# Uniqueness Theorem

by JaredPM
Tags: theorem, uniqueness
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Yes, of course if you change the initial conditions, you will have different solutions! What the "existance and uniqueness" theorem for initial value problems says is that a given (well behaved) differential equation, with specific initial conditions will have a unique solution. Specifically, the basic "existance and uniqueness theorem" for first order equations, as given in most introductory texts, says "If f(x, y) is continuous in x and y and "Lipschitz" in y in some neighborhood of $(x_0, y_0)$ then the differential equation dy/dx= f(x,y) with initial value $y(x_0)= y_0$ has a unique solution in some neighborhood of $x_0$". (A function, f(x), is said to be "Lipschitz" in x on a neighborhood if there exists some constant C so that |f(x)- f(y)|< C|x- y| for all x and y in that neighborhood. One can show that all functions that are differentiable in a given neighborhood are Lipschitz there so many introductory texts use "differentiable" as a sufficient but not necessary condition.) We can extend that to higher order equations, for example $d^2y/dx^2= f(x, y, dy/dx)$ by letting u= dy/dx and writing the single equation as two first order equations, dy/dx= u and du/dx= f(x, y, u). We can then represent those equations as a single first order vector equation by taking $V=$ so that $dV/dx= =$. Of course, we now need a condition of the form $V(x_0)= [itex] is given which means that we must be given values of y and its derivative at the same value of [itex]x_0$, not two different values. For example, the very simple equation $d^2y/dx^2+ y= 0$ with the boundary values y(0)= 0, $y'(\pi/2)= 0$ does NOT have a unique solution. Again, the basic "existance and uniqueness theorem" for intial value problems does NOT say that there exist a unique solution to a differential equation that will work for any initial conditions. It says that there exists a unique solution that will match specific given initial conditions.