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Consider the following PDE:$$0=u_t+u_{yy}+u_{xx}+(x-y)u_y+y^{-\frac{3}{2}}u^2+1,$$ with $t \in [0,T], $ and a terminal condition $u_T=-1$ for all $x$ and $y.$ The domain for $x$ and $y$ can be bounded if needed, and the original problem allows to add more boundary conditions on the first derivatives if that can help.

Ideally I would like to find a solution of course, however if I can manage to find an existence / unicity result, even in a weak sense, I would be more than happy. I am open to fixed-point theorems, bootstraps arguments to show regularity ...

It would be highly appreciated if the answer contains a brief explanation, although I'd be satisfied with just the references.

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  • $\begingroup$ And the domain contains $y=0$? $\endgroup$
    – username
    Commented Apr 5, 2022 at 19:53
  • $\begingroup$ Yes 0 can be in the domain without any issue. $\endgroup$
    – FDR
    Commented Apr 5, 2022 at 21:56
  • $\begingroup$ But does it have to be? If ti doesn't then $y^{-3/2}$ is a nice bounded function, so the problem is very standard, specially because it is the inverse time direction to that of the heat equation. So you could just solve the heat equation (with your nonlinear term which is very nice) with initial data $-1$, call that $v$, and then $u(t)=v(T-t)$ $\endgroup$
    – username
    Commented Apr 6, 2022 at 13:43
  • $\begingroup$ Thank you for your answer. No it doesn't. And it is actually better if $0$ is not in the domain of $y$. So the term $y^{-\frac{3}{2}}$ is indeed bounded. Can you please elaborate on what you mean by solving the heat equation with the nonlinear term ? is it in the weak sense ? can you point towards a method ? $\endgroup$
    – FDR
    Commented Apr 6, 2022 at 13:59
  • $\begingroup$ A possible approach to solve such an equation is by a gradient expansion but I do not know if this can help. $\endgroup$
    – Jon
    Commented Apr 8, 2022 at 8:34

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The idea I was trying to convey in the comments is the following. Let us consider the equation

$$0=u_t+u_{yy}+u_{xx}+(x-y)u_y+y^{-\frac{3}{2}}u^2+1.$$

I assume at the leading order the solution of the equation

$$0=u^{(0)}_t+y^{-\frac{3}{2}}u^{(0)2}+1.$$

This has a known exact solution. Then, the next-to-leading order equation can be computed by taking

$$u(t,x,y)=\sum_{n=0}^\infty u^{(n)}(t,x,y).$$

I assume there could be an ordering parameter such that some kind of convergence exists for the above series. This point is crucial as, being not proven convergence, we cannot claim existence of such a solution. This will give to the next-to-leading order

$$u^{(1)}_t+2y^{-\frac{3}{2}}u^{(0)}u^{(1)}=-u^{(0)}_{yy}-u^{(0)}_{xx}-(x-y)u^{(0)}_y.$$

Note that now the problem is a linear one and one could use Green function techniques to get higher order terms.

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  • $\begingroup$ Do you agree that for the higher orders, using your notations, you should have $$u_{t}^{(k)}+2y^{-\frac{3}{2}}\sum_{i+j=k}^{+\infty}u^{(i)}u^{(j)}=u_{yy}^{(k-1)}+u_{xx}^{(k-1)}+(x-y)u_{y}^{(k-1)}.$$ This is somewhat similar to the approach in my comment. You solve for an initial function, and then by iterations solve for the rest. The difficult part is to prove that the series converge. In my case, with power of $T-t$, I can manage when $t$ is very close to $T.$ $\endgroup$
    – FDR
    Commented Apr 9, 2022 at 14:54
  • $\begingroup$ You are correct. Yes, proof of convergence could be complicated. $\endgroup$
    – Jon
    Commented Apr 9, 2022 at 15:42
  • $\begingroup$ Using your method, we have $$u_{0}\left(t,x,y\right)=y^{3/4}\tanh\left(arctanh\left(-\alpha y^{-3/4}\right)-y^{-\frac{3}{4}}\left(t-T\right)\right)$$ and then for every $k>0$ that $$\partial_{t}u_{k}+2y^{-\frac{3}{2}}u_{0}u_{k}=-\partial_{yy}u_{k-1}-\partial_{xx}u_{k-1}-(x-y)\partial_{y}u_{k-1}-\sum_{\begin{array}{c} i+j=k\\ i,j<k \end{array}}^{+\infty}u_{i}u_{j}=f_{k}(t,x,y)$$ giving the simple solution $$u_{k}\left(t,x,y\right)=-\int_{t}^{T}f_{k}(s,x,y)\exp\left(\int_{t}^{s}y^{-3/2}u_{0}(u,x,y)du\right)ds.$$ $\endgroup$
    – FDR
    Commented Apr 10, 2022 at 13:03
  • $\begingroup$ My question is now that we know how to write the series, how one proves convergence ? And even upon succeeding, how one can say that this proves existence of a solution ? Fyi, a simple implicit scheme without a lot of efforts gives a nice smooth looking solution for this. $\endgroup$
    – FDR
    Commented Apr 10, 2022 at 13:10
  • $\begingroup$ In principle, you are proving existence by construction but without convergence this is not possible. I have in mind something like en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem. $\endgroup$
    – Jon
    Commented Apr 10, 2022 at 18:53

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