Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 !

share|improve this question
    
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

1 Answer 1

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 $f\in C^2$ ).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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