3 Applications of P.D.E.(A.P.D.E.) · 2019-08-15 · 3 Applications of P.D.E.(A.P.D.E.) Solutions of one dimensional wave equation { One dimensional equation of heat conduction {

3 Applications of P.D.E.(A.P.D.E.)

Solutions of one dimensional wave equation – Onedimensional equation of heat conduction – Steady state solutionof two-dimensional equation of heat equation (Insulated edgesexcluded) – Fourier series solutions in cartesian coordinates.

3.0.1 Introduction

Many engineering and physical problems when formulated in themathematical language give rise to partial differential equations. Besidesthese, partial differential equations also play an important role in thetheory of elasticity, Hydraulics, etc.,Since the general solution of a partial differential equation in a region R

contains arbitrary constants or arbitrary functions, the unique solution ofa partial differential equation corresponding to a physical problem willsatisfy certain other conditions at the boundary of the region R. Theseare known as boundary conditions. When these conditions arespecified for the time t = 0, they are known as initial conditions. Suchproblems are called Boundary value problems .In the application of ordinary linear differential equations, we first find

the general solution and then determine the arbitrary constants from theinitial values. But the same method is not applicable to problemsinvolving partial differential equation.Most of the boundary value problems involution linear partial

differential equation can be solved by the Method of Separation ofVariables. In this method, right from the beginning, we try to find theparticular solutions of the partial differential equation which satisfy all orsome of the boundary conditions and then adjust them till the remainingconditions are also satisfied. A combination of these particular solutionsgives the solution of the problem.

137

138 Unit IV - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS (B.V.P.)

3.1 Method of separation of variables

Let z be the dependent variable and x, y be the independent variables[i.e., z = z(x, y)]. Assume the solution to be the product of twofunctions, one of them a function of x alone and the other functionof y alone. Then the solution of the partial differential equations isconverted into the solution of ordinary differential equations. Thismethod is explained in the following examples.

3.1.1 Examples of Method of separation of variables

Example 3.1. Solve by the method of separation of variables the partial

differential equation∂2z

∂x2− 2

∂z

∂x+∂z

∂y= 0.

Solution : Given partial differential equation∂2z

∂x2− 2

∂z

∂x+∂z

∂y= 0 (1)

Here z is the function of x and y.So let z = XY . (2)be the solution of (1)[X is a function of x only; Y is a function of y only.]

From (2)∂z

∂x= X ′Y ;

∂2z

∂x2= X ′′Y and

∂z

∂y= XY ′

∴ (1) ⇒ X ′′Y − 2X ′Y +XY ′ = 0Separating the variables, we get

X ′′ − 2X ′

X= − Y ′

YLHS is a function of x only and RHS is a function of y only. Since the

variables x and y are independent, each side is a constant.

∴X ′′ − 2X ′

X= − Y ′

Y= k (say)

X ′′ − 2X ′ = kX (3)Y ′ + kY = 0 (4)

Equation (3) and (4) are ordinary differential equations.For (3), the A.E. is m2 − 2m− k = 0⇒ m = 1±

√1 + k

∴ Solution of (3) is X = Ae(1+√1+k)x +Be(1−

√1+k)x.

For (4), the A.E. is m+ k = 0⇒ m = −1∴ Solution of (4) is Y = Ce−ky.

Substituting the values of X and Y in (2), we get

MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 139

z = XY =[Ae(1+

√1+k)x +Be(1−

√1+k)x

]Ce−ky

=[A1e

(1+√1+k)x +B1e

(1−√1+k)x

]e−ky

Where A1 = AC and B1 = BC.Which is the required general solution of the given P.D.E.The constants A1, B1 and k can be determined so as to satisfy the given

boundary conditions.

Example 3.2. Solve by the method of separation of variables the partial

differential equation∂u

∂x= 2

∂u

∂t+ u where u(x, 0) = 6e−3x.

Solution : Given partial differential equation∂u

∂x= 2

∂u

∂t+ u (1)

Here u is the function of x and t.So let u = XT . (2)be the solution of (1)X is a function of x only; T is a function of t only, be a solution of the

given equation(1)

Then∂u

∂x= X ′T and

∂u

∂t= XT ′

The given equation can be written as

⇒ X ′T = 2XT ′ +XT = X (2T ′ + T )

Separating the variables, we get

X ′

X=

2T ′ + T

T

LHS is a function of x only and RHS is a function of t only. Since thevariables x and t are independent, each side is a constant.

∴X ′

X=

2T ′ + T

T= k (say)

HenceX ′

X= k (3)

andT ′

T=

1

2(k − 1) (4)

Integrating Equations (3) and (4), we get


log x = kx+ log a and log T =1

2(k − 1)t+ log b

∴ logX

a= kx and log

T

b=

1

2(k − 1)t

X

a= ekx and

T

b= e

12 (k−1)t

X = aekx and T = be12 (k−1)t

Substituting the values of X and T in (2), we getu = XT = aekx.be

12 (k−1)t = abkxe

12 (k−1)t (5)

This is the required general solution of the given P.D.E.Given u(x, 0) = 6e−3x.From (5), u(x, 0) = 6e−3x.∴ abekx = 6e−3x ⇒ ab = 6 and k = −3Hence u(x, t) = 6e−3xe

12 (−3−1)t = 6e−3x−2t this is the required solution.

Example 3.3. Find a solution of the equation∂2u

∂x2=∂u

∂y+ 2u in the form

u = X(x)Y (y). Solve the equation subject to the conditions u = 0 and∂u

∂x= 2e−y when x = 0 for all values of y.

Solution : Given partial differential equation∂2u

∂x2=∂u

∂y+ 2u (1)

Here u is the function of x and y.So let u = XY . (2)be the solution of (1)[X is a function of x only; Y is a function of y only.]

From (2)∂u

∂x= X ′Y ;

∂2u

∂x2= X ′′Y and

∂u

∂y= XY ′

∴ (1) ⇒ X ′′Y = XY ′ + 2XYSeparating the variables, we get

X ′′

X=Y ′

Y+ 2

LHS is a function of x only and RHS is a function of y only. Since thevariables x and y are independent, each side is a constant.

∴X ′′

X=Y ′

Y+ 2 = k (say)

∴X ′′

X= k and

Y ′

Y+ 2 = k

X ′′ − kX = 0 (3)


Y ′

Y= k − 2 (4)

Equation (3) is an ordinary differential equation.

For (3), the A.E. is m2 − k = 0

⇒ m = 1±√1 + k

∴ Solution of (3) is X = Ae

(√k)x+Be

(−√k)x.

For (4) , integrating w.r.to y, we get

⇒ log Y = (k − 2)y + log a

∴Y

a= e(k−2)y ⇒ Y = ae(k−2)y

∴ Solution of (4) is Y = ae(k−2)y.Substituting the values of X and Y in (2), we get

u = XY =

[Ae

(√k)x+Be

(−√k)x]ae(k−2)y

=

[A1e

(√k)x+B1e

(−√k)x]e(k−2)y (5)

Where A1 = Aa and B1 = Ba.This is the required general solution of the given P.D.E.

To find the Particular solutionGiven that u = 0 when x = 0 (i.e.,) u(0, y) = 0.From (5), u(0, y) = (A1 +B1) e

(k−2)y.∴ (A1 +B1) e

(k−2)y = 0 ⇒ A1 +B1 = 0 ⇒ B1 = −A1

∴ u(x, y) = A1[e(√k ) x − e( −

√k ) x]e(k−2)y (6)

Again we have the condition∂u

∂x= 2e−y when x = 0.

From (6),

∂u

∂x= A1

√k[e(

√k) x + e(−

√k)x]e(k−2)y = A1

√k2 cosh(

√kx)e(k−2)y

= 2A1

√k cosh(

√kx)e(k−2)y

x = 0,∂u

∂x= 2A1

√k e(k−2)y

∴√ke(k−2)y = 2e−y

A1

√ke(k−2)y = e−y

⇒ k − 2 = −1 ⇒ k = 1 and A1

√k = 1 ⇒ A1 =

1√k= 1

Hence


u(x, y) = [e(√k ) x − e( −

√k ) x]e(k−2)y = [ex − e−x]e−y

= 2e−y sinhx

This is the required solution.

Example 3.4. Using the method of separation of variables, solve∂2u

∂x2− ∂u

∂y= u subject to u(0, t) = 0 = u(π, t).

Solution : Given partial differential equation∂2u

∂x2− ∂u

∂y= u (1)

Here u is the function of x and t.So let u = XT . (2)be the solution of (1)[Where X is a function of x only; T is a function of t only.]

From (2)∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T

∴ (1) can be written asThe given equation can be written as X ′′T −XT ′ = XT

Separating the variables, we get

X ′′

X= 1 +

T ′

T


∴X ′′

X= 1 +

T ′

T= −k2 (say)

∴X ′′

X= −k2 and 1 +

T ′

T= −k2

X ′′ + k2X = 0 (3)T ′

T= −

(1 + k2

)(4)

Equation (3) is an ordinary differential equation.

For (3), the A.E. is m2 + k2 = 0

⇒ m = ±ik

∴ Solution of (3) is X = A cos(kx) +B sin(kx).For (4) , integrating w.r.to t, we get

⇒ log T = −(1 + k2

)t+ log a

∴T

a= e−

(1+k2

)t ⇒ T = ae−

(1+k2

)t


∴ Solution of (4) is T = ae−(1+k2

)t.

Substituting the values of X and T in (2), we get

u = XT = [A cos(kx) +B sin(kx)][ae−

(1+k2

)t]

= [A cos(kx) +B sin(kx)][e−(1+k2

)t]

(5)

Where A1 = Aa and B1 = Ba.This is the required general solution of the given P.D.E.

To find the Particular solutionGiven that u(0, t) = 0.

From (5), u(0, t) = [A1 cos(0) +B1 sin(0)][e−(1+k2

)t].

[A1 cos(0) +B1 sin(0)][e−(1+k2

)t]= 0

⇒ [A1] e−(1+k2

)t = 0

∴ A1 = 0[∵ e−

(1+k2

)t 6= 0

](5) ⇒ u(x, t) = XT = B1 sin(kx)e

−(1+k2

)t (6)

Given that u(π, t) = 0.

From (6), u(π, t) = B1 sin(kπ)[e−(1+k2

)t].

∴ [B1 sin(kπ)][e−(1+k2

)t]= 0

⇒ sin kπ = 0 = sinnπ for n = 1, 2, 3, . . .

⇒ k = n

∴ (6) ⇒ u(x, t) = Bn sin(nπ)e−(1+n2

)t

This is the required solution.

Example 3.5. Solve 2x∂z

∂x− 3y

∂z

∂x= 0 by method of separation of

variables.

Solution : Given partial differential equation 2x∂z

∂x− 3y

∂z

∂x= 0 (1)

Here z is the function of x and y.So let z = XY . (2)be the solution of (1)[Where X is a function of x only; Y is a function of y only.]

From (2)∂z

∂x= X ′Y and

∂z

∂y= XY ′


∴ (1) ⇒ 2xX ′Y − 3yXY ′ = 0Separating the variables, we get

2xX ′

X=

3yY ′

Y

LHS is a function of x only and RHS is a function of y only. Since thevariables x and y are independent, each side is a constant.

∴2xX ′

X=

3yY ′

Y= k (say)

∴X ′

X=

k

2x(3)

Y ′

Y=

k

3y(4)

Equation (3) is integrating, we get

⇒ logX =k

2log x+ logA = log x

k2 + logA = logAx

k2

∴ X = Ax(k/2)

Equation (4) is integrating, we get

⇒ log Y =k

3log y + logB = log y

k3 + logB = logBy

k3

∴ Y = By(k/3)

∴ Solution of (4) is T = ae−(1+k2

)t.

Substituting the values of X and Y in (2), we get

u = XY = Ax(k/2)By(k/3)

= Cx(k/2)y(k/3) (5)

Where C = AB.This is the required general solution of the given P.D.E.

3.1.2 Exercises & Solutions

By the method of separation of variables solve the followingequations.

1. 2∂2u

∂x2− ∂u

∂y= 0

[Ans: u =

[Ae

(√k)x+Be

(−√k)x]e2ky

]2.∂u

∂x=∂u

∂y;u(0, y) = 8e−3y + 4e−5y

[Ans: u = 8e−12x−3y + 4e−20x−5y

]


3.∂u

∂x+ 4 =

∂u

∂t;u = 4e−3x when t = 0

[Ans: u = 4e−3x−2t

]4. 3

∂u

∂x+ 2

∂u

∂y= 0; u(x, 0) = 4e−x

[Ans: u = 4e−

32 (y−x)

]3.2 One dimensional wave equation - Equation of vibrating string :

Consider an elastic string of length `, which is tightly stretched andfixed at its two end points. Take one fixed end as the origin O(0, 0) andthe other end E(`, 0), the string in the equilibrium position as x-axis andy-axis through the origin and perpendicular to x-axis. We require thedifferential equation governing the motion when the string is set vibratingin the vertical plane.

We make the following assumptions:

1. The motion takes place entirely in one plane. [i.e., xy plane]

2. In this vertical plane each point of the string moves in a directionperpendicular to the equilibrium position of the string.

3. Tension T of the stretched string before fixing it at the end points isconstant at all times and at all points of the deflected string.

4. Tension T in the string to be so large that gravity may be neglectedin comparison with it.

5. The effect of friction is negligible.

6. The string is perfectly flexible. [It can transmit only tension but notbending shearing forces]

7. The slope of the deflection curve at all points and at all times is sosmall.

8. m-Mass per unit length of the string.

When the string is set vibrating in the xy plane, the displacement y ofany point of the string is a function of x, its distance from origin, andthe time t. Consider a small portion of the string lying between P andQ where P (x, y) and Q(x + δx, y + δy) are two neightbouring points onthe string. Let the angles made by the tangents at P and Q be θ andθ + δθ respectively with the x-axis. The only forces acting on PQ are


the tensions T1 and T2 along the tangents at P and Q. Since the motionof string is considered in the vertical plane, (ie. xy plane) the horizontalcomponents of tension must be constant.

T1 cos(θ) = T2 cos(θ + δθ) = T (a constant) (1)For vertical motion of PQ, be resultant force is given by

T2 sin(θ ++δθ)− T1 sin(θ) (2)By Newton’s second law of motion, the above resultant force (2) = mass

of the element X its acceleration in the y direction

T2 sin(θ ++δθ)− T1 sin(θ) = mδx∂2y

∂t2(3)

T2 sin(θ + δθ)

T− T1 sin(θ)

T=mδx∂2y

∂t2

T

T2 sin(θ + δθ)

T2 cos(θ + δθ)− T1 sin(θ)

T1 cos(θ)=mδx∂2y

∂t2

T

tan(θ + δθ)− tan(θ) =mδx∂2y

∂t2

T(4)

tan(θ) and tan(θ + δθ) are the slopes of the tangents at P and Q.

∴

(dy

dx

)x

= tan(θ) and ∴

(dy

dx

)x+δx

= tan(θ + δθ)

Hence (4) (4) ⇒ m

T

∂2y

∂t2=

1

δx

[(dy

dx

)x+δx

−(dy

dx

)x

]As δx→ 0, we have

m

T

∂2y

∂t2=∂2y

∂x2

Let a2 =T

m

Then∂2y

∂t2= a2

∂2y

∂x2Which is the equation of the vibrating string, also

known as one dimensional wave equation .

One dimensional wave equation

The one dimensional wave equation is∂2y

∂t2= a2

∂2y

∂x2, where dependent

variable y is function of independent variables x and t, i.e., y(x, t) and

a2 =T

m=

Tension

Mass per unit length of the string= positive.


SinceT

m= positive, it is denoted by a2 ( and it is not denoted by a ).

Solution of one dimensional wave equation:∂2y

∂t2= a2

∂2y

∂x2(1)

Here y is a function of x and t.So assume the solution of (1) is of the form y = XT (2)Where X is a function of x alone and T is a function of t alone.

From (2),∂2y

∂x2= X ′′T and

∂2y

∂t2= XT ′′

Substituting these in (1), we get

XT ′′ = a2X ′′T

By separating the variables,

X ′′

X=

1

a2T ′′

T


X ′′

X=

1

a2T ′′

T= k (say)

Hence X ′′ = kX and T ′′ = a2kT

X ′′ − kX = 0 (3)and

T ′′ − a2kT = 0 (4)The nature of the solutions of (3) and (4) depends on the value of k.

Case I : Let k be positive , say k = p2

∴ (3) and (4) becomeX ′′ − p2X = 0 (3)

andT ′′ − a2p2T = 0 (4)

Now we get two ordinary differential equations.The Auxiliary equations are

m2 − p2 = 0 and m2 − a2p2 = 0

∴ m = ±p m = ±apSo X = A1e

px + A2e−px and T = A3e

apt + A4e−apt


y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


Case II : Let k be positive , say k = −p2∴ (3) and (4) become

X ′′ + p2X = 0 (3)and

T ′′ + a2p2T = 0 (4)Now we get two ordinary differential equations.The Auxiliary equations are

m2 + p2 = 0 and m2 + a2p2 = 0

m2 = −p2 m2 = −a2p2

∴ m = ±ip m = ±iapSo X = A5 cos px+ A6 sin px and T = A7 cos apt+ A8 sin aptSubstituting the values of X and T in (2), we get

y(x, t) = (A5 cos px+ A6 sin px) (A7 cos apt+ A8 sin apt)

Case III : Let k = 0∴ (3) and (4) become

X ′′ = 0 (3)and

T ′′ = 0 (4)Solutions are X = A9x+ A10 and T = A11t+ A12


y(x, t) = (A9x+ A10) (A11t+ A12)

Thus the various possible solutions of the wave equation are

(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)

(2)y(x, t) = (A5 cos px+ A6 sin px) (A7 cos apt+ A8 sin apt)

(3)y(x, t) = (A9x+ A10) (A11t+ A12)

3.2.1 Zero Velocity Problems

Example 3.6. A tightly stretched string of length ` is fastened at both ends.Motion is started by displacing the string into the form y = k

(`x− x2

),

from which it is released at time t = 0. Find the displacement at any pointon the string at a distance x from one end at time t. [UQ]

Solution : The displacement function y(x, t) of the string is the solution

of wave equation∂2y

∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension

Mass


Solving the equation (I) by method of separation of variable, we get thefollowing three possible solutions.

(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.Boundary conditions

(i)y(x = 0, t) = 0,∀t > 0

(ii)y(x = `, t) = 0,∀t > 0Initial conditions

(iii)∂y

∂(x, t = 0) = 0,∀x ∈ (0, `)

(iv)y(x, t = 0) = f(x) = k(`x− x2

),∀x ∈ (0, `)

The suitable solution which satisfies above boundary conditions isy(x, t) = (c1 cos px+ c2 sin px) (c3 cos apt+ c4 sin apt) (1)

Applying condition (i) in equation (1), we getWe have (i) ⇒ y(x = 0, t) =0,∀t > 0

(1) ⇒ (c1 + 0) (c3 cos pat+ c4 sin pat) =0

Here (c3 cos pat+ c4 sin pat) 6=0 (∵ it is defined ∀t > 0)

∴ c1 = 0

∴ Now (1) ⇒ y(x, t) = c2 sin px (c3 cos pat+ c4 sin pat) (2)Applying condition (ii) in equation (2), we getWe have (ii) ⇒ y(x = `, t) =0,∀t > 0

(2) ⇒ (c2 sin p`) (c3 cos pat+ c4 sin pat) =0


∴ c2 6=0 (∵ it gives trivial solution)

∴ sin p` =0

= sinnπ

∴ p =nπ

`

∴ Now (2) ⇒ y(x, t) = c2 sin(nπx

`

)(c3 cos

nπ

àt+ c4 sin

nπ

àt)

(3)

Before applying condition (iii) in (3), differentiate (3) w.r.t. ‘t’∂y

∂t(x, t) = c2 sin

nπx

`

[−c3

(nπà)sin

nπ

àt+ c4

(nπà)cos

nπ

àt]

(4)


Applying condition (iii) in equation (4), we get

We have (iii) ⇒ ∂y

∂t(x, t = 0) = 0,∀x ∈ (0, `)

(4) ⇒ c2 sinnπx

`

[c4

(nπà)]

= 0

Here, c2 6= 0 (∵ it gives trivial solution)

sinnπx

`6= 0 (∵ it is defined ∀x ∈ (0, `))

nπ

à 6= 0 (∵ all are constants)

∴ c4 = 0

∴ Now (3) ⇒ y(x, t) = c2 sin(nπ`

)c3 cos

(nπàt)

= c2c3 sin(nπ`

)cos(nπàt)

= cn sin(nπ`

)cos(nπàt)

where cn = c2c3∴ By superposition principle( i.e., adding all such solutions ), the most

general solution is

y(x, t) =∞∑n=1

cn sin(nπx

`

)cos(nπa

`t)

(5)

Applying condition (iv) in equation (5), we get

We have (iv) ⇒ y(x, t = 0) = f(x),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cn sin(nπx

`

)= f(x) = k

(`x− x2

)(6)

To find cn,Expand f(x) in a half range Fourier sine series in the interval (0, `).

f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx

From (6) and (7), we have cn = bn


∴ cn =2

`

`∫0

f(x) sinnπx

`dx

=2

`

`∫0

k(`x− x2) sinnπx

`dx

=2k

`

[(0 + 0− 2`3

n3π3(−1)n

)−(0 + 0− 2`3

n3π3

)](sinnπ = 0 , cosnπ = (−1)n)

=4k`2

n3π3[1− (−1)n]

∴ cn =

0, if n is even8k`2

n3π3, if n is odd

∴ The most general solution (5) reduces to

y(x, t) =∞∑

n=odd

8k`2

n3π3sin(nπ`x)cos(nπa

`t)

This is the required displacement function y(x, t).

Example 3.7. A string is stretched and fastened to two points at a distance

‘`’ apart. Motion is started by displacing the string in the form y = k sinπx

`from which it is released at time t = 0. Show that the displacement of anypoint on the string at a distance x from one end at time t is given by

k sinπx

`cos

πax

`.



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension

MassSolving the equation (I) by method of separation of variable, we get the

following three possible solutions.

(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.


Boundary conditions(i)y(x = 0, t) = 0,∀t > 0


(iii)∂y

∂(x, t = 0) = 0,∀x ∈ (0, `)

(iv)y(x, t = 0) = f(x) = k sinπx

`,∀x ∈ (0, `)





∴ c1 =0∴ Now (1) ⇒ y(x, t) = c2 sin px (c3 cos pat+ c4 sin pat) (2)

Applying condition (ii) in equation (2), we getWe have (ii) ⇒ y(x = `, t) =0,∀t > 0




∴ sin p` =0

= sinnπ

∴ p =nπ

`∴ Now (2) ⇒ y(x, t) = c2 sin

(nπx`

)(c3 cos

nπ

àt+ c4 sin

nπ

àt)

(3)

Before applying condition (iii) in (3), differentiate (3) w.r.t. ‘t’∂y

∂t(x, t) = c2 sin

nπx

`

[−c3

(nπà)sin

nπ

àt+ c4

(nπà)cos

nπ

àt]

(4)


We have (iii) ⇒ ∂y

∂t(x, t = 0) = 0,∀x ∈ (0, `)

(4) ⇒ c2 sinnπx

`

[c4

(nπà)]

= 0

Here, c2 6= 0 (∵ it gives trivial solution)

sinnπx

`6= 0 (∵ it is defined ∀x ∈ (0, `))

nπ

à 6= 0 (∵ all are constants)

∴ c4 = 0


∴ Now (3) ⇒ y(x, t) = c2 sin(nπ`

)c3 cos

(nπàt)

= c2c3 sin(nπ`

)cos(nπàt)

= cn sin(nπ`

)cos(nπàt)

where cn = c2c3∴ By superposition principle( i.e., adding all such solutions ), the most

general solution is

y(x, t) =∞∑n=1

cn sin(nπx

`

)cos(nπa

`t)

(5)


We have (iv) ⇒ y(x, t = 0) = f(x),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cn sin(nπx

`

)= f(x) = k sin

πx

`(6)


f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


k sinnπx

`=

∞∑n=1

cn sin(nπx

`

)k sin

nπx

`= c1 sin

(πx`

)+ c2 sin

(2πx

`

)+ c3 sin

(3πx

`

)+ . . .

Comparing the like coefficients, we get

∴ c1 = k, c2 = c3 = c4 = · · · = 0


∴ y(x, t) = k sinπx

`cos

πaxt

`

This is the required displacement function y(x, t).


Example 3.8. [Bisection problem]A taut string of length ‘`’ is fastened at both ends. The midpoint of thestring is taken to a height ‘’ and then released from rest in that position.Find the displacement of any of the string at any subsequent time.

Solution : This is of Zero Velocity example format as previous.The displacement function y(x, t) of the string is the solution of wave

equation∂2y

∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)


(i)y(x = 0, t) = 0,∀t > 0


(iii)∂y

∂(x, t = 0) = 0,∀x ∈ (0, `)

(iv)y(x, t = 0) = f(x),∀x ∈ (0, `), where f(x) is to be find later.The suitable solution which satisfies above boundary conditions is

y(x, t) = (c1 cos px+ c2 sin px) (c3 cos apt+ c4 sin apt) (1)As in previous example, the solution of (I) satisfying theboundary conditions (i),(ii) and (iii) is

y(x, t) = cn sinnπx

`cos

nπat

`(where cn = c2c3)

∴ By superposition principle( i.e., adding all such solutions ), the mostgeneral solution is

y(x, t) =∞∑n=1

cn sin(nπx

`

)cos(nπa

`t)

(5)


We have (iv) ⇒ y(x, t = 0) = f(x),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cn sin(nπx

`

)= f(x) (6)



f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


cn =2

`

`∫0

f(x) sinnπx

`dx (8)

where f(x) is not yet find, we will find f(x) as followsTaking OA = `, length of the string with B as its point of bisection.

The initial position of the string is given in the figure.

To find y = f ( x )=

{Equation of OC, 0 < x < `/2Equation of CA, `/2 < x < `

Equation of the line OC joining (0, 0) & (`/2, h) is given by

y − y1y2 − y1

=x− x1x2 − x1

y − 0

h− 0=x− 0`2 − 0

∴ y =2h

`, 0 ≤ x ≤ `

2

Equation of the line CA joining (`/2, h) & (`, 0) is given by


y − y1y2 − y1

=x− x1x2 − x1

y − h

0− h=x− `

2

`− `2

⇒ y − h

−h=x− `

2`2

−yh+ 1 =

x`2

− 1 ⇒ y = 2h− 2hx

`

∴ y =2h

`[`− x] ,

`

2≤ x ≤ `

∴ The required equation of the string is

y = f(x) =

2h

`x, 0 ≤ x ≤ `

22h

`[`− x],

`

2≤ x ≤ `

∴ (8) ⇒ cn =2

`

`2∫

0

2hx

`sin

nπx

`dx +

`∫`2

2h (`− x)

`sin

nπx

`dx

=4h

`2

`2∫

0

x sinnπx

`dx +

`∫`2

(`− x) sinnπx

`dx

=4h

`2

[x

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)] `

2

0

+

[(`− x)

(− cos nπx`

nπ`

)− (−1)

(− sin nπx

`(nπ`

)2)]`

`2

=4h

`2

{[− `2

2nπ

(cos

nπ

2

)+

(`

nπ

)2 (sin

nπ

2

)]− [0 + 0]

+ [0 + 0]−

[−`2

2nπ

(cos

nπ

2

)−(`

nπ

)2 (sin

nπ

2

)]}=

8h

n2π2sin

nπ

2∴ The most general solution (5) reduces to

i.e., (5) ⇒ y(x, t) =∞∑n=1

8h

n2π2

[sin

nπ

2

]sin(nπx

`

)cos(nπa

`t)

=8h

π2

∞∑n=1

1

n2

[sin

nπ

2

]sin(nπx

`

)cos(nπa

`t)


This is the required displacement solution y(x, t).

Example 3.9. [Section problem]A uniform string is fixed at the ends x = 0 and x = ` are fixed. One end istaken at the origin and at a distance ‘b’ from this end the string is displaceda distance ‘h’ transversely and is released from rest in that position. Findthe displacement of any of the string at any subsequent time.


equation∂2y

∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)


(i)y(x = 0, t) = 0,∀t > 0


(iii)∂y

∂(x, t = 0) = 0,∀x ∈ (0, `)




`cos

nπat

`(where cn = c2c3)


y(x, t) =∞∑n=1

cn sin(nπx

`

)cos(nπa

`t)

(5)



We have (iv) ⇒ y(x, t = 0) = f(x),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cn sin(nπx

`

)= f(x) (6)


f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


cn =2

`

`∫0

f(x) sinnπx

`dx (8)

where f(x) is not yet find, we will find f(x) as followsTaking OA = `, length of the string with B as its points of bisection.

The initial position of the string is given by the equations to the linesOB and BA.


{Equation of OC, 0 < x < bEquation of CA, b < x < `

Equation of the line OC joining (0, 0) & (b, h) is given by


y − y1y2 − y1

=x− x1x2 − x1

y − 0

h− 0=x− 0

b− 0

∴ y =h

bx, 0 ≤ x ≤ b

∴ y =h

bx, 0 ≤ x ≤ b

Equation of the line CA joining (b, h) & (`, 0) is given by

y − y1y2 − y1

=x− x1x2 − x1

y − h

0− h=x− b

`− b⇒ y − h

−h=x− b

`− b(y − h)(`− b) = h(x− b) ⇒ y`− yb− h`+ hb = −hx+ hb

y(`− b) = h(`− x) ⇒ y(b− `) = h(x− `)

∴ y =h

b− `[x− `], b ≤ x ≤ `


y = f(x) =

h

bx , 0 ≤ x ≤ b

h

b− `[x− `] , b ≤ x ≤ `

∴(8)⇒


cn=2

`

b∫0

hx

bsin

nπx

`dx+

`∫b

h (x− `)

b− `sin

nπx

`dx

=2h

`

{1

b

[x

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)]b

0

+1

b− `

[(x− `)

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)]`

b

=2h

`

{1

b

[−x `

nπ

(cos

nπx

`

)+

(`

nπ

)2 (sin

nπx

`

)]b

0

+1

b− `

[− (x− `)

`

nπ

(cos

nπx

`

)+

(`

nπ

)2 (sin

nπx

`

)]`b

=2h

`

{1

b

[− b`

nπ

(cos

nπb

`

)+

(`

nπ

)2(sin

nπb

`

)]− [0 + 0]

+1

b−`

[[0 + 0]−

[−(b−`) `

nπ

[cos

nπb

`

]+

[`

nπ

]2[sin

nπb

`

]]]}

=2h

`

{−`nπ

[cos

nπb

`

]+

[`

nπ

]21

b

[sin

nπb

`

]+

`

nπ

[cos

nπb

`

]

− 1

b− `

[`

nπ

]2 [sin

nπb

`

]}

=2h

`

[(`

nπ

)21

b

(sin

nπb

`

)− frac1b− `

(`

nπ

)2(sin

nπb

`

)]

=2h

`

[(`

nπ

)2(sin

nπb

`

)(1

b− 1

b− `

)]

=−2h`2

b(b− `)n2π2

(sin

nπb

`

)∴ The most general solution (5) reduces to

i.e., (4) ⇒ y(x, t) =−2h`2

b(b− `)π2

∞∑n=1

1

n2

(sin

nπb

`

)sin(nπx

`

)cos(nπa

`t)

This is the required displacement solution y(x, t).

Example 3.10. [Trisection problem]The points of trisection of a string are pulled aside through a distance ‘h’


on opposite sides of the position of equilibrium, and the string is releasedfrom rest. Derive an expression for the string at any subsequent time andshow that the middle point of the string always remains at rest.


equation∂2y

∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)


(i)y(x = 0, t) = 0,∀t > 0


(iii)∂y

∂(x, t = 0) = 0,∀x ∈ (0, `)




`cos

nπat

`(where cn = c2c3)


y(x, t) =∞∑n=1

cn sin(nπx

`

)cos(nπa

`t)

(5)


We have (iv) ⇒ y(x, t = 0) = f(x),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cn sin(nπx

`

)= f(x) (6)



f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


cn =2

`

`∫0

f(x) sinnπx

`dx (8)

where f(x) is not yet find, we will find f(x) as followsTaking OA = `, length of the string with B and C as its points of trisection.The initial position of the string is given in the figure.


Equation of OD, 0 < x < `/3Equation of DE, `/3 < x < 2`/3Equation of EA, 2`/3 < x < `

Equation of the line OD joining (0, 0) & (`/3, h) is given by

y − y1y2 − y1

=x− x1x2 − x1

y − 0

h− 0=x− 0`3 − 0

∴ y =3hx

`, 0 ≤ x ≤ `

3Equation of the line DE joining (`/3, h) & (2`/3,−h) is given by

y − y1y2 − y1

=x− x1x2 − x1

⇒ y − h

−h− h=x− `

32`3 − `

3

⇒ y − h

−2h=x− `

3`3

y

−2h+

1

2=

x`3 − 1

⇒ y

−2h=

3x

`− 3

2⇒ y =

−6hx

`+ 3h


∴ y =3h

`[`− 2x],

`

3≤ x ≤ 2`

3

Equation of the line EA joining (2`/3,−h) & (`, 0) is given by

y − y1y2 − y1

=x− x1x2 − x1

⇒ y + h

0 + h=x− 2`

3

`− 2`3

⇒ y + h

h=x− 2`

3`3

y

h+ 1 =

x`3 − 2

⇒ y =3hx

`− 3h ⇒ y =

3h

`[x− `]

∴ y =3h

`[`− 2x],

`

3≤ x ≤ 2`

3


y = f(x) =

3hx

`, 0 ≤ x ≤ `

33h

`[`− 2x] ,

`

3≤ x ≤ 2`

33h

`[x− `] ,

2`

3≤ x ≤ `

∴(8)⇒


cn =2

`

`3∫

0

3hx

`sin

nπx

`dx+

2`3∫

`3

3h (`− 2x)

`sin

nπx

`dx

+

`∫2`3

3h (x− `)

`sin

nπx

`dx

=6h

`2

`3∫

0

x sinnπx

`dx+

2`3∫

`3

(`− 2x) sinnπx

`dx+

`∫2`3

(x− `) sinnπx

`dx

=6h

`2

{[x

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)] `

3

0

+

[(`− 2x)

(− cos nπx`

nπ`

)− (−2)

(− sin nπx

`(nπ`

)2)] 2`

3

`3

+

[(x− `)

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)]`

2`3

=6h

`2

{[−x `

nπ

(cos

nπx

`

)+

(`

nπ

)2 (sin

nπx

`

)] `3

0

+

[− (`− 2x)

`

nπ

(cos

nπx

`

)− (2)

(`

nπ

)2 (sin

nπx

`

)] 2`3

`3

+

[− (x− `)

`

nπ

(cos

nπx

`

)+

(`

nπ

)2 (sin

nπx

`

)]`2`3


=6h

`2

{[− `2

3nπ

(cos

nπ

3

)+

(`

nπ

)2 (sin

nπ

3

)]− [0 + 0]

+

[`2

3nπ

(cos

2nπ

3

)− (2)

(`

nπ

)2(sin

2nπ

3

)]

−

[− `2

3nπ

(cos

nπ

3

)− (2)

(`

nπ

)2 (sin

nπ

3

)]

+[0 + 0]−

[`2

3nπ

(cos

2nπ

3

)+

(`

nπ

)2(sin

2nπ

3

)]}

=6h

`2

[3

(`

nπ

)2 (sin

nπ

3

)− 3

(`

nπ

)2(sin

2nπ

3

)]

=18h

n2π2

[(sin

nπ

3

)−(sin

2nπ

3

)]=

18h

n2π2

[sin

nπ

3− sin

(nπ − nπ

3

)]=

18h

n2π2

[sin

nπ

3−(sinnπ cos

nπ

3− cosnπ sin

nπ

3

)]=

18h

n2π2

[sin

nπ

3+ cosnπ sin

nπ

3

](sinnπ = 0)

=18h

n2π2

[sin

nπ

3(1 + cosnπ)

]=

18h

n2π2

[sin

nπ

3[1 + (−1)n]

]∴ cn =

{36h

n2π2

[sin

nπ

3

], if n is even

0 , if n is odd∴ The most general solution (5) reduces toi.e., (5)⇒

y(x, t) =∞∑

n=even

36h

n2π2

[sin

nπ

3

]sin(nπ`x)cos(nπa

`t)

=36h

π2

∞∑n=even

1

n2

[sin

nπ

3

]sin(nπ`x)cos(nπa

`t)

=36h

π2

∞∑n=1

1

(2n)2

(sin

2nπ

3

)sin

(2nπ

`x

)cos

(2nπa

`t

)


=9h

π2

∞∑n=1

1

n2

(sin

2nπ

3

)sin

(2nπ

`x

)cos

(2nπa

`t

)∴ y(x, t) =

9h

π2

∞∑n=1

1

n2

(sin

2nπ

3

)sin

(2nπ

`x

)cos

(2nπa

`t

)(9)

This is the required displacement solution y(x, t).By putting x = `/2 at M in (9), the displacement at the midpoint is

y

(x =

`

2, t

)=9h

π2

∞∑n=1

1

n2

(sin

2nπ

3

)sin

(2nπ

`

(`

2

))cos

(2nπa

`t

)=9h

π2

∞∑n=1

1

n2

(sin

2nπ

3

)sin(nπ) cos

(2nπa

`t

)=0 (∵ sinnπ = 0)

Hence there is no displacement at x = `/2,∀t.i.e., at midpoint M , the string is at rest.

Example 3.11. A uniform elastic strings of length 60 cm is subjected toa constant tension of 2 kg. If the ends are fixed and the initialdisplacement y = 60x− x2, 0 < x < 60 while the initial velocity is zero.Find the displacement function y(x, t).


equation∂2y

∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.In this problem ` = 60 and tension T = 2 kg.Boundary conditions

(i)y(x = 0, t) = 0,∀t > 0



(iii)∂y

∂(x, t = 0) = 0,∀x ∈ (0, `)

(iv)y(x, t = 0) = f(x) = 60x− x2 =(`x− x2

),∀x ∈ (0, `).


As in previous example, the solution of (I) satisfying theboundary conditions (i),(ii) and (iii) is


`cos

nπat

`(where cn = c2c3)


y(x, t) =∞∑n=1

cn sin(nπx

`

)cos(nπa

`t)

(5)


We have (iv) ⇒ y(x, t = 0) = f(x) =(`x− x2

),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cn sin(nπx

`

)= f(x) (6)


f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


cn =2

`

`∫0

f(x) sinnπx

`dx


=2

`

`∫0

(`x− x2

)sin

nπx

`dx

=2

`

[(0 + 0− 2`3

n3π3(−1)n

)−(0 + 0− 2`3

n3π3

)]∵ sinnπ = 0, cosnπ = (−1)n

=4`2

n3π3[1− (−1)n]

∴ cn =

0, if n is even8`2

n3π3, if n is odd


y(x, t) =∞∑

n=odd

8`2

n3π3sin(nπx

`

)cos(nπa

`t)

Put ` = 60

y(x, t) =∞∑

n=odd

28800

n3π3sin(nπx

`

)cos(nπa

`t)

This is the required solution y(x, t).

3.2.2 [Non - Zero Velocity Problem]

Example 3.12. A tightly stretched string with fixed end points x = 0 andx = ` is initially at rest in its equilibrium position. If it is set vibrating bygiving to each of its points a velocity λx(`− x), find y(x, t).



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.




(iii)y(x, t = 0) = 0,∀x ∈ (0, `)

(iv)∂y

∂(x, t = 0) = f(x) = λ

(`x− x2

),∀x ∈ (0, `)





∴ c1 =0∴ Now (1) ⇒ y(x, t) = c2 sin px (c3 cos pat+ c4 sin pat) (2)

Applying condition (ii) in equation (2), we getWe have (ii) ⇒ y(x = `, t) =0,∀t > 0




∴ sin p` =0

= sinnπ

∴ p =nπ

`∴ Now (2) ⇒ y(x, t) = c2 sin

(nπx`

)(c3 cos

nπ

àt+ c4 sin

nπ

àt)

(3)

Applying condition (iii) in (3), we getWe have (iii) ⇒ y(x, t = 0) = 0

∴ (3) ⇒[c2 sin

nπx

`(c3)]= 0

c2 6= 0 (∵ it gives trivial solution)

sinnπx

`6= 0 (∵ it is defined for all x)

∴ c3 = 0

∴ Now (3) ⇒ y(x, t) = c2 sin(nπ`x)c4 sin

(nπàt)

= c2c4 sin(nπ`x)sin(nπàt)

= cn sin(nπ`x)sin(nπàt)

where cn = c2c3


∴ By superposition principle(i.e., adding all such above solutions ), themost general solution is

y(x, t) =∞∑n=1

cn sin(nπx

`

)sin(nπa

`t)

(4)

Before applying condition (iv), differentiate (4) w.r.t. ‘t’

∂y

∂t(x, t) =

∞∑n=1

cn sin(nπx

`

) nπa`

cos(nπa

`t)

(5)


We have (iv) ⇒ ∂y

∂t(x, t = 0) = f(x),∀x ∈ (0, `)

(5) ⇒∞∑n=1

cnnπa

`sin(nπx

`

)= f(x) = λ

(`x− x2

)(6)

To find cn,Expand f(x) in a half range Fourier sine series in the interval (0, `)

f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx

From (6) and (7), we have bn = cnnπa

`

cnnπa

`=2

`

`∫0

f(x) sinnπx

`dx

=2

`

`∫0

λ(lx− x2) sinnπx

`dx

=2λ

`

[(0 + 0− 2`3

n3π3(−1)n

)−(0 + 0− 2`3

n3π3

)](∵ sinnπ = 0 , cosnπ = (−1)n)

=4λ`2

n3π3[1− (−1)n]

∴ cn =

0 , if n is even8λ`3

n4π4, if n is odd


∴ The most general solution is

i.e., (4) ⇒ y(x, t) =8λ`3

π4

∞∑n=odd

1

n4sin(nπx

`

)sin(nπa

`t)

This is the required displacement y(x, t).

Example 3.13. A string of length ` is tightly stretched and fixed at itsends at the points (0, 0) and (2`, 0) of the xy plane. It is made to vibratetransversely in the xy plane by giving to each of its points a transversevelocity ν in the xy plane, where ν is given by

ν =

{kx ; 0 < x ≤ `

k(2`− x) ; ` < x ≤ 2`Find the expression for the transverse

displacement of the string at any time t.



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.Let 2` = L.


(ii)y(x = L, t) = 0,∀t > 0Initial conditions

(iii)y(x, t = 0) = 0,∀x ∈ (0, L)

(iv)∂y

∂t(x, t = 0) = f(x) = v =

{kx ; 0 < x ≤ L/2

k(L− x) ;L/2 < x ≤ LThe suitable solution which satisfies above boundary conditions is

y(x, t) = (c1 cos px+ c2 sin px) (c3 cos apt+ c4 sin apt) (1)Applying condition (i) in equation (1) we getWe have (i) ⇒ y(x = 0, t) =0,∀t > 0



∴ c1 =0


∴ Now (1) ⇒ y(x, t) = c2 sin px (c3 cos pat+ c4 sin pat) (2)Applying condition (ii) in equation (2), we getWe have (ii) ⇒ y(x = L, t) =0,∀t > 0

(2) ⇒ (c2 sin pL) (c3 cos pat+ c4 sin pat) =0



∴ sin pL =0

= sinnπ

∴ p =nπ

L∴ Now (2) ⇒ y(x, t) = c2 sin

(nπLx)(

c3 cosnπ

Lat+ c4 sin

nπ

Lat)

(3)

Applying condition (iii) in (3), we getWe have (iii) ⇒ y(x, t = 0) = 0,∀x ∈ (0, L)

∴ (3) ⇒[c2 sin

nπx

L(c3)]= 0

c2 6= 0 (∵ it gives trivial solution)

sinnπx

L6= 0 (∵ it is defined for all x)

∴ c3 = 0

∴ Now (3) ⇒ y(x, t) = c2 sin(nπLx)c4 sin

(nπLat)

= c2c4 sin(nπLx)sin(nπLat)

= cn sin(nπLx)sin(nπLat)

where cn = c2c3∴ By superposition principle(i.e., adding all such above solutions ), the

most general solution is

y(x, t) =∞∑n=1

cn sin(nπLx)sin(nπaLt)

(4)


∂y

∂t(x, t) =

∞∑n=1

cn sin(nπLx) nπa

Lcos(nπaLt)

(5)



We have

(iv)⇒ ∂y

∂t(x, t = 0)=f(x),∀x ∈ (0, L)

(3)⇒∞∑n=1

cnnπa

Lsin(nπxL

)=f(x)=ν=

{kx ; 0<x≤ L

2

k(L−x) ; L2 <x≤L(6)

To find cn,Expand f(x) in a half range Fourier sine series in the interval (0, L)

f(x) =∞∑n=1

bn sinnπx

L(7)

where bn =2

L

L∫0

f(x) sinnπx

Ldx


L


∴ cnnπa

L=2

L

L∫0

f(x) sinnπx

Ldx

=2

L

L/2∫0

kx sinnπx

Ldx+

L∫L/2

k(L− x) sinnπx

Ldx

=2k

L

L/2∫0

x sinnπx

Ldx+

L∫L/2

(L− x) sinnπx

Ldx

=2k

L

[x

(− cos nπxL

nπL

)− (1)

(− sin nπx

L(nπL

)2)]L/2

0

+

[(L− x)

(− cos nπxL

nπL

)− (−1)

(− sin nπx

L(nπL

)2)]L

L/2

=2k

L

[−x L

nπ

(cos

nπx

L

)+

(L

nπ

)2 (sin

nπx

L

)]L/20

+

[−(L− x)

L

nπ

(cos

nπx

L

)−(L

nπ

)2 (sin

nπx

L

)]LL/2

=2k

L

{[− L2

2nπ

(cos

nπx

2

)+

(L

nπ

)2 (sin

nπx

2

)]− [0 + 0]

+[0 + 0]−

[−L2

2nπ

(cos

nπx

2

)−(L

nπ

)2 (sin

nπ

2

)]}=4kL

n2π2sin

nπ

2

∴ cnnπa

L=4kL

n2π2sin

nπ

2

⇒ cn =4kL2

n3π3asin

nπ

2

∴ y(x, t) =∞∑n=1

[4kL2

n3π3asin

nπ

2

]sin(nπLx)sin(nπaLt)

=4kL2

π3a

∞∑n=1

1

n3

[sin

nπ

2


Replace L = 2`



i.e., (4) ⇒ y(x, t) =16k`2

π3a

∞∑n=1

1

n3

[sin

nπ

2

]sin(nπ2`x)sin(nπa

2`t)


Example 3.14. A tightly stretched string with fixed end points x = 0 and

x = ` is initially in a position with velocity ν given by ν = y0 sin3(πx`

).

Find the displacement y(x, t) in terms of Fourier coefficients of f(x).



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)


(i)y(x = 0, t) = 0,∀t > 0


(iii)y(x, t = 0) = 0,∀x ∈ (0, `)

(iv)∂y

∂t(x, t = 0) = f(x) = ν = y0 sin

3(πx`

)The suitable solution which satisfies above boundary conditions is

y(x, t) = (c1 cos px+ c2 sin px) (c3 cos apt+ c4 sin apt) (1)Applying condition (i),(ii) and (iii) as in previous problem, weget∴ By superposition principle(i.e., adding all such above solutions ), the


y(x, t) =∞∑n=1

cn sin(nπx

`

)sin(nπa

`t)

(4)


∂y

∂t(x, t) =

∞∑n=1

cn sin(nπx

`

) nπa`

cos(nπa

`t)

(5)




∂t(x, t = 0) = f(x)

(5) ⇒∞∑n=1

cnnπa

`sin(nπx

`

)= f(x) = ν = y0 sin

3(πx`

)(6)


f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


`To find cn,

y0 sin3(πx`

)=

∞∑n=1

bn sin(nπx

`

)y0

[3 sin

(πx`

)−sin 3

(πx`

)4

]=b1 sin

[π`x]+b2 sin

[2π

`x

]+b3 sin

[3π

`x

]+. . .[

∵ sin3A =3 sinA− sin 3A

4

]3y04

sin[πx`

]−y0

4sin

[3πx

`

]=b1 sin

[π`x]+b2 sin

[2π

`x

]+b3 sin

[3π

`x

]+. . .

Comparing the like coefficients, we get

∴ b1=3y04

and b3=−y04, bn=0 for n 6= 2, 4, 5, 6

∴ c1=b1`

πa=3y04

`

3πaand c3=b3

`

3πa=−y04

`

3πa, cn=0 for n 6= 2, 4, 5, 6[

∵ bn = cnnπa

`⇒ cn = bn

`

nπa

]∴ The most general solution is

i.e.,(4)⇒y(x, t)=c1 sin[π`x]sin[πa`t]+ c3 sin

[3π

`x

]sin

[3πa

`t

]=

3y04

`

πasin[π`x]sin[πa`t]+−y04

`

3πasin

[3π

`x

]sin

[3πa

`t

]=

3y0`

4πasin[π`x]sin[πa`t]− y0`

12πasin

[3π

`x

]sin

[3πa

`t

]This is the required displacement y(x, t).


Example 3.15. A string is stretched between two fixed points at a distanceof 4` apart and the point of the string are given initial velocity V where

V =

cx

`; in 0 < x ≤ 2`

c

`(4`− x) ; in 2` < x ≤ 4`

Find the displacement of the string at

any time.



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.Let 4` = L⇒ 2` = L/2 and k = c/`


(ii)y(x = L, t) = 0,∀t > 0Initial conditions

(iii)y(x, t = 0) = 0,∀x ∈ (0, L)

(iv)∂y

∂t(x, t = 0) = f(x) = V =

{kx ; 0 < x ≤ L/2

k(L− x) ;L/2 < x ≤ LThe suitable solution which satisfies above boundary conditions is

y(x, t) = (c1 cos px+ c2 sin px) (c3 cos apt+ c4 sin apt) (1)Applying condition (i),(ii) and (iii) as in previous problem, weget∴ By superposition principle(i.e., adding all such above solutions ), the


y(x, t) =∞∑n=1

cn sin(nπLx)sin(nπaLt)

(4)


∂y

∂t(x, t) =

∞∑n=1

cn sin(nπLx) nπa

Lcos(nπaLt)

(5)




∂t(x, t = 0) = f(x)

∞∑n=1

cnnπa

Lsin(nπxL

)= f(x) = V =

{kx; 0 < x ≤ L/2

k(2`− x); L/2 < x < L

(6)

To find cn,Expand f(x) in a half range Fourier sine series in the interval (0, L)

f(x) =∞∑n=1

bn sinnπx

L7) (()

where bn =2

L

L∫0

f(x) sinnπx

Ldx


L


cnnπa

L=2

L

L∫0

f(x) sinnπx

Ldx

=2

L

L/2∫0

kx sinnπx

Ldx+

L∫L/2

k(L− x) sinnπx

Ldx

=2k

L

L/2∫0

x sinnπx

Ldx+

L∫L/2

(L− x) sinnπx

Ldx

=2k

L

[x

(− cos nπxL

nπL

)− (1)

(− sin nπx

L(nπL

)2)]L/2

0

+

[(L− x)

(− cos nπxL

nπL

)− (−1)

(− sin nπx

L(nπL

)2)]L

L/2

=2k

L

[−x L

nπ

(cos

nπx

L

)+

(L

nπ

)2 (sin

nπx

L

)]L/20

+

[−(L− x)

L

nπ

(cos

nπx

L

)−(L

nπ

)2 (sin

nπx

L

)]LL/2

=2k

L

{[− L2

2nπ

(cos

nπx

2

)+

(L

nπ

)2 (sin

nπx

2

)]− [0 + 0]

+[0 + 0]−

[−L2

2nπ

(cos

nπx

2

)−(L

nπ

)2 (sin

nπ

2

)]}=4kL

n2π2sin

nπ

2

∴ cnnπa

L=4kL

n2π2sin

nπ

2

⇒ cn =4kL2

n3π3asin

nπ

2

∴ y(x, t) =∞∑n=1

[4kL2

n3π3asin

nπ

2


=4kL2

π3a

∞∑n=1

1

n3

[sin

nπ

2


Replace L = 4` and k = c/`



i.e., (4) ⇒ y(x, t) =64c`

π3a

∞∑n=1

1

n3

[sin

nπ

2

]sin(nπ4`x)sin(nπa

4`t)


Example 3.16. A string is stretched between two fixed points at a distanceof ` apart and the point of the string are given initial velocity V where:

V =

cx

`; 0 < x ≤ `/2

c

`(`− x) ; `/2 < x ≤ `

x being the distance from an end point.

Find the displacement of the string at any time.



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)


(i)y(x = 0, t) = 0,∀t > 0


(iii)y(x, t = 0) = 0,∀x ∈ (0, `)

(iv)∂y

∂t(x, t = 0) = f(x) = V =

c

`x ; 0 < x ≤ `/2

c

`(`− x) ; `/2 < x ≤ `


Applying condition (i),(ii) and (iii) as in previous problem, weget∴ By superposition principle(i.e., adding all such above solutions ), the


y(x, t) =∞∑n=1

cn sin(nπx

`

)sin(nπa

`t)

(4)



∂y

∂t(x, t) =

∞∑n=1

cn sin(nπx

`

) nπa`

cos(nπa

`t)

(5)


We have (iv)⇒ ∂y

∂t(x, t=0)=f(x)

∞∑n=1

cnnπa

`sin[nπx`

]=f(x)=V =

c

`x ; 0<x≤`/2

c

`(`−x) ; `/2<x≤`

(6)


f(x) =∞∑n=1

bn sinnπx

`(7)

where bn =2

`

`∫0

f(x) sinnπx

`dx


`


cnnπa

`=2

`

`∫0

f(x) sinnπx

`dx

=2

`

`/2∫0

c

`x sin

nπx

`dx+

`∫`/2

c

`(`− x) sin

nπx

`dx

=2c

`2

`/2∫0

x sinnπx

`dx+

`∫`/2

(`− x) sinnπx

`dx

=2c

`2

[x

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)]`/2

0

+

[(`− x)

(− cos nπx`

nπ`

)− (−1)

(− sin nπx

`(nπ`

)2)]`

`/2

=2c

`2

[−x `

nπ

(cos

nπx

`

)+

(`

nπ

)2 (sin

nπx

`

)]`/20

+

[−(`− x)

`

nπ

(cos

nπx

`

)−(`

nπ

)2 (sin

nπx

`

)]``/2

=2c

`2

{[− `2

2nπ

(cos

nπx

2

)+

(`

nπ

)2 (sin

nπx

2

)]− [0 + 0]

+[0 + 0]−

[−`2

2nπ

(cos

nπx

2

)−(`

nπ

)2 (sin

nπ

2

)]}=

4c

n2π2sin

nπ

2

∴ cnnπa

`=

4c

n2π2sin

nπ

2

⇒ cn =4c`

n3π3asin

nπ

2

∴ y(x, t) =∞∑n=1

[4c`

n3π3asin

nπ

2

]sin(nπx

`

)sin(nπa

`t)

=4c`

π3a

∞∑n=1

1

n3

[sin

nπ

2

]sin(nπx

`

)sin(nπa

`t)



3.2.3 [Initial displacement and Initial Velocity problem]

Example 3.17. A tightly stretched string with fixed end points x = 0 andx = ` is initially in the position y = f(x). It is set vibrating by giving allits points a velocity ∂y

∂t = g(x) at time t = 0. Find y(x, t) in the form of aFourier series.



∂t2= a2

∂2y

∂x2. (I)

where a2 =T

M=

Tension



(1)y(x, t) =(A1e

px + A2e−px) (A3e

apt + A4e−apt)


(3)y(x, t) = (A9x+ A10) (A11t+ A12)


(i)y(x = 0, t) = 0,∀t > 0


(iii)∂y

∂(x, t = 0) = f(x),∀x ∈ (0, `)

(iv)y(x, t = 0) = g(x),∀x ∈ (0, `)The suitable solution which satisfies above boundary conditions is

y(x, t) = (c1 cos px+ c2 sin px) (c3 cos apt+ c4 sin apt) (1)Applying condition (i) in equation (1) we getWe have (i) ⇒ y(x = 0, t) = 0

(c1 + 0) (c3 cos pat+ c4 sin pat) = 0

Here (c3 cos pat+ c4 sin pat) 6= 0 (∵ itisdefined∀t)

∴ c1 = 0

∴ Now (1) ⇒ y(x, t) = c2 sin px (c3 cos pat+ c4 sin pat) (2)Applying condition (ii) in equation (2) we getWe have (ii) ⇒ y(x = `, t) = 0

(c2 sin p`) (c3 cos pat+ c4 sin pat) = 0

Here (c3 cos pat+ c4 sin pat) 6= 0 (∵ it is defined ∀t)c2 6= 0 (∵ it gives trivial solution)


∴ sin p` = 0

= sinnπ

∴ p =nπ

`

∴ Now (2) ⇒ y(x, t) = c2 sin(nπx

`

)(c3 cos

nπ

àt+ c4 sin

nπ

àt)


y(x, t) =∞∑n=1

sin(nπx

`

) [Cn cos

(nπa`t)+Dn sin

(nπa`t)]

(3)

where Cn = c2c3 and Dn = c2c4Applying condition (iii) in (3), we getWe have (iii) ⇒ y(x, t = 0) = f(x)

∞∑n=1

cn sin(nπx

`

)= f(x)

Which is the half range sine series for f(x) in (0, `).

∴ Cn =2

`

`∫0

f(x) sinnπx

`dx

Before applying condition (iv) in (3) differentiate (3) w.r.t. ‘t’

∂y

∂t(x, t) =

∞∑n=1

sin(nπx

`

)[−nπa`

Cn sin(nπa

`t)+nπa

`Dn cos

(nπa`t)]

(4)

Applying condition (iv) in equation (4),we getWe have (iv) ⇒ ∂y

∂t (x, t = 0) = g(x)

∞∑n=1

Dnnπa

`sin(nπx

`

)= g(x)

Which is also the half range sine series for g(x) in (0, `)

∴ Dnnπa

`=

2

`

`∫0

g(x) sinnπx

`dx⇒ Dn =

2

nπa

`∫0

g(x) sinnπx

`dx

Substituting the values of Cn and Dn in equation (3) we get the solutionof the wave equation satisfying the given boundary conditions.



1. Solve completely the equation∂2y

∂t2= a2

∂2y

∂x2, representing the vibration

of string of length `, fixed at both ends, given that y(x = 0, t) = 0,

y(x = `, t) = 0, y(x, 0) = f(x) and∂y

∂t(x, t = 0) = 0 , 0 < x < `.y(x, t) = ∞∑

n=1

bn sinnπx

`cos

nπat

`, where bn =

2

`

`∫0

f(x) sinnπa

`dx

2. A string is stretched between two fixed points at a distance of 60 cm

apart and the point of the string are given initial velocity V where:

V =

λx

30; 0 < x ≤ 30

λ

60(60− x) ; 30 < x ≤ 60

x being the distance from an end

point. Find the displacement of the string at any time.[y(x, t) =

240λ

π3a

∞∑n=1

1

n3sin

nπ

2sin

nπx

60sin

nπat

60

]3. A string is stretched fastened at two points x = 0 and x = `. Motion

is stated by displacing the string into the form of the curvey(x, 0) =

(`x− x2

)and also by imparting a constant velocity vc to

every point of the string in this position at t = 0. Find thesubsequent displacement of the string as a function of x and t.{y(x, t) =

4`3

π3

∞∑n=odd

1

n3sin

nπx

`sin

nπat

`

+4vc`

π3a

∞∑n=odd

1

n2sin

nπx

`sin

nπat

`

}

3.3 One Dimensional heat flow equation

IntroductionThe partial differential equation of the one dimensional heat flow equation

is∂u

∂t= a2

∂2u

∂x2

where

a2 =k

ρc= diffusivity ( the unit is m2/sec) of the substance.

=Thermal conductivity

(Density)(Specific heat)= positive and u is a function of x and t.


Temperature gradient

The potential derivative∂u

∂xis the rate of change of temperature w.r.t.

distance is called the temperature gradient.

Solution of one dimensional heat flow equation

The one dimensional heat flow equation is∂u

∂t= a2

∂2u

∂x2(1)

where u is a function of x and t.Consider the trial solution of u(x, t) in (1) as

u(x, t) = XT (2)

= X(x)T (t)where X is a function x alone and T is a function of t alone.Now differentiating (2) w.r.t. x & y partially, we get

∂u

∂x= X ′T

∂u

∂t= XT ′

∂2u

∂x2= X ′′T

∴ The equation (1) ⇒ X ′T = a2X ′′TBy variable separable method, we getX ′′

X=

1

a2T ′

T(3)

In (3), LHS is a function X alone and RHS is a function of T alone. HereX and T are independent, so each ratio is equal to a constant.

X ′′

X=

1

a2T ′

T= k (say)

X ′′

X= k

1

a2T ′

T= k

X ′′ − kX = 0 (4) T ′ − ka2T = 0 (5)

Now there are 3 cases for the value of k.

Case I: Let k be positive , say k = p2, then

(4) ⇒ X ′′ − p2X = 0 (5) ⇒ T ′ − p2a2T = 0

The corresponding auxiliary equations (A.E.) are

m2 − p2 = 0 m− p2a = 0∴ m = ±p ∴ m = p2a2

∴ Solution of X is ∴ Solution of T is

X = C1epx + C2e

−px T = C3ep2a2t


∴ The trial solution (2) is

u(x, t) = XT

=(C1e

px + C2e−px)C3e

p2a2t

Case II: Let k be negative , say k = −p2, then

(4) ⇒ X ′′ + p2X = 0 (5) ⇒ T ′ + p2a2T = 0


m2 + p2 = 0 m+ p2a = 0∴ m = ±pi ∴ m = −p2a2∴ Solution of X is ∴ Solution of T is

X=C4 cos px+C5 sin px T =C6e−p2a2t


u(x, t) = XT

= (C4 cos px+ C5 sin px)C6e−p2a2t

Case III: Let k = 0, then

(4) ⇒ X ′′ = 0 (5) ⇒ T ′ = 0


m2 = 0 m = 0∴ m = 0( twice ) ∴ Solution of T is T = C9

∴ Solution of X is X = C7 + C8x


u(x, t) = XT

= (C7 + C8x)C9

∴ The three possible solution of (1) are

(1)u(x, t) =(C1e

px + C2e−px)C3e

a2p2t

(2)u(x, t) = (C4 cos px+ C5 sin px)C6e−a2p2t

(3)u(x, t) = (C7 + C8)A9

Initial and Boundary conditions of one dimensional heat flowequation

The one dimensional heat flow equation is∂u

∂t= a2

∂2u

∂x2where u(x, t) is

the temperature at time t at a point distant x from the left end of the rod.The boundary conditions are


(1)u(x = 0, t) = k◦1C ∀t(1)u(x = `, t) = k◦2C ∀t

(` being the length of the one dimensional rod)The initial condition is(iii)u(x, t = 0) = f(x), 0 < x < `

Steady state conditions of one dimensional heat flow equationsWhen steady state conditions exist the heat flow equation is independent

of time t. i.e.,

∂u

∂t= 0

∴ The heat flow equation becomes∂2u

∂x2= 0 (or)

d2u

dx2= 0

Since u is a function of x alone. Integrating twice we get u(x) = c1x+c2Thermally insulated endsIf an end of heat conducting body is thermally insulated, it means that noheat passes through that section i.e., the temperature gradient is zero at

that point i.e.,∂u

∂x= 0

3.3.1 Types of boundary value problems in one dimensional heat flow

(I) (i)u(x = 0, t) = 0 (ii)u(x = `, t) = 0[Both the ends are at ’0’ temperature]

(II) Steady state conditions and zero boundary conditions

(III) Steady state conditions and non-zero boundary conditions

(IV) (i)∂u

∂x(x = 0, t) = 0 (ii)u(x = `, t) = 0

[The end x = 0 is thermally insulated]

(V) (i)u(x = 0, t) = 0 (ii)∂u

∂x(x = `, t) = 0

[The end x = ` is thermally insulated]

(VI) (i)∂u

∂x(x = 0, t) = 0 (ii)

∂u

∂x(x = `, t) = 0

[Both the ends are thermally insulated]


3.3.2 Examples of TYPE (I)

[Both the ends are at ‘0’ temperature]

Example 3.18. A rod of length ` with insulated sides is initially at auniform temperature f(x). Its ends are suddenly cooled to 0◦C and arekept at that temperature. Find the temperature function u(x, t).

Solution : The temperature function u(x, t) satisfies the one dimensional

heat equation is∂u

∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=

Thermal conductivity

(Density)(Specific heat)Solving the equation (I) by method of separation of variable, we get the


(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t

(2)u(x, t) = (A4 cos px+ A5 sin px)A6e−a2p2t

(3)u(x, t) = (A7 + A8x)A9

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.From the problem, the boundary and initial conditions are

(i)u(x = 0, t) = 0,∀t > 0

(ii)u(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x),∀x ∈ (0, `)

The suitable solution which satisfies above boundary conditions is

u(x, t) = (A cos px+B sin px) e−a2p2t (1)

Applying condition (i) in equation (1) we get

We have (i) ⇒ u(x = 0, t) = 0

(1) ⇒ Ae−a2p2t = 0

Here e−a2p2t 6= 0 (∵ it is defined for all t)

∴ A = 0

∴ Now (1) ⇒ u(x, t) = B sin pxe−a2p2t (2)Applying condition (ii) in equation (2) we getWe have


(ii) ⇒ u(x = `, t) = 0

(2) ⇒ B sin pxe−a2p2t = 0


B 6= 0 (∵ it gives trivial solution )

∴ sin p` = 0

= sinnπ

∴ p =nπ

`

∴ Now (2) ⇒ u(x, t) = B sinnπx

è−a2 n2π2

`2t


u(x, t) =∞∑n=1

Bn sinnπx

è−a2 n2π2

`2t (3)


We have (iii) ⇒ u(x, t = 0) = f(x)

∴ (3) ⇒∞∑n=1

Bn sinnπx

`= f(x) (4)

To find Bn,Expand f(x) in a half range Fourier sine series in the interval (0, `)

f(x) =∞∑n=1

bn sinnπx

`(5)

where bn =2

`

`∫0

f(x) sinnπx

`dx

From (4) and (5), we have Bn = bn =2

`

`∫0

f(x) sinnπx

`dx

Substituting this value of Bn in (3), we get the required solution.

Example 3.19. A rod of length ` cm long with insulated lateral surface isinitially at temperature ‘u0’ at an inner point distance x cm from one end.If both ends are kept at zero temperature, find the temperature functionu(x, t).



∂t= a2

∂2u

∂x2. (I)


where a2 =k

ρc=




(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.For this problem, the boundary and initial conditions are

(i)u(x = 0, t) = 0,∀t > 0

(ii)u(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x) = u0,∀x ∈ (0, `)

The suitable solution which satisfies above boundary conditions isu(x, t) = (A cos px+B sin px) e−a2p2t (1)

Applying condition (i) in equation (1) we getWe have

(i) ⇒ u(x = 0, t) = 0

(1) ⇒ Ae−a2p2t = 0


∴ A = 0


(ii) ⇒ u(x = `, t) = 0




∴ sin p` = 0

= sinnπ

∴ p =nπ

`


è−a2 n2π2

`2t

∴ By superposition principle(i.e., adding all such above solutions ), the



u(x, t) =∞∑n=1

Bn sinnπx

è−a2 n2π2

`2t (3)

Applying condition (iii) in equation (3), we getWe have (iii) ⇒ u(x, t = 0) = f(x) = u0

∴ (3) ⇒∞∑n=1

Bn sinnπx

`= f(x) = u0 (4)


f(x) =∞∑n=1

bn sinnπx

`(5)

where bn =2

`

`∫0

f(x) sinnπx

`dx

From (4) and (5), we have Bn = bn

∴ Bn =2

`

`∫0

f(x) sinnπx

`dx

=2

`

`∫0

u0 sinnπx

`dx =

2u0`

`∫0

sinnπx

`dx

=2u0`

[− cos nπx`

nπ`

]`0

=2u0nπ

[1− (−1)n]

∴ Bn =

{ 4u0nπ

, if n is odd

0, if n is even

∴ The required most general solution (3) is

u(x, t) =∞∑

n=odd

4u0nπ

sinnπx

è

−a2n2π2t`2

=4u0π

∞∑n=1

1

(2n− 1)sin

(2n− 1)πx

è

−a2(2n−1)2π2t

`2

Example 3.20. A rod of length ` is heated so that its ends A and B are atzero temperature. If initially its temperature is u = cx(`− x)/`2, find thetemperature function u(x, t).




∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=




(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9


(i)u(x = 0, t) = 0,∀t > 0

(ii)u(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x) =c(`x− x2

)`2

,∀x ∈ (0, `)



(i) ⇒ u(x = 0, t) = 0

(1) ⇒ Ae−a2p2t = 0


∴ A = 0


(ii) ⇒ u(x = `, t) = 0




∴ sin p` = 0

= sinnπ

∴ p =nπ

`



è−a2 n2π2

`2t


u(x, t) =∞∑n=1

Bn sinnπx

è−a2 n2π2

`2t (3)


We have (iii) ⇒ u(x, t = 0) = f(x) =c(`x− x2

)`2

∴ (3) ⇒∞∑n=1

Bn sinnπx

`= f(x) =

c(`x− x2

)`2

(4)


f(x) =∞∑n=1

bn sinnπx

`(5)

where bn =2

`

`∫0

f(x) sinnπx

`dx



∴ Bn =2

`

`∫0

f(x) sinnπx

`dx

=2

`

`∫0

c(`x− x2

)`2

sinnπx

`dx

=2c

`3

`∫0

(`x− x2

)sin

nπx

`dx

=2c

`3

[[`x−x2

] [− cos nπx`

nπ`

]−[`−2x]

[− sin nπx

`(nπ`

)2]+(−2)

[cos nπx

`(nπ`

)3]]`

0

=2c

`3

{− `

nπ

[`x−x2

] [cos

nπx

`

]+

(`

nπ

)2

[`−2x][sin

nπx

`

]−2

(`

nπ

)3 [cos

nπx

`

]}`

0

=2c

`3

{[−0 + 0− 2

(`

nπ

)3

(−1)n

]−

[−0 + 0− 2

(`

nπ

)3]}

=2c

`3

{2

(`

nπ

)3

[1− (−1)n]

}=

4c

n3π3[1− (−1)n]

∴ Bn =

{ 8c

n3π3, if n is odd

0, if n is even


u(x, t) =∞∑

n=odd

8c

n3π3sin

nπx

è

−a2n2π2t`2

=8c

π3

∞∑n=1

[1

(2n− 1)

]3sin

(2n− 1)πx

è

−a2(2n−1)2π2t

`2

3.3.3 Examples of TYPE (II)

Steady state conditions and zero boundary conditions


Example 3.21. A rod of length ` has its ends A and B kept at 0◦C and120◦C respectively until steady state conditions prevail. If the temperatureat B is reduced to 0◦C and kept so while that of A is maintained, find theresulting temperature distribution u(x, t) taking origin at A.



∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=




(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.First let us find the temperature distribution at any distance x, before

the end A and B are reduced to zero. Prior to the temperature changeat the ends A and B, when t = 0, the heat flow was independent of time(steady state conditions). When the temperature u depends only on x andnot on t,When steady state conditions prevail(I) reduces to

∂u

∂t= 0 ⇒ ∂2u

∂x2= 0 (II)

The general solution of (II) is u = ax+ b (III)where a and b are arbitrary constants.Given that u = 0 when x = 0∴ From (III), we get 0 = a(0) + b⇒ b = 0Also given that u = 120 when x = `

From (III),a`+ b = 120

a`+ 0 = 120

a =120

`x

∴ (III) becomes u =120

`x, which is initial temprature distribution.

After steady state conditionsWhen the temperature at A and B are reduced to 0◦C, the state is not

more steady state. For this transient state, the boundary conditions are


(i)u(x = 0, t) = 0,∀t > 0

(ii)u(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x) =120

`x, 0 < x < `



(i) ⇒ u(x = 0, t) = 0

(1) ⇒ Ae−a2p2t = 0


∴ A = 0


(ii) ⇒ u(x = `, t) = 0




∴ sin p` = 0

= sinnπ

∴ p =nπ

`


è−a2 n2π2

`2t


u(x, t) =∞∑n=1

Bn sinnπx

è−a2 n2π2

`2t (3)


We have (iii) ⇒ u(x, t = 0) = f(x) =120

`x, 0 < x < `

∴ (3) ⇒∞∑n=1

Bn sinnπx

`= f(x) =

120

`x (4)



f(x) =∞∑n=1

bn sinnπx

`(5)

where bn =2

`

`∫0

f(x) sinnπx

`dx


∴ Bn =2

`

`∫0

f(x) sinnπx

`dx

=2

`

`∫0

120

`x sin

nπx

`dx

=240

`2

[x

(− cos nπx`

nπ`

)− (1)

(− sin nπx

`(nπ`

)2)]`

0

=240

`2

[[−` `nπ

(−1)n + 0

]− (0 + 0)

]∴ Bn =

240

nπ(−1)n+1


u(x, t) =∞∑n=1

240

nπ(−1)n+1 sin

nπx

è

−a2n2π2t`2

Example 3.22. An insulated rod of length 30 cm has its ends A and Bkept at 20◦C and 80◦C respectively until steady state conditions prevail.The temperature at each end is then suddenly reduced to 0◦C and kept so.Find the resulting temperature distribution u(x, t) taking origin at A.



∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=




(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9



the end A and B are reduced to zero. Prior to the temperature change atthe ends A and B, when t = 0, the heat flow was independent of time(steady state conditions). When the temperature u depends only on xand not on t,When steady state conditions prevail(I) reduces to

∂u

∂t= 0 ⇒ ∂2u

∂x2= 0 (II)

The general solution of (II) is u = ax+ b (III)Where a and b are arbitrary constants.Given that u = 20 when x = 0∴ From (III), we get 20 = a(0) + b⇒ b = 20Also given that u = 80 when x = 30From (III), 30a+ b = 80

30a+ 20 = 80

a = 2∴ (III) becomes u = 2x+ 20, which is initial temprature distribution.

After steady state conditionsWhen the temperature at A and B are reduced to 0◦C, the state is not

more steady state. For this transient state, the boundary conditions are

(i)u(x = 0, t) = 0,∀t > 0

(ii)u(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x) = 2x+ 20, 0 < x < 30



(i) ⇒ u(x = 0, t) = 0

(1) ⇒ Ae−a2p2t = 0


∴ A = 0



(ii) ⇒ u(x = `, t) = 0


Here e−a2p2t 6= 0 (∵ it is defined for all


∴ sin p` = 0

= sinnπ

∴ p =nπ

`


è−a2 n2π2

`2t


u(x, t) =∞∑n=1

Bn sinnπx

è−a2 n2π2

`2t (3)

Applying condition (iii) in equation (3), we getWe have (iii) ⇒ u(x, t = 0) = f(x) = 2x+ 20, 0 < x < 30

∴ (3) ⇒∞∑n=1

Bn sinnπx

`= f(x) = 2x+ 20 (4)

To find Bn,Expand f(x) in a half range Fourier sine series in the interval (0, ` = 30)

f(x) =∞∑n=1

bn sinnπx

`(5)

where bn =2

`

`∫0

f(x) sinnπx

`dx



∴ Bn =2

`

`∫0

f(x) sinnπx

`dx =

2

`

`∫0

(2x+ 20) sinnπx

`dx

=2

`

[(2x+ 20)

(− cos nπx`

nπ`

)− (2)

(− sin nπx

`(nπ`

)2)]`

0

=2

`

[−(2`+ 20)

`

nπ(−1)n + 0

]−[−(20)

`

nπ+ 0

]=

2

`

[−(2`+ 20)

`

nπ(−1)n + (20)

`

nπ

]= 2

[−80

(−1)n

nπ+

20

nπ

](∵ ` = 30)

∴ Bn =40

nπ[1− 4 (−1)n]


u(x, t) =∞∑n=1

40

nπ[1− 4 (−1)n] sin

nπx

30e

−a2n2π2t900

3.3.4 Examples of TYPE (III)

Steady state conditions and non-zero boundary conditions

Example 3.23. Two ends A and B of rod of length 20 cm have thetemperatures at 30◦C and 80◦C respectively until steady state conditionsprevail. Then the temperature at the ends A and B changed to 40◦C and60◦C respectively. Find the resulting temperature distribution u(x, t)taking origin at A.



∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=




(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9



the end A and B are reduced to zero. Prior to the temperature changeat the ends A and B, when t = 0, the heat flow was independent of time(steady state conditions). When the temperature u depends only on x andnot on t,When steady state conditions prevail(I) reduces to

∂u

∂t= 0 ⇒ ∂2u

∂x2= 0 (II)

The general solution of (II) is u = ax+ b (III)Where a and b are arbitrary constants. Given that u = 30 when x = 0∴ From (III), we get 30 = a(0) + b⇒ b = 30Also given that u = 80 when x = 20From (III),

20a+ b = 80

20a+ 30 = 80

a =5

2

∴ (III) becomes u =5

2x+ 30 (IV)

After steady state conditions with non-zero boundaryconditionsWhen the temperature at A and B are reduced to 40◦C and 60◦C, the

state is not more steady state. For this transient state, the boundaryconditions are

(i)u(x = 0, t) = 40,∀t > 0

(ii)u(x = `, t) = 60,∀t > 0 where ` = 20

(iii)u(x, t = 0) = f(x) =5

2x+ 30, 0 < x < `

Here we have non zero boundary values. In such case, thetemperature function u(x, t) is given by

u(x, t) = us(x) + ut(x, t) (V)Where us(x) is a solution of (I) involving x only and satisfying the

boundary conditions (i) and (ii). ut(x, t) is a function defined by (V)satisfying (I).i.e., us(x) is a steady state solution of (I) and ut(x, t) may therefore


regarded as transient solution which decreases with increase of time t.us(x) satisfies (I).

To find us(x)The general solution of (I) is us(x) = a1x+ b1 (VI)Where a1 and b1 are arbitrary constants.

By the condition (i), we have u = 40 when x = 0From (VI), we get us(0) = b1 = 40By the condition (ii), we have u = 60 when x = `From (VI), us(`) = a1`+ b1 = 60

a1`+ 40 = 60

a1 =20

`

∴ (VI) becomes us(x) =20

`x+ 40 (VII)

To find ut(x, t) [zero boundary conditions]Hence the boundary conditions for the transient solution ut(x, t) by

using (V) are

(iv)ut(x = 0, t) = u(0, t)− us(0) = 40− 40 = 0,∀t > 0

(v)ut(x = `, t) = u(`, t)− us(`) = 60− 60 = 0,∀t > 0

(vi)ut(x, t = 0) = f(x) = u(x, 0)− us(x)

=

(5

2x+ 30

)−(20

`x+ 40

)=

(5

2x+ 30

)− (x+ 40) (∵ ` = 20)

=3

2x− 10, 0 < x < ` = 20

The suitable solution which satisfies above boundary conditions isut(x, t) = (A cos px+B sin px) e−a2p2t (1)

Applying condition (iv) in equation (1) we getWe have

(iv) ⇒ ut(x = 0, t) = 0

(1) ⇒ Ae−a2p2t = 0


∴ A = 0

∴ Now (1) ⇒ ut(x, t) = B sin pxe−a2p2t (2)Applying condition (v) in equation (2) we getWe have


(v) ⇒ ut(x = `, t) = 0

(2) ⇒ B sin pè−a2p2t = 0



∴ sin p` = 0

= sinnπ

∴ p =nπ

`


è−a2 n2π2

`2t


ut(x, t) =∞∑n=1

Bn sinnπx

è−a2 n2π2

`2t (3)

Applying condition (vi) in equation (3), we get

We have (vi) ⇒ ut(x, t = 0) = f(x) =3

2x− 10, 0 < x < ` = 20

∞∑n=1

Bn sinnπx

`= f(x) =

3

2x− 10 (4)

To find Bn,Expand f(x) in a half range Fourier sine series in the interval (0, ` = 20)

f(x) =∞∑n=1

bn sinnπx

`(5)

where bn =2

`

`∫0

f(x) sinnπx

`dx


∴ Bn =2

`

`∫0

f(x) sinnπx

`dx =

2

`

`∫0

(3

2x− 10

)sin

nπx

`dx

=2

`

[(3

2x− 10

)(− cos nπx`

nπ`

)−(3

2

)(− sin nπx`(

nπ`

)2)]`

0

=2

`

[(3`

2− 10

)[(`

nπ

)(− cosnπ)

]+ 0−

(10`

nπ+ 0

)]=

1

10

[−400

nπ(−1)n − 200

nπ

](∵ ` = 20)

∴ Bn = − 20

nπ[1 + 2(−1)n]



ut(x, t) =∞∑n=1

{− 20

nπ[1 + 2(−1)n]

}sin

nπx

è

−a2n2π2t`2

Hence u(x, t) = x+ 40− 20

π

∞∑n=1

{[1 + 2(−1)n]

n

}sin

nπx

è

−a2n2π2t`2

Example 3.24. * Two ends A and B of rod of length `cm have thetemperatures at 0◦C and 100◦C respectively until steady state conditionsprevail. Then the temperature at the ends A and B changed to 50◦C and150◦C respectively. Find the resulting temperature distribution u(x, t)taking origin at A. {

u(x, t) =100x

`+ 50− 200

π

∞∑n=odd

1

nsin

nπx

è

−a2n2π2t`2

}

3.3.5 Examples of TYPE (IV)

The end x = 0 is thermally insulated

Example 3.25. Find the solution of heat equation∂u

∂t= a2

∂2u

∂x2, such that

(i)∂u

∂x= 0 when x = 0 (ii) u = 0 when x = ` (iii) u = u0 when t = 0 for

all x between 0 and `.



∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=




(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9


(i)∂u

∂x(x = 0, t) = 0,∀t > 0

(ii)u(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x) = u0, 0 < x < `



Before applying the condition (i) in equation (1) we have todifferentiating equation (1) with respect to x∂u

∂x(x, t) = (−Ap sin px+Bp cos px) e−a2p2t (II)

Applying condition (i) in equation (II) we get

We have (i) ⇒ ∂u

∂x(x = 0, t) = 0

(II) ⇒ Bpe−a2p2t = 0


p = 0

∴ B = 0

∴ Now (1) ⇒ u(x, t) = A cos pxe−a2p2t (2)Applying condition (ii) in equation (2) we get

We have (ii) ⇒ u(x = `, t) = 0

(2) ⇒ A cos pè−a2p2t = 0


A 6= 0 (∵ it gives trivial solution )

∴ cos p` = 0

p` = an odd multiple ofπ

2

∴ p =

(2n− 1

`

)π

2, n = 1, 2, 3, . . .

∴ Now (2) ⇒ u(x, t) = A cos(2n− 1)πx

2è−a2 (2n−1)2π2

4`2t


u(x, t) =∞∑n=1

An cos(2n− 1)πx

2è−a2 (2n−1)2π2

4`2t (3)


∴ (3) ⇒∞∑n=1

An cos(2n− 1)πx

2`= f(x) = u0 (4)

To find An,Expand f(x) in a half range Fourier cosine series in the interval (0, `)


f(x) =∞∑n=1

an cosnπx

2`(5)

where an =2

`

`∫0

f(x) cosnπx

2`dx

From (4) and (5), we have An = an

∴ An =2

`

`∫0

f(x) cosnπx

2`dx =

2

`

`∫0

u0 cosnπx

`dx

=2u0`

`∫0

cosnπx

2`dx =

2u0`

[sin nπx

2`nπ2`

]`0

∴ An =4u0`

sinnπ

2

∴ The required most general solution

u(x, t) =∞∑

n=odd

4u0nπ

sinnπ

2cos

nπx

2è

−a2n2π2t4`2

=4u0π

∞∑n=1

sin (2n−1)π2

(2n− 1)cos

(2n− 1)πx

2è

−a2(2n−1)2π2t

4`2

3.3.6 Examples of TYPE VI

Both the ends are thermally insulated

Example 3.26. Find the solution of one dimensional diffusion equation∂u

∂t= a2

∂2u

∂x2satisfying the boundary conditions:

(i)u is bounded as t→ ∞

(ii)

[∂u

∂x

]x=0

= 0,∀t

(iii)

[∂u

∂x

]x=`

= 0,∀t

(iv)u(x, 0) = x(`− x),∀0 < x < `



∂t= a2

∂2u

∂x2. (I)

where a2 =k

ρc=


(Density)(Specific heat)



(1)u(x, t) =(A1e

px + A2e−px)A3e

a2p2t


(3)u(x, t) = (A7 + A8x)A9


(i)∂u

∂x(x = 0, t) = 0,∀t > 0

(ii)∂u

∂x(x = `, t) = 0,∀t > 0

(iii)u(x, t = 0) = f(x) = x(`− x), 0 < x < `


Before applying the condition (i) in equation (1) we have differentiatingequation (1) with respect to x∂u

∂x(x, t) = (−Ap sin px+Bp cos px) e−a2p2t (II)

Applying condition (i) in equation (II) we getWe have

(i) ⇒ ∂u

∂x(x = 0, t) = 0

(II) ⇒ Bpe−a2p2t = 0


p 6= 0

∴ B = 0

∴ Now (1) ⇒ u(x, t) = A cos pxe−a2p2t (2)Before applying the condition (ii) in equation (2) we have differentiating

equation (2) with respect to x∂u

∂x(x, t) = (−Ap sin px) e−p2a2t (III)

Applying condition (ii) in equation (III) we getWe have


(ii) ⇒ ∂u

∂x(x = `, t) = 0

(2) ⇒ −Ap sin pè−a2p2t = 0


A 6= 0 (∵ it gives trivial solution )

p 6= 0

∴ sin p` = 0

= sinnπ

∴ p =nπ

`

∴ Now (2) ⇒ u(x, t) = A cosnπx

è

−a2n2π2

`2t


u(x, t) = A0 +∞∑n=1

An cosnπx

è

−a2n2π2

`2t (3)


∴ (3) ⇒ A0 +∞∑n=1

An cosnπx

`= f(x) = x(`− x) (4)

To find A0 and An,Expand f(x) in a half range Fourier cosine series in the interval (0, `)

f(x) =a02+

∞∑n=1

an cosnπx

`(5)

where

a0 =2

`

`∫0

f(x)dx, an =2

`

`∫0

f(x) cosnπx

`dx

From (4) and (5), we have A0 =a02 , An = an


∴ a0 =2

`

`∫0

f(x)dx =2

`

`∫0

x(`− x)dx

=2

`

`∫0

(`x− x2)dx =2

`

[`x2

2− x3

3

]`0

=2

`

[(`3

2− `3

3

)− (0− 0)

]=

2

``3(1

2− 1

3

)= 2`2

(1

6

)=`2

3


∴ an =2

`

`∫0

f(x) cosnπx

`dx =

2

`

`∫0

x(`− x) cosnπx

`dx

=2

`

`∫0

(`x− x2) cosnπx

`dx

=2

`

[(`x− x2)

(sin nπx

`nπ`

)− (`− 2x)

(− cos nπx

`(nπ`

)2)

+(−2)

(− sin nπx

`(nπ`

)3)]`

0

=2

`

[(`x− x2)

`

nπsin

nπx

`+ (`− 2x)

(`

nπ

)2

cosnπx

`

+2

(`

nπ

)3

sinnπx

`

]`0

=2

`

[(0 + (−`)

(`

nπ

)2

(−1)n + 0

)−

(0 + `

(`

nπ

)2

+ 0

)]

=2

`

[−`(`

nπ

)2

(−1)n − `

(`

nπ

)2]

=2

`

[−`(`

nπ

)2

[(−1)n + 1]

]

∴ an =

−4

(`

nπ

)2

if n is even

0 if n is odd

∴ A0 =a02

=

(a2/3

)2

=a2

6

∴ An =

−4

(`

nπ

)2

if n is even

0 if n is odd∴ The required most general solution (3) is

u(x, t) =a2

6+

∞∑n=even

−4`2

n2π2cos

nπx

è

−a2n2π2t`2

=a2

6− −4`2

π2

∞∑n=even

1

n2cos

nπx

è

−a2n2π2t`2


Example 3.27. * Find the solution of one dimensional diffusion equation∂u

∂t= a2

∂2u

∂x2satisfying the boundary conditions:

(i)u is bounded as t→ ∞

(ii)

[∂u

∂x

]x=0

= 0,∀t

(iii)

[∂u

∂x

]x=`

= 0,∀t

(iv)u(x, 0) = kx, ∀0 < x < `{u(x, t) =

k`

2− −4`k

π2

∞∑n=odd

1

n2cos

nπx

è

−a2n2π2t`2

}


1. A rod of length ` is heated so that its ends A and B are at zerotemperature. If initially its temperature is u = (`x − x2), find thetemperature function u(x, t).[

Ans: u(x, t) =8`2

π2

∞∑n=1

1

(2n− 1)2sin

(2n− 1)πx

è

−a2(2n−1)2π2t

`2

]2. The equation for the conduction of heat along a bar of length ‘`’ is∂u

∂t= a2

∂2u

∂x2neglecting radiation. Find the temperature u(x, t) if the

ends of the bar are maintained at zero temperature and if initially thetemperature is T at the center of the bar and falls uniformly to zeroat its ends.[

Ans: u(x, t) =8T

π2

∞∑n=1

(−1)n+1

(2n− 1)2sin

(2n− 1)πx

è

−a2(2n−1)2π2t

`2

]

3. Determine the solution of one dimensional heat equation∂u

∂t= a2

∂2u

∂x2subject to the conditions u(0, t) = 0 = u(1, t) andu(x, 0) = 3sinnπx; 0 < x < 1, t > 0.[

Ans: u(x, t) = 3∞∑n=1

sinnπxe−n2π2t

]4. The ends A and B of a rod 20 cm long have the temperatures at 0◦C

and 80◦C until steady state conditions prevail. If the temperature atB is reduced to 0◦C and kept so while that of A is maintained, findthe temperature distribution in the rod at any subsequent time.

MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 213[Ans: u(x, t) =

200

π

∞∑n=1

(−1)n+1

nsin

nπx

è

−a2n2π2t`2

]5. A bar, 10 cm long, with insulated sides, has its ends A and B kept

at 20◦C and 40◦C respectively, until steady state conditions prevail.The temperature at the ends A is then suddenly raised to 50◦C and atthe same instant that at B is lowered to 10◦C. Find the subsequenttemperature function u(x, t).[

Ans: u(x, t) = 50− 4x− 60

π

∞∑n=1

1

nsin

nπx

5e

−a2n2π2t25

]6. A bar, 10 cm long, with insulated sides, has its ends A and B kept

at 50◦C and 100◦C respectively, until steady state conditions prevail.The temperature at the ends A is then suddenly raised to 90◦C and atthe same instant that at B is lowered to 60◦C. Find the temperatureat a distance x from one end at time t.[

Ans: u(x, t) = −3x+ 90− 80

π

∞∑n=1

1

nsin

nπx

5e

−a2n2π2t25

]


3.4 Two dimensional heat flow equations

IntroductionIn unsteady state, the two dimensional heat flow equation is

∂u

∂t= a2

(∂2u

∂x2+∂2u

∂y2

), where a2 is the diffusivity of the material and

temperature function is u(x, y, t).

In steady state (i.e., time is independent),∂u

∂t= 0 i.e.,

∂2u

∂x2+∂2u

∂y2(or)

Laplace equation in two dimensional heat flow equation and temperaturefunction is u(x, y).Note : Laplace equation in three dimensional is

∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= 0 ⇒ uxx + uyy + uzz = 0

Solution of the Laplace equation in two dimensional plates

The one dimensional heat flow equation is∂2u

∂x2+∂2u

∂y2= 0 (1)

where u is a function of x and y.Consider the trial solution of u(x, y) in (1) as

u(x, y) = XY (2)

= X(x)Y (y)where X is a function x alone and Y is a function of y alone.

Now differentiating (2) w.r.t. x & y partially, we get

∂u

∂x= X ′Y

∂u

∂y= XY ′

∂2u

∂x2= X ′′T

∂2u

∂y2= XY ′′

∴ The equation (1) ⇒ X ′′Y −XY ′′ = 0By variable separable method, we getX ′′

X= −Y

′′

Y(3)

In (3), LHS is a function X alone and RHS is a function of Y alone. HereX and Y are independent, so each ratio is equal to a constant.

i.e.,X ′′

X= −Y

′′

Y= k (say)

X ′′

X= k −Y

′′

Y= k

X ′′ − kX = 0 (4) Y ′′ + kY = 0 (5)


Now there are 3 cases for the value of k.

Case I: Let k be positive , say k = p2, then

(4) ⇒ X ′′ − p2X = 0 (5) ⇒ Y ′′ + p2Y = 0


m2 − p2 = 0 m2 + p2 = 0∴ m = ±p ∴ m = ±pi∴ Solution of X is ∴ Solution of Y is

X = C1epx + C2e

−px Y = C3 cos py + C4 sin py


u(x, y) = XY

=(C1e

px + C2e−px)(C3 cos py + C4 sin py)

Case II: Let k be negative , say k = −p2, then

(4) ⇒ X ′′ + p2X = 0 (5) ⇒ Y ′′ − p2Y = 0


m2 + p2 = 0 m2 − p2 = 0∴ m = ±pi ∴ m = ±p∴ Solution of X is ∴ Solution of Y is

X=C5 cos px+C6 sin px Y =C7epy + C8e

−py


u(x, t) = XY

= (C5 cos px+ C6 sin px)(C7e

py + C8e−py)

Case III: Let k = 0, then

(4) ⇒ X ′′ = 0 (5) ⇒ Y ′′ = 0


m2 = 0 m2 = 0∴ m = 0( twice ) ∴ Y = 0( twice )∴ Solution of X is X=C9+C10x ∴ Solution of Y is Y =C11+C12y


u(x, y) = XY

= (C9 + C10x) (C11 + C12y)


∴ The three possible solution of (1) are

(1)u(x, t) =(C1e


(2)u(x, t) = (C5 cos px+ C6 sin px)(C7e

py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)

Types of boundary value problems in 2D heat flow equations

# Finite Plates # Infinite Plates

(I) f(x) 6= 0 parallel to X axis (IV) f(x) 6= 0 parallel to X axis

(III) f(y) 6= 0 parallel to Y axis (V) f(y) 6= 0 parallel to Y axis

(III) f(x) 6= 0 parallel to X axis&f(y) 6= 0 parallel to Y axis

Hierarchy to take 4 boundary conditions for 2D heat equations

1. Take the (iv) boundary condition for u 6= 0.

2. Take the (iii) boundary condition for opposite to u 6= 0.

3. Take the (i) boundary condition for remaining axis among X or Y .

4. Take the (ii) boundary condition for the side opposite to (i) boundarycondition.

Suitable Solution according to X or Y axis

u(x, y) 6= 0 Suitable Solution

Parallel to X axis u(x, y) = (A cos px+B sin px)(Cepy +De−py

)Parallel to Y axis u(x, y) =

(Aepy +Be−py

)(C cos px+D sin px)

3.4.1 Examples of two dimensional finite plates

3.4.2 Examples of Type I : f(x) 6= 0 parallel to X axis

Example 3.28. Solve for the steady state temperature at any point of arectangular plate of sides a and b insulated on the lateral surface and satisfyu(0, y) = 0, u(a, y) = 0, u(x, b) = 0 and u(x, 0) = x(a− x).

Solution :


Let u(x, y) be the temperature distribution satisfying the two dimensionalheat flow equation in steady state conditions is

∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for 0 < y < b

(ii)u(x = a, y) = 0 for 0 < y < b

(iii)u(x, y = b) = 0 for 0 < x < a

(iv)u(x, y = 0) = f(x) = x(a− x)

=(ax− x2

)for 0 < x < a

The suitable solution which satisfies above boundary conditions isu(x, y) = (A cos px+B sin px)

(Cepy +De−py

)(1)


We have (i) ⇒ u(x = 0, y) = 0

(1) ⇒ A(Cepy +De−py

)= 0

Here Cepy +De−py 6= 0 (∵ it is defined ∀y)∴ A = 0


∴ Now (1) ⇒ u(x, y) = B sin px(Cepy +De−py

)(2)

Applying condition (ii) in equation (2) we get

We have (ii) ⇒ u(x = a, y) = 0

(2) ⇒ B sin pa(Cepy +De−py

)= 0

Here Cepy +De−py 6= 0 (∵ it is defined ∀y)∴ B 6= 0 (∵ it gives trivial solution )

sin pa = 0

= sinnπ

∴ pa = nπ

i.e., p =nπ

a

∴ Now (2) ⇒ u(x, y) = B sin(nπx

a

)(Ce

nπya +De−

nπa y)

(3)


We have (iii) ⇒ u(x, y = b) = 0

(3) ⇒ B sin(nπx

a

)(Ce

nπa b+De−

nπa b)= 0

Here sin(nπx

a

)6= 0 (∵ it is defined ∀x)


∴ Cenπa b +De−

nπa b = 0

De−nπa b = −Ce

nπa b

D = −C enπa b

e−nπa b

∴ (3) ⇒ u(x, y) = B sin(nπx

a

)(Ce

nπa y − C

enπa b

e−nπa be−

nπya

)= BC sin

(nπxa

)(enπa (y−b) − e

nπa (b−y)

e−nπa b

)= BCe

nπa b sin

(nπxa

)(e−

nπa (b−y) − e

nπa (b−y)

)= −BCe

nπa b sin

(nπxa

)(e

nπa (b−y) − e−

nπa (b−y)

)= −2BCe

nπa bsin

(nπxa

)sinh

nπ

a(b− y)



u(x, y) =∞∑n=1

Bn sin(nπx

a

)sinh

nπ

a(b− y) (4)


We have (iv) ⇒ u(x, y = 0) = f(x) =(ax− x2

)(4) ⇒

∞∑n=1

Bn sin(nπx

a

)sinh

nπ

ab = f(x)

∴∞∑n=1

An sin(nπx

a

)= f(x) (5)

An = Bn sinhnπ

ab

To find An,Expand f(x) in a half range Fourier sine series in the interval (0, a)

f(x) =∞∑n=1

bn sinnπx

a(6)

where bn =2

a

a∫0

f(x) sinnπx

adx

From (5) and (6), we have An = bn


An =2

a

a∫0

f(x) sinnπx

adx

=2

a

a∫0

(ax− x2

)sin

nπx

adx

=2

a

[(ax−x2)

(− cos nπxa

nπa

)−(a−2x)

(− sin nπx

a(nπa

)2)+(−2)

(cos nπx

a(nπa

)3)]a

0

=2

a

[(ax−x2) a

nπsin

nπx

a+(a− 2x)

( a

nπ

)2cos

nπx

`+2

(`

nπ

)3

sinnπx

`

]a0

=2

a

{[0 + 0− 2

( a

nπ

)3(−1)n

]−[0 + 0− 2

( a

nπ

)3]}=

2

a

[2( a

nπ

)3][1− (−1)n]

=4a2

n3π3[1− (−1)n]

∴ An =

8a2

n3π3if n is odd

0 if n is even

∴ Bn =An

sinh nπba

=

8a2

n3π3sinh

nπb

aif n is odd

0 if n is even


u(x, y) =∞∑

n=odd

8a2

n3π3sinh

nπb

asin

nπx

asinh

nπ

a(b− y)

=8a2

π3

∞∑n=odd

1

n3sinh

nπb

asin

nπx

asinh

nπ

a(b− y)

=8a2

π3

∞∑n=1

sin (2n−1)πxa

(2n− 1)3 sinh nπba

sinh(2n− 1)π

a(b− y)

Example 3.29. Find the steady state temperature distribution in arectangular plate of sides 20 cm and 10 cm insulated on the lateralsurface and maintained at zero temperature along three of its edges and attemperature 100◦C along the remaining longer edge.

Solution :



∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for 0 < y < b

(ii)u(x = 20, y) = 0 for 0 < y < b

(iii)u(x, y = b) = 0 for 0 < x < 20

(iv)u(x, y = 0) = f(x) = 100 for 0 < x < 20


(Cepy +De−py

)(1)


We have (i) ⇒ u(x = 0, y) = 0


)= 0



)(2)



We have (ii) ⇒ u(x = 20, y) = 0

(2) ⇒ B sin 20p(Cepy +De−py

)= 0

Here Cepy +De−py 6= 0 (∵ it is defined ∀y)B 6= 0 (∵ it gives trivial solution )

∴ sin 20p = 0

= sinnπ

∴ 20p = nπ

i.e., p =nπ

20

∴ Now (2) ⇒ u(x, y) = B sin(nπ20x)(Ce

nπy20 +De−

nπy20

)(3)


We have (iii) ⇒ u(x, y = 0) = 0

(3) ⇒ B sin(nπx

20

)(C +D) = 0

Here sin(nπ20x)6= 0 (∵ it is defined ∀x)


∴ C +D = 0

i.e., D = −C


20

)(Ce

nπy20 − Ce−

nπy20

)= BC sin

(nπx20

)(e

nπy20 − e−

nπy20

)= 2BC sin

(nπx20

)sinh

nπy

20

= Cn sin(nπ20x)sinh

nπ

20y

where Cn = 2BC∴ By superposition principle(i.e., adding all such above solutions ), the


u(x, y) =∞∑n=1

Cn sin(nπx

20

)sinh

nπy

20(4)



We have (iv) ⇒ u(x, y = 10) = f(x)

(4) ⇒∞∑n=1

Cn sin(nπ20x)sinh

nπ

2= f(x)

∴∞∑n=1

Bn sin(nπ20x)= f(x) (5)

where Bn = Cn sinhnπ

2To find Bn,Expand f(x) in a half range Fourier sine series in the interval (0, 20)

f(x) =∞∑n=1

bn sinnπx

20(6)

where bn =2

20

20∫0

f(x) sinnπx

20dx


Bn =2

20

20∫0

100 sinnπx

20dx = 10

20∫0

sinnπx

20dx

= 10

[− cos nπx20

nπ20

]200

= −10

(20

nπ

)[cosnπ − 1] = −200

nπ[(−1)n − 1]

∴ An =

{ 400

nπif n is odd

0 if n is even

∴ Cn =Bn

sinh nπ2

=

400

nπ

1

sinh nπ2

if n is odd

0 if n is even∴ The required most general solution (4) is

u(x, y) =∞∑

n=odd

400

nπ

1

sinh nπ2

sinnπx

20sinh

nπ

20y

=400

π

∞∑n=1

sin (2n−1)πx20

(2n− 1) sinh (2n−1)π2

sinh(2n− 1)π

20y


Example 3.30. Solve∂2u

∂x2+∂2u

∂y2= 0 which satisfies the conditions

u(0, y) = u(a, y) = u(x, 0) = 0 and u(x, b) = sin πxa .

Solution : Let u(x, y) be the temperature distribution satisfying thetwo dimensional heat flow equation in steady state conditions is

∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for 0 < y < b

(ii)u(x = a, y) = 0 for 0 < y < b

(iii)u(x, y = 0) = 0 for 0 < x < a

(iv)u(x, y = b) = f(x) = sinπx

afor 0 < x < a


(Cepy +De−py

)(1)

Applying condition (i),(ii),(iii) as in previous problem, we get∴ By superposition principle(i.e., adding all such above solutions ), the


u(x, y) =∞∑n=1

Cn sin(nπx

a

)sinh

nπy

a(4)

Applying condition (iv) in equation (4), we getWe have (iv) ⇒ u(x, y = b) = f(x)

(4) ⇒∞∑n=1

Cn sin(nπx

a

)sinh

nπb

a= f(x) = sin

πx

a

To find Bn,Expand the summation, we get

sinπx

a= C1 sin

πx

asinh

πb

a+ C2 sin

2πx

asinh

2πb

a+ · · ·

Comparing the like coefficients, we have


1 = C1 sinhπb

aand C2 = C3 = · · · = 0

∴ C1 =1

sinh πba

The required most general solution (4) is

u(x, y) =1

sinh πba

sinπx

asinh

πy

a

3.4.3 Egs. of Type II(f(y) 6= 0 parallel to Y axis) and III(f(x) 6= 0 & f(y) 6= 0)

Example 3.31. Solve Find the steady state temperature distribution at apoint of a rectangular plate, if, two adjacent edges are kept at 0◦C andother at 100◦C.

Solution : Let X & Y axes with 0◦C and X = a & Y = b with 100◦C.Then


∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)

out of these solutions, we have to select that solution which suits thephysical nature of the problem and the boundary conditions.Let the temperature along the adjacent edges AB and BC be 100◦C and

the temperature along the other edges OC and OA be 0◦C.The Boundary conditions are


(i)u(x = 0, y) = 0 for 0 < y < b

(ii)u(x, y = 0) = 0 for 0 < x < a

(iii)u(x = a, y) = 100 for 0 < y < b

(iv)u(x, y = b) = 100 for 0 < x < a

Where u1(x, y) and u2(x, y) are solutions of (I) and further u1(x, y) is thetemperature at the edge BC kept at 100◦C and the other three sides at0◦C while u2(x, y) is the temperature at the edge AB maintained at 100◦Cand the other three edges at 0◦C.Boundary conditions for the function u1(x, y) and u2(x, y) as A and B

are

A B

(i) u1(x=0, y) = 0 in 0<y<b (v) u2(x, y=0) = 0 in 0<x<a

(ii) u1(x=a, y) = 0 in 0<y<b (vi) u2(x, y=b) = 0 in 0<x<a

(iii) u1(x, y=0) = 0 in 0<x<a (vii) u2(x=0, y) = 0 in 0<y<b

(iv) u1(x, y=b) = 100 in 0<x<a (viii) u2(x=a, y) = 100 in 0<y<b

The suitable solution which satisfies above boundary conditions(A)is

u1(x, y) = (A cos px+B sin px)(Cepy +De−py

)(1)



u(x, y) =∞∑n=1

Cn sin(nπx

a

)sinh

nπy

a(4)



We have (iv) ⇒ u(x, y = b) = f(x) = 100

(4) ⇒∞∑n=1

Cn sin(nπx

a

)sinh

nπb

a= f(x) = 100

∴∞∑n=1

Bn sin(nπx

a

)= f(x) = 100 (5)

where Bn = Cn sinhnπb

aTo find Bn,Expand f(x) in a half range Fourier sine series in (0, a), we get

f(x) =∞∑n=1

bn sinnπx

a(6)

where bn =2

a

a∫0

f(x) sinnπx

adx


Bn =2

a

a∫0

100 sinnπx

adx =

200

a

a∫0

sinnπx

adx

=200

a

[− cos nπxa

nπa

]a0

= −(200

nπ

)[cosnπ − 1] = −200

nπ[(−1)n − 1]

∴ Bn =

{ 400

nπif n is odd

0 if n is even

∴ Cn =Bn

sinh nπba

=

400

nπ

1

sinh nπba

if n is odd

0 if n is even∴ The required solution (4) is

u1(x, y) =∞∑

n=odd

400

nπ

1

sinh nπba sin nπx

a sinh nπya

=400

π

∞∑n=1

sin (2n−1)πxa

(2n− 1) sinh (2n−1)πba

sinh(2n− 1)π

ay

This is the required solution of u1(x, y).Similarly, the suitable solution which satisfies above boundary


conditions(B)is

u2(x, y) =(Aepx +Be−px

)(C cos py +D sin py) (7)

Similarly by applying condition (v),(vi),(vii) and (viii), we get therequired solution of u2(x, y) as

u2(x, y) =∞∑

n=odd

400

nπ

1

sinh nπab sinh nπa

b sin nπyb sinh nπx

b

=400

π

∞∑n=1

sin (2n−1)πyb

(2n− 1) sinh (2n−1)πab

sinh(2n− 1)π

bx

This is the required solution of u2(x, y).∴ Finally, The required most general solution =u(x, y) = u1(x, y) + u2(x, y).

3.5 Examples of two dimensional infinite plates

3.5.1 Examples of Type (IV) : f(x) 6= 0 parallel to X axis

Example 3.32. An infinitely long plane uniform plate is bounded by twoparallel edges x = 0 and x = ` and an end at right angles to them. Thebreadth of this edge y = 0 is ` and is maintained at temperature f(x).All the other three edges are at temperature zero. Find the steady statetemperature.

Solution :

Let u(x, y) be the temperature at any point P (x, y) in the steady state.Then u(x, y) satisfying the two dimensional heat flow equation in steady


state conditions is∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for all y

(ii)u(x = `, y) = 0 for all y

(iii)u(x, y → ∞) = 0 for 0 < x < `

(iv)u(x, y = 0) = f(x) for 0 < x < `

Now u(0, y) = 0 i.e., [(i) condition] is not satisfied by (1) unless C1 = 0and C2 = 0 which leads to the trivial solution u = 0. u = 0 as y → ∞[condition (iii)] is not satisfied by (3). Hence we choose (2) as the solutionof the differential equation. (i.e.,)The suitable solution which satisfies above boundary conditions is

u(x, y) = (A cos px+B sin px)(Cepy +De−py

)(1)


We have (i) ⇒ u(x = 0, y) = 0


)= 0



)(2)


We have (ii) ⇒ u(x = `, y) = 0

(2) ⇒ B sin p`(Cepy +De−py

)= 0


∴ sin p` = 0

= sinnπ

∴ p =nπ

`



`

)(Ce

nπy` +De−

nπy`

)(3)


We have (iii) ⇒ u(x, y → ∞) = 0

(3) ⇒ B sin(nπx

`

) (Ce∞ +De−∞) = 0

Here sin(nπx

`



∴ C = 0 (∵ e∞ = ∞)


`

)(De−

nπy`

)= BD sin

(nπx`

)(e−

nπy`

)= E sin

nπx

è−

nπy`

where E = BD∴ By superposition principle(i.e., adding all such above solutions ), the


u(x, y) =∞∑n=1

En sin(nπx

`

)e−

nπy` (4)


We have (iv) ⇒ u(x, y = 0) = f(x)

(4) ⇒∞∑n=1

En sinnπx

`= f(x) (5)

To find En,Expand f(x) in a half range Fourier sine series in the interval (0, `)

f(x) =∞∑n=1

bn sinnπx

`(6)

where bn =2

`

`∫0

f(x) sinnπx

`dx

From (5) and (6), we have En = bn.Substituting this value in (4), we get the required most general solution.

Example 3.33. A long rectangular plate has its surfaces insulated and thetwo long sides as well as one of the short sides are maintained at 0◦C.Find an expression for the steady state temperature u(x, y) if the short sidey = 0 is 50 cm long and is kept at 60◦C.


Solution :

Let u(x, y) be the temperature at any point P (x, y) in the steady state.Then u(x, y) satisfying the two dimensional heat flow equation in steadystate conditions is

∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for all y

(ii)u(x = `, y) = 0 for all y

(iii)u(x, y → ∞) = 0 for 0 < x < `

(iv)u(x, y = 0) = f(x) = 60 for 0 < x < `, where ` = 50


(Cepy +De−py

)(1)


We have (i) ⇒ u(x = 0, y) = 0


)= 0




)(2)


We have (ii) ⇒ u(x = `, y) = 0

(2) ⇒ B sin p`(Cepy +De−py

)= 0


∴ sin p` = 0

= sinnπ

∴ p =nπ

`


`

)(Ce

nπy` +De−

nπy`

)(3)


We have (iii) ⇒ u(x, y → ∞) = 0

(3) ⇒ B sin(nπx

`

) (Ce∞ +De−∞) = 0

Here sin(nπx

`



∴ C = 0 (∵ e∞ = ∞)


`

)(De−

nπy`

)= BD sin

(nπx`

)(e−

nπy`

)= E sin

nπx

è−

nπy`



u(x, y) =∞∑n=1

En sin(nπx

`

)e−

nπy` (4)


We have (iv) ⇒ u(x, y = 0) = f(x)

(4) ⇒∞∑n=1

En sinnπx

`= f(x) (5)

To find En,Expand f(x) in a half range Fourier sine series in the interval (0, ` = 50)


f(x) =∞∑n=1

bn sinnπx

`(6)

where bn =2

`

`∫0

f(x) sinnπx

`dx

From (5) and (6), we have En = bn

En =2

`

`∫0

f(x) sinnπx

`dx =

2

`

`∫0

60 sinnπx

`dx

=120

`

`∫0

sinnπx

`nxdx

=120

`

[−cosnπx`nπ`

]`0

dx

=−120

nπ[cosnπ − 1] = −120

nπ[(−1)n − 1]

∴ En =

{ 240

nπif n is odd

0 if n is even


u(x, y) =∞∑

n=odd

240

nπsin

nπx

è−

nπy`

=240

π

∞∑n=1

1

nsin

(2n− 1)πx

è−

(2n−1)πy` , where ` = 50

Example 3.34. An infinite long plate bounded by two parallel edges andan end at right angles to them. The breadth is π. This end is maintainedat a constant temperature u0 at all the points and the edges are at zerotemperature. Find the steady state temperature at any point (x, y) of theplate.

Solution :



∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for all y

(ii)u(x = π, y) = 0 for all y

(iii)u(x, y → ∞) = 0 for 0 < x < π

(iv)u(x, y = 0) = f(x) = u0 for 0 < x < π


(Cepy +De−py

)(1)


We have (i) ⇒ u(x = 0, y) = 0


)= 0




)(2)


We have (ii) ⇒ u(x = π, y) = 0

(2) ⇒ B sin pπ(Cepy +De−py

)= 0


∴ sin pπ = 0

= sinnπ

∴ p = n

∴ Now (2) ⇒ u(x, y) = B sinnx(Ceny +De−ny

)(3)


We have (iii) ⇒ u(x, y → ∞) = 0

(3) ⇒ B sinnx(Ce∞ +De−∞) = 0

Here sinnx 6= 0 (∵ it is defined ∀x)B 6= 0 (∵ it gives trivial solution )

∴ C = 0 (∵ e∞ = ∞)

∴ Now (3) ⇒ u(x, y) = B sinnx(De−ny

)= BD sinnx

(e−ny

)= E sinnxe−ny



u(x, y) =∞∑n=1

En sinnxe−ny (4)


We have (iv) ⇒ u(x, y = 0) = f(x) = u0

(4) ⇒∞∑n=1

En sinnx = f(x) = u0 (5)

To find En,Expand f(x) in a half range Fourier sine series in the interval (0, π)

f(x) = u0 =∞∑n=1

bn sinnx (6)


where bn =2

π

π∫0

f(x) sinnxdx


En =2

π

π∫0

f(x) sinnxdx

=2

π

π∫0

u0 sinnxdx

=2u0π

[− cosnx

n

]π0

dx

=−2u0nπ

[cosnπ − 1] = −2u0nπ

[(−1)n − 1]

∴ En =

{ 4u0nπ

if n is odd

0 if n is even


u(x, y) =∞∑

n=odd

4u0nπ

sinnxe−ny

=4u0π

∞∑n=1

1

nsin[(2n− 1)x]e−(2n−1)y

Example 3.35. A rectangular plate with insulated surfaces is 10 cm wideand so long compared to its width that it may be considered infinite inlength without introducing an appreciable error. If the temperature alongthe short edge y = 0 is given by

u(x, 0) =

{20x, 0 < x < 5

20(10− x), 5 < x < 10

while the two long edges x = 0 and x = 10 as well as the other short edgeare kept at 0◦C, find the temperature function u(x, y) in steady state.

Solution :



∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)


(i)u(x = 0, y) = 0 for all y

(ii)u(x = 10, y) = 0 for all y

(iii)u(x, y → ∞) = 0 for 0 < x < 10

(iv)u(x, y = 0) = f(x) =

{20x, 0 < x < 5

20(10− x), 5 < x < 10


(Cepy +De−py

)(1)



u(x, y) =∞∑n=1

En sin(nπ10x)e−

nπy10 (4)


Applying condition (iv) in equation (4), we getWe have (iv) ⇒ u(x, y = 0) = f(x)

(4) ⇒∞∑n=1

En sin(nπx

`

)= f(x) =

{20x, 0 < x < 5

20(10− x), 5 < x < 10(5)

To find En,Expand f(x) in a half range series in (0, 10), we have

f(x) =∞∑n=1

bn sinnπx

10(6)

where bn =2

10

10∫0

f(x) sinnπx

10dx


En =2

10

10∫0

f(x) sinnπx

10dx

=1

5

5∫0

20x sinnπx

10dx+

10∫5

20(10− x) sinnπx

10dx

= 4

5∫0

x sinnπx

10dx+

10∫5

(10− x) sinnπx

10dx

= 4

[x10

nπ

(− cos

nπx

10

)− (1)

(10

nπ

)2 (− sin

nπx

10

)]50

+

[(10− x)

10

nπ

(− cos

nπx

10

)− (−1)

(10

nπ

)2 (− sin

nπx

10

)]105

= 4

{[− 50

nπ

(cos

nπ

2

)+

100

n2π2

(sin

nπ

2

)]− [0 + 0]

+ [0 + 0]−[−50

nπ

(cos

nπ

2

)− 100

n2π2

(sin

nπ

2

)]}∴ En =

800

n2π2sin

nπ

2



u(x, y) =∞∑n=1

800

n2π2sin

nπ

2sin

nπx

10e−

nπy10

=800

π2

∞∑n=1

1

n2sin

nπ

2sin

nπx

10e−

nπy10

3.5.2 Examples of Type V : f(y) 6= 0 parallel to Y axis

Example 3.36. An infinitely long rectangular plate with insulated surfaceis 10 cm wide. The two long edges and one short edge are kept at zerotemperature, while the other short edge x = 0 is kept at temperature

u =

{20y, 0 < y < 5

20(10− y), 5 < y < 10. Find the temperature function u(x, y) in

steady state.

Solution :


∂2u

∂x2+∂2u

∂y2= 0 (I)


(1)u(x, t) =(C1e



py + C8e−py)

(3)u(x, t) = (C9 + C10x) (C11 + C12y)



(i)u(x, y = 0) = 0 for all x

(ii)u(x, y = 10) = 0 for all x

(iii)u(x→ ∞, y) = 0 for 0 < y < 10

(iv)u(x = 0, y) = f(y) =

{20y, 0 < y ≤ 5

20(10− y), 5 < y ≤ 10

The suitable solution which satisfies above boundary conditions isu(x, y) =

(Aepx +Be−px

)(C cos py +D sin py) (1)


We have (i) ⇒ u(x, y = 0) = 0

(1) ⇒(Aepx +Be−px

)C = 0

Here Aepx +Be−px 6= 0 (∵ it is defined ∀x)∴ C = 0

∴ Now (1) ⇒ u(x, y) =(Aepx +Be−px

)D sin py (2)


We have (ii) ⇒ u(x, y = 10) = 0

(2) ⇒(Aepx +Be−px

)D sin 10p = 0

Here Aepx +Be−px 6= 0 (∵ it is defined ∀x)D 6= 0 (∵ it gives trivial solution )

∴ sin 10p = 0

= sinnπ

∴ p =nπ

10

∴ Now (2) ⇒ u(x, y) =(Ae

nπx10 +Be−

nπx10

)D sin

nπy

10(3)


We have (iii) ⇒ u(x→ ∞, y) = 0

(3) ⇒(Ae∞ +Be−∞)D sin

nπy

10= 0

(Ae∞)D sinnπy

10= 0

Here sinnπy

106= 0 (∵ it is defined ∀y)


∴ A = 0 (∵ e∞ = ∞)


∴ Now (3) ⇒ u(x, y) =(Be

nπx10

)(Dsinnπy

10

)= BD sin

nπy

10

(e−

nπx10

)= E sin

nπy

10e−

nπx10



u(x, y) =∞∑n=1

En sinnπy

10e−

nπx10 (4)


We have (iv) ⇒ u(x = 0, y) = f(y)

(4) ⇒∞∑n=1

En sinnπy

10= f(y) (5)

To find En,Expand f(y) in a half range Fourier sine series in the interval (0, 10)

f(y) =∞∑n=1

bn sinnπy

10(6)

where bn =2

10

10∫0

f(y) sinnπy

10dy



En =1

5

5∫0

20y sinnπy

10dy +

10∫5

20(10− y) sinnπy

10dy

= 4

5∫0

y sinnπy

10dy +

10∫5

(10− y) sinnπy

10dy

= 4

[y10

nπ

(− cos

nπy

10

)− (1)

(10

nπ

)2 (− sin

nπy

10

)]50

+

[(10− y)

10

nπ

(− cos

nπy

10

)− (−1)

(10

nπ

)2 (− sin

nπy

10

)]105

= 4

{[− 50

nπ

(cos

nπ

2

)+

100

n2π2

(sin

nπ

2

)]− [0 + 0]

= + [0 + 0]−[−50

nπ

(cos

nπ

2

)− 100

n2π2

(sin

nπ

2

)]}∴ En =

800

n2π2sin

nπ

2


u(x, y) =∞∑n=1

800

n2π2sin

nπ

2sin

nπy

10e−

nπx10

=800

π2

∞∑n=1

1

n2sin

nπ

2sin

nπy

10e−

nπx10


1. A plate is in the form of the semi-infinite strip 0 ≤ x ≤ ∞, 0 ≤ y ≤ `.The surface of the plate and the edge y = l are insulated. If thetemperature along the edge y = 0 and the short edge at infinity andkept at temperature 0◦C while the temperature along the other shortedge is kept at temperature T ◦C find the steady state temperaturedistribution in the plate.

2. Write down the boundary conditions for finding the steady statetemperature distribution in the unit square defined by 0 ≤ x, y ≤ 1.If the edges y = 1 is kept at temperature f(x) and the other threeedges are insulated.


3. Find the steady state temperature at any point of a square plate if thetwo adjacent edges are kept at 0◦C and the others at 60◦C.

3.6 Assignment IV[Applications of Partial Differential Equations]

1. A string is stretched and fastened to two points ` apart. Motion isstarted by displacing the string into the form y = k(`x − x2) fromwhich it is released at time t = 0. Find the displacement of any pointof the string at a distance x from one end at any time t.

2. A tightly stretched string of length ‘2`’ has its ends fastened at x =0, x = 2`. The mid point of the string is then taken to height ‘b’ andthen released from rest in that position. Find the lateral displacementof a point of the string at time ‘t’ from the instant of release.

3. A tightly stretched string of length ‘`’ is initially at rest in itsequilibrium position and each of its points is given the velocity

V0 sin3(πx`

)Find the displacement of y(x, t).

4. A string is stretched between two fixed points at a distance 2` apartand the points of the string are given initial velocities ν where

ν =

cx

ìn 0 ≤ x ≤ `

c

`(2`− x) in ` ≤ x ≤ 2`

x being the distance from one end

point. Find the displacement of the string at any subsequent time.

5. The ends of A and B of a rod ` c.m. long have their temperaturekept at 30◦C and 80◦C, until steady state conditions prevail. Thetemperature of the end B is suddenly reduced to 60◦C and that ofA is increased to 40◦C. Find the temperature distribution in the rodafter time t.

6. Find the solution of the one dimensional diffusion equation satisfyingthe boundary conditions:

(i) u is bounded as t→ ∞ (ii)

[∂u

∂x

]x=0

= 0 for all t

(iii)

[∂u

∂x

]x=a

= 0 for all t (iv) u(x, 0) = x(a− x), 0 < x < a

7. Find the steady state temperature distribution in a rectangular plateof sides a and b insulated at the lateral surface and satisfying theboundary conditions u(0, y) = u(a, y) = 0 for 0 ≤ y ≤ b; u(x, b) = 0and u(x, 0) = x(a− x) for 0 ≤ x ≤ a.


8. A square plate is bounded by the lines x = 0, y = 0, x = 20 and y = 20.Its faces are insulated. The temperature along the upper horizontaledge is given by while the other two edges are kept at 0circC. Findthe steady state temperature distribution in the plate.

9. A rectangular plate with insulated surface is 10 cm wide and so longcompared to its width that it may be considered infinite in lengthwithout introducing appreciable error. The temperature at short edge

y = 0 is given by

{20x, 0 ≤ x ≤ 5

20(10− x), 5 ≤ x ≤ 10and all the other three

edges are kept at 0◦C. Find the steady-state temperature at any pointof the plate.

10. An infinitely long rectangular plate with insulated surface is 10 cmwide. The two long edges and one short edge are kept at zerotemperature, while the other short edge x = 0 is kept at temperature

u =

{20y, 0 < y ≤ 5

20(10− y), 5 < y ≤ 10. Find the temperature function

u(x, y) in steady state.

11. A rectangular plate with insulated surface is 20 cm wide and so longcompared to its width rectangular plate with insulated surface is 20cm wide and so long compared to its width that it may be consideredinfinite in length without introducing an appreciable error. If thetemperature of the short edge x = 0 is given by and the two longedges as well as the other short edge are kept at 0◦C. Find the steadystate temperature distribution in the plate.

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3 Applications of P.D.E.(A.P.D.E.) · 2019-08-15 · 3 Applications of P.D.E.(A.P.D.E.) Solutions of one dimensional wave equation { One dimensional equation of heat conduction {