Register to reply

Uniqueness Theorem

by JaredPM
Tags: theorem, uniqueness
Share this thread:
Dec24-12, 08:51 PM
P: 15
Can someone give me a qualitative example of the uniqueness theorem of a first order linear differential equation? I have read the definition, but I am not 100% positive of what it means in regards to an initial value problem.
Im confused about what a unique solution is when/if you change the initial value. Would it not have many different solutions?
Phys.Org News Partner Science news on
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
Dec25-12, 07:20 AM
Sci Advisor
PF Gold
P: 39,317
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 [itex](x_0, y_0)[/itex] then the differential equation dy/dx= f(x,y) with initial value [itex]y(x_0)= y_0[/itex] has a unique solution in some neighborhood of [itex]x_0[/itex]".

(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 [itex]d^2y/dx^2= f(x, y, dy/dx)[/itex] 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 [itex]V= <y, u>[/itex] so that [itex]dV/dx= <dy/dx, du/dx>= <u, f(x,y,u)>[/itex]. Of course, we now need a condition of the form [itex]V(x_0)= <y(x_0), u(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[/itex], not two different values.

For example, the very simple equation [itex]d^2y/dx^2+ y= 0[/itex] with the boundary values y(0)= 0, [itex]y'(\pi/2)= 0[/itex] 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.

Register to reply

Related Discussions
Uniqueness theorem Advanced Physics Homework 1
Existence and Uniqueness Theorem Differential Equations 18
Existence and Uniqueness Theorem Calculus & Beyond Homework 0
Uniqueness theorem Calculus 2
Second uniqueness theorem Classical Physics 5