Energy estimates to show existence of a PDE

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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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. – Willie Wong Jun 28 '12 at 10:27
@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.) – TagWoh Jun 28 '12 at 10:33
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 $H^1$). – Willie Wong Jun 28 '12 at 11:00
@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). – TagWoh Jun 28 '12 at 13:28
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. – Willie Wong Jun 28 '12 at 16:10

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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Thanks very much Willie! – TagWoh Jun 30 '12 at 8:03