Some Aspects of Solutions of Partial Differential Equations · K. Sakthivel Department of Mathematics Indian Institute of Space Science & Technology(IIST) Trivandrum - 695 547, Kerala

Some Aspects of Solutions of Partial DifferentialEquations

K. SakthivelDepartment of Mathematics

Indian Institute of Space Science & Technology(IIST)Trivandrum - 695 547, Kerala

[email protected]

Periyar University, SalemFebruary 22, 2013

K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 1 / 24

Plan of the Talk

1 Formulation of Diffusion Model

2 Solution by Fourier Method

3 Nonlinear PDEs - Conservation Equations - Method ofCharacteristics

4 Weak Solutions for PDEs

5 Weak Solutions by Variational Methods


Formulation of Diffusion ModelConsider

A motionless liquid filling a straight tube or pipeA chemical substance, say die, which is diffusing through the liquid

Let u(z, t) be the concentration of the die at position z of the pipe attime t .

On the region x to x + ∆x , the mass of the dye is

M(t) =

∫ x+∆x

xu(z, t)dz, and so

dMdt

=

∫ x+∆x

x

∂u∂t

(z, t)dz.

Fick’s Law: Flux goes from region of higher concentration to theregion of lower concentration. The rate of motion is propositional to theconcentration gradient.

dMdt

=

∫ x+∆x

x

∂u∂t

(z, t)dz = Net Change of Concentration

= k∂u∂x

(x + ∆x , t)− k∂u∂x

(x , t).


Formulation of Diffusion Model.....Now, ∫ x+∆x

x

∂u∂t

(z, t)dz = k [∂u∂x

(x + ∆x , t)− ∂u∂x

(x , t)],

where k−proportionality constant.

Applying Mean Value Theorem (!!!) on x < ζ < x + ∆x , we get

∂u∂t

(ζ, t) = k[∂u∂x

(x + ∆x , t)− ∂u∂x

(x , t)]/∆x .

Taking ∆x → 0, we have

∂u∂t

(x , t) = k∂2u∂x2 (x , t).

In the multidimensional case, we get

ut = k∆u + f (x , t), where f is a “source" or “sink" of the dye.



x

∂u∂t

(z, t)dz = k [∂u∂x

(x + ∆x , t)− ∂u∂x

(x , t)],



∂u∂t

(ζ, t) = k[∂u∂x

(x + ∆x , t)− ∂u∂x

(x , t)]/∆x .


∂u∂t

(x , t) = k∂2u∂x2 (x , t).


ut = k∆u

+ f (x , t), where f is a “source" or “sink" of the dye.



x

∂u∂t

(z, t)dz = k [∂u∂x

(x + ∆x , t)− ∂u∂x

(x , t)],



∂u∂t

(ζ, t) = k[∂u∂x

(x + ∆x , t)− ∂u∂x

(x , t)]/∆x .


∂u∂t

(x , t) = k∂2u∂x2 (x , t).


ut = k∆u + f (x , t), where f is a “source" or “sink" of the dye.


Fourier Transform and its PropertiesLet f ∈ L1(R). The Fourier transform of f is defined as

f (ω) := F [f (x)] =1√2π

∫ ∞−∞

e−iωx f (x)dx .

Inverse Fourier Transform:

f (x) := F−1 [f (ω)] =1√2π

∫ ∞−∞

eiωx f (ω)dω.

Some Properties: If f is a continuous piecewise differentiable functionwith f , fx , fxx ∈ L1(R) and lim

|x |→∞f (x) = 0, then

F [fx (x)] = iωF [f (x)] and F [fxx (x)] = −ω2F [f (x)] !!!.

Fourier transform is linear, that is, f ,g,∈ L1(R) and a,b ∈ R, then

F [af + bg] = aF [f ] + bF [g].


Fourier Transform and its PropertiesLet f ∈ L1(R). The Fourier transform of f is defined as

f (ω) := F [f (x)] =1√2π

∫ ∞−∞

e−iωx f (x)dx .

Inverse Fourier Transform:

f (x) := F−1 [f (ω)] =1√2π

∫ ∞−∞

eiωx f (ω)dω.

Some Properties: If f is a continuous piecewise differentiable functionwith f , fx , fxx ∈ L1(R) and lim

|x |→∞f (x) = 0, then

F [fx (x)] = iωF [f (x)] and F [fxx (x)] = −ω2F [f (x)] !!!.

Fourier transform is linear, that is, f ,g,∈ L1(R) and a,b ∈ R, then

F [af + bg] = aF [f ] + bF [g].


Fourier Transform and its Properties.....

Note that F [f (x)g(x)] 6= F [f (x)]F [g(x)], whereas

F [(f ∗ g)(x)] = F [f (x)]F [g(x)],

where (f ∗ g) is the “convolution" of f and g defined as

(f ∗ g)(x) :=1√2π

∫ ∞−∞

f (x − y)g(y)dy = (g ∗ f )(x).

1-D Heat Conduction Model (in an infinite rod):

ut (x , t) = kuxx (x , t), −∞ < x <∞, 0 < t <∞u(x ,0) = φ(x), −∞ < x <∞.

Applying Fourier transform

F [ut (x , t)] = kF [uxx (x , t)], F [u(x ,0)] = F [φ(x)]


Fourier Transform and its Properties.....

Note that F [f (x)g(x)] 6= F [f (x)]F [g(x)], whereas

F [(f ∗ g)(x)] = F [f (x)]F [g(x)],

where (f ∗ g) is the “convolution" of f and g defined as

(f ∗ g)(x) :=1√2π

∫ ∞−∞

f (x − y)g(y)dy = (g ∗ f )(x).

1-D Heat Conduction Model (in an infinite rod):

ut (x , t) = kuxx (x , t), −∞ < x <∞, 0 < t <∞u(x ,0) = φ(x), −∞ < x <∞.

Applying Fourier transform

F [ut (x , t)] = kF [uxx (x , t)], F [u(x ,0)] = F [φ(x)]


1-D Heat Conduction Model

Recall that

F [ut (x , t)] =1√2π

∫ ∞−∞

e−iξx ∂u∂t

(x , t)dx

=∂

∂t

[ 1√2π

∫ ∞−∞

e−iξx u(x , t)dx]

=∂u∂t

(ξ, t).

Taking φ(ξ) = F [φ(x)], we get the following “ODE":

dudt

(t) = −kξ2u(t) with data u(0) = φ(ξ).

Solution to the transformed ODE: F [u(x , t)] = u(t) = φ(ξ)e−ktξ2.

Note that:

(f ∗ g)(x) = F−1[F [f (x)]F [g(x)]]

and F [e−a2x2] =

1√2a

e−ξ2/4a2

.


1-D Heat Conduction Model

Recall that

F [ut (x , t)] =1√2π

∫ ∞−∞

e−iξx ∂u∂t

(x , t)dx

=∂

∂t

[ 1√2π

∫ ∞−∞

e−iξx u(x , t)dx]

=∂u∂t

(ξ, t).

Taking φ(ξ) = F [φ(x)], we get the following “ODE":

dudt

(t) = −kξ2u(t) with data u(0) = φ(ξ).

Solution to the transformed ODE: F [u(x , t)] = u(t) = φ(ξ)e−ktξ2.

Note that:

(f ∗ g)(x) = F−1[F [f (x)]F [g(x)]]

and F [e−a2x2] =

1√2a

e−ξ2/4a2

.


Solution of 1-D Heat Conduction ModelTaking inverse Fourier transform

u(x , t) = F−1[φ(ξ)e−ktξ2] =

1√2kt

F−1[F [φ(x)]F [e−x2/4kt ]

]= φ(x) ∗

[ 1√2kt

e−x2/4kt]

=1√

4πkt

∫ ∞−∞

φ(y)e−(x−y)2/4ktdy .

Fundamental Solution (or Heat Kernel) of 1-D Heat Equation:

Φ(x , t) =1√4πt

e−x2/4t , k = 1, t > 0.

Multidimensional case:

Ψ(x , t) =1

(4πt)d/2 e−|x |2/4t , x ∈ Rd , t > 0,

where x = (x1, x2, · · · , xd ) and |x | =√

x21 + x2

2 + · · ·+ x2d .


Solution of 1-D Heat Conduction ModelTaking inverse Fourier transform

u(x , t) = F−1[φ(ξ)e−ktξ2] =

1√2kt

F−1[F [φ(x)]F [e−x2/4kt ]

]= φ(x) ∗

[ 1√2kt

e−x2/4kt]

=1√

4πkt

∫ ∞−∞

φ(y)e−(x−y)2/4ktdy .

Fundamental Solution (or Heat Kernel) of 1-D Heat Equation:

Φ(x , t) =1√4πt

e−x2/4t , k = 1, t > 0.

Multidimensional case:

Ψ(x , t) =1

(4πt)d/2 e−|x |2/4t , x ∈ Rd , t > 0,

where x = (x1, x2, · · · , xd ) and |x | =√

x21 + x2

2 + · · ·+ x2d .


Properties of Fundamental SolutionFor t > 0, the function Ψ(x , t) > 0 is an infinitely differentiablefunction of x and t . (Recall Smoothing Property)

Ψt (x , t) = ∆Ψ(x , t), x ∈ Rd , t > 0 and∫Rd

Ψ(x , t)dx = 1, t > 0.

For any continuous and bounded function g(x) :

limt→0

∫Rd

Ψ(x , t)g(x)dx = g(0).

Since Ψ(x , t) is the probability density of a Gaussian random variable,∫Rd

Ψ(x , t)dx = 1, for all t > 0 !!!.

Moreover, by Lebesgue dominated convergence theorem∫Rd

Ψ(x , t)g(x)dx

=1

(4πt)d/2

∫Rd

e−|x |2/4tg(x)dx =

1πd/2

∫Rd

e−|y |2g(y√

4t)dx

=1

(π)d/2

∫Rd

e−|y |2g(0)dx = g(0) as t → 0.


Properties of Fundamental SolutionFor t > 0, the function Ψ(x , t) > 0 is an infinitely differentiablefunction of x and t . (Recall Smoothing Property)

Ψt (x , t) = ∆Ψ(x , t), x ∈ Rd , t > 0 and∫Rd

Ψ(x , t)dx = 1, t > 0.

For any continuous and bounded function g(x) :

limt→0

∫Rd

Ψ(x , t)g(x)dx = g(0).

Since Ψ(x , t) is the probability density of a Gaussian random variable,∫Rd

Ψ(x , t)dx = 1, for all t > 0 !!!.

Moreover, by Lebesgue dominated convergence theorem∫Rd

Ψ(x , t)g(x)dx

=1

(4πt)d/2

∫Rd

e−|x |2/4tg(x)dx =

1πd/2

∫Rd

e−|y |2g(y√

4t)dx

=1

(π)d/2

∫Rd

e−|y |2g(0)dx = g(0) as t → 0.


Conservation Equations - Traffic Flow ModelConsider a stretch of highway on which cars are moving from left toright. Assume that no exit or entrance of ramps.

u(x , t)−density of cars at xf (x , t)−flux of cars at x (cars per minute passing the point x)

Change in the number of cars in [a,b] =ddt

∫ b

au(x , t)dx

Using flux, change in the number of cars in [a,b] =

= f (a, t)− f (b, t) = −∫ b

a

∂f∂x

(x , t)dx ,

by fundamental theorem of calculus. The last two integrals yield∫ b

a

∂u∂t

(x , t)dx = −∫ b

a

∂f∂x

(x , t)dx .

The interval length [a,b] is arbitrary,

ut (x , t) + fx (x , t) = 0. (Conservation Equation)


Traffic Flow Model

The amount of cars passing a given point is generally a function ofdensity u, that is, f (u) : for example, f (u) = u2 (quadratic flow rate).

From conservation equation and chain rule:

ut + 2uux = 0.

Consider the model

ut + 2uux = 0, −∞ < x <∞, 0 < t <∞

with the initial density of cars !!!

u(x ,0) = u0(x) =

1 x ≤ 01− x 0 < x < 10 x ≥ 1


Traffic Flow Model

The amount of cars passing a given point is generally a function ofdensity u, that is, f (u) : for example, f (u) = u2 (quadratic flow rate).

From conservation equation and chain rule:

ut + 2uux = 0.

Consider the model

ut + 2uux = 0, −∞ < x <∞, 0 < t <∞

with the initial density of cars !!!

u(x ,0) = u0(x) =

1 x ≤ 01− x 0 < x < 10 x ≥ 1


Solution by Method of CharacteristicsLet x(t) be a curve in x − t plane and u(x(t), t) is the function of x andt . How does u change from the perspective of x(t)?

By chain ruledudt

= uxdxdt

+ ut ;

but taking traffic flow model into account

ut + 2uux = 0⇔ dx

dt = 2ududt = 0

So

dxdt

= 2u ⇒ x = 2u(x , t)t + x0→ Characteristic equation

Along such characteristic u is constant. From the initial density of cars,the characteristic curve starting from (x0,0) is

x = 2u(x0,0)t + x0


Solution by Method of Characteristics .....Case 1: For x0 ≤ 0, u(x0,0) = 1, the characteristics are

x = 2t + x0 or t =12

(x − x0).

Case 2: For 0 < x0 < 1 u(x0,0) = 1− x0, the characteristics are

x = 2(1− x0)t + x0 or t =12

x − x0

1− x0.

Case 3: For 1 ≤ x0 <∞, u(x0,0) = 0, the characteristics are verticallines given by x = x0.

Observations:Note that characteristics run together starting at t = 1/2.So except for short time t = 1/2, the solution is not continuous.When characteristics run together, we have a phenomenon of“Shock Waves" (discontinuous solutions)Shock waves occurs due to the flux that grows very large as afunction of density u.



x = 2t + x0 or t =12

(x − x0).


x = 2(1− x0)t + x0 or t =12

x − x0

1− x0.

Case 3: For 1 ≤ x0 <∞, u(x0,0) = 0, the characteristics are verticallines given by x = x0.Observations:

Note that characteristics run together starting at t = 1/2.

So except for short time t = 1/2, the solution is not continuous.When characteristics run together, we have a phenomenon of“Shock Waves" (discontinuous solutions)Shock waves occurs due to the flux that grows very large as afunction of density u.



x = 2t + x0 or t =12

(x − x0).


x = 2(1− x0)t + x0 or t =12

x − x0

1− x0.


Note that characteristics run together starting at t = 1/2.So except for short time t = 1/2, the solution is not continuous.

When characteristics run together, we have a phenomenon of“Shock Waves" (discontinuous solutions)Shock waves occurs due to the flux that grows very large as afunction of density u.



x = 2t + x0 or t =12

(x − x0).


x = 2(1− x0)t + x0 or t =12

x − x0

1− x0.


Note that characteristics run together starting at t = 1/2.So except for short time t = 1/2, the solution is not continuous.When characteristics run together, we have a phenomenon of“Shock Waves" (discontinuous solutions)

Shock waves occurs due to the flux that grows very large as afunction of density u.



x = 2t + x0 or t =12

(x − x0).


x = 2(1− x0)t + x0 or t =12

x − x0

1− x0.


Note that characteristics run together starting at t = 1/2.So except for short time t = 1/2, the solution is not continuous.When characteristics run together, we have a phenomenon of“Shock Waves" (discontinuous solutions)Shock waves occurs due to the flux that grows very large as afunction of density u.


Notion of Weak Solution for PDEsTest Functions : C∞0 (Ω)Let Ω ⊂ Rd be a bounded domain. The set of all functions beloging toC∞(Ω) and compactly supported in Ω.

Support of a function ϕ(x) :

supp(ϕ) = x ∈ Ω : ϕ(x) 6= 0.

Example (Compact Support) Suppose d = 1.

ϕ(x) =

e

−1x2−1 if |x | < 1.

0 elsewhere.

Example (Weak Differentiability): Suppose d = 1 and Ω = (−1,1).The function u(x) = |x |, x ∈ Ω is not differentiable in the classicalsense. But it has a weak derivative!!!

Du =

−1 if x < 01 if x > 0.


Notion of Weak Solution for PDEsTest Functions : C∞0 (Ω)Let Ω ⊂ Rd be a bounded domain. The set of all functions beloging toC∞(Ω) and compactly supported in Ω.

Support of a function ϕ(x) :

supp(ϕ) = x ∈ Ω : ϕ(x) 6= 0.

Example (Compact Support) Suppose d = 1.

ϕ(x) =

e

−1x2−1 if |x | < 1.

0 elsewhere.

Example (Weak Differentiability): Suppose d = 1 and Ω = (−1,1).The function u(x) = |x |, x ∈ Ω is not differentiable in the classicalsense. But it has a weak derivative!!!

Du =

−1 if x < 01 if x > 0.


Weak Differentiability......

For any ϕ ∈ C∞0 (Ω), integrating by parts, we get:∫ 1

−1u(x)Dϕ(x)dx

=

∫ 0

−1u(x)Dϕ(x)dx +

∫ 1

0u(x)Dϕ(x)dx

= −∫ 0

−1Du(x)ϕ(x)dx + uϕ

∣∣∣0−1−∫ 1

0Du(x)ϕ(x)dx + uϕ

∣∣∣10

= −∫ 1

−1Du(x)ϕ(x)dx − [u(0)]ϕ(0).

Note that ϕ(1) = ϕ(−1) = 0 and [u(0)] = u(0+)− u(0−) = 0 since u iscontinuous at x = 0.


Sobolev Space H1(Ω)

H1(Ω) := v ∈ L2(Ω),∇v ∈ L2(Ω).

The space H1(Ω) is a separable Hilbert space equipped with innerproduct

(u, v)H1(Ω) =

∫Ω

uvdx +

∫Ω∇u · ∇vdx .

Norm on H1(Ω):

‖u‖2H1(Ω) =

∫Ω|u|2dx +

∫Ω|∇u|2dx .

Space H10 (Ω) := u ∈ H1(Ω) : u|∂Ω = 0.


Weak Solution for Poisson Equation

Let Ω ⊆ Rd be a bounded domain with smooth boundary.

−4u(x) = f (x) in Ωu(x)|∂Ω = 0

(1)

where f (x) is the external force; x = (x1, x2, ..., xd ) and 4 is theLaplace operator

4 =∂2

∂x21

+∂2

∂x22

+ ....+∂2

∂x2d.

Weak and Classical solutions : Assume u ∈ C2(Ω) and ϕ ∈ C∞0 (Ω).Multiplying Poisson equation by ϕ and integrating on Ω,

−∫

Ω4u(x)ϕ(x)dx =

∫Ω

f (x)ϕ(x)dx .


Weak and Classical solutions.....Green’s identity:

−∫

Ω4u(x)ϕ(x)dx = −

∫∂Ω

∂u∂ν

(x)ϕ(x)ds +

∫Ω∇u · ∇ϕdx

where ν is the unit normal vector outward to ∂Ω. Further,∫Ω∇u · ∇ϕdx =

∫∂Ω

u∂ϕ

∂νds −

∫Ω

u4ϕ(x)dx

We get the following identities :∫Ω∇u · ∇ϕdx =

∫Ω

f (x)ϕ(x)dx (2)

−∫

Ωu∆ϕ(x)dx =

∫Ω

f (x)ϕ(x)dx (3)

Thus, if u ∈ C2(Ω) is a solution of Poisson equation then for anyϕ ∈ C∞0 (Ω) the above identities hold.


Weak and Classical solutions.....

Conversely for any ϕ ∈ C∞0 (Ω),u ∈ C2(Ω) satisfies (3), we also get∫Ω

(−4u − f )ϕ(x)dx = 0.

Since ϕ is arbitrary, −4u − f = 0 in Ω, i.e, u is a solution of Poissonequation.

Note : The integral (2) makes sense if u ∈ H1(Ω) whereas for (3) weonly need u ∈ L2(Ω).

Weak Solution: A function u ∈ H1(Ω) is said to be a weak solution of(1), if the following identity holds:∫

Ω∇u · ∇ϕdx =

∫fϕdx for any ϕ ∈ C∞0 (Ω) or H1

0 (Ω)

since C∞0 (Ω) is dense in H10 (Ω).


Optimization Result from Linear AlgebraConsider a system Ax = b, where A is n × n matrix, symmetric andpositive definite and b ∈ Rn.

Recall that for x , y ∈ Rn

x · y =< x , y >= xT y =n∑

i=1

xiyi .

A vector x ∈ Rn is a solution of Ax = b iff it is global minimizer of

φ(x) =12< Ax , x > − < b, x > .

Assume that φ has a global minimizer !!. For any y ∈ Rn, ε ∈ R thevariation Φ(ε) := φ(x + εy) achieves its minimum at ε = 0.

⇒ Φ′(0) =ddεφ(x + εy)|ε=0 = 0.


Optimization Result from Linear Algebra .....

Now

Φ′(ε) =ddε

[12< A(x + εy), x + εy > − < b, x + εy >

]= < A(x + εy), y > − < b, y >

But Φ′(0) = 0 implying that

< Ax , y > − < b, y >=< Ax − b, y >= 0.

Since y is arbitrary, x is a solution of Ax = b.


Variational Approach for Poisson EquationConsider a functional

J[v ] =12

∫Ω|∇v |2dx −

∫Ω

fvdx .

If u ∈ H10 (Ω) is an extremal of the functional J[v ] in H1

0 (Ω), then u is aweak solution of the Dirichlet problem (1).

Suppose u ∈ H10 (Ω) is an extremum of J[v ](need a proof !!), then for

any ϕ ∈ H10 (Ω) and ε ∈ R,

F (ε) := J[u + εϕ] =12

∫Ω|∇(u + εϕ)|2dx −

∫Ω

f (u + εϕ)dx .

But F achieves its minimum at ε = 0, F ′(0) = dJdε (u + εϕ)|ε=0 = 0. Note

that

F ′(ε) =

∫Ω

(∇u + ε∇ϕ)∇ϕdx −∫

Ωfϕdx

SoF ′(0) = 0⇒

∫Ω∇u · ∇ϕdx =

∫Ω

fϕdx .

Therefore, u is a weak solution of the Poisson equation.K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 22 / 24

Uniqueness of Weak SolutionsSuppose u1 and u2 are two weak solutions of the Poisson equation.For any ϕ ∈ H1

0 (Ω), we have∫Ω∇u1 · ∇ϕdx =

∫Ω

fϕdx and∫

Ω∇u2 · ∇ϕdx =

∫Ω

fϕdx

Taking u := u1 − u2, we arrive at∫Ω∇u · ∇ϕdx = 0, ϕ ∈ H1

0 (Ω).

Choosing ϕ = u,∫

Ω|∇u|2dx = 0.

Applying the Poincare inequality:∫Ω|u|2dx ≤ C(Ω)

∫Ω|∇u|2dx ,

so that ∫Ω|u|2dx = 0⇒ u = 0 a.e. in Ω.


Thank You!


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Some Aspects of Solutions of Partial Differential Equations · K. Sakthivel Department of Mathematics Indian Institute of Space Science & Technology(IIST) Trivandrum - 695 547, Kerala