I'm studying the second order elliptic partial differential equations in the 'Partial Differential Equations, EVANS'. The section 6.2.3 begins with defining the adjoint operator $L^*$ of the operator $$Lu=-\sum_{i,j} (a^{ij}u_{x_i})_{x_j}+ \sum _i b^iu_{x_i}+cu$$
by
$$ L^*v = -\sum_{i,j} (a^{ij}v_{x_j})_{x_i}- \sum _i b^iv_{x_i}+(c-\sum_i b^i_{x_i})u.$$
This definition follows from the simple calculation if we use the inner product on $L^2(U)$. But I think that we have to use the inner product on $H^1_0$, i.e. $(u, v)=\int_U uv+Du \cdot Dvdx$ because we want to find solutions in the Sobolev space $H^1_0$. Please give any advise for me. Thanks.
[Math] The adjoint operator of the second order partial differential operator.
partial differential equations
Best Answer
No, the use of $L^2$ inner product is appropriate here. In this subject it's not unusual to borrow structure from different function spaces at once. E.g., later in the section you will see statements like "$u_j\rightharpoonup u$ weakly in $H^1_0$ and also $u_j\to u$ strongly in $L^2$".
This use of $L^2$ product is consistent with the way in which we associate a bilinear form on $H_0^1$ to an elliptic operator. I'll use Laplacian $L=-\Delta$ for illustration, which corresponds to the bilinear form $B[u,v]=\int_\Omega \nabla u\cdot \nabla v $. How does this correspondence come about? For smooth compactly supported functions, it's integration by parts: $$ (Lu,v) = \int_\Omega -\Delta u \, v = \int_\Omega \nabla u\cdot \nabla v $$ Note that the inner product on the left is from $L^2$. The main point is that when we stick two derivatives into an $L^2$ inner product, the result is a bilinear form that is naturally defined on $H_0^1$.