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 March 31st, 2019, 08:43 PM #1 Newbie   Joined: Mar 2019 From: Russia Posts: 2 Thanks: 0 Partial Differential Equation Help please, I need to solve this differential equation $\displaystyle x\frac{\partial^2 U}{\partial x^2}+y\frac{\partial^2 U}{\partial y^2}=aU$ in Matlab (where "a" is a constant parameter, it can be taken by any), I wanted to use the Partial Differential Equation Toolbox, but I ran into a problem, the elliptic equation in this Toolbox is represented in a vector form, namely -div(c*grad(u))+a*u=f. Please help me convert my equation to this form and tell me how it can be done or at least name the sources of information from which I can learn this knowledge. I really need to solve this equation in Matlab,so please tell me how it can be done.
 April 1st, 2019, 07:20 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 600 Thanks: 366 Math Focus: Dynamical systems, analytic function theory, numerics You probably can't convert your equation to this form. Your equation is nonlinear and second equation you have written is linear. In the unlikely even that it is possible to convert it then it will be some nonlinear change of coordinates which is non-obvious. Most importantly, there is no canonical or algorithmic way to change your equation to this form. Thanks from topsquark
 April 2nd, 2019, 06:20 AM #3 Newbie   Joined: Mar 2019 From: Russia Posts: 2 Thanks: 0 Thanks for the answer, in that case could you tell me how to solve it in Matlab? Is there any way to solve this equation using built-in matlab methods, without algorithm processing and writing code from scratch?
 April 2nd, 2019, 02:19 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,484 Thanks: 2041 Your equation is linear; I don't know why SDK thought it isn't. Thanks from SDK
April 2nd, 2019, 10:59 PM   #5
Senior Member

Joined: Sep 2016
From: USA

Posts: 600
Thanks: 366

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by skipjack Your equation is linear; I don't know why SDK thought it isn't.
Thanks for noticing this. I misread the equation and thought $y$ was the solution variable.

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