Generating a $C_0$-semigroup on $L^2$

Question

Consider the linear operator $$A : H^4(\mathbb{R}; \mathbb{R}) \to L^2(\mathbb{R};\mathbb{R})$$ defined by $u\mapsto -(1-\partial_{xx}^2)^2$. Show that $A$ generates a $C_0$-semigroup on $L^2$.

I believe I was suggested to use Fourier transforms, so I found that \begin{align}F(Au)(\omega)&=F((-1+2\partial_{xx}^2-\partial_{xxxx}^4)u)(\omega)=(-1-2\omega^2-\omega^4)F(u)(\omega)\\ &=-(\omega^2+1)^2F(u)(\omega).\end{align}

But from here on I am totally lost. I know that I should use some theorem like Hille-Yosida or Lumer–Phillips, but I have no idea how to combine it with Fourier transforms.

Thanks in advance and happy New Year!

Answer 1

The $C_0$ semigroup would be $$ T(t)u = F^{-1}(e^{-t(1+\omega^2)}F(u)), \;\;\; t \ge 0, $$ where $F$, $F^{-1}$ are the Fourier transform and inverse Fourier transform. You can directly verify that $T$ is a $C_0$ semigroup.

Answer 2

I know that I should use some theorem like Hille-Yosida or Lumer–Phillips, but I have no idea how to combine it with Fourier transforms.

In general, Fourier transforms are used in this context (when allowed) to prove existence and uniqueness of solution for the resolvent equation $(\lambda -A)u=f$, which is needed in the application of the Hille-Yosida Theorem (for all $\lambda>0$) or Lumer-Phillips Theorem (for some $\lambda>0$).

Details:

If you want to apply the Hille-Yosida Theorem, you have to show (among other things) that each $\lambda>0$ belongs to $\rho(A)$.

In your case, the operator $A:D(A)\subset X\to X$ is defined by $$D(A)= H^4(\mathbb R;\mathbb R),\qquad X=L^2(\mathbb R;\mathbb R), \qquad Au=-u+2u_{xx}-u_{xxxx}.$$

So, fixed $\lambda>0$, you have to show that: given $f\in L^2(\mathbb R;\mathbb R)$, there exists a unique $u\in H^4(\mathbb R;\mathbb R)$ such that $$\lambda u+u-2u_{xx}+u_{xxxx}=f.\tag{1}$$

Proof of uniqueness:

Let $u$ be a solution in $H^4(\mathbb R;\mathbb R)$ of $(1)$. Then, taking the Fourier Transform, we conclude that $$u=\left(\frac{\hat{f}}{\lambda +1+2x^2+ x^4}\right)^\vee.$$ This shows that $(1)$ have at most one solution in $H^4(\mathbb R;\mathbb R)$. $\square$

Proof of existence:

Define $$u=\left(\frac{\hat{f}}{\lambda +1+2x^2+ x^4}\right)^\vee.\tag{2}$$

Since $\hat{f}\in L^2(\mathbb R;\mathbb R)$ and $\left|\frac{\hat{f}}{\lambda +1+2x^2+ x^4}\right|\leq |\hat{f}|$, it follows that $\frac{\hat{f}}{\lambda +1+2x^2+ x^4}\in L^2(\mathbb R;\mathbb R)$. Thus $u$ is well-defined and belongs to $L^2(\mathbb R;\mathbb R)$. Taking the Fourier transform of $(2)$, we obtain $$(\lambda+1+2x^2+x^4)\hat{u}=\hat{f}\in L^2.\tag{3}$$ and thus $(1+x^4)\hat{u}\in L^2$ which implies that $u\in H^4(\mathbb R;\mathbb R)$. From $(3)$, $$\lambda \hat{u}+\hat{u}-2(\mathbf i x)^2\hat{u}+(\mathbf ix)^4\hat{u}=\hat{f}$$ and then, since $u\in H^4(\mathbb R;\mathbb R)$, it follows that $$\lambda u+u-2u_{xx}+u_{xxxx}=f.$$

This shows that $(1)$ have a solution in $H^4(\mathbb R;\mathbb R)$. $\square$

Addendum:

From $(2)$, $$\|(\lambda-A)^{-1}f\|_{L^2}=\|u\|_{L^2}=\|\hat{u}\|_{L^2}=\left\|\frac{\hat{f}}{\lambda +1+2x^2+ x^4}\right\|_{L^2}\leq \frac{1}{\lambda}\|\hat{f}\|_{L^2}=\frac{1}{\lambda}\|f\|_{L^2}$$ and thus $$\|(\lambda-A)^{-1}\|_{\mathcal{L}}\leq \frac{1}{\lambda}\tag{4}$$ which is an estimate that also have to be proved in the application of Hille-Yosida.