How to Modify a Finite Difference Method Code for a Non-Zero Boundary Condition?

  • Thread starter fahad20
  • Start date
In summary, the code provided is a Fortran program for solving a given ordinary differential equation problem using the finite difference method. The program prompts the user for input values such as the number of stations, step size, and boundary conditions. It then calculates the elements of the coefficient matrix [C] and the vector {R} using the functions CIJ and RI. The program then solves the resulting matrix equation using Gaussian elimination and outputs the solution in the form of the vector {Y}. The code provided also includes a subroutine for solving the matrix equation using Gaussian elimination. To solve a different problem with different boundary conditions, the code will need to be modified accordingly.
  • #1
fahad20
8
0
hello...i m trying to solve this ordinary differential equation problem..but it is giving errors..pleasezz help

problem: d^2 T/Dr^2 + (1/r) DT/Dr=1+ (r/R)^2

T'(0)=0 and T(R)=0


solve this problem for T(r),where R=1.0

SOLUTION I HAVE CUM UP WITH:

put R=1 in this equation...

now let v=T'
T''=V'

so

DT/Dr=v

Dv/dr=1+(T**2/V**2)-(V**2/T)

V(0)=0 and T(R)=0

and now here's the code...



dimension p(100,4),r(100),xn(100),xt(100)

write(*,10)

10 format(1x,'input number of equation:')

read(*,*)N
write(*,12)

12 format(1x,'enter the initial value of the independent variable:')

read(*,*)T

write(*,14)

14 format(1x,'enter the increment of the independent variable:')

read(*,*)dt

write(*,16)

16 format(1x,'enter the ending value of the independent variable:')

read(*,*)tend

write(*,18)

18 format(1x,'enter the initial value of the dependent variables:')

do 20 i=1,n

20 read(*,*)r(i)

write(*,22)

22 format(9x,'t',5x,'r(1) through r(n)')

write(*,24) T,(r(i),i=1,n)

24 format(f10.3,5e14.5)

26 call rkn(x,n,dt,t,p,xt,xn)

t=t+dt

do 30 i=1,n

30 r(i)=xn(i)

write(*,24)t,(r(i),i=1,n)

if (t.ge.tend) goto 50

goto 26

50 write(*,52)

52 format(1x,'r-k parameters are:')

do 54 i=1,n

54 write(*,56) (p(i,j),j=1,4)

56 format(4e15.5)

end

function f(r,t,i,n)
dimension r(n)

goto (10,20),i

10 f=r(1)

return

20 f=1+(r(2)**2/r(1)**2)-(r(1)**2/r(2))


return

end


subroutine rkn(xin,n,dt,t,p,xt,xout)

! solving n first-order ode d(xin)/dt=f(xin,t)

!knowing xin at t,this sub.finds xout at t+dt

!p are the runge-kutta 4th-order parameters
dimension p(n,4),xin(n),xout(n),xt(n)

do 5 i=1,n

5 p(i,1)=dt*f(xin,t,i,n)

do 15 j=2,3

do 10 i=1,n

10 xt(i)=xin(i)+p(i,j-1)/2.

do 15 i=1,n

15 p(i,j)=dt*f(xt,t+dt/2.,i,n)

do 20 i=1,n

20 xt(i)=xin(i)+p(i,3)

do 25 i=1,n

p(i,4)=dt*f(xt,t+dt,i,n)


25 xout(i)=xin(i)+(p(i,1)+2.*(p(i,2)+p(i,3))+p(i,4))/6.

return

end




but it is giving eror..please correct this..thank u
 
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  • #2


You made several errors, I have bolded them for your below.

fahad20 said:
hello...i m[/b trying to solve this ordinary differential equation problem..but it is giving errors..pleasezz help

problem: d^2 T/Dr^2 + (1/r) DT/Dr=1+ (r/R)^2

T'(0)=0 and T(R)=0solve this problem for T(r),where R=1.0

SOLUTION I HAVE CUM UP WITH:

put R=1 in this equation...

now let v=T'
T''=V'

so

DT/Dr=v

Dv/dr=1+(T**2/V**2)-(V**2/T)

V(0)=0 and T(R)=0

and now heres the code...
...but it is giving eror..please correct this..thank u


I didn't bother checking sentence structure or punctuation. If you'd like, I can correct those for you as well.
 
  • #3


fahad20 said:
but it is giving errors..

What kind of errors?

If the program doesn't even compile, what error messages does the compiler give you?

If the program compiles but doesn't run correctly, show us

(a) some sample input

(b) the output that the program should produce

(c) the output that the program actually produces
 
  • #4


fahad20 said:

Homework Statement



solve for T(r)

Homework Equations





The Attempt at a Solution



hello...i m trying to solve this ordinary differential equation problem..but it is giving errors..pleasezz help

problem: d^2 T/Dr^2 + (1/r) DT/Dr=1+ (r/R)^2

T'(0)=0 and T(R)=0


solve this problem for T(r),where R=1.0

SOLUTION I HAVE CUM UP WITH:

put R=1 in this equation...

now let v=T'
T''=V'

so

DT/Dr=v

Dv/dr=1+(T**2/V**2)-(V**2/T)

V(0)=0 and T(R)=0

and now here's the code...



dimension p(100,4),r(100),xn(100),xt(100)

write(*,10)

10 format(1x,'input number of equation:')

read(*,*)N
write(*,12)

12 format(1x,'enter the initial value of the independent variable:')

read(*,*)T

write(*,14)

14 format(1x,'enter the increment of the independent variable:')

read(*,*)dt

write(*,16)

16 format(1x,'enter the ending value of the independent variable:')

read(*,*)tend

write(*,18)

18 format(1x,'enter the initial value of the dependent variables:')

do 20 i=1,n

20 read(*,*)r(i)

write(*,22)

22 format(9x,'t',5x,'r(1) through r(n)')

write(*,24) T,(r(i),i=1,n)

24 format(f10.3,5e14.5)

26 call rkn(x,n,dt,t,p,xt,xn)

t=t+dt

do 30 i=1,n

30 r(i)=xn(i)

write(*,24)t,(r(i),i=1,n)

if (t.ge.tend) goto 50

goto 26

50 write(*,52)

52 format(1x,'r-k parameters are:')

do 54 i=1,n

54 write(*,56) (p(i,j),j=1,4)

56 format(4e15.5)

end

function f(r,t,i,n)
dimension r(n)

goto (10,20),i

10 f=r(1)

return

20 f=1+(r(2)**2/r(1)**2)-(r(1)**2/r(2))


return

end


subroutine rkn(xin,n,dt,t,p,xt,xout)

! solving n first-order ode d(xin)/dt=f(xin,t)

!knowing xin at t,this sub.finds xout at t+dt

!p are the runge-kutta 4th-order parameters
dimension p(n,4),xin(n),xout(n),xt(n)

do 5 i=1,n

5 p(i,1)=dt*f(xin,t,i,n)

do 15 j=2,3

do 10 i=1,n

10 xt(i)=xin(i)+p(i,j-1)/2.

do 15 i=1,n

15 p(i,j)=dt*f(xt,t+dt/2.,i,n)

do 20 i=1,n

20 xt(i)=xin(i)+p(i,3)

do 25 i=1,n

p(i,4)=dt*f(xt,t+dt,i,n)


25 xout(i)=xin(i)+(p(i,1)+2.*(p(i,2)+p(i,3))+p(i,4))/6.

return

end




but it is giving eror..please correct this..thank u

You have several do loops that don't have continue statements. The loops should look something like this:
Code: 
do 10 i = 1, n
     sum = sum + i
     write(*,*) 'i =', i
     write(*,*) 'sum =', sum
10  continue

Your code is very hard to read with statement labels on input and output statements that you probably don't need, and with gotos here and there.
 
  • #5


i have this code for a 2nd ODE..but with boundary conditions.

this code is for ..

d^2 Z/dR^2 + (1/Z) dZ/dR=-p/T

Z(Ri)=0 and Z(Ro)=0

this is a finite difference method code in fortran..

please c...





!Fortran Version

! Program OdeBcpFD – Ordinary differential equation, boundry-value Problem

! Solved by finite-difference method . UP to 99 stations.

DIMENSION X(99),C(99,99),R(99)

WRITE (*,2)

2 FORMAT(1X,' Program OdeBvpFD finite difference solution of O.D.E. Boundary-Value Problem.'/)

Write (*,4)

4 Format(1X,'have you Edited The functions CIJ And RI for generating the elements of [C] and {R} in [C] {Y}={R}?')

READ (*,*) I1

Write (*,6)

6 format(1x,'enter the number of (in-between,excluding boundaries)stations and stepsize:')


read(*,*)n,dx

write(*,8)
8 Format (1X,'Enter the First (left Boundary) X Value :' )

READ (*,*) X1

! Calculate [C] and {R}

Do 10 I=1,N

X (I) =X1+I*DX

R (I) =RI(x(i),dx)

DO 10 J=1,N

10 c(i,j)=cij(x(i),i,j,dx)

CALL GAUSS (C,N,99,R)




! {Y} is in {R}.

WRITE (*,12) (R(k),K=1,N)

12 Format(1x,'The Solution is : '//5(5E16.5/))

STOP

END

Function RI (X,DX)

RI=-5*DX **2 /100.

Return

End

FUNCTION CIJ (X,I,J,DX)

CIJ=100.

IF (I.EQ.j) CIJ=-2.

IF (I.EQ.(j-1)) CIJ=1.-DX/2./X

IF (I.EQ.(J+1) ) CIJ = 1.+DX/2./X

RETURN

END

SUBROUTINE GAUSS ( C , N, M, R)



!

!SOLVES MATRIX EQUATION C(N,N) *X(N)=R(N) BY GAUSSIAN ELIMINATION.

!X and V share same storage. C is dimensioned (M,M) in the calling program.

!

DIMENSION C(M,M),r(N)

!

N1=N-1

DO 25 K=1,N1

KP1=K+1

!

!NORMALIZATION

!

DO 10 J=KP1,N

10 C(K,J)=C(K,J)/C(K,K)

r(K)=r(K)/C(K,K)

!

!ELIMINATION

!

DO 25 I=KP1,N

DO 20 J=KP1,N

20 C(I,J)=C(I,J)-C(I,K)*C(K,J)

25 r(I)=r(I)-C(I,K)*r(K)

!

!BACKWARD SUBSTITUTION

!

r(N)=r(N)/C(N,N)

DO 35 I=1 ,N1

IR=N-I

IR1=IR+1

DO 35 J=IR1,N

35 r(IR)=r(IR)-C(IR,J)*r(J)

RETURN

END



now i have this problem...

d^2 T/dr^2 + (1/r) dT/dr=0

and B.C's are T(1)100 and T(2)=0

how will this code change..especially ..how to cater for these two B.Cs with r non-zero...

thank u ..
 
  • #7


Is this a programming course? Are you required to use that program?

If, after taking v= T', you multiply the original equation, [itex]v'+ v/r= 1+ r^2[/itex] by r, you get [itex]rv'+ v= (rv)'= r+ r^3[/itex] and can solve for v just by integrating both sides:
[itex]rv= (1/2)r^2+ (1/4)r^4+ C[/itex] so that [itex]v= T'= (1/2)r+ (1/4)r^3+ C/r[/itex] and T can be found by integrating again.
 
  • #8


please tryy to understnd...

the code i have is for ..

d^2 Z/dR^2 + (1/Z) dZ/dR=-p/T

Z(Ri)=0 and Z(Ro)=0

(both B.C's are ZERO)

this is a finite difference method code in fortran..


now i have this problem...

d^2 T/dr^2 + (1/r) dT/dr=0

in which the B.C's are not zero... ----> T(1)100 and T(2)=0

i.e ar r=1,T=100 and at r=2,T=0

how will this code change for above problem..
 

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