# How to change ode45 from solution for range of values, to solution just for one value?

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I G il 9 Gen 2019
Commentato: madhan ravi il 14 Gen 2019
I have next function file:
function [f,R]=fun_z3(z,p)
beta=1;
ri=0.7;
R=ri-z*(ri-1);
f=zeros(4,size(p,2));
f(1,:)=-32.*beta./(R.^4.*p(1,:));
f(2,:)=(-8*f(1,:)./R-f(1,:).*p(2,:))./p(1,:);
f(3,:)=(-p(2,:).*f(2,:)-8.*f(2,:)./R-8.*f(1,:)./(R.*R.*p(1,:))-f(1,:).*p(3,:))./p(1,:);
f(4,:)=(-f(2,:).*p(3,:)-f(3,:).*p(2,:)+8.*(-f(3,:)./R- (f(2,:)./p(1,:)-p(2,:).*f(1,:)./(p(1,:).*p(1,:)))./(R.*R)) -f(1,:).*p(4,:))./p(1,:);
where I call it with
[zv, pv] = ode45(@fun_z3, [1 0], [1; 0; 0; 0]);
Now, I need to calculate pv, but just in case where zv=0 (in upper file zv=1:0), but for series of beta values. When I change boundaries for z, it means [1 0] to [0 0] or [0], I got some errors. Is it possible implement series of beta values just as range of values?
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### Risposta accettata

madhan ravi il 9 Gen 2019
Just parameterize function (lookup doc):
beta=....range of values
for beta = beta
ode45(@(z,p)fun_z3(z,p,beta), [1 0], [1; 0; 0; 0])]; % function call
end
function [f,R]=fun_z3(z,p,beta) % function definition
ri=0.7;
R=ri-z*(ri-1);
f=zeros(4,size(p,2));
f(1,:)=-32.*beta./(R.^4.*p(1,:));
f(2,:)=(-8*f(1,:)./R-f(1,:).*p(2,:))./p(1,:);
f(3,:)=(-p(2,:).*f(2,:)-8.*f(2,:)./R-8.*f(1,:)./(R.*R.*p(1,:))-f(1,:).*p(3,:))./p(1,:);
f(4,:)=(-f(2,:).*p(3,:)-f(3,:).*p(2,:)+8.*(-f(3,:)./R- (f(2,:)./p(1,:)-p(2,:).*f(1,:)./(p(1,:).*p(1,:)))./(R.*R)) -f(1,:).*p(4,:))./p(1,:);
end
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I G il 14 Gen 2019
Modificato: I G il 14 Gen 2019
If I would like to change boundaries, in way that z is going from 0 to 1, now it is from 1 to 0, for what value of beta are my initial conditions, which are now [1; 0; 0; 0]?
Because in case when I am looking from back, from 0, I got different number of values for every cell of p, and that is not ok for me? What would I do to have constant number - constant dimension for p as soluton? Now it is 93x4 values, 97x4 values, 101x4 values, 105x4 values.
madhan ravi il 14 Gen 2019
It would be [0 1] but honestly I have no idea why the p number of values vary in every iteraration.

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