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I want to prove that there is a solution to the problem $$u_t = u^{n}u_{xx} + u^m$$ $$u|_{t=0} = u_0 \in C^{1+\alpha}$$ in the domain $S^1 \times (0,T)$ and $n$ and $m$ are positive integers, and $u_0 > 0$. Take $n$ to be even and $m$ to be odd ($n = 2$ and $m = 3$ for example).

I found a thread on Mathoverflow in which it was suggested that one looks at $$\partial_t u_a = (a+|u_a|^n)\partial^2_x u_a + |u_a|^m$$

and we know the solution to this exists since it is strongly parabolic and so there are references to show it exists. I guess I need to find energy estimates involving $u_a$ that are independent of $a$ in a Hilbert space so that there is a weakly-convergent subsequence (as $a \to 0$) and I want to show that the limit of this subsequence solves the original PDE involving just $u$, right?

How do I find these energy estimates? I only know one method of multiplying by a test function and integrating by parts but that doesn't look good here. Any ideas?

All I have at the moment (from a book) is that $$\sup_{(0,T)}\lVert u_a \rVert_{H^l} < \infty$$ which doesn't help since I don't know if there is a dependence on $a$.

Also, passing to the limit in the equation involving $u_a$ doesn't look easy either.

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  • $\begingroup$ should there, or should there not, be absolute value signs in $u^n$ and $u^m$? If initial value of $u$ is negative and $n$ odd, no amount of regularisation will bring you from a reverse-time parabolic equation to a forward time one. $\endgroup$ – Willie Wong Jun 28 '12 at 10:27
  • $\begingroup$ @WillieWong In the original PDE there are no absolute value signs. I should have mentioned that the initial value $u_0 > 0$. I'll edit my post. (Also, take $n$ to be even and $m$ to be odd.) $\endgroup$ – TagWoh Jun 28 '12 at 10:33
  • $\begingroup$ In $L^2$ and $H^1$ at least (for the latter, take a derivative of the equation and integrate against $u_x$) your equation is norm decreasing (ditto the regularised equation). In other words, the map solution map $U_a(t)$ for the regularised equations have $H^1$ norms bounded by one independent of $a$. This should suffice demonstrating the existence of a limit. (strong in $L^2$, weak in $H^1$). $\endgroup$ – Willie Wong Jun 28 '12 at 11:00
  • $\begingroup$ @WillieWong Thanks for replying. Am I right that you mean I should show $\frac{d}{dt}\lVert u(t) \rVert_{L^2(S^1)} \leq 0$, so that $\lVert u(t) \rVert_{L^2(S^1)} \leq const$ where the constant depends on $u_0$ only (I think)? (And similarly for the $H^1$ norm). If so, for the $L^2$ norm, I tried multiplying the equation by $u$ and integrating. I get an $\int_{S^1} u^4 > 0$ term on the RHS which is annoying. And I get something similar ($\int_{S^1} u^2 u_{x}^2$) with your suggestion to get the $H^1$ norm (which I believe is: take derivative wrt. $x$, multiply by $u_x$ and integrate). $\endgroup$ – TagWoh Jun 28 '12 at 13:28
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    $\begingroup$ ah, oops. That was a mistake on my part. Your operator is indeed indefinite. (Somehow in my brain I had it with a minus sign in front of the semilinear term.) In your case however you can collect $$ \partial_x u^3 \partial_x u = 3 u^2 (\partial_x u)^2 = 3/4 (\partial_x u^2)^2$$ and $u^4 = (u^2)^2$. With luck (if you have some control on the mean value of $u^2$) you can use Wirtinger's to get a sign. But you are right: choosing $u_0 = \text{const}$ it is quite clear you don't have dissipation in general. $\endgroup$ – Willie Wong Jun 28 '12 at 16:10
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Actually, one can get a uniform growth rate on the $H^1$ norm as follows. Multiply the equation by $u$ and integrating by parts you get

$$ \frac12 \partial_t \|u\|_2^2 = \int (au + u^3)u_{xx} \mathrm{d}x + \|u\|_4^4 \leq \|u\|_4^4 \leq C \|u\|^2_{H^1} \|u\|_2^2 $$

where we used the fact that the first integral after the equality sign is negative semidefinite, and the 1-dimensional Sobolev inequality for the last implication.

Similarly since

$$ u_{xt} = \partial_x( (a + u^2) u_{xx}) + 3 u^2 u_x $$

we multiply by $u_x$ and integrate to get

$$ \frac12 \partial_t \|u_x\|_2^2 = - \int (a+u^2) u_{xx} u_{xx} + 3 u^2 u_x^2 \mathrm{d}x \leq 3 \|u\|_{\infty}^2 \|u_x\|_2^2 \leq C \|u\|_{H^1}^4 $$

Combining the two we get that

$$ \partial_t \|u\|_{H^1}^2 \leq C \|u\|_{H^1}^4 $$

which is a Ordinary differential inequality which you can solve. Note that the constant $C$ above only depends on the Sobolev constant in 1 dimension on the circle and not on $a$. This gives you an uniform upperbound to the growth rate of $u_a$ in $H^1$ (at least for small times: there can still be blowup in finite time, but the time of existence is bounded below by something depending only on the $H^1$ norm of the initial data). From this you can grab a weakly converging subsequence.

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  • $\begingroup$ In your answers you have $ \int (au + u^3)u_{xx} \, \mathrm{d}x$, does it mean a double integral, i.e. $ \int \int (au + u^3)u_{xx}\, \mathrm{d}x \,\mathrm{d}t$? $\endgroup$ – JDoeDoe Feb 21 '17 at 10:17
  • $\begingroup$ @JDoeDoe: no, it does not. $\endgroup$ – Willie Wong Feb 21 '17 at 15:00
  • $\begingroup$ So extracting a weakly convergence subsequence implies local existence? Why? $\endgroup$ – energy Feb 10 '20 at 15:02
  • $\begingroup$ The OP asks about the energy estimate. I answered about the energy estimate. There's generally some more work to be done after. $\endgroup$ – Willie Wong Feb 10 '20 at 15:19
  • $\begingroup$ So for then we need Lipschitz Condition on the function right? And also with the energy estimates if the initial condition satisfies that energy of both solutions are equal then we will have the result right? $\endgroup$ – energy Feb 12 '20 at 13:28

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