5

Suppose I have a PDE, for example the Fokker-Planck one, in which I am mostly interested: $$ \frac{\partial}{\partial t}u(x,t)=-\frac{\partial}{\partial x}(\mu(x,t)u(x,t))+\frac{1}{2}\frac{\partial^2}{\partial x^2}(D(x,t)u(x,t)). $$ Denote the solution $u$ as $u^\mu$ to strength the fact that, in particular, it depends on $\mu$.

Where can I find results that guarantee the continuity of the PDE solution with respect to the coefficients? That is, results guaranteeing that $$ \mu_n(x,t)\to \mu(x,t) \implies u^{\mu_n}(x,t)\to u^{\mu}(x,t), $$ under some conditions and appropriate definitions for the above convergences (the simpler, the better).

Hans
  • 9,804
epsilone
  • 441
  • 4
    It depends very much on the type of PDE. The keyword is "stability". For linear evolution equations, look into "semigroup theory" and consult books such as Pazy, Goldstein, or Engel-Nagel (roughly, in order of comprehensiveness). For linear elliptic equations, the standard reference is Gilbarg-Trudinger's book. For nonlinear elliptic equations, I know of this recent book: https://www.amazon.fr/Compactness-Stability-Nonlinear-Elliptic-Equations/dp/3037191341 – Giuseppe Negro May 19 '16 at 09:03
  • @GiuseppeNegro: Helpful comment. +1 What if $D(x,t)$ is a constant and I would like to know the continuity of $u(x,t)$ with respect to $D$? Is there a straightforward answer? – Hans May 10 '18 at 08:04
  • @Hans: see below. – Giuseppe Negro May 10 '18 at 10:17

1 Answers1

1

WORK IN PROGRESS (This is a community wiki to record progress in this problem).

Easy case. If $\mu$ and $D$ are constants, there is an explicit representation of the solution: $$u(t, x)=\sqrt{\frac{1}{2\pi Dt}}\int_{-\infty}^\infty \exp\left(\frac{(x-y-\mu t)^2}{Dt}\right)\, u_0(y)\, dy.$$ From this formula one can easily establish all the continuity properties one wants.

Constant $D$ case. We assume that $D$ is a constant and we ask whether $$\lim_{D\downarrow 0} u(t, x) = w(t, x),$$ where $\partial_t w = -\partial_x (\mu(t, x)w(t, x))+\frac{D}2\partial^2_x w,$ and $w(0, x)=u(0, x)$. (The convergence is intended in an appropriate sense to be specified).

If $\mu=\mu(x)$ the answer is affirmative: see Hans comment below for a transformation that reduces the problem to the constant $\mu$ case. The abstract framework of semigroup theory also applies; see Pazy, Semigroups of linear operators and applications to PDEs, section 3.4: "The Trotter approximation theorem".

If $\mu=\mu(t)$ the answer is affirmative in the following sense. Taking the spatial Fourier transform termwise, the equation reduces to $\partial_t \hat{u} = \mu(t)(-i\xi \hat{u}(t, \xi))-\frac{D}2\xi^2\hat{u}(t, \xi).$ This is a family of ODEs indexed by $\xi$, and by Gronwall's lemma arguments, as $D\to 0$ its solutions converge uniformly for $t$ in compact intervals. (There is some work to be done here, to show that there is uniformity in $\xi$. This seems true, though).

If $\mu=\mu(t, x)$ is analytic in time, the problem can be reduced to the study of the sequence of PDEs $$\partial_t u_n=-t^n \partial_x(\mu_n(x)u_n(t, x))+\frac{D}2\partial_x^2 u_n.$$

Giuseppe Negro
  • 32,319
  • Thank you for the answer. You misunderstood me, though. I am well aware of the analytic solution of the standard heat equation and thus its analyticity with respect to its constant coefficient. I specifically NOT specializing $\mu(x,t)$ to constant. I am leaving $\mu(x,t)$ as a general function as it is, maybe specializing to be an analytic function if that gives any nice result. As such, I am particularly interested in the continuity of $u(x,t)$ with respect to $D$ at $D=0$. – Hans May 10 '18 at 15:06
  • Well, this formula already shows a non-trivial phenomenon: $$\lim_{D\to 0} u(t, x) = u_0(x-\mu t), $$ where in the right-hand side you see the solution to the transport equation $\partial_t u + \mu\partial_x u=0$. It should be proven that this remains true if $\mu=\mu(t, x)$. This shouldn't be hard. – Giuseppe Negro May 10 '18 at 15:23
  • Are you still talking about a constant $\mu$ which is NOT I want? By "this formula" are you referring to the standard constant coefficient Gaussian integral you wrote in your answer? – Hans May 10 '18 at 16:29
  • Let me rephrase. If $\mu$ is constant, we see from the Gaussian integral that $\lim_{D\to 0} u(t, x)$ is the solution to the equation $\partial_t u + \mu \partial_x u=0$. I conjecture that, even if $\mu=\mu(t, x)$, this is still true. (With some assumptions on $\mu$, of course). – Giuseppe Negro May 10 '18 at 16:43
  • There is nothing to say for the constant case because we have the Gaussian convolution integral and it is obviously analytic with respect to every constant coefficient. As a matter of fact, I know the analytic solution for $\mu(x,t) = ax+b$ and thus is easy to deduce the regularity for that case. I have the same conjecture. The proof is the hard part and is what I am after. – Hans May 10 '18 at 17:04
  • 1
    For $\mu(x,t)=\mu(x)$ i.e. depending only on x, we can make a transformation $x=t+g(y)$ for some function $g$ so as to turn the PDE into the standard heat constant coefficient heat equation $\frac{\partial v(t,y)}{\partial t}=\frac12D\frac{\partial^2 v(t,y)}{\partial y^2}$. The hard problem is when $\mu(t,x)$ is a nontrivial function of both $t$ and $x$. – Hans May 10 '18 at 20:15
  • @Hans: I made the answer CW to keep recording any progress on your question. BTW, it is probably the case that some theorem in abstract semigroup theory applies here. For the autonomous equation (i.e., $\mu=\mu(x)$), that's quite easy and I added a reference (the Trotter approximation theorem). There are generalizations of this theorem to various time-dependent cases, but it is a literature I am not familiar with. – Giuseppe Negro May 11 '18 at 13:23
  • Thank you, Giuseppe. Can you derive your assertion for when $\mu(t,x)$ is analytic in $t$? Is this using the method in the proof of the Cauchy–Kovalevskaya theorem? – Hans May 14 '18 at 18:25
  • That looks like a bad idea because ${t=0}$ is characteristic for the operator $\partial_t-\partial^2_x$. However, I know very little about these things, so I don't really know. – Giuseppe Negro May 14 '18 at 18:42