# Unique soultion of a set of nonlinear differential equations.

1. Jun 1, 2008

### arroy_0205

Is there any theorem or result which tells us whether a given set of nonlinar coupled differential equation (ordinary/partial) will have unique solution set? I need to know the answer for a second order ODE set. I understand there may be some difficulty since in this case the integration constants may come in such a way that a complete answer to this question may not be possible, but how far can one go?

2. Jun 1, 2008

### HallsofIvy

The solution to a given set of differential equations is NEVER unique. The solution to a set of differential equations, satisfying a given additional conditions may be unique.

In particular, the usual "existence and uniqueness" theorem is this: If (t0,X0) is a point in Rn+1 (that is, t0 is a real number, X0 is a vector in Rn) and, in some neighborhood of (t0,X0), F(t, X), where t is a real number and X is a function with values in Rn, is continuous in both variables and satisfies a "Lischitz condition" (see below) in X, then there exist a unique solution to dX/dt= F(t,X) in some neighborhood of (t0, X0). Since X(t) is in Rn so is dX/dt and so must be F(t,X). If you write each component of F(t,X) as fn(t, X), you have the system of equations dXn/dt= fn(t, x1, x2, ....xn). Requiring that they satisfy X(t0)= X0 means that the values of all the xn must be given at the same Xn, what is called an "initial value problem".

Second or higher order problems can be handled in the same way by defining xn+1[/sup]= dx1/dt, xn+2= dx2/dt, etc. so that each second derivative becomes a the first derivative of a new variable and you have a 2n first order equations.

If you are not given all the values at the same t0 (i.e. not an "initial value problem") then the question is much harder.

For example, the differential equation $d^2x/dt^2= -x$ has the general solution x(t)= C1cos(t)+ C2sin(t) and it is easy to find a unique solution for the initial value problem y(0)= A, y'(0)= B for any numbers A or B. On the other hand, there exist an infinite number of solutions to that equation that satisfy y(0)= 0, y($\pi$)= 0 while there is NO solution to that equation satisfying y(0)= 0, y($\pi$)= 1.

3. Jun 2, 2008