Solving coupled differential equations

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)?

[David Goodmanson]

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.

Answer 1

[Torsten]

MATLAB is not able to get a solution. Take the one from WolframAlpha:

https://www.wolframalpha.com/input?i=y%27%28t%29+%3D+-f%28t%29*y%28t%29-gamma*x%28t%29%2C+x%27%28t%29%3Df%28t%29*x%28t%29

[Jon]

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..

Answer 2

[Sam Chak]

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

[Sam Chak]

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]

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