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within my thesis, I am struggeling with the following PDE:

$u_t+a(x,y)u_{xx}+b(x,y)u_{xy}+c(x,y)u_{yy}+d(x,y)u_{x}+e(x,y)u_{y}+f(x,y)u=0$ $u(T,x,y)=1,$

where $a,b,c,d,e,f$ are polynomials and the domain is the set $\{(t,x,y):t\in[0,T],x+y=1,x\geq0,y\geq0\}$.

I showed, however, that there exists a unique solution and I would like to have this solution explicitly, either via Maple/Matlab or by hand. My problem hereby is that this PDE is not in the right form for Matlab, Maple is not able to solve it (here my post in mapleprime) and I don't know how to handle this by hand since I never discretized a PDE in two spatial variables. Besides that, substituting $x$ by $1-y$ also does not work since there are mixed derivatives.

Do you have any advice for me? I would be so grateful.

Thanks a lot for your help.

Best regards,

utcyp

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  • $\begingroup$ do you have some additional constraints on the boundary of your triangle? You could use a uniform grid and do some finite differences. $\endgroup$
    – user35593
    Commented Aug 7, 2018 at 17:23
  • $\begingroup$ Hey user35593, thanks for your reply. I only have these constraints. Actually, my domain should be a line segment. Therefore, I guess that it is not possible to discretize it. $\endgroup$
    – utcyp
    Commented Aug 8, 2018 at 8:58
  • $\begingroup$ Ok I didnt see that. But then how do you define the derivative by x and/or y if your function is only defined on the x+y=1 line? How did you prove existence and uniqueness? For me it seems the problem is not well-posed. $\endgroup$
    – user35593
    Commented Aug 8, 2018 at 16:58
  • $\begingroup$ Hey, in order to show the existence and uniqueness I used Thm. 10 of Ch. 2.9 from Krylov in a $n$-dimensional space where I don't get these difficulties since I then have a open set (the $n$-dimensional probability simplex). Hence, the problem should be well-posted in more dimensions, isn't it? $\endgroup$
    – utcyp
    Commented Aug 9, 2018 at 6:33
  • $\begingroup$ The statements about uniqueness seems to be incorrect. As an example, make a change of variables (x,y): turn on $\pi/4$. The equation will have the same form. Put $a\equiv=-1$, $b\equiv c\equiv\ldots0$. Then the solution of the heat is not unique since no conditions on the segment's boundary are posed. $\endgroup$
    – Andrew
    Commented Sep 8, 2018 at 5:55

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