I have a coupled second order DE; how to solve them with boundary conditions; How to get get F1 and F2

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equation 1: d^2f1/dx^2-Q/K1*F1+P/K1=0
equation 2 d^2f2/dx^2-Q/K1*F2+P/K2=0 Boundary conditions F1(0)=0 F2(l)=P F1(l-u)=F2(l-u) dF1/dx(l-u)=dF2/dx(l-u) PS All variables l, u, P,Q and K1 are known

Antworten (1)

Torsten
Torsten am 28 Sep. 2017
Use "bvp4c" with the "multipoint boundary value problem" facility:
https://de.mathworks.com/help/matlab/ref/bvp4c.html#bt5uooc-23
https://de.mathworks.com/help/matlab/math/boundary-value-problems.html#brfhdsd-1
Best wishes
Torsten.
  2 Kommentare
Torsten
Torsten am 2 Okt. 2017
Bearbeitet: Torsten am 2 Okt. 2017
function dydx = f(x,y,region)
P = ...;
Q = ...;
K1 = ...;
K2 = ...;
dydx = zeros(2,1);
dydx(1) = y(2);
% The definition of dydx(2) depends on the region.
switch region
case 1 % x in [0 l-u]
dydx(2) = y(1)*Q/K1-P/K1
case 2 % x in [l-u l]
dydx(2) = y(1)*Q/K1-P/K2;
end
function res = bc(YL,YR)
P=...;
res = [YL(1,1) % y(0) = 0
YR(1,1) - YL(1,2) % Continuity of F(x) at x=l-u
YR(2,1) - YL(2,2) % Continuity of dF/dx at x=l-u
YR(1,end) - P]; % y(l) = P
You should be able to add initial conditions and call "bvp4c" following the example in my above link.
Of course, you will have to give values to the parameters used in the function routines.
Best wishes
Torsten.

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