I have this problem: $$ \left\{\begin{matrix} u_{xx}= u_{t}+u && 0<x< +\infty & t>0\\ u(x,0)=0 &&& x>0&\\ u_{x}(0,t)=f(t)&&& t\geq 0 \end{matrix}\right. $$ I need to solve this problem with sine and cosine transforms, the definition used for sine and cosine transforms are the following ones:
$Fc(f)(w)=\int_{0}^{+\infty} f(x) cos(wx)dx$
$Fs(f)(w)=\int_{0}^{+\infty} f(x) sin(wx)dx$
and the antitransforms are:
$Fc^{-1}(f)(w)= \frac{2}{\pi} \int_{0}^{+\infty} \hat{f}(w) cos(wx)dw$
$Fs^{-1}(f)(w)= \frac{2}{\pi} \int_{0}^{+\infty} \hat{f}(w) sin(wx)dw$
I tried to think about taking the sine transform on both sides of the differential equation, then:
$\hat{u}(w,t)=\int_{0}^{+\infty} u(x,t) sin(wx)dx$
$\hat{u}_{xx}(w,t)=-(w)^{2}\hat{u}(w,t)$
$\hat{u}_{t}(w,t)=\frac{d}{dt}\hat{u}(w,t)$
So, when I replace that in the differential equation:
$-(w)^{2}\hat{u}(w,t)=\frac{d}{dt}\hat{u}(w,t) + \hat{u}(w,t)$
The solution is:
$\hat{u}(w,t)=A(w)e^{(-1-w²)t}$
But then if $t=0$
$\hat{u}(w,0)=\int_{0}^{+\infty} u(x,0) sin(wx)dx=0$
$\hat{u}(w,0)=A(w)=0$ (the trivial solution)
What should I do? There's no difference if I take the cosine transform.
Hint
As you have mentioned in your comments, we have: $$Fc(f_x)=wFs(f)-f(0)$$ $$Fs(f_x)=-wFc(f)$$ so: $$Fc(f_{xx})=wFs(f_{x})-f_{x}(0)=-w^2 Fc(f)-f_{x}(0)$$ $$Fs(f_{xx})=-wFc(f_{x})=-w^2 Fs(f)+w f(0)$$ and in your equation you obtain: $$Fs(u_{xx})(w,t)=-w^2 Fs(u)(w,t)+w \times 0$$ $$Fc(u_{xx})(w,t)=-w^2 Fc(u)(w,t)(w,t)-\color{red}{ f(t)}$$ the red term is the one that generate an non zero solution.