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Apr 9, 2022 at 12:49 answer added Jon timeline score: 1
Apr 9, 2022 at 9:03 comment added FDR @username Any help would bighly appreciated, as you seem to think the problem is an easy one ?
Apr 8, 2022 at 14:30 comment added FDR Thank you for your comment Jon, I will give it a try, but it could take some time..
Apr 8, 2022 at 14:27 comment added Jon Let me say that here the idea is very simple. Formally, insert a small parameter $\epsilon$ on the perturbation part, apply a perturbation series, and set $\epsilon$ to 1 in the end. You will get a non-trivial series but, as I said, this cannot grant existence and could be a too "physicist"-like approach for your taste. The book I cited is for mathematicians anyway.
Apr 8, 2022 at 14:22 comment added FDR Oh I was more thinking about a reference where this kind of nonlinearity is addressed with your method. Is there a specific chapter in the book I should look into ? I'm not a physicist so reading it might be hard for me.. especially if the chances it works are low.
Apr 8, 2022 at 14:18 comment added Jon This is an approach widely used in fluid mechanics. A good reference for mathematicians is books.google.it/…. See discussion starting from page 38.
Apr 8, 2022 at 13:59 comment added FDR I'm afraid I am not familiar with the method. Could you point me towards a a reference please ?
Apr 8, 2022 at 13:52 comment added Jon I meant $\partial_t u_0+y^{-\frac{3}{2}}u_0^2+1=0$ at the leading order and the remain as a perturbation. This has an exact solution in terms of a tanh. A good non-trivial starting point. Anyway, for existence you need at least convergence but beyond a perturbation approach it is difficult to see a solution.
Apr 8, 2022 at 9:53 comment added FDR An existence and unicity argument is more than enough for my problem..
Apr 8, 2022 at 9:19 comment added FDR $$\begin{cases} u_{0} & =-1\\ u_{1} & =\partial_{yy}u_{0}+\partial_{xx}u_{0}+(x-y)\partial_{y}u_{0}+y^{-\frac{3}{2}}u_{0}^{2}+1\\ & =y^{-\frac{3}{2}}+1\\ u_{2} & =\partial_{yy}u_{1}+\partial_{xx}u_{1}+(x-y)\partial_{y}u_{1}+y^{-\frac{3}{2}}u_{1}^{2}+1\\ ... nu_{n}= & \partial_{yy}u_{n-1}+\partial_{xx}u_{n-1}+(x-y)\partial_{y}u_{n-1}+y^{-\frac{3}{2}}u_{n-1}^{2}+y^{-\frac{3}{2}}\sum_{k+m=n}^{+\infty}u_{k}(x,y)u_{m}(x,y) \end{cases}$$ which gives a polynomial representation of the solution. But that doesn't say it exists right ?
Apr 8, 2022 at 9:19 comment added FDR Hi Jon and thank you for your comment. I tried this, maybe that is what you mean by gradient expansion. I first substitute $u\left(t,x,y\right)=\sum_{0}^{\infty}u_{n}\left(x,y\right)\left(T-t\right)^{n}$ and by identifying the powers of $t$ we get the system:
Apr 8, 2022 at 8:34 comment added Jon A possible approach to solve such an equation is by a gradient expansion but I do not know if this can help.
Apr 6, 2022 at 13:59 comment added FDR 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 ?
Apr 6, 2022 at 13:43 comment added username 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 $u(t)=v(T-t)$
Apr 5, 2022 at 21:56 comment added FDR Yes 0 can be in the domain without any issue.
Apr 5, 2022 at 19:53 comment added username And the domain contains $y=0$?
S Apr 5, 2022 at 10:52 review First questions
Apr 5, 2022 at 11:43
S Apr 5, 2022 at 10:52 history asked FDR CC BY-SA 4.0