# Finding behaviour of ODE solutions without solving for them

1. May 8, 2009

### tun

Hi all,

I am swiftly learning that past the undergraduate level we can't really solve anything explicitly, unless we are really lucky with the problem. In my research I have an ODE which is nonlinear and nonautonomous and is awkward to analyse.

My question is this: given some information about the solution to the ODE, is it possible to deduce an approximate formula for the solution or some more information about it? In my particular case, from simulation I expect the solution to be a periodic function in the asymptotic limit; can I use this to say much more?

Cheers,

Tom

2. May 10, 2009

### HallsofIvy

That depends a lot on the form of the equation.

For example if I have a non-linear differential equation of the form dy/dx= (y-a)(y-b), the first thing I can do is look for the "equilibrium" solutions: the constant solutions. If y is a constant, the dy/dx= 0 so we must have (y-a)(y-b)= 0 which means either y= a or y= b. It is easy to check that y= a (for all x) and y= b (for all x) satisfy the equation.

I also know, from the fundamental existence and uniqueness theorem for intial value problems that two distinct solutions cannot cross. If they did, taking the point at which they cross as intial value, we would have two distinct solutions to an initial valud problem- which cannot happen.

Further, assuming that a< b, suppose for some x0, y(x0)< a< b, then both y- a and y- b are negative so their product, and thus dy/dx is positive: y is increasing. Since y(x0)< a and the solution cannot cross that line, y(x) must approach a in the limit as x goes to infinity and, there being no equilibrium solution less than a, must go to negative infinity as x goes to negative infinity.

If, for some x0, y(x0)> b> a, then both y- a and y- b are postive so their product, and dy/dx is again positive. Now y goes to infinity as x goes to infinity and approaches b as x goes to negative infinity.

If, for some x0, a< y(x0)< b, then x- a is positive and x- b is negative so their product, and dy/dx, is negative. Since y cannot cross y= a or y= b, y must approach a as x goes to infinity and must approach b as x goes to negative infinity.

Finally, notice that if an initial value of y is close to a, whether above or below it, the values of y will go toward a. We say that y= a is a stable equilibrium solution. If an initial value of y is close to b, if above the solution tends to infinity, if below to a, but in any case away from b. We say that y= b is an unstable equilibrium solution.

There's a lot more that can be said for more complicated equations but thinking about "stable" and "unstable" "equilibrium" solutions is a start.

3. May 11, 2009

### tun

Thanks for replying. I understand the example that you gave, but unfortunately mine is more complicated! My equation is:
$$\ddot{\psi} + 4g_{2}\dot{\psi} + 4g_{1}\cos(t)\sin(\psi) = 0$$
I can write this as two first order equations for $$\psi_1 = \psi$$ and $$\psi_2 = \dot{\psi}$$:
$$\dot{\psi}_1 = \psi_2$$
$$\dot{\psi}_2 = -4[g_{2}\dot{\psi}_2 + g_{1}\cos(t)\sin(\psi_1)]$$

then think about the things that you suggested. There are equilibria at psi_2 = 0 and psi_1 = 0, pi, -pi, etc. Then things get a bit tricky due to the cos(t) term.

I can see that, since cos(t)sin(psi) is between -1 and 1, if g2*psi2 is large and negative then psi2 will be increasing, and if g2*psi2 is large and positive then psi2 will be decreasing. However, close to the equilibrium point psi2 = 0 the nonlinear term must become important, so I'm not sure how to assess the stability. Also I take it that this method can't tell you about behaviour when the solution doesn't decay away to a constant?

Cheers,

Tom