Solving ODE

[fraud profile]

Hi,

I am using lsode to solve a simple ODE on Octave 2.1.72.
The ode is:
Solve: y''(t) + y(t) =0
such that:
y(0)=3 & y(pi/2)=3

My code for this is:

-------- file f.m ---------
function xdot = f (x, t)
  xdot = zeros (2,1);

  xdot(1) =  x(2);
  xdot(2) =  -x(1);
endfunction
---- end of f.m -----

-------- file ode.m ---------
 l = linspace (0, pi/2, 20);

y_min=[3;3];
    psi = lsode( "f", y_min, l );
 
plot(l,psi(:,1));
---- end of ode.m -----

I am getting correct answer with this. But lsode requires y_min which is actually the value of x(1) and x(2) at t=0. I have the value of x(1) at t=0 and t=pi/2.(I got the value of x(2) at t=0 by solving this on Matlab and then solved this by adding y_min=[3;3]; here). How can I use the given boundary conditions so that I don't need the values of x(2) to solve the problem.

Thank you

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[LUK ShunTim]

fraud profile wrote:

> Hi,
>
> I am using lsode to solve a simple ODE on Octave 2.1.72. <http://2.1.72.>
> The ode is:
> Solve: y''(t) + y(t) =0
> such that:
> y(0)=3 & y(pi/2)=3
>
> My code for this is:
>
> -------- file f.m ---------
> function xdot = f (x, t)
>   xdot = zeros (2,1);
>
>   xdot(1) =  x(2);
>   xdot(2) =  -x(1);
> endfunction
> ---- end of f.m -----
>
> -------- file ode.m ---------
>  l = linspace (0, pi/2, 20);
>
> y_min=[3;3];
>     psi = lsode( "f", y_min, l );
>  
> plot(l,psi(:,1));
> ---- end of ode.m -----
>
> I am getting correct answer with this. But lsode requires y_min which is
> actually the value of x(1) and x(2) at t=0. I have the value of x(1) at
> t=0 and t=pi/2.(I got the value of x(2) at t=0 by solving this on Matlab
> and then solved this by adding y_min=[3;3]; here). How can I use the
> given boundary conditions so that I don't need the values of x(2) to
> solve the problem.
>
> Thank you
>

Hello,

Maybe be you can google for "shooting method" if you want to use lsode
or other initial value problem solvers for a boundary value problem.

Alternatively, you can try the odebvp package from octave-forge.

Regards,
ST
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[Carlo de Falco-2]

Hi,

On 08/mag/08, at 01:08, fraud profile wrote:

> Hi,
>
> I am using lsode to solve a simple ODE on Octave 2.1.72.
> The ode is:
> Solve: y''(t) + y(t) =0
> such that:
> y(0)=3 & y(pi/2)=3

LSODE is intended for solving intial value problems,
what you have is a boundary value problem.

if you install the BIM package from octave-forge you can solve your
equation by:

Nnodes = 100;
Nelements = Nnodes-1;
x = linspace(0, 1, Nnodes)';
y(Nnodes, 1) = y_pihalf = 3;
y(1, 1) = y_0 = 3;

A =  BIM1Alaplacian(x, -ones(Nelements, 1));
A += BIM1Areaction(x, ones(Nelements, 1), -ones(Nnodes, 1));

y(2:end-1) = A(2:end-1,2:end-1) \ ( - A(2:end-1, [1 end]) * [y_0;  
y_pihalf]);
plot(x,y)

c.


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[tiago.charters.azevedo]

The prolem that you describe
Solve: y''(t) + y(t) =0
such that:
y(0)=3 & y(pi/2)=3

seems to be boundary value problem (BVP), not an initial conditions problem.
The function lsode solves an initial conditions problem not an  boundary value
problem.

I suggest using the odebvp package which solves a linear BVP. You can get it
from octave-forge. It should work on you octave version

http://octave.sourceforge.net/doc/f/lfdif.html

tca

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[Carlo de Falco-2]

On 08/mag/08, at 10:43, Carlo de Falco wrote:

> Hi,
>
> On 08/mag/08, at 01:08, fraud profile wrote:
>
>> Hi,
>>
>> I am using lsode to solve a simple ODE on Octave 2.1.72.
>> The ode is:
>> Solve: y''(t) + y(t) =0
>> such that:
>> y(0)=3 & y(pi/2)=3
>
> LSODE is intended for solving intial value problems,
> what you have is a boundary value problem.
>
> if you install the BIM package from octave-forge you can solve your
> equation by:
>
> Nnodes = 100;
> Nelements = Nnodes-1;
> x = linspace(0, 1, Nnodes)';
> y(Nnodes, 1) = y_pihalf = 3;
> y(1, 1) = y_0 = 3;
>
> A =  BIM1Alaplacian(x, -ones(Nelements, 1));
> A += BIM1Areaction(x, ones(Nelements, 1), -ones(Nnodes, 1));

sorry, to be consistent with your problem this line should be:
A += BIM1Areaction(x, ones(Nelements, 1), ones(Nnodes, 1));

> y(2:end-1) = A(2:end-1,2:end-1) \ ( - A(2:end-1, [1 end]) * [y_0;
> y_pihalf]);
> plot(x,y)

c.
>

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