Finding behaviour of ODE solutions without solving for them

  • Thread starter tun
  • Start date
  • Tags
    Ode
In summary: Your equation is nonlinear and nonautonomous. This makes it difficult to solve explicitly. However, it is possible to deduce an approximate formula for the solution or some more information about it.
  • #1
tun
6
0
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
 
Physics news on Phys.org
  • #2
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
Thanks for replying. I understand the example that you gave, but unfortunately mine is more complicated! My equation is:
[tex]\ddot{\psi} + 4g_{2}\dot{\psi} + 4g_{1}\cos(t)\sin(\psi) = 0[/tex]
I can write this as two first order equations for [tex]\psi_1 = \psi [/tex] and [tex]\psi_2 = \dot{\psi}[/tex]:
[tex] \dot{\psi}_1 = \psi_2 [/tex]
[tex] \dot{\psi}_2 = -4[g_{2}\dot{\psi}_2 + g_{1}\cos(t)\sin(\psi_1)] [/tex]

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
 

1. What is the purpose of finding the behavior of ODE solutions without solving for them?

Finding the behavior of ODE solutions without solving for them allows us to understand the general trends and patterns of the solution without having to go through the tedious process of solving the equation. This can save time and resources, and provide valuable insights into the underlying dynamics of the system.

2. How can we determine the behavior of ODE solutions without solving for them?

There are several methods for determining the behavior of ODE solutions without solving for them. These include graphical analysis, numerical methods such as Euler's method or Runge-Kutta methods, and analytical techniques such as stability analysis and phase plane analysis.

3. Can we accurately predict the behavior of ODE solutions without solving for them?

While it is not always possible to accurately predict the exact behavior of ODE solutions without solving for them, these methods can provide a good approximation and give us a general understanding of the solution's behavior. The accuracy of the predictions will depend on the complexity of the system and the chosen method of analysis.

4. What are some practical applications of finding the behavior of ODE solutions without solving for them?

The ability to determine the behavior of ODE solutions without solving for them has many practical applications in various fields such as physics, engineering, biology, and economics. It can help in predicting the behavior of physical systems, designing control systems, and understanding the dynamics of biological processes, among others.

5. Are there any limitations to finding the behavior of ODE solutions without solving for them?

One limitation of this approach is that it may not always provide an accurate representation of the solution, especially for highly nonlinear systems. Additionally, it may be challenging to analyze systems with multiple variables and parameters. Also, the accuracy of the predictions may decrease as the time frame of the solution increases.

Similar threads

  • Differential Equations
Replies
16
Views
862
Replies
3
Views
774
  • Differential Equations
Replies
7
Views
2K
Replies
2
Views
2K
  • Differential Equations
Replies
6
Views
2K
  • Differential Equations
Replies
8
Views
2K
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
9
Views
2K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
4
Views
2K
Back
Top