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Solve time-dependent ODE using the result from another time-dependent ODE

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Hello everyone,
I needed to solve two ODEs of the following forms:
dRdS = 1 - B(S)*R(S) - A(S)*(R(S)^2); (1)
dWdS = - A(S)*R(S)*W(S) + R(S)*P(S); (2)
where B(S) , A(S) and P(S) are all deterministic functions depending on S, defined as follows:
Bs = linspace(78.809,80,100);
As = linspace(78.809,80,100);
Ps = linspace(78.809,80,100);
A = (2./(vm.*(As.^2))) .* ((sigma^2*vm)/(dv^2) + abs(alpha-beta*vm)/dv + r + 1/dtau);
B = -(2*(r-q)*Bs)./(vm*(Bs.^2));
P = (2./(vm.*Ps.^2)).*( ((-(sigma^2)*vm)/2)*(2*C/(dv^2)) - (abs(alpha-beta*vm)/2)*(2*C/dv) - (1/dtau)*C );
all the parameters above are defined beforehand.
I solved (1) by first writing a function dRdS:
function dRdS = dRdS(S,R,Bs,B,As,A)
B = interp1(Bs,B,S);
A = interp1(As,A,S);
dRdS = 1 - B.*R - A.*R*R;
end
and solved it using ode45:
S_span = [78.809 80];
IC = 0;
opts = odeset('RelTol',1e-2,'AbsTol',1e-4);
[S, R] = ode45(@(S,R) dRdS(S,R,Bs,B,As,A), S_span, IC, opts);
where I obtained a 53x2 matrix of [S, R].
I then moved on to write a function dWdS to solve (2):
function dWdS = dWdS(S,W,Bs,B,As,A,Ps,P,R)
B = interp1(Bs,B,S);
A = interp1(As,A,S);
P = interp1(Ps,P,S);
dWdS = -A.*R.*W-R.*P;
end
and using ode45:
IC_W = 0;
[S, W] = ode45(@(S,W) dWdS(S,W,Bs,B,As,A,Ps,P,Rm), S_span, IC_W);
While the ODE (1) was successfully solved, I encountered error message while solving ODE (2) that says:
"Error using odearguments (line 95)
@(S,W)DWDS(S,W,BS,B,AS,A,PS,P,R) returns a vector of length 53, but the length of initial
conditions vector is 1. The vector returned by @(S,W)DWDS(S,W,BS,B,AS,A,PS,P,R) and the
initial conditions vector must have the same number of elements."
In light of this error, I defined IC_W = zeros(53), but encountered another error: "Arrays have incompatible sizes for this operation."
So, I'd like to ask what exactly might have gone wrong in this implementation? If I'm adopting a wrong method for solving ODE (2), could someone enlighten me with a better method for solving such ODE which have the result from another ODE as part of the coefficient?
Sorry about the long question, and thanks so much!
  4 comentarios
Mingze Yin
Mingze Yin el 12 de Dic. de 2021
@Torsten Thanks for answering! Could you introduce to me a little more on how to solve both equations in one call? Sorry im actually not very familiar with the ode45 solver...
Mingze Yin
Mingze Yin el 12 de Dic. de 2021
@Jan Thanks for answering! Allow me to spend some time understanding and implementing the ideas you shared :)

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Torsten
Torsten el 12 de Dic. de 2021
function main
IC = [0;0];
S_span = [78.809 80];
opts = odeset('RelTol',1e-4,'AbsTol',1e-4);
[S, y] = ode45(@fun, S_span, IC, opts);
R = y(:,1);
W = y(:,2);
end
function dy = fun(S,y)
R = y(1);
W = y(2);
vm = ...;
sigma=...;
dv = ...;
alpha=...;
beta=...;
dtau=...;
r=...;
q=...;
C=...;
A = (2./(vm.*(S.^2))) .* ((sigma^2*vm)/(dv^2) + abs(alpha-beta*vm)/dv + r + 1/dtau);
B = -(2*(r-q)*S)./(vm*(S.^2));
P = (2./(vm.*S.^2)).*( ((-(sigma^2)*vm)/2)*(2*C/(dv^2)) - (abs(alpha-beta*vm)/2)*(2*C/dv) - (1/dtau)*C );
dRdS = 1 - B.*R - A.*R*R;
dWdS = -A.*R.*W-R.*P;
dy = [dRdS;dWdS];
end
  6 comentarios
Torsten
Torsten el 14 de Dic. de 2021
Use bvp4c instead of ode45 or adjust the initial condition v_start for V at s=s_start such that you receive the desired terminal value v_terminal_desired for V at s=s_terminal. This can be done by using "fzero" to solve
v_terminal(v_start) - v_terminal_desired = 0.
Pavitra Vaishali
Pavitra Vaishali el 12 de Abr. de 2023
Hi I have same kind of problem, although I get martices as answer. Please, if anyone may help me with it

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