# College Math Teaching

## October 1, 2014

### Osgood’s uniqueness theorem for differential equations

I am teaching a numerical analysis class this semester and we just started the section on differential equations. I want them to understand when we can expect to have a solution and when a solution satisfying a given initial condition is going to be unique.

We had gone over the “existence” theorem, which basically says: given $y' = f(x,y)$ and initial condition $y(x_0) = y_0$ where $(x_0,y_0) \in int(R)$ where $R$ is some rectangle in the $x,y$ plane, if $f(x,y)$ is a continuous function over $R$, then we are guaranteed to have at least one solution to the differential equation which is guaranteed to be valid so long as $(x, y(x)$ stays in $R$.

I might post a proof of this theorem later; however an outline of how a proof goes will be useful here. With no loss of generality, assume that $x_0 = 0$ and the rectangle has the lines $x = -a, x = a$ as vertical boundaries. Let $\phi_0 = f(0, y_0)x$, the line of slope $f(0, y_0)$. Now partition the interval $[-a, a]$ into $-a, -\frac{a}{2}, 0, \frac{a}{2}, a$ and create a polygonal path as follows: use slope $f(0, y_0)$ at $(0, y_0)$, slope $f(\frac{a}{2}, y_0 + \frac{a}{2}f(0, y_0))$ at $(\frac{a}{2}, y_0 + \frac{a}{2}f(0, y_0))$ and so on to the right; reverse this process going left. The idea: we are using Euler’s differential equation approximation method to obtain an initial piecewise approximation. Then do this again for step size $\frac{a}{4},$

In this way, we obtain an infinite family of continuous approximation curves. Because $f(x,y)$ is continuous over $R$, it is also bounded, hence the curves have slopes whose magnitude are bounded by some $M$. Hence this family is equicontinuous (for any given $\epsilon$ one can use $\delta = \frac{\epsilon}{M}$ in continuity arguments, no matter which curve in the family we are talking about. Of course, these curves are uniformly bounded, hence by the Arzela-Ascoli Theorem (not difficult) we can extract a subsequence of these curves which converges to a limit function.

Seeing that this limit function satisfies the differential equation isn’t that hard; if one chooses $t, s \in (-a.a)$ close enough, one shows that $| \frac{\phi_k(t) - \phi_k(s)}{(t-s)} - f(t, \phi(t))| 0$ where $|f(x,y_1)-f(x,y_2)| \le K|y_1-y_2|$ then the differential equation $y'=f(x,y)$ has exactly one solution where $\phi(0) = y_0$ which is valid so long as the graph $(x, \phi(x) )$ remains in $R$.

Here is the proof: $K > 0$ where $|f(x,y_1)-f(x,y_2)| \le K|y_1-y_2| < 2K|y_1-y_2|$. This is clear but perhaps a strange step.
But now suppose that there are two solutions, say $y_1(x)$ and $y_2(x)$ where $y_1(0) = y_2(0)$. So set $z(x) = y_1(x) -y_2(x)$ and note the following: $z'(x) = y_1(x) - y_2(x) = f(x,y_1)-f(x,y_2)$ and $|z'(x)| = |f(x,y_1)-f(x,y_2)| 0$. A Mean Value Theorem argument applied to $z$ means that we can assume that we can select our $x_1$ so that $z' > 0$ on that interval (since $z(0) = 0$).

So, on this selected interval about $x_1$ we have $z'(x) < 2Kz$ (we can remove the absolute value signs.).

Now we set up the differential equation: $Y' = 2KY, Y(x_1) = z(x_1)$ which has a unique solution $Y=z(x_1)e^{2K(x-x_1)}$ whose graph is always positive; $Y(0) = z(x_1)e^{-2Kx_1}$. Note that the graphs of $z(x), Y(x)$ meet at $(x_1, z(x_1))$. But $z'(x) 0$ where $z(x_1 - \delta) > Y(x_1 - \delta)$.

But since $z(0) = 0 z'(x)$ on that interval.

So, no such point $x_1$ can exist.

Note that we used the fact that the solution to $Y' = 2KY, Y(x_1) > 0$ is always positive. Though this is an easy differential equation to solve, note the key fact that if we tried to separate the variables, we’d calculate $\int_0^y \frac{1}{Kt} dt$ and find that this is an improper integral which diverges to positive $\infty$ hence its primitive cannot change sign nor reach zero. So, if we had $Y' =2g(Y)$ where $\int_0^y \frac{1}{g(t)} dt$ is an infinite improper integral and $g(t) > 0$, we would get exactly the same result for exactly the same reason.

Hence we can recover Osgood’s Uniqueness Theorem which states:

If $f(x,y)$ is continuous on $R$ and for all $(x, y_1), (x, y_2) \in R$ we have a $K > 0$ where $|f(x,y_1)-f(x,y_2)| \le g(|y_1-y_2|)$ where $g$ is a positive function and $\int_0^y \frac{1}{g(t)} dt$ diverges to $\infty$ at $y=0$ then the differential equation $y'=f(x,y)$ has exactly one solution where $\phi(0) = y_0$ which is valid so long as the graph $(x, \phi(x) )$ remains in $R$.