How to numerically solve system of equations and differential equations simultaneously?

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I CHUN LIN
I CHUN LIN 2023-4-19
评论： I CHUN LIN 2023-4-24
Hi all,
I have 3 variables A, B, C, where A and B can be solved by system of equations. For example,
A + B = C + 6
A - B = 2*C
While C should be solved by differential equation,
diff(C,t) == 6*A
This is just a simple example, actually my equations are very complicated. I have tried:
  1. Using "solve". I think the equations are so complicated that the empty solutions appear. So I tend to find the numerical solution.
  2. Using "fsolve". The problem is that there is an undefined variable C. Because of the error "FSOLVE requires all values returned by functions to be of data type double.", I can't use symbolic.
  3. Using "vpasolve". There are similar problems with "fsolve". The error is "Symbolic parameters not supported in nonpolynomial equations.".
Are there other methods I should try? Thank you for any advice.

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采纳的回答

Askic V
Askic V 2023-4-19
编辑：Askic V 2023-4-19
Ran in:
Does this make sense to you (in your simple example)?
syms A B C(t)
eqn1 = diff(C,t) == 6*A;
C_sol = dsolve(eqn1)
C_sol = 
eqn2 = A + B == C_sol + 6;
eqn3 = A - B == 2*C_sol;
[A,B] = equationsToMatrix([eqn2, eqn3], [A, B]);
X = linsolve(A,B)
X = 
  3 个评论
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Askic V
Askic V 2023-4-19
It indeed shows no error, but if this is applicable to your complicated example, (since you stated you want numerical approach)?
I CHUN LIN
I CHUN LIN 2023-4-19
Thank you @Askic V. Regarding this, the output shows "struct with fields: [0×1 sym]". I'm still trying to debug. I think my equations shouldn't be no solution.

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更多回答（2 个）

Sam Chak
Sam Chak 2023-4-19
Ran in:
Hi @I CHUN LIN
It seems that simple Substitution-and-Elimination method produces the solution
.
Thus, the linear ODE becomes
which indicates that it is possible to find an explicit solution of the differential equation analytically.
syms C(t)
eqn = diff(C, t) == 9*(C + 2);
S = dsolve(eqn)
S = 
  1 个评论
I CHUN LIN
I CHUN LIN 2023-4-19
Thank you @Sam Chak, I know what you mean. However, if A and B are very complicated, I can not use the Substitution-and-Elimination method, but the numerical method. So I want to find whether matlab provides numerical tools (ex. vpasolve, fzero...) which contain symbolic variables.

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Torsten
Torsten 2023-4-19
编辑：Torsten 2023-4-19
Ran in:
MATLAB's ode solvers allow a mixture of differential and algebraic equations as in your case.
These systems are called differential-algebraic equations. For an example, see
Solve Robertson Problem as Semi-Explicit Differential Algebraic Equations (DAEs)
under
https://de.mathworks.com/help/matlab/ref/ode15s.html
Or as a solution for your simple example:
M = [0 0 0;0 0 0; 0 0 1];
tspan = [0 1];
fun = @(t,y) [y(1)+y(2)-y(3)-6;y(1)-y(2)-2*y(3);6*y(1)];
options = odeset('Mass',M,'MStateDependence','none','MassSingular','yes','RelTol',1e-7,'AbsTol',1e-7);
% Starting values for A and B are taken arbitrary ;
% They will be adjusted according to the algebraic equations 1 and 2 by
% ode15s to y0(1) = 4.5 and y0(2) = 2.5 (see below)
y0=[0 0 1];
[T,Y] = ode15s(fun,tspan,y0,options);
plot(T,Y)
Y(1,1)
ans = 4.5000
Y(1,2)
ans = 2.5000
Y(1,3)
ans = 1
grid on
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Torsten
Torsten 2023-4-21
编辑：Torsten 2023-4-21
Ran in:
I stick to your program structure of first solving the algebraic equations and inserting the result into the differential equations.
This won't usually work if the algebraic equations are difficult.
You should try my method from above for more complicated systems (i.e. solving algebraic and differential equations all together using ode15s).
syms x1 x2 x3
a = 1;
b = 2;
c = 3;
d = 4;
Eq1 = (a+1i*b)*x1 + 1i/2*x2*c - 1i/2*conj(x2)*d == 0;
Eq2 = (-a+1i*c)*x1 + 1i/2*d*x2 + 1i/2*b*x3 + a*conj(x3) == 0;
Solution = solve([Eq1, Eq2], [x1 x2]);
x1 = matlabFunction(Solution.x1);
x2 = matlabFunction(Solution.x2);
options = odeset('RelTol',1e-10,'AbsTol',1e-10);
fun = @(x,t) 1i*c*x1(x(1));
tspan = [0:0.05: 1];
y0 = 1;
[T,X3] = ode15s(fun,tspan,y0,options);
figure(1)
plot(T,[real(X3),imag(X3),abs(X3)])
figure(2)
plot(T,[real(x1(X3)),imag(x1(X3)),abs(x1(X3))])
figure(3)
plot(T,[real(x2(X3)),imag(x2(X3)),abs(x2(X3))])

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