# Proving an operator is surjective for Lumer-Phillips Theorem application.

I am trying to show that the following $\lambda I - \mathcal{L}_0$ is subjective for all $\lambda > 0$. The purpose of this is to use the Lumer-Phillips Theorem. The operator $\mathcal{L}_0$ is given by $(-c_\mu^2 + \gamma \partial_{xx} - \nu \partial_x)$, where $c_{\mu}$ is a constant. The domain of this operator is $D_{\mathcal{L}_0}= H_0^1(\mathbb{R},\mathbb{C})\cap H^2(\mathbb{R},\mathbb{C})$. Note that the linear operator maps in $\mathcal{L_0}:D_\mathcal{L_0} \subset X \longrightarrow X$, where $X$ is given by $L^2(\mathbb{R},\mathbb{C})$.

I thought that it was best to show that the range of this operator is dense and closed, but I am stuggling to show either of these things.

To show the range is dense:

So far, my working has been let $y \in X$ and let $u \in D_{\mathcal{L}}$. For the range of $\lambda I - \mathcal{L}_0$ to be dense, we must stow that for all $x \in D_{\mathcal{L}_0}$, $v \in X$, $\langle (\lambda I - \mathcal{L}_0) x, v \rangle = 0$ only when $v = 0$. I cannot thing what to choose $x$ as to give a case when $v=0$.

Also, I have been thinking of wanting to use the other version of the Lumer-Phillips theorem, because both the operator and the adjoint are dissipative. In which case, how would I show $L_0$ is a closed operator.

Kindest regards,

Catherine

Let us apply the idea explained here.

Take $\lambda>0$. You want to show that: given $f\in L^2(\mathbb R;\mathbb C)$, there exists $u\in H_0^1(\mathbb{R},\mathbb{C})\cap H^2(\mathbb{R},\mathbb{C})$ such that $$\lambda u+c_\mu^2u - \gamma u_{xx} + \nu u_x=f\tag{1}$$ where I'm assuming that $c_\mu^2$, $\gamma$, and $\nu$ are positive constants.

Taking the Fourier Transform, we conclude that $$u=\left(\frac{\hat{f}}{\lambda +c_\mu^2 + \gamma x^2 + \mathbf i \nu x}\right)^\vee.\tag{2}$$ This shows that any solution $u$ of $(1)$ has the form $(2)$. It remains to prove that the function $u$ given by $(2)$ is indeed a solution of the equation in $H_0^1(\mathbb{R},\mathbb{C})\cap H^2(\mathbb{R},\mathbb{C})$.

Proof:

Since $\hat{f}\in L^2(\mathbb R;\mathbb C)$ and $$\left|\frac{\hat{f}}{\lambda +c_\mu^2 + \gamma x^2 + \mathbf i \nu x}\right|^2=\frac{|\hat{f}|^2}{(\lambda +c_\mu^2 + \gamma x^2)^2 +\nu^2 x^2}\leq \frac{|\hat{f}|^2}{(\lambda+c_\mu^2)^2},$$ it follows that $$\frac{\hat{f}}{\lambda +c_\mu^2 + \gamma x^2 + \mathbf i \nu x}\in L^2(\mathbb R;\mathbb C).$$

Thus, $u$ given by $(2)$ is well-defined and belongs to $L^2(\mathbb R;\mathbb C)$. Taking the Fourier transform of $(2)$ (which can be done because $u\in L^2$), we obtain $$(\lambda +c_\mu^2 + \gamma x^2 + \mathbf i \nu x)\hat{u}=\hat{f}\in L^2\tag{3}$$ and thus $(1+x^2)\hat{u}\in L^2$ which implies that $$u\in H^2(\mathbb R;\mathbb C).\tag{4}$$ From $(3)$, $$\lambda \hat{u}+c_\mu^2\hat{u} - \gamma (\mathbf i x)^2 \hat{u} + \nu (\mathbf i x)\hat {u}=\hat{f}$$ and then, by $(4)$, $$\lambda u+c_\mu^2u - \gamma u_{xx} + \nu u_x=f.$$

This shows that $(2)$ is a solution of $(1)$ in $H^2(\mathbb R;\mathbb C)=H_0^1(\mathbb{R},\mathbb{C})\cap H^2(\mathbb{R},\mathbb{C})$. $\square$