how to solve a set of differential equation systems like this?

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Daniel Niu
Daniel Niu 2022-11-16
编辑： Torsten 2022-11-18
Dear friend,
How to solve a ordinary differential equation systems like above using MATLAB?
a=b=c=1
Your help would be highly appreciated.
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John D'Errico
John D'Errico 2022-11-16
编辑：John D'Errico 2022-11-16
@Sam Chak
Actually, the point where you identify a singularity is an interesting one, since in order for a singularity to exist, one would need to have
x^2 + (c*t-y)^2 == 0
And that implies two things MUST happen, x==0, AND y = c*t. Note that both differential equations have a corresponding term in the denominator. But then look carefully. In the numerator of both equations, they have a similar term on top. As such, that fraction is actually going to have a finite limit at that point. At the exact point of interest, the result is effectively a Nan, but everywhere else, it is well defined. Not really any different from the fraction x/x.

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Torsten
Torsten 2022-11-17
编辑：Torsten 2022-11-17
Ran in:
ar = 0.01;
af = 0.001;
x0 = 3;
y0 = -3;
Lr = 0;
Lf = 0;
z0 = [x0; y0; Lr; Lf];
[T,Z] = ode45(@(t,z)fun(t,z,ar,af),[0 11.5],z0);
plot(Z(:,1),Z(:,2),Z(:,3),zeros(size(T)))
ylim([-3 0.5])
function dz = fun(t,z,ar,af)
x = z(1);
y = z(2);
Lr = z(3);
Lf = z(4);
sr = exp(-ar*Lr);
sf = exp(-af*Lf);
dz = zeros(4,1);
dz(1) = sf* (Lr-x)/sqrt((Lr-x)^2+(0-y)^2);
dz(2) = sf* (0-y)/sqrt((Lr-x)^2+(0-y)^2);
dz(3) = sr;
dz(4) = norm([dz(1) dz(2)]);
end
  2 个评论
Daniel Niu
Daniel Niu 2022-11-18
Dear Torsten,
Sorry to bother you again. I want to know if the rabbit is not moving horizontal but from (0,0) to (-100,100)
how to modify the code?
Thank you!
Torsten
Torsten 2022-11-18
编辑：Torsten 2022-11-18
Ran in:
Uninteresting case since both fox and rabbit will run on the same straight line.
ar = 0.01;
af = 0.001;
x0r = 0;
y0r = 0;
x0f = 3;
y0f = -3;
Lr0 = 0;
Lf0 = 0;
z0 = [x0r; y0r; x0f; y0f; Lr0; Lf0];
[T,Z] = ode45(@(t,z)fun(t,z,ar,af),[0 11.5],z0);
plot(Z(:,1),Z(:,2),Z(:,3),Z(:,4))
function dz = fun(t,z,ar,af)
xr = z(1);
yr = z(2);
xf = z(3);
yf = z(4);
Lr = z(5);
Lf = z(6);
sr = exp(-ar*Lr);
sf = exp(-af*Lf);
dz = zeros(6,1);
dz(1) = sr* (-1/sqrt(2));
dz(2) = sr* (1/sqrt(2));
dz(3) = sf* (xr-xf)/sqrt((xr-xf)^2+(yr-yf)^2);
dz(4) = sf* (yr-yf)/sqrt((xr-xf)^2+(yr-yf)^2);
dz(5) = norm([dz(1) dz(2)]);
dz(6) = norm([dz(3) dz(4)]);
end

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更多回答（1 个）

Torsten
Torsten 2022-11-17
Hint:
The length L(t) of a curve in parametric form (x(t),y(t)) where x(t) and y(t) are given by differential equations
dx/dt = f(x,y)
dy/dt = g(x,y)
can be computed by additionally solving a differential equation for L:
dL/dt = sqrt(f(x,y).^2+g(x,y).^2)
with initial condition
L(0) = 0.
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Torsten
Torsten 2022-11-17
编辑：Torsten 2022-11-17
r1 is part of your ODE system:
dz1/dt = s_f* (r1-z1)/sqrt((r1-z1)^2+(r2-z2)^2)
dz2/dt = s_f* (r2-z2)/sqrt((r1-z1)^2+(r2-z2)^2)
And I think the speed of the rabbit will also decrease with time...
Sam Chak
Sam Chak 2022-11-17
编辑：Sam Chak 2022-11-17
Hi @Daniel Niu
Your original "Fox-chasing-Rabbit" example assumes that the Rabbit is moving at a constant velocity . That's why the solution for the Rabbit's position is .
If the Rabbit's velocity is time-varying, or is reactively dependent on the position of the Fox {, }, then naturally there exists , and I think that is what @Torsten tried to tell you.

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