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I'd like to know the asymptotic ($ t\to\infty $) behavior of

$$ \partial_tu = -y\partial_{y}u + x^2 + y x \partial_xu - (1 + \partial_xu)^2 $$

with $ u(0, x, y) = 0 $.

DSolve doesn't give anything. What else could I try?

If I can't solve for $ u $ directly, can I at least get some approximation of $ \tilde{u}(x, y) $, assuming that $ u(t, x, y) = kt + \tilde{u}(x, y) $?

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    $\begingroup$ Nitpick: $t \to \infty$ doesn't necessarily mean "stationary". $\partial_t u \to 0$ is "stationary", but it's not a given that $u$ approaches this value from arbitrary initial conditions. $\endgroup$ – Michael Seifert Jul 18 at 17:29
  • $\begingroup$ Also, would a numerical solution from NDSolve be acceptable? If so, you'll probably need spatial boundary conditions; what are they? $\endgroup$ – Michael Seifert Jul 18 at 17:29
  • $\begingroup$ @MichaelSeifert I agree stationary is the wrong word. Asymptotic would be better, I don't believe $\partial_tu\to 0$ $\endgroup$ – Bananach Jul 18 at 18:47
  • $\begingroup$ @MichaelSeifert that's the problem. I don't know boundary conditions except for sub exponential growth at infinity. From my experience with the heat equation that should be enough, theoretically, but it might not be good enough for NDSolve $\endgroup$ – Bananach Jul 18 at 21:24
  • $\begingroup$ Even for the conventional heat equation, you still need to specify spatial boundary conditions. What is the domain of $x$ and $y$ values you're interested in? How does $u$ behave at the "edges" of this domain? (Note that the domain in question may be "all of $\mathbb{R}^2$", though this is harder for NDSolve to handle without a few tricks.) If you don't know the boundary conditions for a non-linear PDE, then I strongly suspect there's no hope of finding a solution. $\endgroup$ – Michael Seifert Jul 19 at 12:09
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I don't think there is an analytic solution, but if "just some terms in an expansion" is fine, then the solution for small $t$ reads $$ \begin{aligned} u(t,x,y)&= \left(x^2-1\right)t+\\ &+x (x y-2)t^2\\ &+\frac{1}{3} \left(x^2 (y (2 y-1)-4)-6 x y+4\right)t^3\\ &+\mathcal O(t^4) \end{aligned} $$

Check:

D[u[t, x, y], t] == -y D[u[t, x, y], y] + D[u[t, x, y], {y, 2}] + x^2 + y x D[u[t, x, y], x] - (1 + D[u[t, x, y], x])^2 /. u -> ((-1 + #2^2) #1 + #2 (-2 + #2 #3) #1^2 + 1/3 (4 - 6 #2 #3 + #2^2 (-4 + #3 (-1 + 2 #3))) #1^3 &)

Series[%, {t, 0, 2}] // Simplify

(* True *)
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  • $\begingroup$ Sorry, I suspect that $u(t,x,y)\sim kt +\tilde{u}(x,y)$ and I meant an expansion of $\tilde{u}(x,y)$. Updated the question $\endgroup$ – Bananach Jul 18 at 15:41
  • $\begingroup$ @Bananach An expansion in powers of what? what is the small parameter here? $\endgroup$ – AccidentalFourierTransform Jul 18 at 17:39
  • $\begingroup$ I don't know. Maybe polynomials? Fourier modes? The result after a finite number of Newton iterations? $\endgroup$ – Bananach Jul 18 at 18:47
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    $\begingroup$ @Bananach I don't think you understood my question. You seem to be after some approximate solution; but approximate with respect to what? What is "small" here? What exactly do you mean by an "expansion of $\tilde u(x,y)$"? What exactly are you looking for? Unless you ask a precise question, you won't get a precise answer... $\endgroup$ – AccidentalFourierTransform Jul 18 at 18:50
  • $\begingroup$ I did understand your question. I am looking for any formula that approximates $\tilde{u}(x,y)$. I gave the example of a Newton algorithm: after any finite number of iterations this might give an analytic expression and I'd be happy with such an expression. Alternatively, one could do a Picard iteration. I also gave the example of Fourier or polynomial expansions. The solution might be analytic and it might be possible to find the first few polynomial coefficients. I cannot say precisely what I want because I don't know what Mathematica can do. Obviously I would be most happy with the exact .. $\endgroup$ – Bananach Jul 18 at 21:00

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