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Consider the following Cauchy problem $$u_t - \Delta u + f(x,u,u_x) = 0, \quad (t,x) \in (0,T) \times \mathbb{R}^n,$$ with initial condition $u(0,x) = g(x), \ x \in \mathbb{R}^n.$

Suppose that $u \in C^2((0,T) \times \mathbb{R}^n)$ is a solution of the initial value problem.

Can we obtain (under some reasonable assumptions) that $$u_t(0,x) - \Delta u(0,x) + f(x,u,u_x)= 0,$$ that is, that the solution $u$ solves the PDE at time $t=0$ too?

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  • $\begingroup$ Of course. This is simply the differential equation evaluated at t=0, so every sufficiently regular solution will satisfy it. Without a specific regularity assumption, the question is too vague. $\endgroup$ Commented Jun 29, 2017 at 2:03
  • $\begingroup$ @MichaelRenardy Why will every sufficiently regular solution satisfy the equation at $t=0$ in addition to the initial condition? We have $u \in C^2((0,T)\times\mathbb{R}^n$ and may assume $f$ Lipschitz in $u$ and $u_x$ and $g$ Lipschitz. $\endgroup$
    – Jun
    Commented Jun 29, 2017 at 11:36
  • $\begingroup$ Michael, the solution is assumed to be $C^2$ only for $t>0$. $\endgroup$ Commented Aug 1, 2017 at 17:07

2 Answers 2

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The question is vague, because the OP does not mention in which sense the PDE is to be satisfied at initial time.

In addition, the assumption that $u$ is $C^2$ for $T>0$ is very weak, if not useless. For let me assume that $f$ is a smooth function with $f(\cdot,0,0)\equiv0$, and that the initial data $g$ is $L^1\cap L^\infty({\mathbb R}^n)$. Then it is classical, because of the smoothing property of the heat semi-group, that $u$ is actually $C^\infty$ for $t>0$ ; hence the assumption is automatically fulfilled. Therefore the question asks whether, if $g\in L^1\cap L^\infty({\mathbb R}^n)$, one has $$u_t(0)=\Delta g-f(\cdot,g,\nabla g).$$ certainly, this does not hold pointwisely, since the right-hand side may not be a function. It holds in the distributional sense ; just pass to the limit (in ${\cal D}'$) in the equation satisfied at $\{s\}\times{\mathbb R}^n$, as $s\rightarrow0+$. But then $u_t(0)$ is just defined as the limit (in ${\cal D}'$) of $u_t(s)$ as $s\rightarrow0+$, and this is a tautology. The answer will be better if we can prove that the distributional limit of $\frac{u(s)-g}s$ equals $\Delta g-f(\cdot,g,\nabla g)$. I guess that this can be done by the usual techniques of semi-groups.

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  • $\begingroup$ Indeed what I need to prove (and I hope it's true) is that $u$ is smooth also at time $0$ under some assumptions and it solves the PDE (not just the initial condition) in the classical sense ("pointwise") at time $t=0$. However I'm not familiar with the "usual techniques of semigroups". $\endgroup$
    – Jun
    Commented Aug 1, 2017 at 19:16
  • $\begingroup$ Also, we may take $f=-u - u_x$ and $g$ smooth. $\endgroup$
    – Jun
    Commented Aug 1, 2017 at 19:31
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This is a well known result from diffusion PDE theory. For example [1] studied the semi-linear equation in form of

$$u_t - \Delta u + u^\gamma = 0, \quad (t,x) \in (0,T) \times \mathbb{R}^n, \gamma>1$$ with initial value $u(0,x) = g(x), \ x \in \mathbb{R}^n.$ And with typical Green function representation we could write (2.1) in [1]. $$u(t)=e^{t\Delta}g+{\displaystyle \int_{0}^{t}e^{(t-s)\Delta}u^{\gamma}ds}$$ with $e^{t\Delta}g(x)={\displaystyle \int_{\mathbb{R}^{n}}Gr_{t}(x-y)dy}$ and $Gr_{t}(x)=\left(4\pi t\right)^{-\frac{n}{2}}e^{-\frac{\|x\|^{2}}{4t}}$. Therefore an intuition that such a solution $u$ exists if $u^\gamma,\gamma>1$satisfy regularities. And if $f(x,u,u_x)$ is somehow regular enough then we can expand it into analytic form in terms of powers of $u,u_t$ and the Green function representation argument applied.

A rigid proof requires more work but I believe that is what I(and possibly @ Michael Renardy) had in mind about regularity on solution to the equation above.

[1]Weissler, Fred B. "Existence and non-existence of global solutions for a semilinear heat equation." Israel Journal of Mathematics 38.1 (1981): 29-40.

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  • $\begingroup$ Yes, I can assume that the we have a Green function representation of the type $$u(t,x)=e^{t\Delta}g+{\displaystyle \int_{0}^{t}e^{(t-s)\Delta}f }ds$$ for the solution of the initial value problem in my question. My main problem is: how does this imply that $u$ solves the PDE at time $0$ too? $\endgroup$
    – Jun
    Commented Aug 1, 2017 at 16:53
  • $\begingroup$ @Jun I assume you can have a direct check for $t=0$ using derivation on Green function representations. Now I have the same question Serre raised below in his answer... $\endgroup$
    – Henry.L
    Commented Aug 1, 2017 at 18:41
  • $\begingroup$ I've added a comment under Denis Serres' question and I hope it makes the question clearer. How would you do the direct verification of the result on the green function representation? $\endgroup$
    – Jun
    Commented Aug 1, 2017 at 19:18
  • $\begingroup$ @Jun If you just want to verify the smoothness you can try to use a Fubini-type theorem or consider the semi-group consists of differential operators of modulus of continuity assumptions like mentioned in Serre's answer. This may or may not hold due to the choice of $f$ as explained in Serre's answer. $\endgroup$
    – Henry.L
    Commented Aug 1, 2017 at 19:22
  • $\begingroup$ Actually, we might as well assume $f = - u - u_x$ and $g$ smooth. Your last comment is not quite clear to me. Could you add some details, maybe updating your answer? $\endgroup$
    – Jun
    Commented Aug 1, 2017 at 19:34

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