Partial Differential Equations

Question

Please help me sir. Question Partial Differential Equations. Find the solution the following heat equation ∂u/dt = 4 ∂^2u/dx^2, 0<x<1, t>0 with boundary conditions u(0,t)=0, u(1,t)=0, t>0 and initial condition u(x,0)=x(1-x), 0<x<1

thank you.

Answer 1

Let's break this problem into steps.

Step 1. Separate your variables.

Let u(x,t) = X(x)T(t). Plug this into your PDE and separate the X and T dependencies. You'll eventually be left with

T'(t)/(4*T(t)) = X''(x)/X(x).

I grouped the coefficient 4 with the T dependence for convenience. We introduce the separation constant s.

T'(t)/(4*T(t)) = X''(x)/X(x) = s.

This gives us two ODE's X''(x) = s*X(x) and T'(t) = 4*s*T(t).

Step 2. Solve the X(x) equation with the appropriate boundary conditions.

It is not hard to show that the boundary conditions u(0,t) = 0 and u(1,t) = 0 imply X(0) = 0 and X(1) = 0, respectively. Thus, we need to find the solution of X''(x) = s*X(x) that satisfies the boundary conditions X(0) = X(1) = 0.

I'll leave it to you to show that, when s >= 0, the only solution to this boundary value problem is X(x) = 0 for all x. This is a trivial solution that we throw out. When s < 0, we get nontrivial solutions. Indeed, when s < 0,

X(x) = C₁*cos(sqrt(-s)*x) + C₂*sin(sqrt(-s)*x)

is the general solution of X''(x) = s*X(x), where C₁ and C₂ are arbitrary constants. If X(0) = 0, then hopefully you can see C₁ = 0. If X(1) = 0, then we have C₂sin(sqrt(-s)) = 0. We could let C₂ = 0, but that would lead us to the trivial solution X(x) = 0 again. Instead, we require sin(sqrt(-s)) = 0. This implies sqrt(-s) = pi*n, where n is any natural number. Thus, s = -(pi*n)^2. Thus, we have infinitely many nontrivial solutions to our boundary value problem:

X_n(x) = C_n*sin(pi*n*x),

where C_n is arbitrary for each n.

Step 3. Obtain the corresponding time dependence T(t) for the solutions of the boundary value problem.

We've seen that X_n(x) = C_n*sin(pi*n*x) are the nontrivial solutions of the boundary value problem. Their corresponding time dependence solves the ODE T'(t) = 4*s*T(t). The general solution of this ODE is

T(t) = D*exp(4*s*t),

where D is an arbitrary constant. For the time dependence of our nontrivial boundary value problem solutions, s = -(n*pi)^2. Thus, the corresponding time dependence of the boundary value problem solutions is

T_n(t) = D_n*exp(-4*(n*pi)^2*t),

where D_n is an arbitrary constant for each n.

Step 4. Superimpose your nontrivial solutions of the PDE.

In Steps 2-3, we found X_n(x) = C_n*sin(pi*n*x) and T_n(t) = D_n*exp(-4*(n*pi)^2*t). Thus, we have the following nontrivial solutions of the PDE:

u_n(x,t) = X_n(x)T_n(t) = C_n*sin(pi*n*x) D_n*exp(-4*(n*pi)^2*t).

Since C_n and D_n are arbitrary, we can fold D_n into C_n so that

u_n(x,t) = C_n*sin(pi*n*x)*exp(-4*(n*pi)^2*t).

We superimpose these solutions to get a new solution

u(x,t) = ∑ u_n(x,t) = ∑C_n*sin(pi*n*x)*exp(-4*(n*pi)^2*t).

Here, the sum goes from n = 1 to n = ∞.

Step 5. Use Fourier series and the initial condition to determine the arbitrary constants in your solution u(x,t).

The initial condition enforces

u(x,0) = x(1-x) = ∑C_n*sin(pi*n*x).

Looking carefully at the second equality, we see that the C_n are simply the Fourier sine series coefficients of the function x(1-x) defined on the interval [0,1]. There is a formula for these coefficients given by

C_n = 2/L*∫₀^Lx*(1-x)*sin(n*pi*x/L) dx.

Here, L is the length of the interval, i.e., L = 1. One can evaluate this integral above by repeated integration by parts. One finds

C_n = (4(-1)ⁿ⁺¹+4)/(pi*n)³.

Thus, the final solution of this problem is

u(x,t) = ∑(4(-1)ⁿ⁺¹+4)/(pi*n)³*sin(pi*n*x)*exp(-4*(n*pi)^2*t).