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A question arised when I was self-studying the PDE book of Evans.

Let $Lu = - \sum _{i,j=1}^n (a^{ij} u_{x_i} ) _{x_j} + \sum_{i=1}^n b^{i} u_{x_i} + cu$ be a uniformly elliptic operator on a bounded domain $U$ with $a^{ij}, c \in L^{\infty} (U)$ and $b^i \in C^1 ( \bar U)$. Here, $a^{ij} = a^{ji}$.

The formal adjoint of $L$, denoted $L^*$, is defined as follows.

$L^* v = - \sum_{i,j=1}^n (a^{ij} v_{x_j})_{x_i} - \sum_{i=1}^n b^i v_{x_i} +(c-\sum_{i=1}^n b^i_{x_i} )v$.

It is easy to check that the associated bilinear form to $L^*$ satisfies $B^* [ u,v] = B[v,u]$, where $B$ is associated to $L$. ($u, v \in H^1_0 (U)$)

Now, here is (a part of) the proof stated in the book. Here, $\gamma>0$ is chosen so that $\beta ||u||^2 _ {H^1_0} \leq B[u,u] + \gamma ||u||^2_{L^2}$ for all $u \in H^1_0 (U)$.

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Now, the author claims that: $v \in L^2(U) $ satisfies $v=K^* v$ if and only if $v \in H^1_0 (U) $ is a weak solution of the system [$L^*v = 0 $ on $U$, $v =0$ on $\partial U$]. (Here, $K^* : L^2 (U) \rightarrow L^2 (U)$ is the adjoint operator of the compact operator $K$.)

How can I prove this claim? (Note: The analogous claim for $L$ and $K$ is indeed obvious.)

J. Doe
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1 Answers1

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In my proof I will let ( , ) to denote the $L^2(U)$ inner product

Let $v∈L^2(U)$

$v = K^*v $ iff

($v$,$u$) = ($K^*v$,$u$) = ($v$,$Ku$) for any $u∈L^2(U)$

$=\gamma$($v$,$L_{\gamma}^{-1}$$u$) by Definition of $K$ in (18)

iff (this holds because we can replace $u$ with $L_\gamma$$u$)

($v$,$L_{\gamma}u$) = $\gamma(v,u)$ for any $u∈L^2(U)$

iff

$0$ = ($v$,$Lu$) = ($L^*v$,$u$) = $B^*$[$v$,$u$] for any $u∈L^2(U)$

iff $[L^∗v=0$ on $U$, $v=0$ on $∂U]$

Hope that can help.

I just guess the solution from the uppest and the lowest logcally equivalence and luckily got the second and third

Pi Pi
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