I am trying to solve the following problem:
Let $\Omega$ = (0, $\infty$), $f$ $\in$ $C$([0, $\infty$) $\times$ $\bar{\Omega}$), $h$ $\in$ $C$([0, $\infty$)). Consider the PDE Problem
$$
\left\{
\begin{array}{ll}
u_t – \Delta u &= f & \text{in } (0, \infty) \times \Omega, \\
u + u_x &= h & \text{on } \partial \Omega, \\
u(0, x) &= g(x) & \text{for } x \in \Omega
\end{array}
\right.
$$
Determine the PDE Problem which the corresponding Green's function must solve and derive Green's formula.
Hint: You may assume $u$ to be smooth, and thus
$$
\int_{\Omega} \delta(x-y)u(t,x)dx = u(t,y).
$$
My approach was to first find the adjoint differential operator to $L$ = $\partial_t$ $-$ $\Delta$ by testing with the desired Green's function $G^{(s,y)}$ for some time $s$ and some point $y$ in $\Omega$, resulting in the following expression:
$$
\int_{\Omega_s} f G^{(s,y)}d(t,x) = \int_{\Omega}u G^{(s,y)} \vert_{t=s}dx – \int_{\Omega}u G^{(s,y)} \vert_{t=0}dx + \int^s_0 \int_{\partial \Omega} u \partial_{\nu}G^{(s,y)} – G^{(s,y)} \partial_{\nu} u \ dS dt + \int_{\Omega_s}(L^* G)u\ d(t,x)
$$
where $L^*$ denotes the adjoint operator $-\partial_t$ $-$ $\Delta$ to $L$. I am pretty sure the system of equations has to look something like this:
$$
\left\{
\begin{array}{ll}
L^*G^{(s,y)} &= f & \text{in } (0, s) \times \Omega, \\
? &= \ ? & \text{on } \partial \Omega, \\
G^{(s,y)} &= \delta(x-y) & \text{for } t = s
\end{array}
\right.
$$
However, I have no idea how the boundary condition for $G$ should look in order to somehow include the function $h$ in the solution. Any help would be appreciated.
Best Answer
I figured it out. By multiplying the PDE for $u$ with $G$ and integrating one gets: $$ \begin{align} \int_{\Omega_s}G^{(s,y)} (u_t - \Delta u) d(t,x) &= \int_{\Omega}G^{(s,y)}(s,x) u(s,x) - G^{(s,y)}(0,x)u(0,x)dx - \int_{\Omega_s} G^{(s,y)}_t u\ d(t,x) \\ &- \int_0^s \int_{\partial \Omega} G^{(s,y)} \frac{\partial u}{\partial \nu} - u \frac{\partial G^{(s,y)}}{\partial \nu} dSdt - \int_{\Omega_s} u \Delta G^{(s,y)}d(t,x), \end{align} $$ by partial integration, and by substituting in the boundary conditions for $u$, and the two conditions for $G^{(s,y)}$ which I included above, one gets $$ \begin{align} \int_{\Omega_s} G^{(s,y)} f\ d(t,x) &= u(s,y) - \int_{\Omega} G^{(s,y)}(0,x)g(x)\ dx + \int_0^s \int_{\partial \Omega}G^{(s,y)} h\ dSdt\\ &- \int_0^s \int_{\partial \Omega} u \Bigl(G^{(s,y)} - \frac{\partial G^{(s,y)}}{\partial \nu} \Bigr)dSdt. \end{align} $$ Thus, the remaining condition for Green's Function must be $$ G^{(s,y)} + G^{(s,y)}_x = 0 \text{ on } (0,s) \times \Omega $$ in order to make the last term containing $u$ in an integral to vanish. One then gets the following Green's Formula: $$ u(s,y) = \int_{\Omega_s} G^{(s,y)} f\ d(t,x) + \int_{\Omega} G^{(s,y)}(0,x)g(x)\ dx - \int_0^s \int_{\partial \Omega}G^{(s,y)} h\ dSdt. $$