EXERCISE
a)Give the solution of the problem:
$u_t(x,t)-u_{xx}(x,t)=2e^tcosx ,x>0 ,t>0$
$u(0,t)=e^t, t\geq 0$
$u(x,0)=cosx,x\geq0$
Use Laplace transformation
b)Solve the problem:
$u_{tt}-u(x,t)=3u_{xx}(x,t) , x\in(0,\infty)$
$u(x,0)=e^{-x}, u_t(x,0)=-2e^{-x}$
$u(0,t)=e^{-2t}, t\geq 0$
Use Fourier transformation
MY ATTEMPT
a) $L(u(x,t))=\int_{0}^{\infty} u(x,t) e^{-st} dx=u(x,s)$ where L is the symbol of laplace transformation
We have that: $u_t(x,t)-u_{xx}(x,t)-2\cdot e^t\cdot cosx=0$
Due to Laplace transformation the equation transforms into: $su(x,s)-u(x,0)-\frac{\partial^2{u(x,s)}}{\partial{x^2}}-\frac{2cosx}{s-1}=0 \iff su(x,s)-cosx-\frac{\partial^2{u(x,s)}}{\partial{x^2}}-\frac{2cosx}{s-1}=0 \iff \frac{\partial^2{u(x,s)}}{\partial{x^2}}-su(x,s)=-cosx-\frac{2cosx}{s-1}$
So the solution of the PDE is:
$u(x,s)=A(s)e^{\sqrt s x}+B(s)e^{-\sqrt s x} + \frac{1}{s+1}cosx$
(Is it right or there are problems in computation?)
So now with the help of L symbol( Laplace transformation formula) we have:
$u(0,t)=e^t\implies u(0,s)=\frac{1}{s-1}$
So we have that:
$u(0,s)=\frac{1}{s-1}=A(s)+B(s)+\frac{1}{s+1} \iff A+B=\frac{2}{(s-1)(s+1)}$
So here I assumed that because of $x\geq 0, u(x,t)\longrightarrow 0, u(x,s)\longrightarrow 0 \Longrightarrow A(s)=0 , B=\frac{2}{(s-1)(s+1)}$
So, we have that:
$u(x,s)=\frac{2}{(s-1)(s+1)}e^{-\sqrt s x} + \frac{1}{s+1}cosx$
So, from here and then I am not sure how to proceed! I know that I have to take that:
$u(x,t)=L^{-1}(u(x,s))=L^{-1}(\frac{2}{(s-1)(s+1)}e^{-\sqrt s x} + \frac{1}{s+1}cosx)$ but the calculation seems very compicated to me.
I think that I am losing something here. I would really appreciate a thorough solution and explanation since I've just started working on these problems.
b)I am not sure what kind of Fourier transformation I have to use. I tried the simple Fourier transformation but I didn't figure it out. May I have to use Fourier cosine or sine transform? I would really appreciate a thorough solution and explanation here too.
Thanks, in advance!