MA 201: Method of Separation of Variables

Finite Vibrating String ProblemLecture - 12

MA201(2018): PDE

IBVP for Vibrating string with no external forces

• We consider the problem in a computational domain

(x , t) ∈ [0, L]× [0,∞)

• The IBVP under consideration consists of the following:

• The governing equation:

utt = c2uxx , (x , t) ∈ (0, L)× (0,∞). (1)

The boundary conditions for all t > 0:

u(0, t) = 0, u(L, t) = 0. (2)

The initial conditions for 0 ≤ x ≤ L are

u(x , 0) = φ(x), ut(x , 0) = ψ(x). (3)

MA201(2018): PDE

Separation of variables method

• The main idea of this method is to convert the given partialdifferential equation into several ordinary differential equations.How?

• The solution is assumed to consist of the product of two or morefunctions.

• The number of functions involved depends on the number ofindependent variables.

• For wave equation u = u(x , t), so we will assume a solution in theform u(x , t) = X (x)T (t), where X is a function of x only and T isa function of T only.

• Substituting this solution in the given equation we will have a pair ofODEs.

• Note that this method can be used only for bounded domains sothat boundary conditions are prescribed.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Recall the wave equation

utt − c2uxx = 0. (4)

• Assume a solution of the form

u(x , t) = X (x)T (t). (5)

Here, X (x) is function of x alone and T (t) is a function of t alone.

• Substituting (5) in equation (4)

XT′′ = c

2X

′′T . (6)

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Separating the variables

X′′

X=

T′′

c2T.

• Here the left side is a function of x and the right side is a functionof t.

• The equality will hold only if both are equal to a constant, say, k .

• We get two differential equations as follows:

X′′ − kX = 0, (7a)

T′′ − c

2kT = 0. (7b)

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Since k is any constant,

it can be zero, or

it can be positive, or

it can be negative.

• Consider all the possibilities andexamine what value(s) of k lead to a non-trivial solution.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)Case I: k = 0

• In this case the equations (7) reduce to

X′′ = 0, and T

′′ = 0

• Giving rise to solutions

X (x) = Ax + B, T (t) = Ct + D.

• Boundary conditions

u(0, t) = u(L, t) = 0,

leads to X (x) = 0. Hence u = X (x)T (t) = 0.

• This case of k = 0 is rejected since it gives rise to trivial solutiononly.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

Case II: k > 0, let k = λ2 for some λ > 0.

• In this case the equations (7) reduce to the equations

X′′ − λ2X = 0, and T

′′ − c2λ2T = 0

• Giving rise to solutions

X (x) = Aeλx + Be

−λx ,

T (t) = Cecλt + De

−cλt .

• Therefore

u(x , t) = (Aeλx + Be−λx)(Cecλt + De

−cλt).

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Using boundary condition u(0, t) = 0,

A+ B = 0, B = −A.

• Using boundary condition u(L, t) = 0,

(AeλL + Be−λL)(Cecλt + De

−cλt) = 0.

• The t part of the solution cannot be zero as it will lead to T (t) = 0 andthen case k > 0 will be rejected straightway.

• Then we must have

A(eλL − e−λL) = 0,

• Which leads to A = 0 as λ 6= 0.

• k > 0 also gives rise to trivial solution:so k > 0 is also rejected.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)Case III: k < 0, let k = −λ2 for some λ > 0.

• In this case equations (7) reduce to the equations

X′′ + λ2X = 0 and T

′′ + c2λ2T = 0

• Giving rise to solutions

X (x) = A cosλx + B sinλx ,

T (t) = C cos(cλt) + D cos(cλt).

• Hence

u(x , t) = (A cosλx + B sinλx)(C cos(cλt) + D sin(cλt)).

• Using boundary condition u(0, t) = 0, A = 0.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Using boundary condition u(L, t) = 0,

B sinλL = 0.

• B 6= 0 as that will lead to a trivial solution.

• Hence we must have

sinλL = 0.

• Which gives us

λ =nπ

L= λn, n = 1, 2, 3, . . . .

• These λn’s are called eigenvalues and note thatcorresponding to each n there will be an eigenvalue.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Accordingly, the solution is

u(x , t) = (A cosλx + B sinλx)(C cos(cλt) + D sin(cλt))

= sinλnx(BC cos(cλnt) + BD sin(cλnt))

= sinnπx

L

[

An cosnπct

L+ Bn sin

nπct

L

]

, λn =nπ

L

= un(x , t).

• The solution corresponding to each eigenvalue is called aneigenfunction

• Thus, un(x , t) is the eigenfunction corresponding to theeigenvalue λn.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)• Since the wave equation is linear and homogeneous, any linearcombination will also be a solution

• Hence, we can expect the solution in the following form:

u(x , t) =

∞∑

n=1

un(x , t)

=∞∑

n=1

sinnπx

L

[

An cosnπct

L+ Bn sin

nπct

L

]

, (8)

• providedi. An and Bn are determined uniquely andii. each of the resulting series for those coefficients converges, andiii. the limit of the series is twice continuously differentiable with respectto x and t so that it satisfies the equation utt − c

2uxx = 0,

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• What is the idea?

• First, we assume that (!)(i) the infinite series converges for some An and Bn

(ii) term-wise differentiation with respect to t is possible andit converges

• Next, we calculate An and Bn using the given initial conditions.

• Once, both the coefficients An and Bn are calculated, we thenprove that the series actually holds following properties

i. the resulting series for those coefficients converge, and

ii. the limit of the series is twice continuously differentiable with respectto x and t, and it satisfies the partial differential equation.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Using the initial condition u(x , 0) = φ(x)

φ(x) =∞∑

n=1

An sinnπx

L. (9)

• This series can be recognized as the half-range sine expansion of afunction φ(x) defined in the range (0, L).

• An can be obtained by multiplying equation (9) by sin nπx

Land

integrating with respect to x from 0 to L.

• Therefore

An =2

L

∫ L

0

φ(x) sinnπx

Ldx , n = 1, 2, 3, . . . (10)

• Here, we have used the fact that

∫

L

−L

sinnπx

Lsin

nπx

Ldx = L.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)• To use the other initial condition ut(x , 0) = ψ(x), we need todifferentiate (8) w.r.t. t to get

ut(x , t) =∞∑

n=1

sinnπx

L

(

nπc

L

) [

−An sinnπct

L+ Bn cos

nπct

L

]

.

• Then

ψ(x) =∞∑

n=1

Bn

nπc

Lsin

nπx

L.

• Similarly

Bn =2

nπc

∫ L

0

ψ(x) sinnπx

Ldx , n = 1, 2, 3, . . . (11)

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)• Now, consider the infinite series

∞∑

n=1

un(x , t) =∞∑

n=1

sinnπx

L

[

An cosnπct

L+ Bn sin

nπct

L

]

, (12)

with

•

An =2

L

∫ L

0

φ(x) sinnπx

Ldx , n = 1, 2, 3, . . . (13)

Bn =2

nπc

∫ L

0

ψ(x) sinnπx

Ldx , n = 1, 2, 3, . . . . (14)

• Clearly, the series satisfies both initial and boundary conditions.

• In fact, it can be proved that it is the solution of the finite vibratingstring problem.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)• Consider following infinite series

w1(x , t) =∞∑

n=1

An cosnπct

Lsin

nπx

L, (15)

w2(x , t) =

∞∑

n=1

Bn sinnπct

Lsin

nπx

L. (16)

• Using the trigonometric identity

sinnπx

Lcos

nπct

L=

1

2sin

nπ

L(x − ct) +

1

2sin

nπ

L(x + ct) (17)

• We write

w1(x , t) =1

2

∞∑

n=1

An sinnπ

L(x − ct) +

1

2

∞∑

n=1

An sinnπ

L(x + ct)

=1

2

∞∑

n=1

An sinnπ

L(ξ) +

1

2

∞∑

n=1

An sinnπ

L(η),

ξ = x − ct ∈ R, η = x + ct ∈ R.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)• We know that the Fourier series

∞∑

n=1

An sinnπξ

Lwith An =

2

L

∫

L

0

φ(x) sinnπx

Ldx

converges to φ(ξ) in ξ ∈ (0, L). But, ξ ∈ R.

• Define the odd periodic extension of φ by

φo(x) = φ(x) if 0 < x < L,

φo(x) = −φ(−x) if − L < x < 0,

φo(x) = φ(x + 2L), for rest x ∈ R.

• From convergence of Fourier series, we conclude that the series

∞∑

n=1

An sinnπξ

Lwith An =

1

L

∫

L

−L

φo(x) sinnπx

Ldx =

2

L

∫

L

0

φo(x) sinnπx

Ldx = An

converges to φo(ξ) in [−L, L]. What about the convergence in (−∞,∞)?

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)• Therefore, the infinite series

∞∑

n=1

An sinnπξ

Lwith An =

2

L

∫

L

0

φ(x) sinnπx

Ldx

converges to φo(ξ) in ξ ∈ (−∞,∞).

• As a consequence, we have

w1(x , t) =1

2φo(ξ) +

1

2φo(η)

• Assume that φ ∈ C2[0, L]. What about the smoothness of φo in

(−∞,∞)?

• Then we obtain

∂2w1

∂t2− c

2 ∂2w1

∂x2= 0.

MA201(2018): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Similarly, under the assumption ψ ∈ C2[0, L], we have

∂2w2

∂t2− c

2 ∂2w2

∂x2= 0.

• Hence,

∂2

∂t2(w1 + w2)− c

2 ∂2

∂x2(w1 + w2) = 0.

• Recall

w1(x , t) =∞∑

n=1

An cosnπct

Lsin

nπx

L,

w2(x , t) =

∞∑

n=1

Bn sinnπct

Lsin

nπx

L.

MA201(2018): PDE

Formal Solution of the Finite Vibrating String Problem

• The solution is given by

u(x , t) = w1(x , t) + w2(x , t)

=∞∑

n=1

An cosnπct

Lsin

nπx

L+

∞∑

n=1

Bn sinnπct

Lsin

nπx

L

• with

An =2

L

∫ L

0

φ(x) sinnπx

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

Bn =2

nπc

∫ L

0

ψ(x) sinnπx

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

• Is the solution unique?

MA201(2018): PDE