I'd like to solve the following problem, but I don't know how to approach it.
(Adjoint dynamics) Suppose that $u$ is a smooth solution of
$$\left\{\begin{align}
u_t+Lu&=0\quad \text{in}\quad U_T\\
u&=0\quad \text{on}\quad \partial U\times[0,T]\\
u&=g\quad \text{on}\quad U\times\{t=0\},
\end{align}\right.$$
where $L$ denotes a second-order elliptic operator, and that $v$ is a smooth solution of the adjoint problem
$$\left\{\begin{align}
v_t-L^*v&=0\quad \text{in}\quad U_T\\
v&=0\quad \text{on}\quad \partial U\times[0,T]\\
v&=h\quad \text{on}\quad U\times\{t=T\}.
\end{align}\right.$$
Show
$$\int_U g(x)v(x,0)\;dx = \int_U u(x,T)h(x)\; dx. $$
How to get the integrals in the last equality? Any idea is appreciated.
(In this kind of problem it is hard to do something if we do not know the right idea. So, I don't have tried so much. I've tried to consider the special case $L=\Delta$, but it didn't help me).
EDIT. Solution suggested by @JoeyZou. Is it ok?
Let $L$ be given by
$$Lw=-\sum_{i,j}(a^{i,j}w_{x_i})w_{x_j}+\sum_{i}b^iw_{x_i}+cw$$
Then, by definition, the adjoint $L^*$ of $L$ is given by
$$L^*w=-\sum_{i,j}(a^{i,j}w_{x_j})w_{x_i}-\sum_{i}b^iw_{x_i}+\left(c+\sum_{i}b^i_{x_i}\right)w$$
So, an integration by parts yields
$$\int_UuL^*v\;dx=\int_UvLu\;dx$$
Now, from the equations we have $0=(u_t+Lu)v$ and $uv_t=uL^*v$. So,
$$\begin{align}
0&=\int_{U_T}u_tv\;d(x,t)+\int_{U_T}vLu\;d(x,t)\\\\
&=\int_U\int_0^Tu_tv\;dt\;dx+\int_0^T\int_UvLu\;dx\;dt\\\\
&=\int_U\left(uv\Bigg|_0^T-\int_0^Tuv_t\;dt\right)dx+\int_0^T\int_UuL^*v\;dx\;dt\\\\
&=\int_Uuv\Bigg|_0^T\;dx-\int_{U_T}uv_t\;d(x,t)+\int_{U_T}uv_t\;d(x,t)\\\\
&=\int_U\Big(u(x,T)v(x,T)-u(x,0)v(x,0)\Big)\;dx
\end{align}$$
and thus
$$\int_Ug(x)v(x,0)\;dx=\int_Uu(x,0)v(x,0)\;dx=\int_Uu(x,T)v(x,T)\;dx=\int_Uu(x,T)h(x)\;dx.$$
Best Answer
Notice that $\int\limits_{U_T}{(u_t+Lu)v\text{ d}x\text{ d}t} = 0$. Use the equalities $$ \int\limits_{0}^{T}{u_t(x,t)v(x,t)\text{ d}t} = u(x,T)h(x)-g(x)v(x,0) - \int\limits_{0}^{T}{u(x,t)v_t(x,t)\text{ d}t} $$ and $$ \int\limits_{U}{Lu(x,t)v(x,t)\text{ d}x} = \int\limits_{U}{u(x,t)L^*v(x,t)\text{ d}x}$$ to conclude.