Simple applications to solve partial differential equations using Fourier transform

Applications using Fourier Transform

APPLICATIONS:

Simple applications to solve partial differential equations using Fourier transform :

Problem 1:

Solve the one‒dimensional wave equation

Subject to the initial conditions

y (x, 0) = 0, ∂u/∂x(x, 0) = e^−x2

Step 1: Fourier transform with respect to x

Define the Fourier transform

Step 2. Solution of the transformed ODE

Step 3: Apply initial conditions

Step 4: Transformed solution

Step 5. Inverse Fourier Transform (Detailed steps)

The transformed solution is

Step 5(a): Write the inverse Fourier transform

Step 5(b) Express the sine function in exponential form

Step 5(c) Substitute and split the integral

Step 5(d): Use the standard Fourier integral result

The known identity is

Step 5(e): Apply the identity :

Let

b = x + ct and b = x ‒ ct

Step 5(f): Combine the integrals

Problem 2:

We solve the diffusion (heat) equation

Subject to the initial condition

 u(x, 0) = δ(x)

where δ(x) is the Dirac delta function.

Solution:

Step 1: Take Fourier transform with respect to x

Step 2: Solve the transformed ODE

This is a first‒order linear ODE :

Step 3: Apply the initial condition :

Taking Fourier transform of the initial condition:

 u(x, 0) = δ(x)

Using the property

F [δ(x)] = 1

we get

Step 4: Inverse Fourier Transform :

Problem 3:

Solve the equation ∂u/∂t = ∂²u/∂x², x > 0, t> 0

subject to the conditions

(i) u = 0, when x = 0, t> 0

(ii) u = , when t = 0

and (iii) u (x, t) is bounded.

Solution:

Since (u)_x=0 is given, taking Fourier sine transform of both sides of the given equation, we have

Now, when t= 0, Fourier sine transform of u (x, t)... (1)

Applying the inverse Fourier sine transform, we have

which is required solution.

Problem 4:

Solve the following heat equation

Solution:

Here the problem is given in 0 < x < ∞. Hence Fourier transform can not be applied. Also the application of Fourier cosine transform requires a derivative value. So let us apply Fourier sine transform on both sides.

d/dt Fs(s, t) = c² [‒s² F_S (s, t) + su(0, t)]

 Fs' (s, t) = ‒c²s²Fs (s, t)         ['.' u (0,t) = 0]

  Fs' (s, t) + c²s²F_S(s, t) = 0

which is a linear equation

The solution is

        ……….(2)

Taking Fourier sine transform of the boundary condition we obtain,

Fs [u (x, 0)] = Fs f(x)]

which is the required solution.

Problem 5:

Solve the wave equation  subject to the initial conditions y(x, 0) = f(x), ‒∞<x< ∞, ∂y/∂t (x, 0) = g(x) and the boundary conditions y (x, t) → 0, as x → ± ∞.

Solution:

Taking Fourier transforms of the equation with respect to x, we get,

Taking Fourier transforms of the initial conditions, we have

Problem 6:

Solve the Laplace equation  = 0, y ≥ 0 subject to the boundary conditions u (x, 0) = f(x), ‒∞<x<∞ and u (x, y)→ 0 as y→ ∞.

Solution:

Fourier transforms of the Laplace equation with respect to x, we get

Problem 7:

Solve the one‒dimensional heat flow equation ∂u/∂t = a² ∂²u/∂x² for a rod with insulated sides extending from ‒∞ to ∞ and with initial temperature distribution given by u(x, 0) = f(x)

Solution:

Assume that u (x, t) → 0 as x → ± ∞

Taking Fourier transforms of the given equation with respect to x, we get

Taking Fourier inverse transforms of (4) by using convolution theorem and using (5)

we get

EXERCISES

1. Solve the equation ∂u/∂t = a² ∂²u/∂x² , −∞ < x < ∞ using transform method, given that u(x,0) =  and u (x, t) → 0 as x → ± ∞.

2. Solve the equations ∂²y/∂t² = c² ∂²y/∂x², satisfying the initial conditions y (x, 0) = f(x). ∂u/∂t (x, 0) = 0 and the boundary conditions y(0,t) = 0, y (x, t) → 0 as x → ∞. Use transform method.

3. Solve the Laplace equation ∂²u/∂x² + ∂²u/∂y² = 0; x≥0, using transform method, given that u(0,y) = f(y), −∞ < y < ∞ and u(x, y) → 0 as x → ∞.

4. Solve the equation ∂u/∂t = a² ∂²u/∂x², x≥0, using transform method, given that u (0,t) = f(t), t ≥ 0, u (x, t) → 0 as x → 0 and u (x, 0) = 0

5. Using transform method, solve the equation ∂u/∂x = a² ∂²u/∂x², x ≥ 0, subject to the boundary conditions ∂u/∂x (0,t) = f(t), t≥ 0 and u (x, t) → 0 as x → ∞ and the initial condition u (x, 0) = 0