# Using the uniqueness theorem for differential equations without initial value?

The uniqueness theorem I am referring to is this one:

Let $f(x,y)$ be a real valued function which is continuous on the rectangle $R = \{(x,y); \lvert x - x_0 \rvert \leq a, \lvert y - y_0 \rvert \leq b \}$. Assume $f$ has a partial derivative with respect to $y$ and that $\frac{\partial f}{\partial y}$ is also continuous on the rectangle $R$. Then there exists an interval $I = [x_0 - h, x_0 + h]$ (with $h \leq a$) such that the initial value problem $y^\prime = f(x,y)$, $y(x_0) = y_0$ has a unique solution $y(x)$ defined on the interval $I$.

Suppose we are considering the differential equation $\frac{dh}{dt} = -h^{1/3}$. I want to determine if, for any $h(x_0) = y_0$ and $h(x_1) = y_1$, with $x_0, x_1, y_0, y_1$ all being positive, if the solutions intersect each other. I know that $h^{1/3}$ and its partial derivative with respect to $h$ are not continuous at $h = 0$. I am confused about how to approach questions like this using the uniqueness theorem, since the uniqueness theorem refers to a given initial value and the uniqueness of the solution that corresponds to the initial value. I am not sure how to use the theorem to compare solutions with different initial values. I have thought about the problem, but all I have is that the solutions will intersect the equilibrium solution $h = 0$ at some point, which is obvious from the solution to the differential equation.

• You want to determine whether any two solutions with all positive initial conditions are guaranteed to intersect? The uniqueness theorem won't help you with that (especially since this equation doesn't even have unique solutions to its IVPs). – Ian Oct 13 '17 at 3:08
• Not sure what you mean Ian, that equation is separable. If two solutions intersected than that would mean that the solution wouldn’t be unique for that initial condition. – DaveNine Oct 13 '17 at 4:23
• OP you have shown that when $h(t)=0$, you are not guaranteed uniqueness for any rectangle there. however for any other rectangle where $h(t) \neq 0$, you have uniqueness, and so this guarantees that no two solutions may intersect. Imagine if two different solutions had the same initial condition — your uniqueness theorem guarantees that they really are the same. – DaveNine Oct 13 '17 at 4:28
• @DaveNine The equation does not have unique solutions. Separation of variables breaks down as a result of divisions by zero. In this case what can happen is that the solution can remain zero for an arbitrarily long period of time before abruptly "switching back" to a $\frac{-2}{3} t^{3/2}$ type behavior. That said, you are right that if there were uniqueness then solutions can't cross...except that the initial times aren't the same, which needs to be taken into account. – Ian Oct 13 '17 at 12:49
• My mistake, you’re definitely right about the solutions. – DaveNine Oct 13 '17 at 14:11