Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 319 / 1

A C0-semigroup on a Banach space $X$ is a family $U(t)\in \mathcal{B}(X)$ such thatfor every $x\in X$. Its infinitesimal generator of a semigroup iswith generator domain

For $M\ge 1$ and $\omega \in \mathbb{R}$, write $A\in \mathcal{G}(M,\omega )$ when $A$ generates a $C_0$-semigroup satisfying $\Vert U(t)\Vert \le M{e}^{\omega t}$. The Hille-Yosida theorem states that this holds exactly when $A$ is closed and densely defined,and, for every real $\lambda >\omega$ and every integer $n\ge 1$,The estimates for every resolvent power, rather than only $n=1$, are essential when $M>1$.

For $x\in X$ and $t>0$, form the Bochner integralThe semigroup property gives, for $h>0$,Strong continuity lets $h\downarrow 0$, yieldingAlso,by strong continuity. Every $x\in X$ is therefore a norm limit of elements of $D(A)$, soThis approximation is the basic Yosida averaging of a semigroup.

Let $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and define the positive self-adjoint operatoron $L^2(\mathbb{T})$. On the Fourier mode $e^{inx}$ it acts by multiplication by $\sqrt{1+n^2}$. ConsequentlyThe energy space is the periodic Sobolev spacewith inner productIf $u=\sum u_n{e}^{inx}$ and $v=\sum v_n{e}^{inx}$, thenwhich is precisely the stated energy norm.

With $v=u_t$, the periodic Klein-Gordon equation becomesFor $AZ$ to belong to $H^1\times L^2$, one needs $v\in H^1$ and $u\in H^2$. Thus

The Lumer-Phillips theorem says that a densely defined operator on a Hilbert space generates a contraction $C_0$-semigroup exactly when it is maximal dissipative:and $\mathrm{Ran}(\lambda I-A)$ is the whole space for some, equivalently every, $\lambda >0$.

For $Z=(u,v)\in D(A)$, self-adjointness of $B$ givesThus both $A$ and $-A$ are dissipative. To check maximality, solveThe equations giveOn Fourier mode $n$, the last operator has multiplier $1+n^2+{\lambda }^2>0$, so it gives a unique $u\in H^2$ and then $v\in H^1$ whenever $(f,g)\in \mathcal{H}$. Hence $\lambda I-A$ is onto; the same calculation applies to $-A$.

The two contraction semigroups generated by $A$ and $-A$ are inverses. They form a strongly continuous unitary group $U(t)$ on the complexification of $\mathcal{H}$, or an orthogonal group on the real space, andfor every $t\in \mathbb{R}$.

SetThen the forced equation is the abstract Cauchy problemFor $Z_0\in \mathcal{H}$ and $F\in C([0,T];\mathcal{H})$, a mild solution of an abstract Cauchy problem is a function $Z\in C([0,T];\mathcal{H})$ satisfying the variation-of-constants formula

Suppose $F(s)\in D(A)$ and both $F$ and $AF$ are continuous. If also $Z_0\in D(A)$, then the closedness of $A$ permits differentiation under the Bochner integral:Thus $Z\in C^1([0,\infty );\mathcal{H})$ and $Z^{\prime }=AZ+F$. The assumption $Z_0\in D(A)$ is necessary here: a unitary group has no smoothing, so the conditions on $F$ alone cannot make $U(t)Z_0$ differentiable for arbitrary $Z_0\in \mathcal{H}$.

For the resulting strong solution, skew symmetry of $A$ gives the energy estimateIntegration yields

Write the nonlinearity asIn one dimension, the Sobolev embedding theorem gives $H^1(\mathbb{T})\hookrightarrow L^{\infty }(\mathbb{T})$. HenceThus $N:\mathcal{H}\to \mathcal{H}$ is locally Lipschitz.

On $C([0,T];\mathcal{H})$ defineOn a ball of radius $R$, unitarity givesChoose $R>\Vert Z_0{\Vert }_{\mathcal{H}}$ and then $T>0$ small enough that the first bound preserves the ball and $CTR<1$. The contraction mapping theorem gives a unique fixed point. Precisely, the local mild solution is

For general energy data $Z_0\in \mathcal{H}$, this solution need not be differentiable in $\mathcal{H}$. If $Z_0\in D(A)=H^2\times H^1$, standard semilinear evolution theory and the smoothness of $N$ give a local classical solution.

The blow-up alternative for a semilinear evolution equation says that the solution continues while its $\mathcal{H}$ norm stays finite, but global existence does not hold for every datum. Spatially constant solutions obeyFor sufficiently large $y(0)>1$ with $y^{\prime }(0)\ge 0$, the solution grows until $y^{\prime \prime }\ge y^2/2$ and blows up in finite time. These constant functions are periodic and belong to every Sobolev space, so they provide finite-time blow-up examples for the original equation.