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Ajuda Scilab >> Equações Diferenciais > bvode

# 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 contains `z(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 the `i`-th differential equation. In the following, `M` will represent the sum of the elements of `m`.

x_low

a scalar: left end of interval

x_up

a scalar: right end of interval

zeta

a vector of size `M`, `zeta(j)` gives `j`-th side condition point (boundary point). One must have ```x_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)` and `fixpnt` 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` then `collpnt` is set to `max( max(m(j))+1, 5-max(m(j)) )`.

subint: ipar(3)

Gives the number of subintervals in the initial mesh. If `ipar(3) = 0` then `bvode` arbitrarily sets `subint = 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 the `tol` argument or to `0`. In the latter case the actual value will automatically be set to `size(tol,'*')`.

ndimf: ipar(5)

Gives the dimension of `fspace` (a real work array). Its value provides a constraint on `nmax` the maximum number of subintervals.

The `ipar(5)` value must respect the constraint `ipar(5)>=nmax*nsizef` where `nsizef=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 on `nmax`, the maximum number of subintervals.

The `ipar(6)` value must respect the constraint `ipar(6)>=nmax*nsizei` where `nsizei= 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

= 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 occupy `fspace(1), ..., fspace(n+1)`. The user needs to supply only the interior mesh points ```fspace(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` and `x_up` (the dimension of `fixpnt`). `ipar(11)` must be set to the dimension of the `fixpnt` argument or to `0`. In the latter case the actual value will automatically be set to `size(fixpnt,'*')`.

ltol

an array of dimension `ntol=ipar(4)`. `ltol(j) = l` specifies that the `j`-th tolerance in the `tol` array controls the error in the `l`-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 the `ltol(j)`-th component of . Thus, the code attempts to satisfy on each subinterval

where is the approximate solution vector and is the exact solution (unknown).

fixpnt

an array of dimension `nfxpnt=ipar(11)`. It contains the points, other than `x_low` and `x_up`, which are to be included in every mesh. The code requires that all side condition points other than `x_low` and `x_up` (see description of `zeta` ) be included as fixed points in `fixpnt`.

fsub

an external used to evaluate the column vector `f=` for any `x` such as `x_low` <= `x` <= `x_up` and for any `z=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:

`void fsub(double *x, double *zu, double *f)`
• And in Scilab:

`function f=fsub(x, zu, parameters)`
dfsub

an external used to evaluate the Jacobian of `f(x,z(u))` at a point `x`. Where `z(u(x))` is defined as for `fsub` and the `N` by `M` array `df` should be filled by the partial derivatives of `f`: 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:

`void dfsub(double *x, double *zu, double *df)`
• And in Scilab:

`function df=dfsub(x, zu, parameters)`
gsub

an external used to evaluate given z= `z = zeta(i)` for `1<=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:

`void gsub(int *i, double *zu, double *g)`
• And in Scilab:

`function g=gsub(i, zu, parameters)`

Note that in contrast to `f` in `fsub`, here only one value per call is returned in `g`.

dgsub

an external used to evaluate the `i`-th row of the Jacobian of `g(x,u(x))`. Where `z(u)` is as for `fsub`, `i` as for `gsub` and the `M`-vector `dg` should be filled with the partial derivatives of `g`, viz, for a particular call one calculates The 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

`void dgsub(int *i, double *zu, double *dg)`
• And in Scilab

`function dg=dgsub(i, zu, parameters)`
guess

An external used to evaluate the initial approximation for `z(u(x))` and `dmval(u(x))` the vector of the `mj`-th derivatives of `u(x)`. Note that this subroutine is used only if `ipar(9) = 1`, and then all `M` components of `zu` and `N` components of `dmval` should be computed for any `x` such as `x_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

`void fsub(double *x, double *zu, double *dmval)`
• 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` with `arg_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. If `guess` is given `iguess` 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  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.

```function [z, lhS]=zstart(x)
z=zeros(5,1);z(5)=1;
lhS=[0;1];
endfunction
zu=bvode(xpoints,N,m,x_low,x_up,zeta,ltol=ltol,tol=tol,guess=zstart)```
• 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` and `x=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=)
xtitle('Two different solutions','x',' ')
subplot(2,1,2)
plot2d(x,z1(1,:),style=)
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=;

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,:))= ')```

• external — objeto Scilab, função ou rotina externa
• ode — Solucionador de equações diferenciais ordinárias
• dassl — Equação diferencial algébrica

### 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

1. 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.

2. G. Bader and U. Ascher, a new basis implementation for a mixed order boundary value ode solver, siam j. scient. stat. comput. (1987).

3. 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.

4. 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.

5. 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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