Matlab gives [ empty sym ] when using dsolve

16 次查看（过去 30 天）
I-Chun
I-Chun 2023-7-5
评论： Steven Lord 2023-7-5
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.

请先登录，再回答此问题。

采纳的回答

Samay Sagar
Samay Sagar 2023-7-5
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.
  3 个评论
显示 1更早的评论
I-Chun
I-Chun 2023-7-5
@Samay Sagar, @John D'Errico
Ah, that makes sense. I should not be attached to finding analytical solution. Thank you very much.
Steven Lord
Steven Lord 2023-7-5
Ran in:
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);
Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.
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')

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Symbolic Computations in MATLAB 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by