Find the lagest function value on the boundary

General Discussion
1

Hello to all of you FreeFem users,

I want to find the largest value of my PDE solution on the boundary of the domain. I know “u.max” gives the largest value of the u on the mesh, but I don’t know how to modify it to get the largest function value on the boundary of the domain.

Thank you in advance.

2

Are you looking for something like this?

mesh Th = square(10,10);
fespace Vh(Th, P2);
Vh u = x*y;
varf onBoundary(u,v) = on(1, 2, 3, 4, u = 1.0);
real[int] onB = onBoundary(0, Vh, tgv = -1);
u[] .*= onB;
cout << u[].max << endl;
plot(u);
3

Hello,

To find the largest value of your PDE solution 𝑢u specifically on the boundary of the domain in FreeFem, you can follow these steps:

  1. Identify Boundary Elements: Use boundary labels to identify the elements that lie on the boundary.
  2. Evaluate Boundary Values: Extract the values of 𝑢u at these boundary elements.
  3. Find Maximum: Compute the maximum value among these boundary values.
    Thanks
4

Thanks for your reply! I understand this strategy but I don’t know what is the syntax for them. For example, how to identify the element that lies on the boundary?

5

Thank you, it works. My understanding of this code is that, when envaluates onBoundary(0,vh, tgv=-1), it returns a vector that is zero on the interior point and is 1 for all boundary points. Therefore, u=u.*onB makes the function value on the interior point to be zero, which stands out from the boundary points. Is that correct understanding? If so I think an addition modification is required if the function value is negative on the boundary. In addition to that, could I ask why tgv is set to be -1 when constructs onB?

6

Your understanding is correct. No modification is needed if the function value on the boundary is negative. A negative number multiplied by 1 is still negative.

See here for an explanation of tgv = -1.