求助：利用Hille-Yoshida定理证明算子生成$C_0$半群

Great question! Let's break down how to verify each condition of the Hille-Yosida theorem for your operator, focusing on the resolvent part you're stuck on. First, let's recap your setup clearly:

You're working with the linear operator on $L^2(\mathbb{R})$ (I'm assuming $L^2$ since you mention decay at infinity and resolvent norms) defined by:
$$A w = \partial_{xx} w - a \partial_x w - b x^2 w$$
with domain $D(A)$ consisting of functions $w \in L^2(\mathbb{R})$ where $w, \partial_x w, \partial_{xx} w$ are all in $L^2(\mathbb{R})$, and $w(x) \to 0$ as $x \to \pm\infty$. Your goal is to show $A$ generates a strongly continuous semigroup so you can construct mild solutions to the associated evolution equation.

1. Is $D(A)$ dense in $L^2(\mathbb{R})$? #

Yes, you're correct here. The space of smooth, compactly supported functions $C_c^\infty(\mathbb{R})$ is a subset of $D(A)$ (these functions vanish at infinity and have all derivatives in $L^2$), and $C_c^\infty(\mathbb{R})$ is well-known to be dense in $L^2(\mathbb{R})$. So this condition is satisfied.

2. Is $A$ a closed operator? #

Also correct! Here's a quick way to confirm: suppose we have a sequence ${w_n} \subset D(A)$ where $w_n \to w$ in $L^2$ and $A w_n \to f$ in $L^2$. We need to show $w \in D(A)$ and $A w = f$.

Using integration by parts (since boundary terms vanish because $w_n \to 0$ at infinity), for any smooth compactly supported test function $\phi$, we have:
$$\langle A w_n, \phi \rangle = \int_{\mathbb{R}} w_n \left( \partial_{xx} \phi + a \partial_x \phi - b x^2 \phi \right) dx$$
Taking the limit as $n \to \infty$, the left side becomes $\langle f, \phi \rangle$, and the right side becomes $\langle w, \partial_{xx} \phi + a \partial_x \phi - b x^2 \phi \rangle$. This tells us $w$ is twice weakly differentiable, with $\partial_{xx} w - a \partial_x w - b x^2 w = f$. Additionally, since $w_n \to w$ in $L^2$ and each $w_n$ vanishes at infinity, $w$ must also vanish at infinity (a standard result for $L^2$ limits). Thus $w \in D(A)$ and $A w = f$, so $A$ is closed.

3. Resolvent Condition: $\lambda > 0$ implies $\lambda \in \rho(A)$ and $| R(\lambda, A) | \leq 1/\lambda$ #

This is the tricky part, but we can use an energy estimate to get the norm bound easily, and then confirm surjectivity/injectivity.

First, let's recall the resolvent equation: for a given $f \in L^2(\mathbb{R})$, we need to find $w \in D(A)$ such that:
$$(\lambda I - A) w = f \implies \lambda w - \partial_{xx} w + a \partial_x w + b x^2 w = f$$

Step 1: Dissipativity Check (Key for Hille-Yosida) #

First, we need $A$ to be dissipative, which means $\langle A w, w \rangle \leq 0$ for all $w \in D(A)$. Let's compute this inner product (assuming $a$ is real, which is standard for such operators):
$$\langle A w, w \rangle = \int_{\mathbb{R}} (\partial_{xx} w - a \partial_x w - b x^2 w) \overline{w} dx$$

• The first term: $\int \partial_{xx} w \overline{w} dx = -\int |\partial_x w|^2 dx$ (integration by parts twice, boundary terms vanish).
• The second term: $\int \partial_x w \overline{w} dx = \frac{1}{2} \int \partial_x |w|^2 dx = 0$ (since $|w|^2$ vanishes at infinity).
• The third term: $-b \int x^2 |w|^2 dx$.

So combining these:
$$\langle A w, w \rangle = - | \partial_x w |^2 - b | x w |^2$$
For this to be non-positive, we need $b \geq 0$. I'll assume $b > 0$ here (if $b \leq 0$, the operator might not be dissipative, so Hille-Yosida doesn't apply).

Step 2: Norm Bound for the Resolvent #

Multiply the resolvent equation by $\overline{w}$ and integrate over $\mathbb{R}$:
$$\lambda | w |^2 - \langle A w, w \rangle = \langle f, w \rangle$$
Substitute the expression we found for $\langle A w, w \rangle$:
$$\lambda | w |^2 + | \partial_x w |^2 + b | x w |^2 = \langle f, w \rangle$$
Take absolute values on both sides:
$$\lambda | w |^2 + | \partial_x w |^2 + b | x w |^2 \leq | f | | w |$$
Since all terms on the left are non-negative, we can drop the positive ones to get:
$$\lambda | w |^2 \leq | f | | w |$$
If $w \neq 0$, divide both sides by $| w |$:
$$\lambda | w | \leq | f | \implies | w | \leq \frac{1}{\lambda} | f |$$
This directly gives the norm bound $| R(\lambda, A) f | = | w | \leq \frac{1}{\lambda} | f |$.

Step 3: $\lambda \in \rho(A)$ (Resolvent Exists) #

• Injectivity: If $f = 0$, then the energy equation becomes $\lambda | w |^2 + | \partial_x w |^2 + b | x w |^2 = 0$. All terms are non-negative, so each must be zero. This implies $w = 0$, so $\lambda I - A$ is injective.
• Surjectivity: To show every $f$ has a solution, we can use the Green's function for the ODE. The homogeneous equation $\partial_{xx} w - a \partial_x w - (b x^2 + \lambda) w = 0$ has two solutions: one decaying at $+\infty$ and one at $-\infty$ (these are parabolic cylinder functions). We can construct a Green's function that combines these to give a solution $w(x) = \int_{\mathbb{R}} G(x,y) f(y) dy$, which will be in $D(A)$ and satisfy the decay condition. Alternatively, if $A$ is self-adjoint (which it is if $a$ is real), then its spectrum is real, and since the potential $b x^2$ tends to $+\infty$, the spectrum is bounded below. For $\lambda > 0$ larger than the lower bound of the spectrum, $\lambda$ is in the resolvent, but our energy estimate shows it holds for all $\lambda >0$.

Conclusion #

All three conditions of the Hille-Yosida theorem are satisfied, so $A$ generates a strongly continuous contraction semigroup on $L^2(\mathbb{R})$. This semigroup gives you the mild solutions you're looking for.

内容的提问来源于stack exchange，提问作者Catherine Drysdale