Dummy argument not agreeing with actual argument when passing function - fortran

I'm trying to implement Newton's method but I'm getting a confusing error message. In my code you'll see I called external with f1 and f2 which I assumes tells the computer to look for the function but it's treating them as variables based on the error message. I've read the stack overflow posts similar to my issue but none of the solutions seem to work. I've tried with and without the external but the issue still persists. Hoping someone could see what I'm missing.
implicit none
contains
subroutine solve(f1,f2,x0,n, EPSILON)
implicit none
real(kind = 2), external:: f1, f2
real (kind = 2), intent(in):: x0, EPSILON
real (kind = 2):: x
integer, intent(in):: n
integer:: iteration
x = x0
do while (abs(f1(x))>EPSILON)
iteration = iteration + 1
print*, iteration, x, f1(x)
x = x - (f1(x)/f2(x))
if (iteration >= n) then
print*, "No Convergence"
stop
end if
end do
print*, iteration, x
end subroutine solve
end module newton
Program Lab10
use newton
implicit none
integer, parameter :: n = 1000 ! maximum iteration
real(kind = 2), parameter :: EPSILON = 1.d-3
real(kind = 2):: x0, x
x0 = 3.0d0
call solve(f(x),fp(x),x0,n, EPSILON)
contains
real (kind = 2) function f(x) ! this is f(x)
implicit none
real (kind = 2), intent(in)::x
f = x**2.0d0-1.0d0
end function f
real (kind = 2) function fp(x) ! This is f'(x)
implicit none
real (kind = 2), intent(in)::x
fp = 2.0d0*x
end function fp
end program Lab10```

You seem to be passing function results to your subroutine and not the functions themselves. Remove (x) when calling solve() and the problem will be resolved. But more importantly, this code is a prime example of how to not program in Fortran. The attribute external is deprecated and you better provide an explicit interface. In addition, what is the meaning of kind = 2. Gfortran does not even comprehend it. Even if it comprehends the kind, it is not portable. Here is a correct portable modern implementation of the code,
module newton
use iso_fortran_env, only: RK => real64
implicit none
abstract interface
pure function f_proc(x) result(result)
import RK
real(RK), intent(in) :: x
real(RK) :: result
end function f_proc
end interface
contains
subroutine solve(f1,f2,x0,n, EPSILON)
procedure(f_proc) :: f1, f2
real(RK), intent(in) :: x0, EPSILON
integer, intent(in) :: n
real(RK) :: x
integer :: iteration
x = x0
do while (abs(f1(x))>EPSILON)
iteration = iteration + 1
print*, iteration, x, f1(x)
x = x - (f1(x)/f2(x))
if (iteration >= n) then
print*, "No Convergence"
stop
end if
end do
print*, iteration, x
end subroutine solve
end module newton
Program Lab10
use newton
integer, parameter :: n = 1000 ! maximum iteration
real(RK), parameter :: EPSILON = 1.e-3_RK
real(RK) :: x0, x
x0 = 3._RK
call solve(f,fp,x0,n, EPSILON)
contains
pure function f(x) result(result) ! this is f(x)
real (RK), intent(in) :: x
real (RK) :: result
result = x**2 - 1._RK
end function f
pure function fp(x) result(result) ! This is f'(x)
real (RK), intent(in) :: x
real (RK) :: result
result = 2 * x
end function fp
end program Lab10
If you expect to pass nonpure functions to the subroutine solve(), then remove the pure attribute. Note the use of real64 to declare 64-bit (double precision) real kind. Also notice how I have used _RK suffix to assign 64-bit precision to real constants. Also, notice I changed the exponents from real to integer as it is multiplication is more efficient than exponentiation computationally. I hope this answer serves more than merely the solution to Lab10.

Related

Computing the Jacobian matrix in Fortran

In Newton's method, to solve a nonlinear system of equations we need to find the Jacobian matrix and the determinant of the inverse of the Jacobian matrix.
Here are my component functions,
real function f1(x,y)
parameter (pi = 3.141592653589793)
f1 = log(abs(x-y**2)) - sin(x*y) - sin(pi)
end function f1
real function f2(x,y)
f2 = exp(x*y) + cos(x-y) - 2
end function f2
For the 2x2 case I am computing the Jacobian matrix and determinant of the inverse of Jacobian matrix like this,
x = [2,2]
h = 0.00001
.
.
! calculate approximate partial derivative
! you can make it more accurate by reducing the
! value of h
j11 = (f1(x(1)+h,x(2))-f1(x(1),x(2)))/h
j12 = (f1(x(1),x(2)+h)-f1(x(1),x(2)))/h
j21 = (f2(x(1)+h,x(2))-f2(x(1),x(2)))/h
j22 = (f2(x(1),x(2)+h)-f2(x(1),x(2)))/h
! calculate the Jacobian
J(1,:) = [j11,j12]
J(2,:) = [j21,j22]
! calculate inverse Jacobian
inv_J(1,:) = [J(2,2),-J(1,2)]
inv_J(2,:) = [-J(2,1),J(1,1)]
DET=J(1,1)*J(2,2) - J(1,2)*J(2,1)
inv_J = inv_J/DET
.
.
How do I in Fortran extend this to evaluate a Jacobian for m functions evaluated at n points?
Here is a more flexible jacobian calculator.
The results with the 2×2 test case are what you expect
arguments (x)
2.00000000000000
2.00000000000000
values (y)
1.44994967586787
53.5981500331442
Jacobian
0.807287239448229 3.30728724371454
109.196300248300 109.196300248300
I check the results against a symbolic calculation for the given inputs of
Console.f90
program Console1
use ISO_FORTRAN_ENV
implicit none
! Variables
integer, parameter :: wp = real64
real(wp), parameter :: pi = 3.141592653589793d0
! Interfaces
interface
function fun(x,n,m) result(y)
import
integer, intent(in) :: n,m
real(wp), intent(in) :: x(m)
real(wp) :: y(n)
end function
end interface
real(wp) :: h
real(wp), allocatable :: x(:), y(:), J(:,:)
! Body of Console1
x = [2d0, 2d0]
h = 0.0001d0
print *, "arguments"
print *, x(1)
print *, x(2)
y = test(x,2,2)
print *, "values"
print *, y(1)
print *, y(2)
J = jacobian(test,x,2,h)
print *, "Jacobian"
print *, J(1,:)
print *, J(2,:)
contains
function test(x,n,m) result(y)
! Test case per original question
integer, intent(in) :: n,m
real(wp), intent(in) :: x(m)
real(wp) :: y(n)
y(1) = log(abs(x(1)-x(2)**2)) - sin(x(1)*x(2)) - sin(pi)
y(2) = exp(x(1)*x(2)) + cos(x(1)-x(2)) - 2
end function
function jacobian(f,x,n,h) result(u)
procedure(fun), pointer, intent(in) :: f
real(wp), allocatable, intent(in) :: x(:)
integer, intent(in) :: n
real(wp), intent(in) :: h
real(wp), allocatable :: u(:,:)
integer :: j, m
real(wp), allocatable :: y1(:), y2(:), e(:)
m = size(x)
allocate(u(n,m))
do j=1, m
e = element(j, m) ! Get kronecker delta for j-th value
y1 = f(x-e*h/2,n,m)
y2 = f(x+e*h/2,n,m)
u(:,j) = (y2-y1)/h ! Finite difference for each column
end do
end function
function element(i,n) result(e)
! Kronecker delta vector. All zeros, except the i-th value.
integer, intent(in) :: i, n
real(wp) :: e(n)
e(:) = 0d0
e(i) = 1d0
end function
end program Console1
I will answer about evaluation in different points. This is quite simple. You just need an array of points, and if the points are in some regular grid, you may not even need that.
You may have an array of xs and array of ys or you can have an array of derived datatype with x and y components.
For the former:
real, allocatable :: x(:), y(:)
x = [... !probably read from some data file
y = [...
do i = 1, size(x)
J(i) = Jacobian(f, x(i), y(i))
end do
If you want to have many functions at the same time, the problem is always in representing functions. Even if you have an array of function pointers, you need to code them manually. A different approach is to have a full algebra module, where you enter some string representing a function and you can evaluate such function and even compute derivatives symbolically. That requires a parser, an evaluator, it is a large task. There are libraries for this. Evaluation of such a derivative will be slow unless further optimizing steps (compiling to machine code) are undertaken.
Numerical evaluation of the derivative is certainly possible. It will slow the convergence somewhat, depending on the order of the approximation of the derivative. You do a difference of two points for the numerical derivative. You can make an interpolating polynomial from values in multiple points to get a higher-order approximation (finite difference approximations), but that costs machine cycles.
Normally we can use auto difference tools as #John Alexiou mentioned. However in practise I prefer using MATLAB to analytically solve out the Jacobian and then use its build-in function fortran() to convert the result to a f90 file. Take your function as an example. Just type these into MATLAB
syms x y
Fval=sym(zeros(2,1));
Fval(1)=log(abs(x-y^2)) - sin(x*y) - sin(pi);
Fval(2)=exp(x*y) + cos(x-y) - 2;
X=[x;y];
Fjac=jacobian(Fval,X);
fortran(Fjac)
which will yield
Fjac(1,1) = -y*cos(x*y)-((-(x-y**2)/abs(-x+y**2)))/abs(-x+y**2)
Fjac(1,2) = -x*cos(x*y)+(y*((-(x-y**2)/abs(-x+y**2)))*2.0D0)/abs(-
&x+y**2)
Fjac(2,1) = -sin(x-y)+y*exp(x*y)
Fjac(2,2) = sin(x-y)+x*exp(x*y)
to you. You just get an analytical Jacobian fortran function.
Meanwhile, it is impossible to solve the inverse of a mxn matrix because of rank mismatching. You should simplify the system of equations to get a nxn Jacobin.
Additionally, when we use Newton-Raphson's method we do not solve the inverse of the Jacobin which is time-consuming and inaccurate for a large system. An easy way is to use dgesv in LAPACK for dense Jacobin. As we only need to solve the vector x from system of linear equations
Jx=-F
dgesv use LU decomposition and Gaussian elimination to solve above system of equations which is extremely faster than solving inverse matrix.
If the system of equations is large, you can use UMFPACK and its fortran interface module mUMFPACK to solve the system of equations in which J is a sparse matrix. Or use subroutine ILUD and LUSOL in a wide-spread sparse matrix library SPARSEKIT2.
In addition to these, there are tons of other methods which try to solve the Jx=-F faster and more accurate such as Generalized Minimal Residual (GMRES) and Stabilized Bi-Conjugate Gradient (BICGSTAB) which is a strand of literature.

How to calculate several different random numbers from normal dist? using fortran

I have to find 'n' random numbers from a normal distribution given the mean and standard deviation. I have figure out how to get a random number, but when try to loop it to get several different random numbers, it gives me the same number x amount of times?
program prac10
implicit none
real :: mu, sigma
integer :: i
!print*, "mean and stdev?"
!read*, mu, sigma
mu=1.1
sigma=0.1
do i=1, 2 # here is the part I think I am stuck on??
call normal(mu,sigma)
enddo
end program
subroutine normal(mu,sigma)
implicit none
integer :: i
integer :: n
real :: u, v, z, randnum
real :: mu, sigma
real :: pi=3.141593
call RANDOM_SEED()
call RANDOM_NUMBER(u)
call RANDOM_NUMBER(v)
z=sqrt(-2*log(u))*cos(2*pi*v)
randnum=mu+sigma*z
print*, randnum
end subroutine
particularly the part where I should be looping/repeating. I used from 1:2, replacing n with 2 for now so that I wouldn't have to input it every time I try to run it
The most important fact is that you must not call RANOM_SEED repeatedly. It is supposed to be called only once.
It is also good to modify the subroutine to generate the number and pass it further. Notice also the change of formatting to make it more readable and the change in the value of pi.
program prac10
implicit none
real :: mu, sigma, x
integer :: i
call RANDOM_SEED()
mu = 1.1
sigma = 0.1
do i = 1, 2
call normal(x,mu,sigma)
print *, x
end do
end program
subroutine normal(randnum,mu,sigma)
implicit none
integer :: i
integer :: n
real :: u, v, z, randnum
real :: mu, sigma
real :: pi = acos(-1.0)
call RANDOM_NUMBER(u)
call RANDOM_NUMBER(v)
z = sqrt(-2*log(u)) * cos(2*pi*v)
randnum = mu + sigma*z
end subroutine

Using Minpack to solve S-curve

I'd like to use Minpack (fortran) to estimate the D parameter in the following generalized form of the S-curve: y = (A - D) / (1 + (x**B/C)) + D
The idea is that in this application, the user provides A [which is always 0 to force passage through (0,0)], B, and C, and from there Minpack will find a value for D that forces passage through (1,y), where y is also supplied by the user but must be <= 1. I was able to accomplish this task with the code below, however, minpack is claiming it hasn't converged when in fact it appears that it has. For example, when running this code and entering the values 1 (at the first prompt) and 0 4 0.1 (at the second prompting), minpack returns info = 2, which according to the comments in lmdif means:
relative error between two consecutive iterates is at most xtol.
I'm tempted to comment out line 63, but am worried that's playing with fire...are there any seasoned minpack users out there who could comment on this? Line 63 is the one that reads:
if (info /= 1) stop "failed to converge"
Am I mis-using Minpack even though it appears to converge (based on my verifying the value in pars)?
module types
implicit none
private
public dp
integer, parameter :: dp=kind(0d0)
end module
module f_vals
DOUBLE PRECISION, SAVE, DIMENSION(:), POINTER:: fixed_vals
end module
module find_fit_module
! This module contains a general function find_fit() for a nonlinear least
! squares fitting. The function can fit any nonlinear expression to any data.
use minpack, only: lmdif1
use types, only: dp
implicit none
private
public find_fit
contains
subroutine find_fit(data_x, data_y, expr, pars)
! Fits the (data_x, data_y) arrays with the function expr(x, pars).
! The user can provide any nonlinear function 'expr' depending on any number of
! parameters 'pars' and it must return the evaluated expression on the
! array 'x'. The arrays 'data_x' and 'data_y' must have the same
! length.
real(dp), intent(in) :: data_x(:), data_y(:)
interface
function expr(x, pars) result(y)
use types, only: dp
implicit none
real(dp), intent(in) :: x(:), pars(:)
real(dp) :: y(size(x))
end function
end interface
real(dp), intent(inout) :: pars(:)
real(dp) :: tol, fvec(size(data_x))
integer :: iwa(size(pars)), info, m, n
real(dp), allocatable :: wa(:)
tol = sqrt(epsilon(1._dp))
!tol = 0.001
m = size(fvec)
n = size(pars)
allocate(wa(m*n + 5*n + m))
call lmdif1(fcn, m, n, pars, fvec, tol, info, iwa, wa, size(wa))
open(222, FILE='D_Value.txt')
write(222,4) pars(1)
4 format(E20.12)
close(222)
if (info /= 1) stop "failed to converge"
contains
subroutine fcn(m, n, x, fvec, iflag)
integer, intent(in) :: m, n, iflag
real(dp), intent(in) :: x(n)
real(dp), intent(out) :: fvec(m)
! Suppress compiler warning:
fvec(1) = iflag
fvec = data_y - expr(data_x, x)
end subroutine
end subroutine
end module
program snwdeplcrv
! Find a nonlinear fit of the form y = (A - D) / (1 + (x**B/C)) + D.
use find_fit_module, only: find_fit
use types, only: dp
use f_vals
implicit none
real(dp) :: pars(1), y_int_at_1
real(dp) :: y(1) = 1.0 ! Initialization of value to be reset by user (y: value of S-curve # x=1)
real(dp) :: A, B, C
integer :: i
allocate(fixed_vals(3)) ! A, B, C parameters
pars = [1._dp] ! D parameter in S-curve function
! Read PEST-specified parameters
write(*,*) ' Enter value that S-curve should equal when SWE=1 (must be <= 1)'
read(*,*) y_int_at_1
if(y_int_at_1 > 1.0) y_int_at_1 = 1
y = y_int_at_1
! Read PEST-specified parameters
write(*,*) ' Enter S-curve parameters: A, B, & C. D parameter to be estimated '
read(*,*) A, B, C
fixed_vals(1) = A
fixed_vals(2) = B
fixed_vals(3) = C
call find_fit([(real(i, dp), i=1,size(y))], y, expression, pars)
print *, pars
contains
function expression(x, pars) result(y)
use f_vals
real(dp), intent(in) :: x(:), pars(:)
real(dp) :: y(size(x))
real(dp) :: A, B, C, D
A = fixed_vals(1)
B = fixed_vals(2)
C = fixed_vals(3)
D = pars(1)
y = (A - D) / (1 + (x**B / C)) + D
end function
end program

Using MKL to solve a non-linear system of equations with an objective function stored in another module

I'm trying to use the MKL trust region algorithm to solve a nonlinear system of equations in a Fortran program. I started from the example provided online (ex_nlsqp_f90_x.f90 https://software.intel.com/en-us/node/501498) and everything works correctly. Now, because I have to use this in a much bigger program, I need the user defined objective function to be loaded from a separate module. Hence, I split the example into 2 separate files, but I'm not able to make it compile correctly.
So here is the code for module which contains user defined data structure and the objective function
module modFun
implicit none
private
public my_data, extended_powell
type :: my_data
integer a
integer sum
end type my_data
contains
subroutine extended_powell (m, n, x, f, user_data)
implicit none
integer, intent(in) :: m, n
real*8 , intent(in) :: x(n)
real*8, intent(out) :: f(m)
type(my_data) :: user_data
integer i
user_data%sum = user_data%sum + user_data%a
do i = 1, n/4
f(4*(i-1)+1) = x(4*(i-1)+1) + 10.0 * x(4*(i-1)+2)
f(4*(i-1)+2) = 2.2360679774998 * (x(4*(i-1)+3) - x(4*(i-1)+4))
f(4*(i-1)+3) = ( x(4*(i-1)+2) - 2.0 * x(4*(i-1)+3) )**2
f(4*(i-1)+4) = 3.1622776601684 * (x(4*(i-1)+1) - x(4*(i-1)+4))**2
end do
end subroutine extended_powell
end module modFun
and here the portion of the main program calling it
include 'mkl_rci.f90'
program EXAMPLE_EX_NLSQP_F90_X
use MKL_RCI
use MKL_RCI_type
use modFun
! user's objective function
! n - number of function variables
! m - dimension of function value
integer n, m
parameter (n = 4)
parameter (m = 4)
! precisions for stop-criteria (see manual for more details)
real*8 eps(6)
real*8 x(n)
real*8 fjac(m*n)
! number of iterations
integer fun
! Additional users data
type(my_data) :: m_data
m_data%a = 1
m_data%sum = 0
rs = 0.0
fun = djacobix(extended_powell,n,m,fjac,x,eps(1),%val(loc(m_data)))
end program EXAMPLE_EX_NLSQP_F90_X
Also djacobix code
INTERFACE
INTEGER FUNCTION DJACOBIX(fcn, n, m, fjac, x, eps, user_data)
USE, INTRINSIC :: ISO_C_BINDING
INTEGER, INTENT(IN) :: n
INTEGER, INTENT(IN) :: m
DOUBLE PRECISION, INTENT(IN) :: eps
DOUBLE PRECISION, INTENT(IN), DIMENSION(*) :: x
DOUBLE PRECISION, INTENT(OUT), DIMENSION(m, *) :: fjac
INTEGER(C_INTPTR_T) :: user_data
INTERFACE
SUBROUTINE fcn(m, n, x, f, user_data)
USE, INTRINSIC :: ISO_C_BINDING
INTEGER, INTENT(IN) :: n
INTEGER, INTENT(IN) :: m
DOUBLE PRECISION, INTENT(IN), DIMENSION(*) :: x
DOUBLE PRECISION, INTENT(OUT), DIMENSION(*) :: f
INTEGER(C_INTPTR_T), INTENT(IN) :: user_data
END SUBROUTINE
END INTERFACE
END FUNCTION
END INTERFACE
When i compile the following errors are generated:
mpiifort -g -t -mkl -I/apps/rhel6/intel/composer_xe_2015.3.187/mkl/include/intel64/lp64 -c modFun.f90
mpiifort -g -t -mkl -I/apps/rhel6/intel/composer_xe_2015.3.187/mkl/include/intel64/lp64 -c main.f90
main.f90(30): error #7065: The characteristics of dummy argument 5 of the associated actual procedure differ from the characteristics of dummy argument 5 of the dummy procedure. [EXTENDED_POWELL]
fun = djacobix(extended_powell,n,m,fjac,x,eps(1),%val(loc(m_data)))
-------------------^
I have the feeling I have to create an interface to override the check on the m_data, but I can't figure out where and how. Can anyone help me with this problem providing a working example?
I guess the reason is that the function djacobix passes the pointer instead of the true value of variable user_data.
You can check the manual at https://software.intel.com/content/www/us/en/develop/documentation/onemkl-developer-reference-c/top/nonlinear-optimization-problem-solvers/jacobian-matrix-calculation-routines/jacobix.html where a sentence shows that "You need to declare fcn as extern in the calling program."

Only one error left in my code,

PROGRAM MPI
IMPLICIT NONE
INTEGER, PARAMETER :: nn=100
DOUBLE PRECISION h, L
DOUBLE PRECISION, DIMENSION (2*nn) :: y, ynew
DOUBLE PRECISION, DIMENSION (nn) :: qnew,vnew
DOUBLE PRECISION, DIMENSION (2*nn) :: k1,k2,k3,k4
INTEGER j, k
INTEGER i
INTEGER n
n=100 !particles
L=2.0d0
h=1.0d0/n
y(1)=1.0d0
DO k=1,2*n ! time loop
CALL RHS(y,k1)
CALL RHS(y+(h/2.0d0)*k1,k2)
CALL RHS(y+(h/2.0d0)*k2,k3)
CALL RHS(y+h*k3,k4)
ynew(1:2*n)=y(1:2*n) + (k1 + 2.0d0*(k2 + k3) + k4)*h/6.0d0
END DO
qnew(1:n)=ynew(1:n)
vnew(1:n)=ynew(n+1:2*n)
DO i=1,n
IF (qnew(i).GT. L) THEN
qnew(i) = qnew(i) - L
ENDIF
END DO
write(*,*) 'qnew=', qnew(1:n)
write(*,*) 'vnew=', vnew(1:n)
END PROGRAM MPI
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
! Right hand side of the ODE
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
SUBROUTINE RHS(y,z)
IMPLICIT NONE
INTEGER, PARAMETER :: nn=100
DOUBLE PRECISION, DIMENSION (2*nn) :: y
DOUBLE PRECISION, DIMENSION (2*nn) :: z
DOUBLE PRECISION, DIMENSION (nn) :: F
DOUBLE PRECISION, DIMENSION (nn) :: g
INTEGER n
INTEGER m
n=100
m=1/n
z(1:n)=y(n+1:2*n)
CAll FORCE(g,F)
z(n+1:2*n)=F(1:n)/m
RETURN
END
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
! Force acting on each particle
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
SUBROUTINE FORCE(g,F)
IMPLICIT NONE
INTEGER, PARAMETER :: nn=100
DOUBLE PRECISION, DIMENSION (nn) :: F
DOUBLE PRECISION, DIMENSION (nn) :: q
DOUBLE PRECISION, DIMENSION (nn) :: g
DOUBLE PRECISION u
INTEGER j, e
INTEGER n
n=100
e=1/n
DO j=2,n+1
CALL deriv((abs(q(j)-q(j-1)))/e,u)
g(j-1)=((y(j)-y(j-1))/(abs(y(j)-y(j-1))))*u
CALL deriv((abs(q(j)-q(j+1)))/e,u)
g(j+1)=((y(j)-y(j+1))/(abs(y(j)-y(j+1))))*u
F(j)=g(j-1)+g(j+1)
END DO
RETURN
END
SUBROUTINE deriv(c,u,n)
IMPLICIT NONE
INTEGER, INTENT(in) :: n
DOUBLE PRECISION, DIMENSION(n), INTENT(IN) :: c
DOUBLE PRECISION, DIMENSION(n), INTENT(OUT) :: u
INTEGER, PARAMETER :: p=2
INTEGER, PARAMETER :: cr=100
INTEGER :: i
DOUBLE PRECISION L
L=2.0d0
DO i= 1,n
IF (c(i) .LE. L) THEN
u(i)=cr*(L*(c(i)**(-p))-L**(1-p))
ELSE IF (c(i) .GT. L) THEN
u(i)=0
END IF
END DO
RETURN
END SUBROUTINE deriv
I am only getting one same error on line 85 and 87. It says:
y has no implicit type at y(j-1) ans at y(j+1).
There's a lot wrong here. We can point out some of the things, but you're going to have to sit down with a book and learn about programming, starting with smaller programs and getting them right, then building up.
Let's look at the last routine in the code you posted above. I've changed the syntax of some of the variable declarations just to make it shorter so more fits on screen at once.
SUBROUTINE deriv(c,u)
IMPLICIT NONE
DOUBLE PRECISION :: deriv, c, u
INTEGER :: p, x, cr, n
L=2.0d0
cr=100
p=2
n=100
DO i= 1,n
IF (c(i).LE. L) THEN
u(c)=cr*(L*c^(-p)-L^(1-p))
ELSE IF (c(i) .GT. L) THEN
u(c)=0
END IF
RETURN
END
So you've made deriv a double precision variable, but it's also the name of the subroutine. That's an error; maybe you meant to make this a function which returns a double precision value; then you're almost there, you'd need to change the procedure header to FUNCTION DERIV(c,u) -- but you never set deriv anywhere. So likely that should just be left out. So let's just get rid of that DOUBLE PRECISION deriv declaration. Also, L, which is used, is never declared, and x, which isn't, is declared.
Then you pass in to this subroutine two variables, c and u, which you define to be double precision. So far so good, but then you start indexing them: eg, c(i). So they should be arrays of double precisions, not just scalars. Looking at the do loop, I'm guessing they should both be of size n -- which should be passed in, presumably? . Also, the do loop is never terminated; there should be an end do after the end if.
Further, the ^ operator you're using I'm assuming you're using for exponentiation -- but in Fortran, that's **, not ^. And that c^(-p) should (I'm guessing here) be c(i)**(-p)?
Finally, you're setting u(c) -- but that's not very sensible, as c is an array of double precision numbers. Even u(c(i)) wouldn't make sense -- you can't index an array with a double precision number. Presumably, and I'm just guessing here, you mean the value of u corresponding to the just-calculated value of c -- eg, u(i), not u(c)?
So given the above, we'd expect the deriv subroutine to look like this:
SUBROUTINE deriv(c,u,n)
IMPLICIT NONE
INTEGER, INTENT(in) :: n
DOUBLE PRECISION, DIMENSION(n), intent(IN) :: c
DOUBLE PRECISION, DIMENSION(n), intent(OUT) :: u
INTEGER, PARAMETER :: p=2, cr=100
DOUBLE PRECISION, PARAMETER :: L=2.0
INTEGER :: i
DO i= 1,n
IF (c(i) .LE. L) THEN
u(i)=cr*(L*c(i)**(-p)-L**(1-p))
ELSE IF (c(i) .GT. L) THEN
u(i)=0
END IF
END DO
RETURN
END SUBROUTINE deriv
Note that in modern fortran, the do loop can be replaced with a where statement, and also you don't need to explicitly pass in the size; so then you could get away with the clearer and shorter:
SUBROUTINE DERIV(c,u)
IMPLICIT NONE
DOUBLE PRECISION, DIMENSION(:), intent(IN) :: c
DOUBLE PRECISION, DIMENSION(size(c,1)), intent(OUT) :: u
INTEGER, PARAMETER :: p=2, cr=100
DOUBLE PRECISION, PARAMETER :: L=2.0
WHERE (c <= L)
u=cr*(L*c**(-p)-L**(1-p))
ELSEWHERE
u=0
ENDWHERE
RETURN
END SUBROUTINE DERIV
But notice that I've already had to guess three times what you meant in this section of code, and this is only about 1/4th of the total of the code. Having us try to divine your intention in the whole thing and rewrite accordingly probably isn't the best use of anyone's time; why don't you proceed from here working on one particular thing and ask another question if you have a specific problem.