Starnge problem

Accelerated Computing HPC Compilers Legacy PGI Compilers
1

I’m writing a rpogram in Fortran for computing Boundary Value Problem ODE.
Here is my code:

module_BVP.f90

module module_BVP

public :: Step, Tridiagonal_Solv, ReadFromFile, WriteToFile, channel

contains

!==========  Step =============================================================================================
!           - a,     - b 
!     - n.
function Step(n, a, b) result(h)
	implicit none

	integer,			 intent(in)	:: n
	double precision,	 intent(in) :: b, a
	real							:: h
	 
	h = (b-a) / (n-1)

end function Step

!======================================================================================================================

!============== Subroutine Values =================

subroutine Tridiagonal_Solv(n, a, b, alpha, beta)
	implicit none 
	integer						:: i, j, n 
	double precision					:: a, b ! 
	real,			  dimension(:)		:: q, w, t, y 
	real,			  dimension(:)		:: sigma, teta
	double precision					:: p, l, z
	double precision,		  intent(in) :: alpha, beta 
	double precision, dimension(n)             :: x, r, vector_B 
	double precision, dimension(n,n)			 :: matrix_A 

	x(0) = a
	x(n) = b
	p = -1 / (Step(n, a, b)*Step(n, a, b))
	z = 2 / (Step(n, a, b)*Step(n, a, b)) + 1
	l = 2 / Step(n, a, b)

	!    x(i)  r(x(i))
	do i = 2, n-1
		x(i) = a + i*Step(n, a, b)
		r(i) = 2*sin(x(i))
	end do

	!    nxn . 
	do i = 1, n     
		do j = 1, n
			matrix_A(i,j) = 0
		end do
	end do

	!         (   )
	matrix_A(1,1) = -3 / (2*Step(n, a, b))
	matrix_A(1,2) = l
	matrix_A(1,3) = -1 / (2*Step(n, a, b))

	!    
	do i = 2, n-1
		matrix_A(i,i-1) = p
		matrix_A(i,i) = z
		matrix_A(i,i+1) = p
		vector_B(i) = r(i)
	end do
	
	!          (   )
	matrix_A(n,n-2) = 1 / (2*Step(n, a, b))
	matrix_A(n, n-1) = l
	matrix_A(n,n) = 3 / (2*Step(n, a, b))

	vector_B(1) = alpha
	vector_B(n) = beta

	y(0) = 0
	y(n+1) = 0
	sigma(0) = 0
	teta(0) = 0
	w(1) = 0
	t(n) = 0

	do i = 2, n-1
		w(i) = matrix_A(i,i-1)
	end do

	do i = 1, n
		q(i) = matrix_A(i,i)
	end do

	do i = 1, n-1
		t(i) = matrix_A(i,i+1)
	end do

	! ,     
	do i = 1, n 
		sigma(i) = -t(i)/(q(i) + w(i)*sigma(i-1))
		teta(i) = (r(i) - w(i)*teta(i-1))/(q(i) + w(i)*sigma(i-1))
	end do

	y(n) = teta(n)

	! ,    
	do i = n-1, 1, -1 
		y(i) = sigma(i)*y(i+1) + teta(i)
	end do

end subroutine Tridiagonal_Solv
!==============================================

!================= Subroutine _Read ==================

subroutine ReadFromFile(n, a, b, alpha, beta)
	implicit none
	integer				 :: n, free, err
	double precision			 :: a, b
	double precision			 :: alpha, beta

	call channel(free)
	open(unit = free, file = "Values.dat", status = "old", action = "read")
		do
			read(free, *, iostat = err) n, a, b, alpha, beta
			if(err == 0) then
				print *, "Reading data..."
				print *
				print *, "     n     a           b            alpha     beta"
				print "(i7, f8.2, f16.10, f11.3, f10.3)", n, a, b, alpha, beta
				print *
				print *, "Data reading complete successfully!"
				exit
			 else if(err < 0) then
				print *, "Error in data reading!"
				exit
			end if
		end do
	close(free)

end subroutine ReadFromFile

!===============================================

!================= Subroutine _Write ========

subroutine WriteToFile(n, x, y)
	implicit none
	integer						   :: free, n, i
	double precision, dimension(n) :: x, y

	call channel(free)
	open(unit = free, file = "Results.dat", status = "replace", action = "write")
		write(free, *) " i      x(i)	     y(i)"
		write(free, *)
		do i = 1, n
			write(free, fmt = "(i3, f13.8, f13.8)"), i, x(i), y(i)
		end do
		print *
		print *, "The results was writen in file 'Results.dat'"
	close(free)

end subroutine WriteToFile

!==============================================

!================== CHANNEL ==========================================================================================

SUBROUTINE channel (NFREE)

      LOGICAL FISOPEN

      INTEGER NFREE

      FISOPEN = .TRUE.
      NFREE = 9
      DO WHILE (FISOPEN)
         NFREE = NFREE + 1
         INQUIRE (NFREE, OPENED = FISOPEN)
      END DO
      RETURN
      END SUBROUTINE channel

!====================================================================================================================

end module module_BVP

BVP.f90

program BVP
use module_BVP

	implicit none

	integer				     :: n
	double precision			     :: a, b
	double precision			     :: alpha, beta
	double precision, dimension(n) :: x, y

	call ReadFromFile(n, a, b, alpha, beta)
	call Tridiagonal_Solv(n, a, b, alpha, beta)
	call WriteToFile(n, x, y)

end program BVP

There is no error when I compile it. But when run it in the Results.dat file there is no results writen. Only zeros in all columns. Why is that happen.

2

Hi ferry2,

You need to pass in your main program’s “x” and “y” variables to Tridiagonal_Solv. Otherwise, the “x” and “y” in Tridiagonal_Solv are local to the subroutine.

Though, I’m a bit surprised the code made it to writing out a file. It should get a segmentation violation in Tridiagonal_Solv since several of your arrays are assumed-shape but not used as arguments. Also, since x and y’s size wont be known until the file is read in, they should be allocatable arrays. Finally, be sure to declare the arrays with a “0” lower bound. In Fortran, the default is 1.

  • Mat
3

Now my code is:

module_BVP.f90

module module_BVP

public :: Step, Tridiagonal_Solv, ReadFromFile, WriteToFile, channel

contains

!========== Функция Step =============================================================================================
! Изчисляваме големината на стъпката по зададени ляв край на интервала - a, десен край на интервала - b 
! и брой на итерациите - n.
function Step(n, a, b) result(h)
	implicit none

	integer,			 intent(in)	:: n
	double precision,	 intent(in) :: b, a
	real							:: h
	 
	h = (b-a) / (n-1)

end function Step

!======================================================================================================================

!============== Subroutine Values =====================================================================================
! Подпрограма, в която изчисляваме всички стойности на x_i
subroutine Tridiagonal_Solv(n, a, b, alpha, beta, x, y)
	implicit none 
	integer										  :: i, j, n ! i и j са броячи за циклите, а n е размерността на матриците и векторите
	double precision							  :: a, b ! а и b са съответно ляв и десен край на интервала
	real, allocatable, dimension(:)				  :: q, w, t ! q, w, t - Три едномерни масива за коефициентите на системата; 
															   
	real, allocatable, dimension(:)				  :: sigma, teta ! Eдномерни масиви за векторите от прогонъчни коефициенти
	double precision							  :: p, l, z ! Променливи, в които се съхраняват изчислени изрази
	double precision							  :: alpha, beta ! alpha и beta са стойностите на граничните условия
	double precision, allocatable, dimension(:)	  :: x, r, vector_B, y ! х е вектор за променливата "х", r е вектор за функцията r(x)
																				  !y - Eдномерен масив за вектора от неизвестни
	double precision, allocatable, dimension(:,:) :: matrix_A ! A e тридиагонална матрица, в която

	allocate(q(n), w(n), t(n), y(0:n+1), sigma(0:n), teta(0:n), x(0:n), r(n), vector_B(n), matrix_A(n,n))

	x(0) = a
	x(n) = b
	p = -1 / (Step(n, a, b)*Step(n, a, b))
	z = 2 / (Step(n, a, b)*Step(n, a, b)) + 1
	l = 2 / Step(n, a, b)

	! Пресмятаме стойностите на x(i) и r(x(i))
	do i = 2, n-1
		x(i) = a + i*Step(n, a, b)
		r(i) = 2*sin(x(i))
	end do

	! Създаваме си нулева nxn матрица. 
	do i = 1, n     
		do j = 1, n
			matrix_A(i,j) = 0
		end do
	end do

	! Записваме елементите стоящи на първия ред на матрицата (от първото гранично условие)
	matrix_A(1,1) = -3 / (2*Step(n, a, b))
	matrix_A(1,2) = l
	matrix_A(1,3) = -1 / (2*Step(n, a, b))

	! Записваме всички останали елементи
	do i = 2, n-1
		matrix_A(i,i-1) = p
		matrix_A(i,i) = z
		matrix_A(i,i+1) = p
		vector_B(i) = r(i)
	end do
	
	! Записваме и елементите стоящи на последния ред на матрицата (от второто гранично условие)
	matrix_A(n,n-2) = 1 / (2*Step(n, a, b))
	matrix_A(n, n-1) = l
	matrix_A(n,n) = 3 / (2*Step(n, a, b))

	vector_B(1) = alpha
	vector_B(n) = beta

	y(0) = 0
	y(n+1) = 0
	sigma(0) = 0
	teta(0) = 0
	w(1) = 0
	t(n) = 0

	do i = 2, n-1
		w(i) = matrix_A(i,i-1)
	end do

	do i = 1, n
		q(i) = matrix_A(i,i)
	end do

	do i = 1, n-1
		t(i) = matrix_A(i,i+1)
	end do

	! Цикъл, в който изчисляваме прогонъчните коефициенти
	do i = 1, n 
		sigma(i) = -t(i)/(q(i) + w(i)*sigma(i-1))
		teta(i) = (r(i) - w(i)*teta(i-1))/(q(i) + w(i)*sigma(i-1))
	end do

	y(n) = teta(n)

	! Цикъл, в който изчисляваме неизвестните
	do i = n-1, 1, -1 
		y(i) = sigma(i)*y(i+1) + teta(i)
	end do

end subroutine Tridiagonal_Solv
!=====================================================================================================================

!================= Subroutine _Read ==================================================================================

subroutine ReadFromFile(n, a, b, alpha, beta)
	implicit none
	integer						 :: n, free, err
	double precision			 :: a, b
	double precision			 :: alpha, beta

	call channel(free)
	open(unit = free, file = "Values.dat", status = "old", action = "read")
		do
			read(free, *, iostat = err) n, a, b, alpha, beta
			if(err == 0) then
				print *, "Reading data..."
				print *
				print *, "     n     a           b            alpha     beta"
				print "(i7, f8.2, f16.10, f11.3, f10.3)", n, a, b, alpha, beta
				print *
				print *, "Data reading complete successfully!"
				exit
			 else if(err < 0) then
				print *, "Error in data reading!"
				exit
			end if
		end do
	close(free)

end subroutine ReadFromFile

!=====================================================================================================================

!================= Subroutine _Write =================================================================================

subroutine WriteToFile(n, x, y)
	implicit none
	integer										:: free, n, i
	double precision, allocatable, dimension(:) :: x, y

	allocate(x(n), y(n))

	call channel(free)
	open(unit = free, file = "Results.dat", status = "replace", action = "write")
		write(free, *) " i      x(i)	     y(i)"
		write(free, *)
		do i = 1, n
			write(free, fmt = "(i3, f13.8, f13.8)"), i, x(i), y(i)
		end do
		print *
		print *, "The results was writen in file 'Results.dat'"
	close(free)

end subroutine WriteToFile

!=====================================================================================================================

!================== CHANNEL ==========================================================================================

SUBROUTINE channel (NFREE)

      LOGICAL FISOPEN

      INTEGER NFREE

      FISOPEN = .TRUE.
      NFREE = 9
      DO WHILE (FISOPEN)
         NFREE = NFREE + 1
         INQUIRE (NFREE, OPENED = FISOPEN)
      END DO
      RETURN
      END SUBROUTINE channel

!====================================================================================================================

end module module_BVP

BVP.f90

program BVP
use module_BVP

	implicit none

	integer										:: n
	double precision							:: a, b
	double precision							:: alpha, beta
	double precision, allocatable, dimension(:) :: x, y

	allocate(x(n), y(n))

	call ReadFromFile(n, a, b, alpha, beta)
	call Tridiagonal_Solv(n, a, b, alpha, beta, x, y)
	call WriteToFile(n, x, y)

end program BVP

But I’m geting a very strange result. The data in result file is:

i x(i) y(i)

1 0.05208331 760.09265128
2 0.00005207 760.09240824
3 0.00000000*************
4**************************
5 0.00000000 0.00000000

4

Hi ferry2,

The stars indicate that your number is too wide for the width provided, however, this is not your main issue.

You want to allocate x and y only once, preferably in your main program. Hence, remove the allocatable clause from x and y’s declaration in WriteToFile and Tridiagonal_Solv as well as removing them from the allocate statements. By allocating x and y in WriteToFile, you’re actually printing our junk values.

Also, since n isn’t known until after you read in your values, you need to wait to allocate x and y until after ReadFromFile is called. Otherwise n’s value is junk.

Since x and y don’t use element zero, you probably want to change “x(0)=a” to “x(1)=a” and remove “y(0)=0”. Also, since the value of Step never changes, why not call it once and store it’s value in a local variable?

  • Mat
5

@mkcolg thankc for the tips. I edit my program and now the problem is that in the “Results.dat” file I’m geting only zeros. Maybe I have a logical error.

module_BVP.f90

	module module_BVP

	public :: Step, Tridiagonal_Solv, ReadFromFile, WriteToFile, channel

	contains

!========== ??????? Step =============================================================================================
! ??????????? ?????????? ?? ???????? ?? ???????? ??? ???? ?? ????????? - a, ????? ???? ?? ????????? - b 
! ? ???? ?? ?????????? - n.
	function Step(n, a, b) result(h)
	implicit none

	integer,			 intent(in)	:: n
	double precision,	 intent(in) :: b, a
	real							:: h
	 
	h = (b-a) / (n-1)

	end function Step

!======================================================================================================================

!============== Subroutine Values =====================================================================================
! ???????????, ? ????? ??????????? ?????? ????????? ?? x_i
	subroutine Tridiagonal_Solv(n, a, b, alpha,beta)
	implicit none 
	
	integer						   :: n, i
	double precision, intent(in)   :: a, b, alpha, beta
	real						   :: h ! Ïðîìåíëèâà, â êîÿòî ùå ïàçèì ñòîéíîñòòà íà ñòúïêàòà
	double precision, dimension(0:5) :: x, r, u, v, w, z, s, t, y


	h = Step(n, a, b)
	
	x(1) = a
	r(1) = alpha
	do i = 2, n-1
		x(i) = a+i*h
		r(i) = 2*sin(x(i))
	end do
	r(n) = beta
	x(n) = b

	do i = 2, n-1
		u(i) = -1/(h*h)
	end do

	u(n) = 2/h
	
	v(1) = -3/(2*h)
	do i = 2, n-1
		v(i) = 2/(h*h) + 1
	end do
	v(n) = 3/(2*h)

	w(1) = 2/h
	do i = 2, n-1
		w(i) = -1/(h*h)
	end do
	
	s(0) = 0
	t(0) = 0
	do i = 1, n
		s(i) = -w(i)/(v(i) + u(i)*s(i-1))
		t(i) = (r(i) - u(i)*t(i-1))/(v(i) + u(i)*s(i-1))
	end do

	y(n) = t(n)
	do i = n-1, 1, -1
		y(i) = s(i)*y(i+1) + t(i)
	end do
	

	end subroutine Tridiagonal_Solv
!=====================================================================================================================

!================= Subroutine _Read ==================================================================================

	subroutine ReadFromFile(n, a, b, alpha, beta)
	implicit none
	integer						 :: n, free, err
	double precision			 :: a, b
	double precision			 :: alpha, beta

	call channel(free)
	open(unit=free,file="Values.dat",status = "old", action = "read")
		do
			read(free, *, iostat = err) n, a, b, alpha, beta
			if(err == 0) then
				print *, "Reading data..."
				print *
		print *, "     n     a           b            alpha     beta"
		print "(i7, f8.2, f16.10, f11.3, f10.3)", n, a, b, alpha, beta
				print *
				print *, "Data reading complete successfully!"
				exit
			 else if(err < 0) then
				print *, "Error in data reading!"
				exit
			end if
		end do
	close(free)

	end subroutine ReadFromFile

!=====================================================================================================================

!================= Subroutine _Write =================================================================================

	subroutine WriteToFile(n, x, y)
	implicit none
	integer										:: free, n, i
	double precision, dimension(:) :: x, y

	call channel(free)
	open(unit=free,file="Results.dat",status="replace",action="write")
		write(free, *) " i      x(i)	     y(i)"
		write(free, *)
		do i = 1, n
			write(free, fmt = "(i3, f13.8, f13.8)"), i, x(i), y(i)
		end do
		print *
		print *, "The results was writen in file 'Results.dat'"
	close(free)

	end subroutine WriteToFile

!=====================================================================================================================

!================== CHANNEL ==========================================================================================

	SUBROUTINE channel (NFREE)

      LOGICAL FISOPEN

      INTEGER NFREE

      FISOPEN = .TRUE.
      NFREE = 9
      DO WHILE (FISOPEN)
         NFREE = NFREE + 1
         INQUIRE (NFREE, OPENED = FISOPEN)
      END DO
      RETURN
      END SUBROUTINE channel

!====================================================================================================================

	end module module_BVP

BVP.f90

	program BVP
	use module_BVP

	implicit none
	integer, parameter			   :: nn = 100
	integer						   :: n
	double precision			   :: a, b
	double precision			   :: alpha, beta 
	double precision, dimension(nn):: x, y

	call ReadFromFile(n, a, b, alpha, beta)
	call Tridiagonal_Solv(n, a, b, alpha,beta)
	call WriteToFile(n, x, y)

	end program BVP
6

Hi ferry2,

I’m assuming your a student, or at least new to programming, so instead of just telling what’s wrong, I wan you to try and figure this out yourself.

Hint: Reread my first post and then look over your main program again.

  • Mat