Let $\;A:W^{2,2}(\mathbb R;\mathbb R^m) \to L^2(\mathbb R;\mathbb R^m)\;$ be the operator:
$\;Au=-u''+D^2G(\bar u)u\;$ where $\;G \in C^3(\mathbb R^m;\mathbb R)\;$ and $\;\bar u:\mathbb R \to \mathbb R^m\;$ is a function such that: $\;\lim_{x \to -\infty} \bar u(x) =l_{-}\;,\;\lim_{x \to +\infty} \bar u(x) =l_{+}\;$
It holds that the quadratic forms $\;D^2G(l_{\pm})\zeta \cdot \zeta \;$ are positive definite. (Here $\;\cdot\;$ denotes the inner product)
I need to show that $\;A\;$ is self adjoint.
Now as I was trying to prove it I found the following:
DEFINITION: $\;P=−\Delta + V\;$ is an operator in $\;L^2(\mathbb R^n)\;$ where $\;V\;$ is a real function in $\;L^1_{loc}(\mathbb R^n)\;$ and $\;\Delta\;$ stands for the Laplacian operator in $\;\mathbb R^n\;$
THEOREM 1:
If $\;H:D(H)→L^2(\mathbb R^n)\;$ where $\;D(H)=\{u\in W^{1,2}(\mathbb R^n)||V|^{1/2}u \in L^2(\mathbb R^n)\;,\;Pu \in L^2(\mathbb R^n)\}\;$ then $\;Hu=Pu \; \forall u\in D(H)\;$ is a self-adjoint operator bounded from below.
Hence, I believe it is sufficient to show that $\;A\;$ can replace $\;H\;$ in the above Theorem.
- $\;u \in W^{1,2}(\mathbb R)\;$ since $\;u \in W^{2,2}(\mathbb R)\;$
- Here $\;V=D^2G(\bar u)\;$. It should be $\;{\vert D^2G(\bar u) \vert}^{1/2} u\in L^2(\mathbb R)\;$. I don't have any idea on how to proceed. I guess I 'll have to use the fact that $\;D^2G(l_{\pm})\;$ are positive definite but I can't see in which way.
Any help would be valuable, because I'm really stuck here. Thanks in advance!