Answer to Question #88293 in Differential Equations for Rwittik

Question

Solve the following boundary value problem
Ut=Uxx,0<x<l,t>0
U(0,t)=U(l,t)=0
U(x,0)=x(l-x),0<=x<=1

Answer 1

Separate variables. Let

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

Then

$$U_t=XT', U_{xx}=X''T$$

Substitute

$$XT'=X''T$$

Dividing this equation by $XT$, we have

$$\frac{T'}{T}=\frac{X''}{X}=-\lambda$$

for some constant $\lambda$. Therefore, if there exists a solution $U(x, t)=X(x)T(t)$ of the heat equation, then $X(x)$ and $T(t)$ must satisfy the equations 

$$\frac{T'}{T}=-\lambda$$

$$\frac{X''}{X}=-\lambda$$

Solve the eigenvalue problem

$$X''+\lambda X=0, X(0)=X(1)=0$$

We have to find nontrivial solutions X of the eigenvalue problem. If λ=0, then $X(x)=\alpha x+\beta$ and $0=X(0)=0+\beta$ implies that $\beta=0$ and $0=X(1)=\alpha$ implies that $\alpha=0$. If $\lambda<0$ then $0=X(0)=\alpha\cosh(0)+\beta\sinh(0)=\alpha$ and $0=X(1)=0+\beta\sinh(1)=\beta\sinh(1)$ shows that also $\beta=0$. We conclude that $\lambda\leq0$  is not an eigenvalue of the problem. Finally, consider $\lambda>0.$ Then $0=X(0)=\alpha\cos(\sqrt{\lambda}\cdot0)+0=\alpha$ and $0=X(1)=\beta\sin(\sqrt{\lambda}\cdot1).$

Since $X$ is nontrivial solution, $\beta=\not0$ and hence $\sin(\sqrt{\lambda}\cdot1)=0.$ Consequently,

$$\lambda=(\frac{n\pi}{1})^2, n\geq1\pi\pi$$

and the corresponding eigenfunction is given by

$$X_n(x)=\sin(\frac{n\pi x}{1})$$

After substituting $\lambda=(\dfrac{n\pi }{1})^2$ , we get the family of solutions

$$T_n(t)=B_ne^{-k({n\pi \over 1})^2t}$$

Thus we have obtained the following sequence of solutions

$$U_n(x, t)=X_n(x)T_n(t)=B_n\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}$$

We obtain more solutions by taking linear combinations of the $U_n$ ’s ( recall the superposition principle) 

$$U(x, t)=\displaystyle\sum_{n=1}^NX_n(x)T_n(t)=\displaystyle\sum_{n=1}^NB_n\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}$$

and then by passing to the limit $N\to \infin,$

$$U(x, t)=\displaystyle\sum_{n=1}^{\infin}X_n(x)T_n(t)=\displaystyle\sum_{n=1}^{\infin}B_n\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}$$

Finally, we consider the initial condition. At $t=0$, we must have

$$U(x, 0)=\displaystyle\sum_{n=1}^{\infin}B_n\sin(\frac{n\pi x}{1})=x(1-x)$$

$$B_n={2 \over 1}\displaystyle\int_{0}^1x(1-x)\sin(\frac{n\pi x}{1})dx$$

$$\int xsin(n\pi x)dx=-{1 \over \pi n}xcos(n\pi x)+{1 \over \pi n}\int cos(n\pi x)dx=$$

$$=-{1 \over \pi n}xcos(n\pi x)+{1 \over \pi^2 n^2}sin(n\pi x)+C_1$$

$$-\int x^2sin(n\pi x)dx={1 \over \pi n}x^2cos(n\pi x)-2{1 \over \pi n}\int xcos(n\pi x)dx$$

$$-2{1 \over \pi n}\int xcos(n\pi x)dx=-2{1 \over \pi^2 n^2}xsin(n\pi x)+2{1 \over \pi^2 n^2}\int sin(n\pi x)dx=$$

$$=-2{1 \over \pi^2 n^2}xsin(n\pi x)-2{1 \over \pi^3 n^3}cos(n\pi x)+C_2$$

$$B_n=2[-{1 \over \pi n}xcos(n\pi x)+{1 \over \pi^2 n^2}sin(n\pi x)]\begin{matrix} 1 \\ 0 \end{matrix}+$$

$$+2[{1 \over \pi n}x^2cos(n\pi x)-{2 \over \pi^2 n^2}xsin(n\pi x)-{2 \over \pi^3 n^3}cos(n\pi x)]\begin{matrix} 1 \\ 0 \end{matrix}=$$

$$=-{2 \over \pi n}cos(n\pi )+{2 \over \pi n}cos(n\pi )-{4 \over \pi^3 n^3}cos(n\pi )+{4 \over \pi^3 n^3}=$$

$$={4(1-(-1)^n) \over \pi^3 n^3}$$

We arrive at the solution

$$U(x, t)=\displaystyle\sum_{n=1}^{\infin}X_n(x)T_n(t)=\displaystyle\sum_{n=1}^{\infin}{4(1-(-1)^n) \over \pi^3 n^3}\sin(\frac{n\pi x}{1})e^{-k({n\pi \over 1})^2t}$$