Solving the Heat Equation on The Quarter Plane Consider the intial boundary value problem 
\begin{align*}
u_t +u &= u_{xx} \\
u(x,0) &= f(x) \\
u(0,t) &= g(t) \\  
\end{align*}
on the domain $x >0$, $t > 0$, where $f$ and $g$ are continuous functions with compact support. Find a solution of this problem. 

First, I made the substitution $v = e^tu$ to reduce the PDE to
\begin{align*}
v_t &= v_{xx} \\
u(x,0) &= f(x) \\
u(0,t) &= h(t), \\  
\end{align*}
where $h(t) = e^tg(t)$. 
I then attempted to decompose the problem as $v = r + w$, where 
\begin{align*}
r_t &= r_{xx} \\
r(x,0) &= f(x) \\
r(0,t) &= 0, \\  
\end{align*}
\begin{align*}
w_t &= w_{xx} \\
w(x,0) &= 0 \\
w(0,t) &= h(t). \\  
\end{align*}
 I know that $$r(x,t) = \int_{0}^{\infty} (\Phi(x-y,t)-\Phi(x+y,t))f(y)dy, $$
where $\Phi$ is the fundamental solution of the heat equation. However, I'm unsure on how to solve for $w$. If this is the correct approach, how can I proceed?
 A: First, take the Laplace transform in $t$ of the heat equation for $w$:
$$
\partial_{xx}W(x,s) = sW(x,s) -W(x,0) = sW(x,s),
$$
where $W(x,s) = \mathcal{L}_t[w(x,t)](s)$. We can discard the growing solution in $x$ since $w(x,t)$ vanishes as $x\rightarrow \infty$, so the solution is $W(x,s) = C(s)e^{-x\sqrt{s}}$ for some function $C$. But $W(0,s)$ is just the Laplace transform of $h$, so we must have $C(s) = \mathcal{L}_t[h(t)](s)$. Inverting the Laplace transform and using the convolution property gives
$$
w(x,t) = \mathcal{L}_s^{-1}\left[H(s)e^{-x\sqrt{s}}\right](t) = \int_0^t h(\tau)\mathcal{L}_s^{-1}\left[e^{-x\sqrt{s}}\right](t-\tau) d\tau.
$$
This is an answer, since $\mathcal{L}_s^{-1}\left[e^{-x\sqrt{s}}\right](t)$ is a well-defined function of $x$ and $t$ even if you don't have a closed form for it. That said, a table of Laplace transforms will give you the closed form:
$$
\mathcal{L}_s^{-1}\left[e^{-x\sqrt{s}}\right](t) = \frac{x}{2\sqrt{\pi}}\frac{\Theta(t)}{t^{3/2}}\exp\left(-\frac{x^2}{4t}\right),
$$
where $\Theta$ is the unit step function. This function is similar to the space kernel: it solves the heat equation, vanishes as $t\rightarrow 0$, has unit $t$ integral for all $x$, and reduces to a delta function in $t$ as $x\rightarrow 0$. This gives a full integral form for $w$:
$$
w(x,t) = \frac{x}{2\sqrt{\pi}}\int_0^t \frac{h(\tau)}{(t-\tau)^{3/2}}\exp\left[-\frac{x^2}{4(t-\tau)}\right]d\tau.
$$
