Question:
Let $u=u(y,t)$. Solve the following PDE (heat equation) in the region $y,t>0$:
\begin{align} & \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial y^2} = \cos(t) \\ & u(0,t) = 0 \\ & u(\infty,t) = \sin (t) \\ & u(x,0) = 0 \end{align}
Attempt:
So first, to make things a little nicer, I let $u(y,t) = v(y,t) + \sin(t)$. The system for $v$ is then given by
\begin{align} & \frac{\partial v}{\partial t} - \frac{\partial^2 v}{\partial y^2} = 0 \\ & v(0,t) = -\sin(t) \\ & v(\infty,t) = 0 \\ & v(x,0) = 0 \end{align}
Now I applied a Laplace Transform with respect to $t$. This gives
\begin{align} & p\bar v - \frac{d^2 \bar v}{dy^2} = 0 \\ & \bar v(0) = -\frac{1}{p^2+1} \\ & \bar v(\infty) = 0 \end{align}
Solving, this gives $\bar v(y) = -\frac{1}{p^2+1}\exp(-\sqrt{p}y)$.
So my question is, is what I have done so far correct?
And, assuming this is correct, how do you invert this to find $v(y,t)$?
Thanks in advance.
