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We had for this PDE: $-Lu=f$ in $\Omega$ (1) and $u|_{\partial \Omega}=0$ in $\partial \Omega$ (2) the following existence theorem:

Let $\Omega \subseteq \mathbb{R}^n$ be open and $L$ be a uniformly elliptic differential operator of second order. Then there's a$\lambda_0 \in \mathbb{R}$ such that $\forall \lambda \in \mathbb{K}$ with $Re \lambda \geq \lambda_0$ and $\forall f \in H^{-1}(\Omega)$, there is exatly one $u \in H^1_0(\Omega)$ with $\lambda u - Lu =f$ in $H^{-1}(\Omega)$. This means that $$ \lambda \int_\Omega u(x) \overline{\varphi(x)}dx - \langle Lu, \varphi \rangle=\langle f, \varphi \rangle~~ (*)$$ for all $\varphi \in H^1_0(\Omega) $. In particular, we have for (1, 2) with $-L$ replaced by $\lambda-L$ a unique weak solution $u \in H^1_0(\Omega)$ if $f \in L^1_{loc}(\Omega) \cap H^{-1}(\Omega).(**)$

We defined $$\langle Lu, \varphi \rangle= -\sum_{j,k=1}^n \int_\Omega a_{j,k}(x) \partial_{x_j}u(x) \overline{ \partial _{x_k} \varphi(x)} dx+ \sum_{j=1}^n \int_\Omega b_{j}(x) \partial_{x_j}u(x) \overline{ \varphi(x)}dx+ \int_\Omega c(x) u(x) \overline{ \varphi(x)}dx $$.

My Questions:

  1. I understand the statement of this existence theorem, but why is this equal to the statement in (*).
  2. In (**): why does this hold and why needs $f$ to be in this set? If it helps, the statement in our proof was: "As $C^{\infty}_0(\Omega) \subseteq H^1_0(\Omega)$, we get that $u$ is a weak solution of (1,2) under the conditions mentioned in the theorem." Apparently it has to follow from the more general statement but I fail to see this.

Thank you for your help!

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    $\begingroup$ Instead of < ... > you can use \langle ... \rangle or \left< ... \right>. That looks better: $\langle Lu, \varphi \rangle$. $\endgroup$ – md2perpe Aug 30 '17 at 15:32
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To your first question: By the existence theorem, you know that for all (suitable) $\lambda \in \mathbb{K}$ and all $f \in H^{-1}(\Omega)$ the equation \begin{equation} \lambda u - Lu = f \quad \text{ in } H^{-1}(\Omega) \end{equation} has a unique solution $u \in H^1_0(\Omega)$. But in what sense do we understand the above equation? To answer this question, we note that $H^{-1}(\Omega)$ is the dual space of $H^1_0(\Omega)$; that is, $f$ and $\lambda u - Lu$ are linear and bounded maps $H^1_0(\Omega) \to \mathbb{K}$. Thus $\lambda u - Lu = f$ in $H^{-1}(\Omega)$ if and only if $\langle \lambda u - Lu, \varphi \rangle = \langle f, \varphi\rangle $ for all $\varphi \in H^1_0(\Omega)$. Here $\langle \cdot, \cdot \rangle$ is the dual paring. This last equation looks almost like $(*)$.

We now need to understand why $\lambda u$ is an element of $H^{-1}(\Omega)$ for some $\lambda \in \mathbb{K}$ and $u \in H^{1}_0(\Omega)$; recall that $\lambda u \in H^{-1}(\Omega)$ iff $\lambda u$ is a linear and bounded map $H^1_0(\Omega) \to \mathbb{K}$. At first glance this is nonsense, because $\lambda u \in H^1_0(\Omega)$ and not a linear functional. However, this function $\textit{induces}$ a linear and bounded map via \begin{equation} \lambda u \colon H^1_0(\Omega) \to \mathbb{K}\, , \langle \lambda u, \varphi \rangle = \lambda \int_{\Omega} u(x) \overline{\varphi(x)} \, \text{d}x\, . \end{equation} Using this observation, we find that $\langle \lambda u - Lu, \varphi \rangle = \langle f , \varphi \rangle$ is equivalent to $(*)$.

To your second question: The equation \begin{equation} \lambda u - Lu = f \quad \text{ in } H^{-1}(\Omega) \end{equation} is nothing but the weak formulation of \begin{align} \lambda u - L u & = 0 \quad \text{ in } \Omega\, ,\\ u & = 0 \quad\text{ on } \partial \Omega \, . \end{align} The integrability condition on $f$, i.e., $f \in L^1_{loc}(\Omega)$ is not necessary for the existence of a weak solution. However, having that $f$ is in $L^1_{loc}(\Omega)$ we can write $\int_{\Omega} f(x) \overline{\varphi(x)} \, \text{d}x$ instead of $\langle f, \varphi \rangle$.

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