Uniqueness Theorem: Qualitative Example of 1st Order Linear DE

[JaredPM]

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?

 

[HallsofIvy]

Yes, of course if you change the initial conditions, you will have different solutions! What the "existence 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 "existence 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= &lt;y, u&gt; so that dV/dx= &lt;dy/dx, du/dx&gt;= &lt;u, f(x,y,u)&gt;. Of course, we now need a condition of the form V(x_0)= &lt;y(x_0), u(x_0)&gt;is given which means that we must be given values of y and its derivative at the &lt;b&gt;same&lt;/b&gt; value of x_0, not two different values.&lt;br /&gt; &lt;br /&gt; For example, the very simple equation d^2y/dx^2+ y= 0 with the &lt;b&gt;boundary&lt;/b&gt; values y(0)= 0, y&amp;amp;#039;(\pi/2)= 0 does NOT have a unique solution.&lt;br /&gt; &lt;br /&gt; Again, the basic &amp;quot;existence and uniqueness theorem&amp;quot; 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 &lt;b&gt;specific&lt;/b&gt; given initial conditions.