Skip to main content

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

2
  • $\begingroup$ Thank you for the response! I'm still a bit confused on how NeumannValue would be implemented here. If I were to solve the PDE in the form you've given, I would use something like op==0+NeumannValue[ ? ]. What goes inside the NeumannValue function to specify the actual boundary conditions? From your answer, I see that in terms of the variables given in the documentation, c = {{-1,0},{0,0}}, a=-y^2, and alpha=beta=gamma=f=0. But the connection between c and the argument of the NeumannValue call that I need to implement is unclear, if such a connection exists. $\endgroup$ Commented Feb 27, 2024 at 17:15
  • $\begingroup$ @AndrewL From yanswer you get -{{-1,0},{0,0}}.Grad[f]=={Derivative[1,0]f][x,y],0}. This resulty has to be multiplied by the unit normal of the boundary to get the NeumanValue. See my modified answer! $\endgroup$ Commented Feb 28, 2024 at 9:54