My question is from Cazenave's book "Semilinear Schrödinger Equation", page 35. I am stuck with Step 2 of the Strichartz's estimates.

The book says that $||\Phi_f(t)||_{L^2}^2=\left(\int_0^t \mathcal{T}(t-s)f(s)ds,\int_0^t \mathcal{T}(t-\sigma)f(\sigma)d\sigma\right)_{L^2}=$

$\underbrace{\int_0^t\int_0^t\left( \mathcal{T}(t-s)f(s), \mathcal{T}(t-\sigma)f(\sigma)\right)_{L^2}d\sigma ds=\int_0^t \int_0^t \left( f(s),\mathcal{T}(s-\sigma)f(\sigma)\right)_{L^2}d\sigma ds=}_?$


where $\Theta_{t,f}(s)$ was defined earlier in the chapter, but it is not important for the purpose of this question.

$\mathcal{T}(t)$ is defined as the group of isometries on $L^2(\Omega)$ generated by the skew-adjoint operator $iA$, where $A$ is the Laplacian with the Dirichlet boundary conditions on $\partial\Omega$. On the other hand there is a theorem which says that $\mathcal{T}(t)\phi(x)=(4\pi it)^{-\frac{N}{2}}\int_{\mathbb{R}^N}e^\frac{i|x-y|^2}{4t}\phi(y)dy$ for suitable functions $\phi$.

I would really appreciate if someone could give me a hand. I actually struggle with the third equality, namely how the identity below is determined:

$\int_{0}^{t}\int_{0}^{t} (\mathcal{T}(t-s)f(s),\mathcal{T}(t-\sigma)f(\sigma))_{L^2} d\sigma ds=\int_{0}^{t}\int_{0}^{t} (f(s),\mathcal{T}(s-\sigma)f(\sigma))_{L^2} d\sigma ds$.

If I am not wrong, I need to prove that

$\int\mathcal{T}(t-s)f(s)\overline{\mathcal{T}(t-\sigma)f(\sigma)}=\int f(s)\overline{\mathcal{T}(s-\sigma)f(\sigma)}$,

where the $L^2$ inner product is defined by $(u,v)_{L^2}$=Re $\int_{\Omega}u(x)\overline{v(x)}dx$.

Thanks in advance!


On the operator level: if $U,V: L^2 \to L^2$, denote by $U^*$ the adjoint operator of $U$, you have that

$$ \langle U f, V g\rangle = \langle f, U^* V g\rangle $$

by definition. (Note, this is outside any time integration.)

Now: $\mathcal{T}(s)^* = \mathcal{T}(-s)$ (as evident from the explicit formula you gave, or from abstract considerations as a group of isometries).

So before even integrating you have

$$ \langle \mathcal{T}(t-s) f(s), \mathcal{T}(t-\sigma) f(\sigma)\rangle = \langle f(s), \mathcal{T}(s-t) \mathcal{T}(t-\sigma) f(\sigma) \rangle $$

Now use the group property and you are done.

| cite | improve this answer | |
  • $\begingroup$ @ Willie Wong: Thanks a lot for the answer! It is really helpful. I have one more question. Sorry if it sounds stupid! Let us use the equality from your post. Since $\mathcal{T}(\cdot)$ is a group of isometries, we have that $\mathcal{T}(s-t)\mathcal{T}(t-\sigma)$ will be another element of the same group. How do we know that this element will be exactly $\mathcal{T}(s-\sigma)$ or it is just a matter of notation? $\endgroup$ – Candidate Jun 13 '16 at 23:20
  • $\begingroup$ That's just the standard notation of a one parameter group. See also en.wikipedia.org/wiki/C0-semigroup ; If you think in terms of the solution operator: solving forward first by time $t-\sigma$ then by time $s - t$ must equal to solving forward by time $t - \sigma + s - t = s-\sigma$... $\endgroup$ – Willie Wong Jun 14 '16 at 1:53
  • $\begingroup$ @ Willie Wong: Thanks for helping me again! Actually my first attempt was to plug everything into the formula for $\mathcal{T}(t)\phi(x)$ and then derive that $\mathcal{T}(s-t)\mathcal{T}(t-\sigma)=\mathcal{T}(s-\sigma)$. This turned out to be not an easy task. $\endgroup$ – Candidate Jun 14 '16 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.