Question

u_tt =c²u_xx=, 0<x<1

u(0,1)=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

Using the variable separable method.

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

Then, the given differential equation becomes

$X(x)T''(t) = c^2X''(x) T(t)$

So we have that

$\dfrac{1}{c^2T} \dfrac{d^2T}{dt^2} = \dfrac{1}{X} \dfrac{d^2X}{dx^2}$

Using the information that

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

We have that the general solution is of the form

$u(x,t) = \sum_{n=1}^{\infty}[A_n \cos(n\pi c)t +B_n \sin (n\pi c)t] [sin(n\pi )x]$

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

$u_t(x,0) = \sum_{n=1}^{\infty} B_n n\pi c \sin (n \pi )x = 0$

With

$A_n = 2 \int_{0}^{1} 2x(1-x) \sin (n \pi )x dx$

And

$B_n = \dfrac{2}{n \pi c} \int_{0}^{1} 0 \sin (n\pi )x dx = 0$

We evaluate the integral

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

$[(2x-2x^2)(\dfrac{-1}{n\pi} \cos n\pi x)]_0^1 + [(2-4x)(\dfrac{1}{(n\pi)^2} \sin n\pi x)]_0^1$

$+ \dfrac{4}{(n\pi)^2} \int_0^1 \sin (n\pi )xdx$

$= \dfrac{4}{(n\pi)^3} [(-1)^{n+1}+1]$

Hence,

$A_n = \dfrac{8}{(n\pi)^3} [(-1)^{n+1}+1]$

But since the value of $B_n = 0$

We have that the solution with respect to the given conditions becomes

$u(x,t) = \sum_{n=1}^{\infty} A_n \cos(n \pi c)t \sin( n \pi )x$

where

$A_n = \dfrac{8}{(n\pi)^3} [(-1)^{n+1}+1]$