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I'm attempting to use the Fast Poisson Solver routines from the oneMKL library. The driver, in Fortran, is provided in the attachment.

The solver works, however the corner values at locations (1,1) and (1,0) seem to be wrong, when compared with the output of a different solver.

What could be the problem?

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I realized it was an error on my side. The boundary arrays are of length nx+1 and ny+1, but I was wrongly only setting the first nx, ny values. Hence, there was by default a 0.0 in the last value.

The boundary routine is supposed to read

```
subroutine fbd(nx,ny,x,y,bd_ax,bd_bx,bd_ay,bd_by)
integer, intent(in) :: nx, ny
real(dp), intent(in) :: x(nx+1), y(ny+1)
real(dp), intent(out) :: bd_ax(ny+1), bd_bx(ny+1)
real(dp), intent(out) :: bd_ay(nx+1), bd_by(nx+1)
! boundaries are given by
! u(x,y) = cos( pi x ) - sin( 2 pi y )
bd_ax = cos(pi*0.0_dp) - sin(2*pi*y) ! x = 0
bd_bx = cos(pi*1.0_dp) - sin(2*pi*y) ! x = 1
bd_ay = cos(pi*x) - sin(2*pi*0.0_dp) ! y = 0
bd_by = cos(pi*x) - sin(2*pi*1.0_dp) ! y = 1
end subroutine
```

Now the solution matches the multigrid solver in the link in my original post.

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Ivan, I am not quite sure which results you are expecting to see here.

running with the current version of oneMKL, I see

1.00000000000000 0.000000000000000E+000 **0.000000000000000E+000 **

.....

1.00000000000000 1.00000000000000 **0.000000000000000E+000**

Which results do you see with another solvers?

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I realized it was an error on my side. The boundary arrays are of length nx+1 and ny+1, but I was wrongly only setting the first nx, ny values. Hence, there was by default a 0.0 in the last value.

The boundary routine is supposed to read

```
subroutine fbd(nx,ny,x,y,bd_ax,bd_bx,bd_ay,bd_by)
integer, intent(in) :: nx, ny
real(dp), intent(in) :: x(nx+1), y(ny+1)
real(dp), intent(out) :: bd_ax(ny+1), bd_bx(ny+1)
real(dp), intent(out) :: bd_ay(nx+1), bd_by(nx+1)
! boundaries are given by
! u(x,y) = cos( pi x ) - sin( 2 pi y )
bd_ax = cos(pi*0.0_dp) - sin(2*pi*y) ! x = 0
bd_bx = cos(pi*1.0_dp) - sin(2*pi*y) ! x = 1
bd_ay = cos(pi*x) - sin(2*pi*0.0_dp) ! y = 0
bd_by = cos(pi*x) - sin(2*pi*1.0_dp) ! y = 1
end subroutine
```

Now the solution matches the multigrid solver in the link in my original post.

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ok, then the thread is closing.

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