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

조회 수: 20 (최근 30일)
I CHUN LIN 2023년 4월 19일
댓글: I CHUN LIN 2023년 4월 24일
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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### 채택된 답변

편집: Askic V 2023년 4월 19일
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 =
##### 댓글 수: 3이전 댓글 1개 표시이전 댓글 1개 숨기기
It indeed shows no error, but if this is applicable to your complicated example, (since you stated you want numerical approach)?
I CHUN LIN 2023년 4월 19일
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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### 추가 답변 (2개)

Sam Chak 2023년 4월 19일
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 =
##### 댓글 수: 1이전 댓글 -1개 표시이전 댓글 -1개 숨기기
I CHUN LIN 2023년 4월 19일
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 2023년 4월 19일
편집: Torsten 2023년 4월 19일
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
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
##### 댓글 수: 4이전 댓글 2개 표시이전 댓글 2개 숨기기
Torsten 2023년 4월 21일
편집: Torsten 2023년 4월 21일
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))])
I CHUN LIN 2023년 4월 24일
Yes! I made it! Thank you @Torsten, you help me a lot.

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