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Fix $u_0\in H^1(\Omega)$ and $f=f(x,y,t)\in L^2(\Omega\times [0,T])$ where $\Omega$ is a sufficiently smooth bounded domain in $\mathbb{R}^2$. Consider the problem of finding $u:\Omega\times[0,T]\to\mathbb{R}$ satisfying the following variational equation $$ \begin{cases} \langle \nabla u, \nabla v\rangle_{L^2(\Omega)}+\langle u,v\rangle_{L^2(\Omega)}=\langle f,v\rangle_{L^2(\Omega)},\;\;\forall v\in H^1(\Omega),\;\forall t\in(0,T], \\\;u(\cdot,0)=u_0, \; \end{cases} $$ It is clear that this variational problem comes from the equation

$$\begin{cases} -\Delta u+u=f, & \text{ on }\Omega\times(0,T],\\ u(0)=u_0. \end{cases}$$

I did not put boundary conditions, but any boundary condition for me works as long as $\int_{\partial\Omega}(\vec{n}\cdot\nabla u)v=0$ for all $v$.

Note: All of the differential operators seen above are spacial operators, they involve no derivatives in time.

Questions:

  • What are appropriate function spaces to look for solutions?
  • Can we have that $u(\cdot,t)\in L^2([0,T])$ for all $t\in[0,T]$?
  • Can we have that $\text{essup}_{t\in[0,T]}\|u(\cdot,t)\|_{L^2(\Omega)}\leq C$ for some constant $C\in\mathbb{R}$ independent of time and space?
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  • $\begingroup$ This is just a family of elliptic problems with t as a parameter, so what is an initial condition doing here? Also, you say you did not impose boundary conditions, and then impose a Neumann condition? $\endgroup$ – Michael Renardy May 16 '20 at 20:50
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As you correctly pointed out, since the operators appearing here are only acting in space your problem amounts to solving an elliptic equation for each time. In particular assigning $u(0)=u_0$ is meaningless: Even if the right-hand side $f$ is continuous in time, and without further compatibility assumptions between $f$ and $u_0$, it is most likely false that the solution of the elliptic problem at time $t=0$ for the datum $f|_{t=0}$ coincides with your "initial datum" $u_0$. Long story short: since time is not really a variable here, but rather a parameter, you need to figure out a framework to make sense of "regular dependence of the data on the parametere $t$", and then the dependence of the solution will follow. More details below.

In order to give a more specific answer to your question, let $K:=((-\Delta)+ 1)^{-1}$ be the inverse operator mapping the datum $f$ to the solution $u=Kf$ of $-\Delta u+u=f$. By standard elliptic regularity $K$ is a continuous operator from $L^2$ to $H^2$. Here I'm assuming that your boundary conditions, which you apparently don't prescribe explicitly, are nice enough so that this standard regularity machinery works. If not, the just replace $H^2(\Omega)$ below by $H^1(\Omega)$.

Answers

  • Bochner space is the key word here. In your first equation you implicitly consider $f(t)=f(.,.,t)$ as an element of $L^2(\Omega)$ for fixed time. (By the way: it seems that you work exclusively in spatial dimension 2, $(x,y)\in\Omega\subset \mathbb R^2$, if so it would be better to edit your question for the sake of clarity.) This is exactly what Bochner spaces do, typically $L^2((0,T);X)$ is the space of squared integrable functions $f:(0,T)\to X$ with values in the Banach space $X$, here you may want to consider of course $X=L^2(\Omega)$. Just a quick comment, though: whether an $L^2$ function $f(x,y,t)$ in space-time (in the classical sense) can actually be considered as such a Bochner function $t\mapsto f(.,.,t)\in L^2(\Omega)$, and conversely, is actually slightly more delicate than what it may seem at first sight. But this is whole different topics so let me not elaborate too much here, just google "Bochner space" or look up this keyword here on Math.MO you'll find quasillions of related questions and answers.

  • For your second question the answer is yes. But in order to define $u(t)$ you need to be able to evaluate $f(t)$ at (almost every) time $t$. This is my first answer to your question, so let's assume that indeed your datum $f\in L^2(0,T;L^2(\Omega))$. In this setup it makes sense to evaluate $f(t)\in L^2(\Omega)$ for a.e. $t\in (0,T)$, hence you can define $$ u(t):= K[f(t)]=(-\Delta +1)^{-1}f(t) \in H^2(\Omega) \qquad \mbox{for a.e. }t. $$ So the answer is yes, and moreover you see that $u(t)$ lies in the better space $H^2(\Omega)$. Actually what this shows is that $u$ can be properly defined as an element of the Bochner space $L^2(0,T;H^2(\Omega))$. Note here that since $u$ is only defined for a.e. time clearly evaluating $u(0)$ does not make sense (you can always redefine $u$ on a negligible set of times and still get the same $L^2(0,T;H^2(\Omega))$ element)

  • With these preliminaries out of the way, the answer to your question 3 should now be pretty obvious: No, you cannot hope for such $L^\infty(0,T;L^2)$, unless you improve the time regularity of $f$. To see this, take any real-valued function of time $\eta=\eta(t)$ such that $\eta\in L^2(0,T)$. Take an arbitrary $F=F(x,y)\in L^2(\Omega)$ and set $f(x,y,t):=\eta(t)F(x,y)$. Then clearly (the operator $-\Delta +1$ does not depend on time!) the solution of your problem is $$ u(t)=\eta(t) KF \qquad\mbox{hence}\qquad \|u(t)\|_{L^2(\Omega)}=|\eta(t)|\, \|KF\|_{L^2(\Omega)} $$ and therefore $\sup\limits_{t\in (0,T)} \|u(t)\|_{L^2(\Omega)} = +\infty$ unless $\eta\in L^\infty(0,T)$. Note here that the same line of thoughts shows actually that the time regularity of $f$ gives the sharp time regularity of the solution, for example $f\in C^k([0,T];L^2)\Rightarrow u\in C^k([0,T];H^2)$ or whatever (you can substitute $C^k$ by whatever regularity scale you wish, roughly speaking).

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  • $\begingroup$ Thank for the answer, this issue with $u(0)$ is problematic for me because as you said, even if assume compatibility between $f$ and $u_0$, $u$ is defined a.e in time. In classical Sobolev theory, we would encode the boundary conditions within the Sobolev itself (and use Lax-Milgram for ex). Can we do something similar with time conditions ie "encode" the initial condition in the Bochner space itself? $\endgroup$ – UserA May 17 '20 at 5:14
  • $\begingroup$ No, you're misunderstanding the problem, I think. You shouldn't view the condition $u(0)=u_0$ as an "initial/boundary" condition being part of the problem. The issue here is not to find a suitable weak formulation encoding the "initial condition", but that the whole problem is overdetermined (unless you impose an additional compatibility condition on $u_0,f$). Think of the smooth case where $f$ is continuous up to the boundary $t=0$: you rproblem is clearly ill-posed, unless $u_0=K f(0)$. So either you need to assume something more on $u_0,f$, or the problem does not make sense. $\endgroup$ – leo monsaingeon May 17 '20 at 12:05
  • $\begingroup$ I think you misunderstood my comment. Suppose that the compatibility condition holds for starters. Im asking: is ir possible to encode such an initial condition in the Bocnher space? I mean: think about the time independent Laplace equation with the Dirichlet boundary condition. In that case, we look for weak solutions in the space $H^1_0$, which encodes the Dirichlet conditon. Can we do the same in a Bochner space ie to have "contain" the initial condition? $\endgroup$ – UserA May 17 '20 at 12:16
  • $\begingroup$ I'm pretty sure I didn't misunderstand your comment! Usual weak formulations encode simultaneously a PDE and a boundary conditions because integration by parts gives rise to an extra term for regular solutions, which you then force to cancel. In the absence of time derivatives in the PDE no such terms arise, and therefore you cannot encode initial condition in this way. The only thing you can reasonably do is force the initial condition by requiring time-continuity, e.g. $u\in C([0,T);L^2)$ or whatever. But then again you need an extra compatibility condition. $\endgroup$ – leo monsaingeon May 17 '20 at 12:32
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    $\begingroup$ Exactly. This is what I tried to explain in my thrid point in my answer aboe: whatever regularity you need/want for $u$ can only come from the same regularity on $f$. $\endgroup$ – leo monsaingeon May 17 '20 at 14:05

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