Question

Determine the temperature distribution u(x, t) in a bar of length L, whose ends are kept at zero

temperature, and whose initial temperature is:

i) u(x, 0) = f(x) = U₀

ii) u(x, 0) = f(x) = k sin πx/L

Answer 1

We shall apply the heat equation. Which is:

$U_t=cU_{xx}$

To solve this, let $U=XT\implies U_t=XT',U_{xx}=X''T$

The equation becomes:

$XT'=cX''T\\ \frac{T'}{cT}=\frac{X''}{X}=-\lambda^2$

Solving the two differential equations, we have that:

$T=Ae^{-c\lambda^2t} \text{ and } X=B\cos \lambda x+ C \sin \lambda x$

Thus, $U(x,t)=XT'=(A_1\cos \lambda x+ A_2 \sin \lambda x)e^{-c\lambda^2t}$

Applying the boundary conditions $U(0,t)=U(L,t)=0$

$U(0,t)= A_1e^{-c\lambda^2t}=0\\ \text{But, exponential does not vanishes}\\ \implies A_1=0\\ U(x,t)=A_2e^{-c\lambda^2t}\sin \lambda x$

Applying the second condition;

$U(L,t)=A_2e^{-c\lambda^2t} \sin \lambda L=0\\ \implies A_2\sin \lambda L=0\\ \text{But, } A_2≠0\\ \implies \sin \lambda L=0\implies \lambda=\frac{n \pi}{L}, n\in \Z\\ \text{Thus, } U(x,t)=A_2e^{-c(\frac{n \pi}{L})^2t}\sin \frac{n \pi}{L}x~~~~~~~~~~~~~~~~(*)$

Now to apply the conditions given in the question.

(i) $U(x,0)=f(x)=U_0$

By superposition principle, (*) becomes

$U(x,t)=∑_{n=0}^\infty​A_n​e^{−c(\frac{n \pi}{L})^2t}\sin \frac{n \pi}{L}​x\\ U(x,0)=\sum_{n=0}^\infty A_n\sin \frac{n \pi}{L} x=U_0$ ​

This is nothing but the Fourier sine series of $U_0$

Where,. $A_n=\frac{2}{L}\int_0^LU_0 \sin\frac{n \pi}{L}x ~dx=\frac{2}{n\pi}U_0[1-(-1)^n]$

Thus $U(x,t)=\sum_{n=0}^\infty \frac{2}{n\pi}U_0[1-(-1)^n] e^{-c(\frac{n \pi}{L})^2t}\sin \frac{n \pi}{L} x\\$

(ii) For the condition $U(x,0)=f(x)=k \sin \frac{ \pi}{L}x$

$U(x,0) \text{ from } (*) is\\ U(x,0)=A_2\sin \frac{n \pi}{L}x= k \sin \frac{\pi}{L}x\\ \text{Comparing both sides}\\ A_2=k, n=1\\ \text{Thus, } U(x,t)=ke^{-c(\frac{\pi}{L})^2}\sin \frac{\pi}{L}x$