Matlab gives [ empty sym ] when using dsolve
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I-Chun
am 5 Jul. 2023
Kommentiert: Steven Lord
am 5 Jul. 2023
I have two differential equations which are complicated. Later I will substitute the solution into another differential equation, so I need to keep the analytical form.
syms Ap(z) Ac(z)
ode1 = diff(Ap) == (Ap*(Ac^2*conj(Ac)^2*7.12e+27i - Ac*conj(Ac)*(2.45e+29 + 4.89e+28i) + ...
Ac*Ap*conj(Ac)*conj(Ap)*7.12e+27i)*1.53e+34i)/(Ac^3*conj(Ac)^3*5.07e+55i + ...
Ac^2*Ap*conj(Ac)^2*conj(Ap)*1.52e+56i - Ac^2*conj(Ac)^2*6.97e+56i + Ac*Ap^2*conj(Ac)*conj(Ap)^2*1.52e+56i + ...
Ac*Ap*conj(Ac)*conj(Ap)*1.39e+57i + Ac*conj(Ac)*6.22e+58i + Ap^3*conj(Ap)^3*5.07e+55i + Ap*conj(Ap)*5.98e+58i);
ode2 = diff(Ac) == -(1.53e+34*Ac*Ap*conj(Ap)*(2.45e+29 + Ac*conj(Ac)*7.12e+27i + ...
Ap*conj(Ap)*7.12e+27i))/(5.07e+55*Ac^3*conj(Ac)^3 - 6.97e+56*Ac^2*conj(Ac)^2 + 5.07e+55*Ap^3*conj(Ap)^3 + ...
6.22e+58*Ac*conj(Ac) + 5.98e+58*Ap*conj(Ap) + 1.39e+57*Ac*Ap*conj(Ac)*conj(Ap) + ...
1.52e+56*Ac*Ap^2*conj(Ac)*conj(Ap)^2 + 1.52e+56*Ac^2*Ap*conj(Ac)^2*conj(Ap));
odes = [ode1; ode2];
cond = [Ap(0) == 1.3105, Ac(0) == 13.1046];
dsolve(odes, cond)
I want to solve "Ap(z)" and "Ac(z)", but I get a empty solution:
ans =
[ empty sym ]
How can I do for a such problem? Thanks very much for any advice.
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Akzeptierte Antwort
Samay Sagar
am 5 Jul. 2023
In cases where the dsolve function in MATLAB returns an empty solution, it means that it was unable to find an analytical solution for the given set of differential equations. This can happen when the equations are too complex or do not have a closed-form solution.
In such situations, you might consider using numerical methods to approximate the solution instead. MATLAB provides a variety of numerical ODE solvers that can handle complex systems of differential equations. You can use the ode45, ode23, or ode15s functions, among others, to numerically solve the equations.
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Steven Lord
am 5 Jul. 2023
To restate John's second point, where do those very large magnitude coefficients come from?
Perhaps if this is a physical problem a change of units might be in order (using meters instead of millimeters or vice versa.)
If those came about from fitting a polynomial curve to data, what degree polynomial did you fit? If you had say 100 data points you could fit a degree 99 curve to that data, but that doesn't mean you should do that.
x = 0:100;
y = sind(x);
p = polyfit(x, y, numel(x)-1);
Note that the plot of p takes on values over 1 when x is larger than say 80. But the sine function's value for real inputs is between -1 and 1 inclusive.
plot(x, polyval(p, x))
Perhaps use a lower degree curve. Note the strong agreement between the actual data (the + symbols) and the fitted curve (the o symbols.)
p2 = polyfit(x, y, 5);
figure
plot(x, y, '+', x, polyval(p2, x), 'o')
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