# 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

  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 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 z (u) . 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 z (u) . Thus, the code attempts to satisfy ( z (v) −z (u) ) ltol (j) ≤tol (j) ⋅ ( z (u) ) ltol (j) +tol (j) ,for j=1 :ntol abs(z(v)-z(u))_{ltol(j)} <= tol(j) cdot abs(z(u))_{ltol(j)} + tol(j), for j=1:ntol on each subinterval

  where v is the approximate solution vector and u 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=` [ f 1 ( x,z ( u (x) ) ) ; f 2 ( x,z ( u (x) ) ) ,⋯, fN ( x,z ( u (x) ) ) ] [f_1(x,z(u(x))); f_2(x,z(u(x))),dotsaxis, f_N(x,z(u(x)))] 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`:

  df
  (
  i
  ,
  j
  )
  =
  ∂
  f
  i
  ∂
  z
  j
  for
  {
  i
  =
  1
  :
  N
  j
  =
  1
  :
  M
  df(i,j)= {partial{f_i}} over {partial{z_j}} ~ for ~ left lbrace binom{i=1:N}{j=1:M} right none

  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 gi ( ζ i ,z ( u ( ζ i ) ) ) g_i(%zeta_i,z(u(%zeta_i))) given z= z ( u ( ζ i ) ) z(u(%zeta_i)) `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

  dg
  (
  i
  ,
  j
  )
  =
  ∂
  g
  i
  ∂
  z
  j
  for
  {
  i
  =
  1
  :
  M
  j
  =
  1
  :
  M
  dg(i,j)= {partial{g_i}} over {partial{z_j}} ~ for ~ left lbrace binom{i=1:M}{j=1:M} right none

  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

{
u
i
(
m
i
)
=
f
i
(
x
,
z
(
u
(
x
)
)
)
{
i
=
1
:
N
x
∈
x
l
,
x
u
g
j
(
ζ
j
,
z
(
u
(
ζ
j
)
)
)
=
0
,
j
=
1
:
M
left lbrace stack{u_i^(m_i)=f_i(x,z(u(x))) ~~left lbrace { binom{i=1:N}{x in left ] a_l, a_u right [} } right none# `# g_j(%zeta_j,z(u(%zeta_j))) = 0~~j=1:M} right none

where

{
M
=
∑
i
=
1
N
m
i
z
(
u
(
x
)
)
=
[
u
1
(
x
)
;
u
1
1
(
x
)
;
⋯
u
1
(
m
1
−
1
)
(
x
)
;
⋯
;
u
N
(
x
)
;
u
N
1
(
x
)
;
⋯
;
u
N
(
m
N
−
1
)
(
x
)
]
x
l
≤
ζ
1
≤
⋯
≤
ζ
M
≤
x
u
left lbrace stack { M= sum from {i=1} to {N} m_i # ~# z(u(x)) = [u_1(x);u_1^{1}(x);dotsaxis u_1^(m_1-1)(x);~ dotsaxis; ~u_N(x);u_N^{1}(x);dotsaxis; u_N^(m_N-1)(x)]# ~# x_l <= %zeta_1 <= dotsaxis <= %zeta_M <=x_u}right none

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:

  d
  4
  d
  x
  4
  u
  (
  x
  )
  =
  1
  −
  6
  .
  x
  2
  .
  d
  3
  d
  x
  3
  u
  (
  x
  )
  −
  6
  .
  x
  .
  d
  2
  d
  x
  2
  u
  (
  x
  )
  x
  3
  d^4 over {d x^4} u(x)={1 -6*x^2*{{d^3} over {d x^3}} u(x)-6*x*{{d^2} over {d x^2}} u(x)} over x^3

  Subjected to the boundary conditions:

  {
  u
  (
  1
  )
  =
  0
  d
  2
  d
  x
  2
  u
  (
  x
  )
  (
  1
  )
  =
  0
  u
  (
  2
  )
  =
  0
  d
  2
  d
  x
  2
  u
  (
  x
  )
  (
  2
  )
  =
  0
  left lbrace stack{ u(1)=0# {{d^2} over {d x^2}} u(x)(1)=0# u(2)=0# {{d^2} over {d x^2}} u(x)(2)=0} right none

  The exact solution of this problem is known to be:

  u
  (
  x
  )
  =
  1
  4
  (
  10
  log
  (
  2
  )
  −
  3
  )
  (
  1
  −
  x
  )
  +
  1
  2
  [
  1
  x
  +
  (
  3
  +
  x
  )
  log
  (
  x
  )
  −
  x
  ]
  u(x)=1 over 4 (10`log(2)-3)`(1-x)+ {1 over 2} [ {1 over x} + {(3+x) ` log(x)-x}]

      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 $\varphi$ is the meridian angle change of the deformed shell and $\psi$ is a stress function. For $\varepsilon=\mu=10^{-3}$ two different solutions may found depending on the starting point

  {
  ε
  4
  μ
  (
  φ
  '
  '
  +
  φ
  '
  x
  −
  φ
  x
  2
  )
  +
  ψ
  (
  1
  −
  φ
  x
  )
  −
  φ
  +
  γ
  x
  (
  1
  −
  x
  2
  2
  )
  =
  0
  μ
  (
  ψ
  '
  '
  +
  ψ
  '
  x
  −
  ψ
  x
  2
  )
  −
  φ
  (
  1
  −
  φ
  2x
  )
  =
  0
  0
  <
  x
  <
  1
  left lbrace stack{{%epsilon^4} over %mu (%phi''+ {%phi'} over x -{%phi} over {x^2} ) +%psi (1- %phi over x ) -%phi+%gamma x(1-{x^2} over 2)=0# %mu(%psi''+{%psi'} over x-%psi over {x^2})-%phi(1-%phi over{2x})=0 # 0<x<1} right none

  Subject to the boundary conditions

  {
  φ
  =
  0
  x
  ψ
  '
  −
  0.3
  ψ
  +
  0.7x
  =
  0
  left lbrace stack{ %phi=0# x`%psi'-0.3%psi+0.7x=0} right none

  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=[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

- link
- external
- ode
- dassl

## 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.