How to find second derivative as output of ode45?
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I G
el 22 de Ag. de 2018
Comentada: Torsten
el 29 de Ag. de 2018
I am using ode45 function to find numerical solution for my system of equations, where I have 4 equations and 4 variables, with command:
sol=ode45(@fun,[1 0],[1; 0; 0; 0])
where time span is going from 1 to 0, and initial conditions are 1,0,0,0.
So, I need to find numerically the first and the second derivative of my values according to time. For the first derivative I have verified code
[~,SXINT]=deval(sol,sol.x);
where in SXINT are stored first derivatives of all 4 variables.
But I don`t know how to find the second derivative numerically. I tried:
[~,SXINT2]=deval(SXINT,sol.x);
but I got error:
SXINT must be a structure returned by a differential equation solver.
How to make that SXINT be in that structure, or is there some other way to find the second derivative, but with the same precision as deval method is with ode45 solver?
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Torsten
el 22 de Ag. de 2018
Editada: Torsten
el 22 de Ag. de 2018
To get the second derivative, the first derivative must be part of the solution "sol".
So you have to solve 8=4*2 equations, namely your original system and new algebraic equations given by
y(5)-dy(1) = 0
y(6)-dy(2) = 0
y(7)-dy(3) = 0
y(8)-dy(4) = 0
This way, "sol" has stored the first-order derivatives of y(1)-y(4) in y(5)-y(8) which can now be differentiated by a call to "deval".
Best wishes
Torsten.
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Torsten
el 29 de Ag. de 2018
The way I usually calculate higher derivatives of ODE variables is simply differentiating the right-hand side of the ODE equations with respect to the independent variable:
If
dyi/dz = fi(z,y1(z),y2(z),...,yn(z)),
then
d^2yi/dz^2 = partial(fi)/partial(z) + sum_j (partial(fi)/partial(yj)*dyj/dz)
E.g. for your first equation
beta = 1.0;
ri = 0.3;
R = ri-z*(ri-1);
dR/dz = -ri;
dp1/dz = -32*beta/(R^4*p(1));
d^2p1/dz^2 = 32*beta*(4*R^3*dRdz*p(1)+R^4*dp1/dz)/(R^4*p(1))^2
Best wishes
Torsten.
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