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bvode
boundary value problems for ODE using collocation method
bvodeS
Simplified call to bvode
Calling Sequence
zu=bvode(xpoints,N,m,x_low,x_up,zeta,ipar,ltol,tol,fixpnt,fsub,dfsub,gsub,dgsub,guess)
zu=bvodeS(xpoints,m,N,x_low,x_up,fsub,gsub,zeta, <optional_args>)
Arguments
- zu
a column vector of size
M
. The solution of the ode evaluated on the mesh given by points. It containsz(u(x))
for each requested points.- xpoints
an array which gives the points for which we want to observe the solution.
- N
a scalar with integer value, number of differential equations (N <= 20).
- m
a vector of size
N
with integer elements. It is the vector of order of each differential equation:m(i)
gives the order of thei
-th differential equation. In the following,M
will represent the sum of the elements ofm
.- x_low
a scalar: left end of interval
- x_up
a scalar: right end of interval
- zeta
a vector of size
M
,zeta(j)
givesj
-th side condition point (boundary point). One must havex_low<=zeta(j) <=zeta(j+1)<=x_up
All side condition points must be mesh points in all meshes used, see description of
ipar(11)
andfixpnt
below.- ipar
an array with 11 integer elements:
[nonlin, collpnt, subint, ntol, ndimf, ndimi, iprint, iread, iguess, rstart,nfxpnt]
- nonlin: ipar(1)
0 if the problem is linear, 1 if the problem is nonlinear
- collpnt: ipar(2)
Gives the number of collocation points per subinterval where
max(m(j)) <= collpnt <= 7
.If
ipar(2)=0
thencollpnt
is set tomax( max(m(j))+1, 5-max(m(j)) )
.- subint: ipar(3)
Gives the number of subintervals in the initial mesh. If
ipar(3) = 0
thenbvode
arbitrarily setssubint = 5
.- ntol: ipar(4)
Gives the number of solution and derivative tolerances. We require
0 < ntol <= M
.ipar(4)
must be set to the dimension of thetol
argument or to0
. In the latter case the actual value will automatically be set tosize(tol,'*')
.- ndimf: ipar(5)
Gives the dimension of
fspace
(a real work array). Its value provides a constraint onnmax
the maximum number of subintervals.The
ipar(5)
value must respect the constraintipar(5)>=nmax*nsizef
wherensizef=4 + 3*M + (5+collpnt*N)*(collpnt*N+M) + (2*M-nrec)*2*M
(nrec
is the number of right end boundary conditions).- ndimi: ipar(6)
Gives the dimension of
ispace
(an integer work array). Its value provides a constraint onnmax
, the maximum number of subintervals.The
ipar(6)
value must respect the constraintipar(6)>=nmax*nsizei
wherensizei= 3 + collpnt*N + M
.- iprint: ipar(7)
output control, may take the following values:
- -1
for full diagnostic printout
- 0
for selected printout
- 1
for no printout
- iread: ipar(8)
- = 0
causes bvode to generate a uniform initial mesh.
- = xx
Other values are not implemented yet in Scilab
- = 1
if the initial mesh is provided by the user. it is defined in
fspace
as follows: the mesh will occupyfspace(1), ..., fspace(n+1)
. The user needs to supply only the interior mesh pointsfspace(j) = x(j), j = 2, ..., n.
- = 2
if the initial mesh is supplied by the user as with
ipar(8)=1
, and in addition no adaptive mesh selection is to be done.
- iguess: ipar(9)
- = 0
if no initial guess for the solution is provided.
- = 1
if an initial guess is provided by the user through the argument
guess
.- = 2
if an initial mesh and approximate solution coefficients are provided by the user in
fspace
(the former and new mesh are the same).- = 3
if a former mesh and approximate solution coefficients are provided by the user in
fspace
, and the new mesh is to be taken twice as coarse; i.e.,every second point from the former mesh.- = 4
if in addition to a former initial mesh and approximate solution coefficients, a new mesh is provided in
fspace
as well (see description of output for further details on iguess = 2, 3 and 4).
- ireg: ipar(10)
- = 0
if the problem is regular
- = 1
if the first relaxation factor is equal to
ireg
, and the nonlinear iteration does not rely on past convergence (use for an extra-sensitive nonlinear problem only)- = 2
if we are to return immediately upon (a) two successive nonconvergences, or (b) after obtaining an error estimate for the first time.
- nfxpnt: ipar(11)
Gives the number of fixed points in the mesh other than
x_low
andx_up
(the dimension offixpnt
).ipar(11)
must be set to the dimension of thefixpnt
argument or to0
. In the latter case the actual value will automatically be set tosize(fixpnt,'*')
.
- ltol
an array of dimension
ntol=ipar(4)
.ltol(j) = l
specifies that thej
-th tolerance in thetol
array controls the error in thel
-th component of . It is also required that:1 <= ltol(1) < ltol(2) < ... < ltol(ntol) <= M
- tol
an array of dimension
ntol=ipar(4)
.tol(j)
is the error tolerance on theltol(j)
-th component of . Thus, the code attempts to satisfy on each subintervalwhere is the approximate solution vector and is the exact solution (unknown).
- fixpnt
an array of dimension
nfxpnt=ipar(11)
. It contains the points, other thanx_low
andx_up
, which are to be included in every mesh. The code requires that all side condition points other thanx_low
andx_up
(see description ofzeta
) be included as fixed points infixpnt
.- fsub
an external used to evaluate the column vector
f=
for anyx
such asx_low
<=x
<=x_up
and for anyz=z(u(x))
(see description below).The external must have the headings:
In Fortran the calling sequence must be:
subroutine fsub(x,zu,f) double precision zu(*), f(*),x
In C the function prototype must be:
And in Scilab:
function f=fsub(x, zu, parameters)
- dfsub
an external used to evaluate the Jacobian of
f(x,z(u))
at a pointx
. Wherez(u(x))
is defined as forfsub
and theN
byM
arraydf
should be filled by the partial derivatives off
:The external must have the headings:
In Fortran the calling sequence must be:
subroutine dfsub(x,zu,df) double precision zu(*), df(*),x
In C the function prototype must be:
And in Scilab:
function df=dfsub(x, zu, parameters)
- gsub
an external used to evaluate given z=
z = zeta(i)
for1<=i<=M.
The external must have the headings:
In Fortran the calling sequence must be:
subroutine gsub(i,zu,g) double precision zu(*), g(*) integer i
In C the function prototype must be:
And in Scilab:
function g=gsub(i, zu, parameters)
Note that in contrast to
f
infsub
, here only one value per call is returned ing
.
- dgsub
an external used to evaluate the
i
-th row of the Jacobian ofg(x,u(x))
. Wherez(u)
is as forfsub
,i
as forgsub
and theM
-vectordg
should be filled with the partial derivatives ofg
, viz, for a particular call one calculatesThe external must have the headings:
In Fortran the calling sequence must be:
subroutine dgsub(i,zu,dg) double precision zu(*), dg(*)
In C the function prototype must be
And in Scilab
function dg=dgsub(i, zu, parameters)
- guess
An external used to evaluate the initial approximation for
z(u(x))
anddmval(u(x))
the vector of themj
-th derivatives ofu(x)
. Note that this subroutine is used only ifipar(9) = 1
, and then allM
components ofzu
andN
components ofdmval
should be computed for anyx
such asx_low
<=x
<=x_up
.The external must have the headings:
In Fortran the calling sequence must be:
subroutine guess(x,zu,dmval) double precision x,z(*), dmval(*)
In C the function prototype must be
And in Scilab
function [dmval, zu]=fsub(x, parameters)
- <optional_args>
It should be either:
any left part of the ordered sequence of values:
guess, dfsub, dgsub, fixpnt, ndimf, ndimi, ltol, tol, ntol,nonlin, collpnt, subint, iprint, ireg, ifail
or any sequence of
arg_name=argvalue
witharg_name
in:guess
,dfsub
,dgsub
,fixpnt
,ndimf
,ndimi
,ltol
,tol
,ntol
,nonlin
,collpnt
,subint
,iprint
,ireg
,ifail
where all these arguments excepted
ifail
are described above.ifail
can be used to display the bvode call corresonding to the selected optional arguments. Ifguess
is giveniguess
is set to 1
Description
These functions solve a multi-point boundary value problem for a mixed order system of ode-s given by
where
The argument zu
used by the external functions
and returned by bvode
is the column vector formed by
the components of z(u(x))
for a given x
.
The method used to approximate the solution u
is collocation at
gaussian points, requiring m(i)-1
continuous derivatives in the i
-th
component, i = 1:N
. here, k
is the number of collocation points (stages)
per subinterval and is chosen such that k .ge. max m(i)
. a
runge-kutta-monomial solution representation is utilized.
Examples
The first two problems below are taken from the paper [1] of the Bibliography.
The problem 1 describes a uniformy loaded beam of variable stifness, simply supported at both end.
It may be defined as follow :
Solve the fourth order differential equation:
Subjected to the boundary conditions:
The exact solution of this problem is known to be:
N=1;// just one differential equation m=4;//a fourth order differential equation M=sum(m); x_low=1;x_up=2; // the x limits zeta=[x_low,x_low,x_up,x_up]; //two constraints (on the value of u and its second derivative) on each bound. //The external functions //These functions are called by the solver with zu=[u(x);u'(x);u''(x);u'''(x)] // - The function which computes the right hand side of the differential equation function f=fsub(x, zu),f=(1-6*x^2*zu(4)-6*x*zu(3))/x^3,endfunction // - The function which computes the derivative of fsub with respect to zu function df=dfsub(x, zu),df=[0,0,-6/x^2,-6/x],endfunction // - The function which computes the ith constraint for a given i function g=gsub(i, zu), select i case 1 then //x=zeta(1)=1 g=zu(1) //u(1)=0 case 2 then //x=zeta(2)=1 g=zu(3) //u''(1)=0 case 3 then //x=zeta(3)=2 g=zu(1) //u(2)=0 case 4 then //x=zeta(4)=2 g=zu(3) //u''(2)=0 end endfunction // - The function which computes the derivative of gsub with respect to z function dg=dgsub(i, z) select i case 1 then //x=zeta(1)=1 dg=[1,0,0,0] case 2 then //x=zeta(2)=1 dg=[0,0,1,0] case 3 then //x=zeta(3)=2 dg=[1,0,0,0] case 4 then //x=zeta(4)=2 dg=[0,0,1,0] end endfunction // - The function which computes the initial guess, unused here function [zu, mpar]=guess(x),zu=0;mpar=0,endfunction //define the function which computes the exact value of u for a given x ( for testing purposes) function zu=trusol(x) zu=0*ones(4,1) zu(1) = 0.25*(10*log(2)-3)*(1-x) + 0.5 *( 1/x + (3+x)*log(x) - x) zu(2) = -0.25*(10*log(2)-3) + 0.5 *(-1/x^2 + (3+x)/x + log(x) - 1) zu(3) = 0.5*( 2/x^3 + 1/x - 3/x^2) zu(4) = 0.5*(-6/x^4 - 1/x/x + 6/x^3) endfunction fixpnt=[ ];//All boundary conditions are located at x_low and x_up // nonlin collpnt n ntol ndimf ndimi iprint iread iguess rstart nfxpnt ipar=[0 0 1 2 2000 200 1 0 0 0 0 ] ltol=[1,3];//set tolerance control on zu(1) and zu(3) tol=[1.e-11,1.e-11];//set tolreance values for these two controls xpoints=x_low:0.01:x_up; zu=bvode(xpoints,N,m,x_low,x_up,zeta,ipar,ltol,tol,fixpnt,... fsub,dfsub,gsub,dgsub,guess) //check the constraints zu([1,3],[1 $]) //should be zero plot(xpoints,zu(1,:)) // the evolution of the solution u zu1=[];for x=xpoints,zu1=[zu1,trusol(x)]; end; norm(zu-zu1)
Same problem using
bvodeS
and an initial guess.The problem 2 describes the small finite deformation of a thin shallow spherical cap of constant thickness subject to a quadratically varying axisymmetric external pressure distribution. Here is the meridian angle change of the deformed shell and is a stress function. For two different solutions may found depending on the starting point
Subject to the boundary conditions
for
x=0
andx=1
N=2;// two differential equations m=[2 2];//each differential equation is of second order M=sum(m); x_low=0;x_up=1; // the x limits zeta=[x_low,x_low, x_up x_up]; //two constraints on each bound. //The external functions //These functions are called by the solver with zu=[u1(x);u1'(x);u2(x);u2'(x)] // - The function which computes the right hand side of the differential equation function f=fsub2(x, zu, eps, dmu, eps4mu, gam, xt), f=[zu(1)/x^2-zu(2)/x+(zu(1)-zu(3)*(1-zu(1)/x)-gam*x*(1-x^2/2))/eps4mu //phi'' zu(3)/x^2-zu(4)/x+zu(1)*(1-zu(1)/(2*x))/dmu];//psi'' endfunction // - The function which computes the derivative of fsub with respect to zu function df=dfsub2(x, zu, eps, dmu, eps4mu, gam, xt), df=[1/x^2+(1+zu(3)/x)/eps4mu, -1/x, -(1-zu(1)/x)/eps4mu, 0 (1-zu(1)/x)/dmu 0 1/x^2 -1/x]; endfunction // - The function which computes the ith constraint for a given i function g=gsub2(i, zu), select i case 1 then //x=zeta(1)=0 g=zu(1) //u(0)=0 case 2 then //x=zeta(2)=0 g=-0.3*zu(3) //x*psi'-0.3*psi+0.7x=0 case 3 then //x=zeta(3)=1 g=zu(1) //u(1)=0 case 4 then //x=zeta(4)=1 g=1*zu(4)-0.3*zu(3)+0.7*1 //x*psi'-0.3*psi+0.7x=0 end endfunction // - The function which computes the derivative of gsub with respect to z function dg=dgsub2(i, z) select i case 1 then //x=zeta(1)=1 dg=[1,0,0,0] case 2 then //x=zeta(2)=1 dg=[0,0,-0.3,0] case 3 then //x=zeta(3)=2 dg=[1,0,0,0] case 4 then //x=zeta(4)=2 dg=[0,0,-0.3,1] end endfunction gam=1.1 eps=1d-3 dmu=eps eps4mu=eps^4/dmu xt=sqrt(2*(gam-1)/gam) fixpnt=[ ];//All boundary conditions are located at x_low and x_up collpnt=4; nsizef=4+3*M+(5+collpnt*N)*(collpnt*N+M)+(2*M-2)*2*M ; nsizei=3 + collpnt*N+M;; nmax=200; // nonlin collpnt n ntol ndimf ndimi iprint iread iguess rstart nfxpnt ipar=[1 k 10 4 nmax*nsizef nmax*nsizei -1 0 0 0 0 ] ltol=1:4;//set tolerance control on zu(1), zu(2), zu(3) and zu(4) tol=[1.e-5,1.e-5,1.e-5,1.e-5];//set tolreance values for these four controls xpoints=x_low:0.01:x_up; zu=bvode(xpoints,N,m,x_low,x_up,zeta,ipar,ltol,tol,fixpnt,... fsub2,dfsub2,gsub2,dgsub2,guess2); scf(1);clf();plot(xpoints,zu([1 3],:)) // the evolution of the solution phi and psi //using an initial guess // - The function which computes the initial guess, unused here function [zu, dmval]=guess2(x, gam), cons=gam*x*(1-x^2/2) dcons=gam*(1-3*x^2/2) d2cons=-3*gam*x dmval=zeros(2,1) if x>xt then zu=[0 0 -cons -dcons] dmval(2)=-d2cons else zu=[2*x;2;-2*x+cons;-2*dcons] dmval(2)=d2cons end endfunction ipar(9)=1;//iguess zu2=bvode(xpoints,N,m,x_low,x_up,zeta,ipar,ltol,tol,fixpnt,... fsub2,dfsub2,gsub2,dgsub2,guess2); scf(2);clf();plot(xpoints,zu2([1 3],:)) // the evolution of the solution phi and psi
An eigenvalue problem:
// y''(x)=-la*y(x) // BV: y(0)=y'(0); y(1)=0 // Eigenfunctions and eigenvalues are y(x,n)=sin(s(n)*(1-x)), la(n)=s(n)^2, // where s(n) are the zeros of f(s,n)=s+atan(s)-(n+1)*pi, n=0,1,2,... // To get a third boundary condition, we choose y(0)=1 // (With y(x) also c*y(x) is a solution for each constant c.) // We solve the following ode system: // y''=-la*y // la'=0 // BV: y(0)=y'(0), y(0)=1; y(1)=0 // z=[y(x) ; y'(x) ; la] function rhs=fsub(x, z) rhs=[-z(3)*z(1);0] endfunction function g=gsub(i, z) g=[z(1)-z(2) z(1)-1 z(1)] g=g(i) endfunction // The following start function is good for the first 8 eigenfunctions. function [z, lhs]=ystart(x, z, la0) z=[1;0;la0] lhs=[0;0] endfunction a=0;b=1; m=[2;1]; n=2; zeta=[a a b]; N=101; x=linspace(a,b,N)'; // We have s(n)-(n+1/2)*pi -> 0 for n to infinity. la0=input('n-th eigenvalue: n= ?');la0=(%pi/2+la0*%pi)^2; z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ystart=list(ystart,la0)); clf() plot(x,[z(1,:)' z(2,:)']) xtitle(['Startvalue = '+string(la0);'Eigenvalue = '+string(z(3,1))],'x',' ') legend(['y(x)';'y''(x)'])
A boundary value problem with more than one solution.
// DE: y''(x)=-exp(y(x)) // BV: y(0)=0; y(1)=0 // This boundary value problem has more than one solution. // It is demonstrated how to find two of them with the help of // some preinformation of the solutions y(x) to build the function ystart. // z=[y(x);y'(x)] a=0;b=1;m=2;n=1; zeta=[a b]; N=101; tol=1e-8*[1 1]; x=linspace(a,b,N); function rhs=fsub(x, z),rhs=-exp(z(1));endfunction function g=gsub(i, z) g=[z(1) z(1)] g=g(i) endfunction function [z, lhs]=ystart(x, z, M) //z=[4*x*(1-x)*M ; 4*(1-2*x)*M] z=[M;0] //lhs=[-exp(4*x*(1-x)*M)] lhs=0 endfunction for M=[1 4] if M==1 z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ystart=list(ystart,M),tol=tol); else z1=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ystart=list(ystart,M),tol=tol); end end // Integrating the ode yield e.g. the two solutions yex and yex1. function y=f(c),y=c.*(1-tanh(sqrt(c)/4).^2)-2;endfunction c=fsolve(2,f); function y=yex(x, c) y=log(c/2*(1-tanh(sqrt(c)*(1/4-x/2)).^2)) endfunction function y=f1(c1), y=2*c1^2+tanh(1/4/c1)^2-1;endfunction c1=fsolve(0.1,f1); function y=yex1(x, c1) y=log((1-tanh((2*x-1)/4/c1).^2)/2/c1/c1) endfunction disp(norm(z(1,:)-yex(x)),'norm(yex(x)-z(1,:))= ') disp(norm(z1(1,:)-yex1(x)),'norm(yex1(x)-z1(1,:))= ') clf(); subplot(2,1,1) plot2d(x,z(1,:),style=[5]) xtitle('Two different solutions','x',' ') subplot(2,1,2) plot2d(x,z1(1,:),style=[5]) xtitle(' ','x',' ')
A multi-point boundary value problem.
// DE y'''(x)=1 // z=[y(x);y'(x);y''(x)] // BV: y(-1)=2 y(1)=2 // Side condition: y(0)=1 a=-1;b=1;c=0; // The side condition point c must be included in the array fixpnt. n=1; m=[3]; function rhs=fsub(x, z) rhs=1 endfunction function g=gsub(i, z) g=[z(1)-2 z(1)-1 z(1)-2] g=g(i) endfunction N=10; zeta=[a c b]; x=linspace(a,b,N); z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,fixpnt=c); function y=yex(x) y=x.^3/6+x.^2-x./6+1 endfunction disp(norm(yex(x)-z(1,:)),'norm(yex(x)-z(1,:))= ')
See Also
Used Functions
This function is based on the Fortran routine
colnew
developed by
U. Ascher, Department of Computer Science, University of British Columbia, Vancouver, B.C. V6T 1W5, Canada
G. Bader, institut f. Angewandte mathematik university of Heidelberg; im Neuenheimer feld 294d-6900 Heidelberg 1
Bibliography
U. Ascher, J. Christiansen and R.D. Russell, collocation software for boundary-value ODEs, acm trans. math software 7 (1981), 209-222. this paper contains EXAMPLES where use of the code is demonstrated.
G. Bader and U. Ascher, a new basis implementation for a mixed order boundary value ode solver, siam j. scient. stat. comput. (1987).
U. Ascher, J. Christiansen and R.D. Russell, a collocation solver for mixed order systems of boundary value problems, math. comp. 33 (1979), 659-679.
U. Ascher, J. Christiansen and R.D. russell, colsys - a collocation code for boundary value problems, lecture notes comp.sc. 76, springer verlag, b. children et. al. (eds.) (1979), 164-185.
C. Deboor and R. Weiss, solveblok: a package for solving almost block diagonal linear systems, acm trans. math. software 6 (1980), 80-87.
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