Implicit Boundary Value Problem
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Is there a way in Matlab to use the bvp4c or bvp5c functions to solve a system consisting of boundary value and algebraic equations where the deifinction of at least one of the derivatives is implicit?
The system looks somewhat like this:
dY(1)/dx=a
dY(2)/dx=f(Y(1),Y(3))
g(Y(2),Y(3))*[dY(1)/dx]+h(Y(2),Y(3))=0
B.C.:
yb(1)=0
ya(2)=ya(3)=0
where a and K are constants and f, g, h and j are functions relating Y(1) and Y(3) to each other. Naturally, the boundary conditions are known.
8 个评论
J. Alex Lee
2020-2-27
Are you sure about the equations themselves? In this particular case, can't you just divide Eq.2 by f and Eq.3 by h to put dY2/dx on the LHS? In which case you have 2 disparate expressions for the same variable Y2...
Saeid
2020-2-27
darova
2020-2-27
What are functions f() and g()? Is it possible to express Y(3)?
g(Y(2),Y(3))*[dY(1)/dx]+h(Y(2),Y(3))=0
Saeid
2020-2-27
darova
2020-2-27
I think it's possible to solve such system of equations.
Can you attach the whole code? Do you have any attempts?
darova
2020-2-27
But as soon as possible
Saeid
2020-2-27
darova
2020-2-27
Please attach the functions
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J. Alex Lee
2020-2-27
0 个投票
Ok, so looks like what you are after is solving a DAE that has BCs at more than one boundary.
Docs say it is ode15s and ode23t that are capable of handling DAE (not ode15i; https://www.mathworks.com/help/matlab/math/solve-differential-algebraic-equations-daes.html)
It seems Matlab does not have built-in functionality to address directly (https://www.mathworks.com/matlabcentral/answers/478652-solving-bvp-for-dae-in-matlab)
This system seems simple enough to use a shooting strategy, I think. You have 3 different boundary locations, so you can treat the problem as if it were an IVP leaving 2 guesses for 2 out of 3 of the Y's at x=0 as guesses (not sure which ones, you are mixing a's, b's, and 1,2,3's), use either of the odeXX (with appropriate handling of the algebraic equation), and you will get some values for the Y's at the 2 other boundaries in the course of integration. Their deviations from your actual BCs will become residuals for a 2-variable root finding.
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