Solving coupled differential equations

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Jon
Jon 2023-4-21
评论： David Goodmanson 2023-4-23
I am trying to solve the following coupled differential equation:
.
The goal is to obtain analytic solutions of and in terms of , and .
However, MATLAB doesn't seem to obtain the analytic solutions, as posted below. But I do think that we can get and in terms of , and . Is there any other way I can try to solve the problem in MATLAB (or analytically)?
  1 个评论
David Goodmanson
David Goodmanson 2023-4-23
Hi Jon,
In general there won't be an analytical solution, even if Int f(t) dt has an analytic expression. For y, you have the second order differential equation
y'' + y'(-f/f + gamma) + f^2 y = 0
(there is a similar expression for x''). This may or may not (and in most cases does not) have an analytic solution.

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回答（2 个）

Torsten
Torsten 2023-4-21
  1 个评论
Jon
Jon 2023-4-21
编辑：Jon 2023-4-21
Thank you Torsten. Unfortunately, the second equation you entered in the link has to be x'(t) = f(t)y(t) (as well as the first equation). But then the WolframAlpha doesn't give the solution (https://www.wolframalpha.com/input?i=y%27%28t%29+%3D+-f%28t%29*x%28t%29-gamma*y%28t%29%2C+x%27%28t%29%3Df%28t%29*y%28t%29), which I believe means that there's no analytic solution..

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Sam Chak
Sam Chak 2023-4-22
Ran in:
Hi @Jon
I'm unsure if the system has a general analytical solution.
In Pure Math, from the properties of a stable 2nd-order ODE, if , and is monotonically increasing, then the system will converge to the origin. Perhaps, the analytical solution exists for a certain type of .
Here is a simple demonstration using 3 different monotonically increasing functions of :
Definition of state variables:
tspan = linspace(0, 20, 2001);
x0 = [1 0];
[t, x] = ode45(@odefcn, tspan, x0);
% Solution Plot
plot(t, x(:,1), 'linewidth', 1.5, 'color', '#528AFA'),
grid on, xlabel('t'), ylabel('x_{1}')
% Phase Portrait
plot(x(:,1), x(:,2), 'linewidth', 1.5, 'color', '#FA477A'),
grid on, xlabel('x_{1}'), ylabel('x_{2}')
function xdot = odefcn(t, x)
ft1 = tanh(t);
ft2 = t;
ft3 = t^3;
gamma = 1; % gamma > 0
xdot = zeros(2, 1);
xdot(1) = ft3*x(2); % <-- change to ft1, or ft2
xdot(2) = - ft3*x(1) - gamma*x(2); % <-- change to ft1, or ft2
end
  2 个评论
Sam Chak
Sam Chak 2023-4-22
Ran in:
Hi @Jon, The trivial solution would be having as a constant function.
syms y(t) x(t) gamma
sym('gamma', 'real');
assume(t > 0)
f(t) = sign(t);
eqns = [diff(y,t) == f(t)*x,
diff(x,t) == - f(t)*y - gamma*x];
sol = dsolve(eqns);
display(sol.y)
ans = 
However, WolframAlpha shows that analytical solutions (click on the ODEs) exist for
and
.
Sam Chak
Sam Chak 2023-4-22
编辑：Sam Chak 2023-4-22
Ran in:
Hi @Jon, only certain functions of have analytical solutions.
Analytical solution:
can also be expressed as
.
Click on the 2nd-order ODE to see the analytical solution and plots on WolframAlpha.
Note: By the way, this ODE can be rewritten in Sturm–Liouville form. Perhaps, if your problem can be transformed to become a Sturm–Liouville problem, then the analytical solution exists.
syms y(t) gamma
gamma = 2;
Dy = diff(y,t);
eqn = diff(y,t,2) + (gamma - 1)*Dy + exp(2*t)*y == 0;
cond = [y(0)==1, Dy(0)==0];
ySol(t) = dsolve(eqn, cond)
ySol(t) = 
Numerical solution:
tspan = linspace(0, 10, 1001);
x0 = [1 0];
[t, x] = ode45(@odefcn, tspan, x0);
% Solution Plot
plot(t, x(:,1), 'linewidth', 1.5, 'color', '#528AFA'),
grid on, xlabel('t'), ylabel('x_{1}')
% Phase Portrait
plot(x(:,1), exp(t).*x(:,2), 'linewidth', 1.5, 'color', '#FA477A'),
ylim([-2 2]), grid on,
xlabel({'$y_{t}$'}, 'interpreter', 'latex', 'fontsize', 16),
ylabel({'$\dot{y}_{t}$'}, 'interpreter', 'latex', 'fontsize', 16)
function xdot = odefcn(t, x)
ft = exp(t);
gamma = 2; % gamma > 0
xdot = zeros(2, 1);
xdot(1) = ft*x(2); % <-- change to ft1, or ft2
xdot(2) = - ft*x(1) - gamma*x(2); % <-- change to ft1, or ft2
end

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