Question

use the method of separation of variables to determine the general solution to the one dimensional heat equation ∂u/∂t=α(∂²u/∂x²)=0,0≤ x≤ l,t≥ 0 subject to the initial condition u(x,0)=g(x). Hence compute the particular solution when g(x)=0

Answer 1

Solution;

The heat equation;

$\frac{\partial u}{\partial t}=\alpha(\frac{\partial^2u}{\partial x^2})=0,0\leq x\leq l,t\geq0$

Let ;

$u(x,t)=X(x)T(t)$

X(x) is some function dependent on x alone while T(t) is some function dependent on t alone.

The factorized function u(x, t) = X(x)T (t) is a solution to the heat equation if and only if;

$X(x)T'(t)=\alpha X''(x)T(t)$

Hence;

$\frac{X''(x)}{X(x)}=\frac1\alpha\frac{T'(t)}{T(t)}$

The two sides are equal. So both sides must be independent of both x and t and hence equal to some constant, say σ. So we have;

$\frac{X''(x)}{X(x)}=\sigma$ ; $\frac1\alpha\frac{T'(t)}{T(t)}=\sigma$

Therefore;

$X''(x)-\sigma X(x)=0$

$T'(t)-\alpha\sigma T(t)=0$

If σ$\neq$ 0, the general solution to the above equations are;

$X(x)=c_1e^{\sqrt{\sigma}x}+c_2e^{-\sqrt{\sigma}x}$

$T(t)=c_3e^{\alpha\sigma t}$

For arbitrary constants c₁, c₂ and c₃.

If σ = 0, the equations simplify to;

$X''(x)=0;T'(t)=0$

Whose general solutions are;

$X(x)=c_1+c_2x$

$T(t)=c_3$

Therefore the general solutions to the heat equation are;

$u(x,t)=(c_1e^{\sqrt{\sigma}x}+c_2e^{-\sqrt{\sigma}x})c_3e^{\alpha\sigma t}$

($\sigma\neq0$ )

$u(x,t)=(c_1+c_2x)c_3$

($\sigma=0$ )

Subject the solutions to the initial condition;

$u(x,0)=g(x)=(c_1e^{\sqrt{\sigma}x}+c_2e^{-\sqrt{\sigma}x})c_3$

The particular solution if g(x)=0;

$g(x)=0=(c_1e^{\sqrt{\sigma}x}+c_2e^{-\sqrt{\sigma}x})c_3$

Meaning;

$c_1,c_2 ,c_3=0$

Hence;

$u(x,t)=0$