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bvp4c or ode45?

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I G il 3 Mag 2018
Commentato: I G il 3 Mag 2018
I am solving two the first order ODEs ('=d/dz, all other variables are known constants):
p0 p0'=-32 beta/R^4
I have next conditions
p0(z=0)=p0i (I can choose value)
It is necessary to find p0'(z=0) and p1'(z=0) with shooting method (literature says like that), according to already mentioned p0(z=1)=1 and p1(z=1)=0. *How to connect this two conditions* and shoot p0'(z=0) for already known p1'(z=0)?
*Are that conditions p0'(z=0) and p1'(z=0) necessary*, because these are the first order equations, is there only one initial condition enough?
Instead of missing conditions, I also need to solve numerically two ODEs, from the beginning of text, with Runge Kutta method. How to connect that solving with shooting? Is it possible to find missing condition with bvp4c, as shooting method, and after that solve equation with ode45?

Risposta accettata

Torsten il 3 Mag 2018

If you know $p_0|_{z=1}=1$ and $p_1|_{z=1}=0$ and you want to know $p_0'|_{z=0}$ and $p_1'|_{z=0}$, you don't need a shooting method.

Just use ODE45 as you already did and define tspan=[1 0]. This way, you integrate back in z-direction from z=1 to z=0. Once you have reached z=0,

$p_0'|_{z=0}=-\dfrac{32 \beta}{R^4}/p_0$ 



Best wishes


  5 Commenti
Torsten il 3 Mag 2018
If I defined y0=[1 0], does it mean that I gave it 1 for first equation, and 0 for the second equation? Or do I need to have, 4 initial conditions for two equations?
It means that you gave p_0(@tspan(1))=1 and p_1(@tspan(1))=0 as initial conditions. And you need two initial conditions for two first-oder differential equations.
Does it recognize it as initial value for variable-p0 or as initial values for derivative of variable-dp0dz, if I defined equation like this: dp0dz=-32*beta/(p0*R^4)?
y0 is interpreted as initial value vector [p_0,p_1] at tspan(1).
I G il 3 Mag 2018
I cannot find mistake in my equations, this is the last question I promise, but ode45 doesn`t work:
function f=fun(z,p)
R=2; sig=1; beta=1;
[zv,pv]=ode45('fun',[1 0],[1; 0])
Is it possible to call term on left side of the first equation in the second equation? What can be wrong here?

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