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Consider the heat equation on $\mathbb{R}_+\times\mathbb{R}^d$

\begin{align*} \partial_t u -\Delta_x u &= f, \\ u(0,x)&=u_0(x). \end{align*}

In the case where $u_0\in L^2(\mathbb{R}^d)$ and $f\in\mathscr{C}^0(\mathbb{R}_+;L^2(\mathbb{R}^d))\cap L^\infty(\mathbb{R}_+;L^2(\mathbb{R}^d))$, there exists a unique $u\in\mathscr{C}^0(\mathbb{R}_+;L^2(\mathbb{R}^d))$ such as $\tilde{u}$ is solution of the above Cauchy problem in the tempered sense that is

\begin{align*} \partial_t \tilde{u} -\Delta_x \tilde{u} = \tilde{f} +\delta_{t=0}\otimes u_0, \end{align*}

where $\tilde{h}$ denotes the extension by $0$ on $\mathbb{R}$. This solution is given by the Duhamel formula for the Fourier transform : for all $t>0$, \begin{align*} \widehat{u(t)}(\xi) = \widehat{u_0}(\xi) e^{-|\xi|^2t} + \int_0^t \widehat{f(s)}(\xi)e^{-(t-s)|\xi|^2} ds. \end{align*}

I know that in the case $f=0$, the heat equation has an instantaneous regularizing effect in the sense that the above solution $u$ belongs to $\mathscr{C}^\infty(\mathbb{R}_+^*\times\mathbb{R}^d)$ regardless of the regularity of $u_0$. What regularity one gets in the general case $f\in\mathscr{C}^0(\mathbb{R}_+;L^2(\mathbb{R}^d))\cap L^\infty(\mathbb{R}_+;L^2(\mathbb{R}^d))$ described before ? The point is that I would like to keep on working on this nice Fourier framework and get regularity for $u$ by means of decay of $\widehat{u(t)}(\xi)$.

Two remarks

  • In the mentionned case $f\in\mathscr{C}^0(\mathbb{R}_+;L^2(\mathbb{R}^d))\cap L^\infty(\mathbb{R}_+;L^2(\mathbb{R}^d))$ it seems to me that instantaneously smoothness is lost. For example the belonging $u(t)\in H^2(\mathbb{R}^d)$ leads to study the integrability (in time) of $ \| |\xi|^2 e^{-\tau|\xi|^2}\|_\infty$ near $0$ which apparently is false (it behaves like $1/\tau$). But maybe I missed something.

  • If one assumes $f\in L^\infty(\mathbb{R}_+;L^1(\mathbb{R}^d))$, then $(s,\xi)\mapsto \widehat{f(s)}(\xi)$ is bounded on $\mathbb{R}_+\times\mathbb{R}^d$ and hence for any $t>0$ and any multi-index $\alpha\in\mathbb{N}^d$ and $|\xi| \geq 1$ (small $\xi$ are handable)

\begin{align*} |\xi^\alpha \widehat{u(t)}(\xi)| &\leq |\xi^\alpha \widehat{u_0}(\xi)| e^{-|\xi|^2t} + \|f\|_{L^\infty(\mathbb{R}_+;L^1(\mathbb{R}^d))} \int_0^t |\xi|^{|\alpha|}e^{-(t-s)|\xi|^2} ds \\ &= |\xi^\alpha \widehat{u_0}(\xi)| e^{-|\xi|^2t} + \|f\|_{L^\infty(\mathbb{R}_+;L^1(\mathbb{R}^d))} \int_0^t |\xi|^{|\alpha|}e^{-(t-s)|\xi|^2} ds\\ &= |\xi^\alpha \widehat{u_0}(\xi)| e^{-|\xi|^2t} + \|f\|_{L^\infty(\mathbb{R}_+;L^1(\mathbb{R}^d))} \frac{1}{|\xi|^{2-\alpha}}(e^{-t|\xi|^2}-1), \end{align*}

which again is not square integrable in $\xi$.

It appears to me that without any regularity assumptions on the $x$ variable for $f$, one can only expect $L^2_t(H^s_x)$ regularity for $u$, but I am not sure that I did not missed something before.

I would be glad to have any comments or advice. Also if one has a nice reference on the Fourier analysis of the nonhomogeneous heat equation and the regularity properties of its solutions in the previous case, it would be very kind to share it !

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What do you mean by the script C? Do you just mean to say that the norm is continuous in the time variable? – Ray Yang Feb 9 '13 at 20:36
Hi. No, I mean, the function is continuous in time, with values in the mentionned space. For instance $\mathscr{C}^0(\mathbb{R}_+;L^2(\mathbb{R}^d))$ is the vector space of continuous functions of the time variable with value in $L^2(\mathbb{R}^d)$ : $\|f(t)-f(s)\|_{L^2(\mathbb{R}^d)}$ goes to $0$ with $|t-s|$. – xounamoun Feb 11 '13 at 9:09

I think the answer is no, since the analogous result doesn't even hold for elliptic equations (consider the discussion here Counterexample for the solvability of $-\Delta u = f$ for $f\in C^2$ ).

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