- Solve
D.E. $u_t-u_{xx}=e^{-4t}cos(t)sin(2x) 0 \leq x \leq \pi, t \geq 0$
B.C. $u(0,t)=0, u(\pi, t)=0$
I.C. $u(x,0)=sin(3x)$
Attempt:
Let $f(x) = sin(3x), a(t) =0$, and $b(t) = 0$. Then $w(x,t)$ will be
$w(x,t) = (0-0) \frac{x}{\pi} + 0 = 0$
$w(0,t) = 0$ $w( \pi, t) = 0$
$u(0, t) = 0$ $u( \pi, t) = 0$
D.E $v_t-v_{xx}k=u_t-ku_{xx}-w_t+kw_{xx} \rightarrow e^{-4t}cos(t)sin(2x)$
B.C $v(0,t) = u(0,t) -w(0,t) = 0-0 = 0$
$v( \pi, t) = u( \pi,t) - w( \pi, t) = 0-0 = 0$
I.C. $v(x,0) = u(x,0) -w(x,0)$
$v(x,0) = sin(3x)$
By Proposition 1
$b_n e^-(\frac{n \pi}{L})^{2kt}sin( \frac{n \pi x}{L})$
Let $n = 3$, $k=1$, and $L = \pi$
$b_3 e^-(\frac{3 \pi}{\pi})^{2t}sin( \frac{3 \pi x}{\pi})$
$b_3 e^{-9t}sin(3x)$
$v_t-v{xx}k = e^{-4t}cos(t)sin(2x) \rightarrow 0 $
$v(0,t) = 0 \rightarrow v(0,t,s) = 0$
$v(\pi, t) = 0 \rightarrow v( \pi,t,s) = 0$
$v(x,0) = 0 \rightarrow v(x,0,s) = e^{-4s}cos(s)sin(2x)$
The next steps involve integrating from 0 to T for $e^{-4s}cos(s)sin(2x)$. I know that the sin(2x) isn't going to be integrated because we are integrating with respect to s. The problem is that my integration by parts is out of whack. I know I have to set $u = cos(s)$ and $du = -sin(s)$. The dv and v would be $ e^{-4s}$ and $\frac{-1}{4}e^{-4s}$
$udv = uv - \int vdu$
$ cos(s)e^{-4s}=\frac{-1}{4}e^{-4s}cos(s) - \int sin(s)\frac{1}{4}e^{-4s} $
But if I integrate from the right hand side I would have a $\frac{1}{16}e^{-4s}$
This seems to be the only issue because the answer has a $sin(s)$ along with the exponent and $sin(2x)$.