Question

Solve the following PDE using the method of separation of variables and Fourier series

u_tt = c²u_xx, 0<x<1, t<0

u(0,t) = 0, t>0

u(1,t) = 0, t>0

u(x,0) = 2x(1-x), 0<=x<=1

u_t(x,0) =0

Answer 1

We consider the wave equation satisfying Dirichlet boundary condition:

$u_{tt}=c^2u_{xx}$ , $0<x<1,\ \ t>0$

$u(0,t)=u(1,t)=0$ , $t\geq 0$

$u(x,0)=2x(1-x),\quad 0\leq x\leq 1$

$u_t(x,0)=0,\quad\quad \quad \quad \ 0\leq x\leq 1$

We look for solutions of the form $u(x,t)=\sum \limits_{n=1}^\infty\big(A_n\cos (\pi nct)+B_n\sin (\pi nct)\big) \sin (\pi nx)$

Setting $t=0$ , we get

$u(x,0)= \sum \limits_{n=1}^\infty A_n\sin (\pi nx)=2x(1-x)$

Coefficients $A_n$ are equal to $A_n=2\int \limits_0^1 2x(1-x)\sin (\pi nx)dx=- \frac{\cos (\pi nx)}{\pi n} 4x(1-x)\bigg|_0^1+\int\limits_0^1 \frac{\cos (\pi nx)}{\pi n}4(1-2x)dx= \int\limits_0^1 \frac{\cos (\pi nx)}{\pi n}4(1-2x)dx= \frac{\sin (\pi nx)}{\pi ^2 n^2 }4(1-2x)\bigg|_0^1 +\frac{8}{\pi^2n^2}\int\limits_0^1 \sin (\pi n x)dx=-\frac{\cos (\pi n x)}{\pi n}\cdot \frac{8}{\pi^2n^2}\bigg|_0^1 =\frac{8}{\pi^3n^3}(1-\cos(\pi n))=\begin{cases} 0, \quad n=2k \\ \frac{16}{\pi^3n^3}, \quad n=2k+1 \end{cases}$

$u_t(x,t)=\sum \limits_{n=1}^\infty \pi n c\big(B_n\cos (\pi nct)-A_n\sin (\pi nct)\big) \sin (\pi nx)$

Setting $t=0$ , we get 

$u_t(x,0)=\sum\limits_{n=1}^\infty \pi n cB_n\sin (\pi n x)=0$

Therefore, $B_n=0\ \ \forall n$ .

Answer: $u(x,t)=\sum\limits_{k=0}^\infty \frac{16}{\pi^3(2k+1)^3}\cos (\pi (2k+1)ct) \sin (\pi(2k+1)x).$