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

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!

• 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 $\lambda>0$). See details in my answer. – Pedro Jan 3 '18 at 21:00

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.

• Wow! So I don't need to use Lumer-Phillips, where we need to prove that $\lambda-A$ is surjective for some $\lambda>0$? And what strategy did you use to come up with that $C_0$-semigroup? – The Phenotype Jan 2 '18 at 1:39
• @ThePhenotype : For me it's just a matter of having seen examples of semigroups with generators that are multiplication operators on some space. The Fourier transform transforms constant coefficient differential operators into multiplications. So these fit the pattern in the Fourier domain. I'm not even clever enough to make up a good lie that would convince you that I'm clever, or I might be tempted to try. :) – DisintegratingByParts Jan 2 '18 at 1:57

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 $H^4(\mathbb R;\mathbb R)$. $\square$

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.
• Awesome! This would indeed prove the surjectivity for the reflexive variant of Lumer-Phillips. I am not sure anymore if using Hille-Yosida can be completed, as it seems impossible (to me) to prove that $A$ is a closed operator. – The Phenotype Jan 3 '18 at 21:37
• @ThePhenotype In the post, we have proved that $(\lambda-A)^{-1}:X\to X$ is bounded. In particular, it is closed and thus its inverse $\lambda-A:D(A)\to X$ is also closed. As $\lambda:D(A)\to X$ is bounded, the operator $A=\lambda-(\lambda-A):D(A)\to X$ is closed, as desired. Note that this argument is general; it does not depend on any particular property of $A$ and can be used provided the estimate $(4)$ is already proved. Alternatively, just invoke the fact that every densely-defined linear operator with non-empty resolvent set is closed. – Pedro Jan 3 '18 at 23:32
• @CatherineDrysdale Here we use the Fourier transform to obtain a formula for a sufficiently regular solution of the equation. This is the main idea. Since we have a formula, we know that any solution has the specific form given by the formula. In other words, we use the Fourier transform to prove a sentence like this: "if $u$ is a solution in the space $X$ then $u$ is given by...". This shows that the problem does not have two differents solutions which proves the uniqueness. – Pedro Feb 1 '18 at 17:54