Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 105 / 3 / b

Write $B=-A=-{\partial }_x^2+1+x^2$. Smooth compactly supported functions lie in $D(A)$ and are dense in L2 space, so $A$ is densely defined.

For $\lambda >0$, use the formon the one-dimensional harmonic oscillator form domain $X$. It is bounded and coercive in the $X$ norm. The Lax-Milgram theorem gives a unique $u\in X$ satisfying $a_{\lambda }(u,v)=⟨g,v⟩$ for every $v\in X$. Thus $-u^{\prime \prime }+(1+\lambda +x^2)u=g$ in distributions.

We must verify the full specified operator domain, not just the form domain. The equation first gives $u\in H_{\mathrm{loc}}^2$. For compactly supported $H^2$ functions, integration by parts gives the graph estimate for the harmonic oscillator, obtained fromApply this to $v={\chi }_{R}u$, with ${\chi }_R=1$ on $[-R,R]$, supported in $[-2R,2R]$, and derivatives bounded by $C/R$ and $C/R^2$. Its right-hand source is ${\chi }_{R}g-2{\chi }_R^{\prime }{u}^{\prime }-{\chi }_R^{\prime \prime }u$, uniformly bounded in L2 space. The estimate and the Fatou lemma give $x^{2}u\in L^2$; the original equation then gives $u^{\prime \prime }\in L^2$, and hence $u\in H^2$. Thus $u\in D(A)$, proving that $\lambda -A$ is onto. The energy identity proves injectivity and the resolvent of the shifted harmonic oscillator estimateTo prove closedness, suppose $u_n\to u$ and $A{u}_n\to v$ in L2 space. For one fixed $\lambda >0$,Therefore $u$ belongs to the range of the resolvent, namely $D(A)$, and $Au=v$.

All hypotheses of the Hille-Yosida theorem now hold; the resolvent bound is stronger than $1/\lambda$. Let $S(t)$ be its contraction semigroup. The exponentially shifted semigrouphas generator $A+I={\partial }_x^2-x^2$. It is continuous in L2 space, takes the initial value ${\psi }_0$, and satisfies $\boxed{\Vert \psi (t){\Vert }_2\le e^t\Vert {\psi }_0{\Vert }_2}$. For ${\psi }_0\in D(A)$, its time derivative exists in L2 space and equals $(A+I)\psi (t)$, so the equation holds as a strong L2 space solution. For general initial data it is the semigroup solution, without an unproved pointwise differentiability claim.