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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 $$\int_Ug(x)v(x,0)\;dx=\int_Uu(x,0)v(x,0)\;dx=\int_Uu(x,T)v(x,T)\;dx=\int_Uu(x,T)h(x)\;dx.$$

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  • $\begingroup$ Without working out the details, it seems that using the equality $\int\limits_{U_T}{(u_t+Lu)v\text{ d}x\text{ d}t} = 0$ and integrating by parts would help. $\endgroup$
    – Joey Zou
    Jan 4, 2016 at 15:24
  • $\begingroup$ Also, is the second condition supposed to read "$v=h$ on $U\times\{t=T\}$" (and not $t=0$)? $\endgroup$
    – Joey Zou
    Jan 4, 2016 at 21:33
  • $\begingroup$ @JoeyZou Yes. I corrected it. $\endgroup$
    – Pedro
    Jan 5, 2016 at 12:48
  • $\begingroup$ @JoeyZou It seems your approach works (see my edit). $\endgroup$
    – Pedro
    Jan 5, 2016 at 16:55
  • $\begingroup$ I'm voting to close this question as the OP found an answer on his own. $\endgroup$
    – Pragabhava
    Jan 5, 2016 at 17:27

1 Answer 1

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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.

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