Mapping properties of backward and forward heat equation

Question

In a previous question on mathoverflow, I asked about the following:

Let $\Delta$ be the Laplacian on some compact interval $I$ of the real line with let's say Dirichlet boundary conditions.

The functional calculus allows me to study $S_t = e^{-\Delta t}$ and $T_t = e^{(\Delta+f)t}.$

Here $T_t$ is a perturbed heat semigroup and $S_t$ the backward heat semigroup.

Michael Renardy essentially showed that for $S_1T_1$ to be a bounded operator on $L^2$, $f$ would have to be even better than analytic and it was not even clear, if $f$ would not even have to be constant in order for $S_1T_1$ to be bounded on $L^2(I).$

This raised essentially the following follow-up question:

• Since $S_1T_1$ is not bounded from $L^2$ to $L^2$, perhaps there are more canonical spaces between which this operator is bounded?-I would not necessarily be interested now in an answer saying that by some general principle it takes test functions into distributions or something along these lines, but would hope that it is possible to shed some light on the mapping properties of this linear operator. In short: Are there any spaces that seem canonical for the operator $S_1T_1$?

Answer 1

This is not an answer, but just an illustration of how bad things can be.

Assume for convenience and concreteness that $I = [0,\pi]$ and so we can decompose the solution $u$ in sine series $$ u(t,x) = \sum_{k = 1}^\infty \hat{u}(t,k) \sin(k x) $$

Consider the case $f(x) = \cos(x)$ which is real analytic. Then $$ f\cdot u = \sum_{k = 1}^\infty \hat{u}(t,k) \sin(kx) \cos(x) = \sum_{k = 1}^\infty \frac12 \hat{u}(t,k) [\sin((k+1)x) + \sin((k-1)x) ] $$ So the equation $$ u_t = (\Delta + f)u $$ becomes $$ \partial_t \hat{u}(t,k) = - k^2 \hat{u}(t,k) + \frac12 \hat{u}(t,k-1) + \frac12 \hat{u}(t,k+1) $$

Since you are interested in the operator $S_1 T_1$, you want to look at $e^{k^2} \hat{u}(1,k)$. So what we can do is look at $v(t,k) = e^{tk^2} \hat{u}(t,k)$ in which case the quantity of interest is $v(1,k)$. You find that $v(t,k)$ solves the differential equation $$ \partial_t v(t,k) = \frac12 e^{ - 2tk - t} v(t,k+1) + \frac12 e^{2tk - t} v(t,k-1) \tag{A}$$

Notice that if $v(0,k)$ is signed (say non-negative for all $k$), then so is $v(t,k)$ for all positive $t$. And in this case we have that

$$ \partial_t v(t,k) \geq \frac12 e^{2tk-t} v(t,k-1) $$

And we see that

\begin{gather} v(t,1) \geq v(0,1) \\ v(t,2) \geq v(0,2) + \frac{1}{3} (e^{3t} - 1) v(0,1) \\ v(t,3) \geq v(0,3) + \frac15 (e^{5t} -1)v(0,2) + [\frac1{24} (e^{8t} - 1) - \frac{1}{15} (e^{5t}-1)] v(0,1) \\ \vdots \end{gather}

To get a feel of how fast this grows: Consider the initial data $u(0,x) = \sin(x)$, where $v(0,1) = 1$ and $v(0,k) = 0$ for all $k \neq 1$.

Then we have $v(t,1) \geq 1$. Reparametrizing $t = \ln(s+1)$ and setting $w(s,k) = v(\ln(s+1),k)$, we have $$ \partial_s w(s,k) \geq (s+1)^{2k-2} w(s,k-1) \geq 2^{k-1} s^{k-1} w(s,k-1).$$ This allows us, by repeatedly integrating, the extremely naive underestimate that $$ w(s,k) \geq 2^{k(k-1)/2} \cdot \left( \prod_{j = 2}^k \frac{1}{j(j+1)/2 - 1} \right) s^{k(k+1)/2 - 1} $$ Noting that $2^{j}$ grows much faster than $j^2$, we therefore have the lower bound that $$ v(1,k) = w(e-1,k) \gtrsim 2^{k^2/4} $$

This means that even though $u_0$ has compact Fourier support (in terms of regularity this is as good as one can expect), we see that $S_1 T_1 u_0$ is so rough that we cannot even consider it a distribution.