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Let $u$ be a classical solution for the nonhomogeneous heat equation in $\mathbb R_+ \times\mathbb R^d$: $$ \begin{cases} \partial_tu(t,x)-\Delta u(t,x) = f(t,x), \\ u(0,x)=u_0(x). \end{cases} $$ Suppose $f(t,\cdot) \in L^q(\mathbb R^n), 1 \le q \le \infty$, for each $t > 0$, and suppose $f$ has the decay behavior $$ \|f(t, \cdot)\|_{L^q} \le \frac{C}{1+t^\gamma}, \quad \gamma >0. $$ Now my question is that, given $\gamma$ and $q$, what is the decay rate of $\|u(t, \cdot)\|_{L^p(\mathbb R^d)}$ for each $1\le p \le \infty$? Probably there is no decay for all values of $p$ in this range, but I am interested in the optimal results. In particular, I would be interested in a reference (book or research paper), where this has been studied in detail.

I tried to derive the results from the representation formula for $u$: $$ u(t,x)=\int_{\mathbb R^d} H(t, x-y)u_0(y)\, dy + \int_0^t\int_{\mathbb R^d} H(t-s, x-y)f(s,y) \, dy \, ds, $$ where $ H(t,x)=(4\pi t)^{-d/2}e^{-\frac{|x|^2}{4t}} $ is the heat kernel. Here the problem comes with the latter term for large dimensions $d$. Simple arguments by using Minkowski's (or Hölder's) inequalities give some decay estimates, but then one ends up with some strange looking integrability conditions in the time integral. In particular, in this manner one only obtains decay for some values of $p$ and I doubt the estimates obtained are optimal in this sense. So I am wondering what is the correct approach and what would be the reference for this.

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Suppose that $u_0=0$, otherwise $L^p$ decay estimates are well-known. Let $\{T(t)\}_{t\geq0}$ denote the heat semigroup, i.e., $T(t)$ for $t>0$ is convolution with $(4\pi t)^{-d/2}\,e^{-|x|^2/(4t)}$. Notice that, for $p\geq q$, $$ \|T(t)\|_{L^q\to L^p} = \left\|\frac1{(4\pi t)^{d/2}}\,e^{-\,\frac{|x|^2}{4t}}\right\|_{L^r} = C_{d,r}\,t^{-\,\frac{d}2\left(1-\frac1r\right)}, $$ where $1+\frac1p=\frac1q+\frac1r$ (here, $p,\,q,\,r\in[1,\infty]$).

Let $p\geq q$ and $\displaystyle \frac1q-\frac1p<\frac2d$. Further let $\displaystyle \|f(t)\|_{L^q} \lesssim (1+t)^{-\gamma}$ for some $\gamma\in \mathbb R$. Then $\displaystyle \|u(t)\|_{L^p} \lesssim \alpha_\gamma(t)(1+t)^{-\,\frac{d}2\left(\frac1q-\frac1p\right)}$, where $$ \alpha_\gamma(t) = \begin{cases} (1+t)^{1-\gamma}, & \gamma<1, \\ \log \left(e+t\right), & \gamma=1, \\ 1, & \gamma>1. \end{cases} $$

Indeed, $u(t)=\int_0^t T(t-s)f(s)\,ds$, so $$ \begin{aligned} \|u(t)\|_{L^p} &\leq \int_0^t \|T(t-s)\|_{L^q\to L^p} \|f(s)\|_{L^q}\,ds \lesssim \int_0^t \frac{(t-s)^{-\,\frac{d}2\left(1-\frac1r\right)}}{(1+s)^\gamma}\,ds \\ & \lesssim t^{1-\,\frac{d}2\left(1-\frac1r\right)}\sideset{_{2\,}}{_1}F\left(1,\gamma;2-\, \frac{d}2\left(1-\frac1r\right);-t\right). \end{aligned} $$ To find the asymptotics of $\displaystyle \sideset{_{2\,}}{_1}F\left(\dots;-t\right)$ as $t\to\infty$, use $$ \begin{aligned} \sideset{_{2\,}}{_1}F\left(a,b;c;-t\right) & = \frac{\Gamma(b-a)\Gamma(c)}{\Gamma(c-a)\Gamma(b)} \,t^{-a}\sideset{_{2\,}}{_1}F\left(a,a-c+1;a-b+1;-\,\frac1t\right) \\ &\qquad +\,\frac{\Gamma(a-b)\Gamma(c)}{\Gamma(c-b)\Gamma(a)}\,t^{-b}\, \sideset{_{2\,}}{_1}F\left(b-c+1,b;-a+b+1;-\,\frac1t\right) \end{aligned} $$ in case $a-b\notin\mathbb Z$; and similarly for $a-b\in\mathbb Z$.

The estimates above are best possible (as $t\to\infty$) provided that $q=1$ or $\gamma\neq1$.

In case $\gamma<1$ or $(q,\gamma)=(1,1)$, this is shown by the example of $$ u(t,x) = \frac1{(4\pi t)^{d/2}}\,e^{-\,\frac{|x|^2}{4t}}\,h(t), \quad f(t,x) = \frac1{(4\pi t)^{d/2}}\,e^{-\,\frac{|x|^2}{4t}}\,h'(t), $$ where $h\in\mathscr C^\infty(\overline{\mathbb R}_+)$, $h(t)=0$ near $t=0$, upon choosing $h$ appropriately. For instance, $\displaystyle h'(t)=t^{-\delta}$ for $t\geq1$ (and $h'(t)=0$ for $0\leq t\leq1/2$) with $\displaystyle \delta=\gamma-\frac{d}2\left(1-\frac1q\right)$ will do.

For $\gamma>1$, it is enough to remark that $\alpha_\gamma(t)\gtrsim 1$, with this lower bound being implied by the Cauchy problem (e.g., take $f$ such that $f=0$ for $t>1$).

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  • $\begingroup$ Thanks for answering my question. So what is this function $F$ and, in particular, how did you obtain the estimate where this function appears the first time? A simple integration would seem to give a restriction for $r$ in terms of $d$, in order the right hand side to be integrable. $\endgroup$ – Juhana Siljander Apr 17 '15 at 11:05
  • $\begingroup$ @Juhana $\sideset{_{2\,}}{_1}{F}(\dots)$ is the hypergeometric function. Otherwise, it is an exact evaluation of this integral (up to a multiplicative constant $C_{d,r}$). The condition on the function under the integral to be integrable is $\displaystyle \frac1q-\frac1p = 1-\frac1r<\frac2d$. $\endgroup$ – ifw Apr 17 '15 at 12:33
  • $\begingroup$ Ok, so this gives a condition on $p$ given $q$ and $d$. Is it known that this is optimal, i.e. for high enough $p$ there is, in general, no decay (independent of the decay rate $\gamma$ of the forcing term)? $\endgroup$ – Juhana Siljander Apr 19 '15 at 12:08
  • $\begingroup$ @Juhana For $\gamma>1$, there is a decay of $\displaystyle (1+t)^{-\frac{d}2\left(\frac1q-\frac1p\right)}$, and this indeed independent of $\gamma>1$, because whatever one might get from a right-hand side $f$ is overruled by the decay one has for the Cauchy problem. I've edited the second part of my answer. $\endgroup$ – ifw Apr 20 '15 at 6:29
  • $\begingroup$ What I meant is that the condition $\frac1q-\frac1p < \frac2d$ seems to give an upper bound for $p$ in terms of $q$ and $d$ (at least if $q$ is small enough). So what happens if $p$ is larger than this number? $\endgroup$ – Juhana Siljander Apr 21 '15 at 9:40

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