- #1
physicsuser2023
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- TL;DR Summary
- I encountered errors while running a Fortran code to calculate potential values inside a cube, with the first error being a reference to an undefined variable and the second error being insufficient storage allocation.
hey everyone. I wrote a code in Fortran to calculate potential values or solve the Laplace equation inside a cube according to the boundary conditions mentioned in the code.
this is my code:
I first entered the dimensions of the cube as 3x3x3, Iteration or MAX COUNTER as 1000, and MAX ERROR as 0.000001. After these inputs, I enter to make the program perform the operation, but it showed me an error: error 112, reference to undefined variable, array element or function result.
I then set the dimensions of the cube to 10x10x10 and the rest of the inputs the same as before. But this time I received another error: ALLOCATE was unable to obtain sufficient storage.
I tested a code similar to the above code for the two-dimensional case and it worked.
what's the solution?
this is my code:
Fortran:
program laplace_cubic
implicit none
REAL*8 :: LX, LY, LZ, DELTA, MAX_ERR, ERR
INTEGER :: NX, NY, NZ, I, J, K, M
INTEGER*8 :: MAX_COUNT
REAL*8, DIMENSION(:,:,:), ALLOCATABLE :: PHI, PHIO
write(*, *) 'LX='
read(*, *) LX
write(*, *) 'LY='
read(*, *) LY
write(*, *) 'LZ='
read(*, *) LZ
write(*, *) 'ENTER MAX COUNTER='
read(*, *) MAX_COUNT
write(*, *) 'ENTER MAX ERROR='
read(*, *) MAX_ERR
DELTA = 0.01
NX = INT(LX/DELTA)
NY = INT(LY/DELTA)
NZ = INT(LZ/DELTA)
ALLOCATE(PHI(0:NX, 0:NY, 0:NZ))
ALLOCATE(PHIO(0:NX, 0:NY, 0:NZ))
CALL INITIALIZE(PHI, NX, NY, NZ, LX, LY, LZ, DELTA)
DO M = 1, MAX_COUNT
PHIO=PHI
ERR=0.0
DO I = 1, NX - 1
DO J = 1, NY - 1
DO K = 1, NZ - 1
PHI(I, J, K) = (PHIO(I + 1, J, K) + PHIO(I - 1, J, K) + PHIO(I, J + 1, K) + &
PHIO(I, J - 1, K) + PHIO(I, J, K + 1) + PHIO(I, J, K - 1)) / 6.0
ERR = ERR + (PHI(I, J, K) - PHIO(I, J, K))**2
END DO
END DO
END DO
IF (ERR < MAX_ERR) EXIT
END DO
OPEN(100, FILE = 'laplaceoutput.TXT')
DO I = 0, NX
DO J = 0, NY
DO K = 0, NZ
WRITE(100, *) I * DELTA, J * DELTA, K * DELTA, PHI(I, J, K)
END DO
END DO
END DO
end program laplace_cubic
SUBROUTINE INITIALIZE(PHI,NX,NY,NZ,LX,LY,LZ,DELTA)
IMPLICIT NONE
REAL,PARAMETER::PI=3.1415
INTEGER::I,J,K
INTEGER,INTENT(IN)::NX,NY,NZ
REAL*8::DELTA,X,Y,Z,LX,LY,LZ
REAL*8,DIMENSION(0:NX,0:NY,0:NZ)::PHI
DO J = 1, NY - 1
DO K = 1, NZ - 1
PHI(0, J, K) = 1.0 ! potential inside the back face (zy plane)
PHI(NX,J,K) = 0.0
END DO
END DO
DO J = 0, NY
Y = J * DELTA
PHI(0, J, 0) = SIN(PI*Y/LY) !sinusoidal potential along y-axis
END DO
DO K = 0, NZ
Z = K * DELTA
PHI(0, 0, K) = SIN(PI*Z/LZ) ! sinusoidal potential along z-axis
END DO
DO J = 1, NY-1
DO I = 1, NX-1
PHI(I, J, 0) = 1.0 ! potential inside the bottom face (xy plane)
PHI(I,J,NZ) = 0.0
END DO
END DO
DO I = 0, NX
X= I * DELTA
PHI(I, 0, 0) = SIN(PI*X/LX) ! sinusoidal potential along x-axis
END DO
DO I = 1, NX-1
DO K = 1, NZ-1
PHI(I, 0, K) = 1.0 ! potential inside the left face (XZ plane)
PHI(I,NY,K) = 0.0
END DO
END DO
DO K = 1, NZ-1
DO J = 1, NY-1
DO I = 1, NX-1
PHI(I, J, K) = 0.5 ! potential inside the cube
END DO
END DO
END DO
END SUBROUTINE INITIALIZEI then set the dimensions of the cube to 10x10x10 and the rest of the inputs the same as before. But this time I received another error: ALLOCATE was unable to obtain sufficient storage.
I tested a code similar to the above code for the two-dimensional case and it worked.
what's the solution?