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# 数学代写|偏微分方程代考Partial Differential Equations代写|Compactness of the resolvent

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## 数学代写|偏微分方程代考Partial Differential Equations代写|Compactness of the resolvent

An elliptic operator $L+\mu I$ of the type studied above is a bounded, invertible linear map from $H_0^1(\Omega)$ onto $H^{-1}(\Omega)$ for sufficiently large $\mu \in \mathbb{R}$, so we may define an inverse operator $K=(L+\mu I)^{-1}$. If $\Omega$ is a bounded open set, then the Sobolev embedding theorem implies that $H_0^1(\Omega)$ is compactly embedded in $L^2(\Omega)$, and therefore $K$ is a compact operator on $L^2(\Omega)$.

The operator $(L-\lambda I)^{-1}$ is called the resolvent of $L$, so this property is sometimes expressed by saying that $L$ has compact resolvent. As discussed in Example $4.14, L+\mu I$ may fail to be invertible at smaller values of $\mu$, such that $\lambda=-\mu$ belongs to the spectrum $\sigma(L)$ of $L$, and the resolvent is not defined as a bounded operator on $L^2(\Omega)$ for $\lambda \in \sigma(L)$.

The compactness of the resolvent of elliptic operators on bounded open sets has several important consequences for the solvability of the elliptic PDE and the spectrum of the elliptic operator. Before describing some of these, we discuss the resolvent in more detail.
From Theorem 4.22 , for $\mu \geq \gamma$ we can define
$$K: L^2(\Omega) \rightarrow L^2(\Omega), \quad K=\left.(L+\mu I)^{-1}\right|_{L^2(\Omega)} .$$
We define the inverse $K$ on $L^2(\Omega)$, rather than $H^{-1}(\Omega)$, in which case its range is a subspace of $H_0^1(\Omega)$. If the domain $\Omega$ is sufficiently smooth for elliptic regularity theory to apply, then $u \in H^2(\Omega)$ if $f \in L^2(\Omega)$, and the range of $K$ is $H^2(\Omega) \cap H_0^1(\Omega)$; for non-smooth domains, the range of $K$ is more difficult to describe.

If we consider $L$ as an operator acting in $L^2(\Omega)$, then the domain of $L$ is $D=\operatorname{ran} K$, and
$$L: D \subset L^2(\Omega) \rightarrow L^2(\Omega)$$
is an unbounded linear operator with dense domain $D$. The operator $L$ is closed, meaning that if $\left{u_n\right}$ is a sequence of functions in $D$ such that $u_n \rightarrow u$ and $L u_n \rightarrow f$ in $L^2(\Omega)$, then $u \in D$ and $L u=f$. By using the resolvent, we can replace an analysis of the unbounded operator $L$ by an analysis of the bounded operator $K$.

## 数学代写|偏微分方程代考Partial Differential Equations代写|The Fredholm alternativ

Consider the Dirichlet problem
$$L u=f \quad \text { in } \Omega, \quad u=0 \text { on } \partial \Omega,$$
where $\Omega$ is a smooth, bounded open set, and
$$L u=-\sum_{i, j=1}^n \partial_i\left(a_{i j} \partial_j u\right)+\sum_{i=1}^n \partial_i\left(b_i u\right)+c u$$

If $u=v=0$ on $\partial \Omega$, Green’s formula implies that
$$\int_{\Omega}(L u) v d x=\int_{\Omega} u\left(L^* v\right) d x$$
where the formal adjoint $L^$ of $L$ is defined by $$L^ v=-\sum_{i, j=1}^n \partial_i\left(a_{i j} \partial_j v\right)-\sum_{i=1}^n b_i \partial_i v+c v$$
It follows that if $u$ is a smooth solution of (4.28) and $v$ is a smooth solution of the homogeneous adjoint problem,
$$L^* v=0 \quad \text { in } \Omega, \quad v=0 \quad \text { on } \partial \Omega$$
then
$$\int_{\Omega} f v d x=\int_{\Omega}(L u) v d x=\int_{\Omega} u L^* v d x=0 .$$
Thus, a necessary condition for (4.28) to be solvable is that $f$ is orthogonal with respect to the $L^2(\Omega)$-inner product to every solution of the homogeneous adjoint problem.

For bounded domains, we will use the compactness of the resolvent to prove that this condition is necessary and sufficient for the existence of a weak solution of (4.28) where $f \in L^2(\Omega)$. Moreover, the solution is unique if and only if a solution exists for every $f \in L^2(\Omega)$.

# 偏微分方程代写

## 数学代写|偏微分方程代考Partial Differential Equations代写|Compactness of the resolvent

$$K: L^2(\Omega) \rightarrow L^2(\Omega), \quad K=\left.(L+\mu I)^{-1}\right|_{L^2(\Omega)} .$$

$$L: D \subset L^2(\Omega) \rightarrow L^2(\Omega)$$

## 数学代写|偏微分方程代考Partial Differential Equations代写|The Fredholm alternativ

$$L u=f \quad \text { in } \Omega, \quad u=0 \text { on } \partial \Omega,$$

$$L u=-\sum_{i, j=1}^n \partial_i\left(a_{i j} \partial_j u\right)+\sum_{i=1}^n \partial_i\left(b_i u\right)+c u$$

$$\int_{\Omega}(L u) v d x=\int_{\Omega} u\left(L^* v\right) d x$$

$$L^* v=0 \quad \text { in } \Omega, \quad v=0 \quad \text { on } \partial \Omega$$

$$\int_{\Omega} f v d x=\int_{\Omega}(L u) v d x=\int_{\Omega} u L^* v d x=0 .$$

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