Toronto Math Forum
MAT2442018F => MAT244Lectures & Home Assignments => Topic started by: Xinyu Jiao on September 25, 2018, 08:58:39 PM

In class, we were given an example where a differential equation can have two solutions given some initial condition. Specifically, the equation was $y' = y^\alpha$ with $0<\alpha<1$, and initial condition $y(0) = 0$. This shows that it's not unique, because it does not satisfy some condition which I do not understand.
My question is, can there be a differential equation (of order 1) such that given an initial condition, can acquire an infinite number of solutions? The answer to this question should be able to shed light as to the mechanism through which the equation acquires more than one solution.

For condition see Section 2.8 of the textbook or this Lecture Note (https://q.utoronto.ca/courses/56504/files/1311954?module_item_id=332283)
Yes, this equation $y'=3 y^{2/3}$ (I modified it for simplicity) has a general solution $y=(xc)^{3}$ but also a special solution $y=0$. Thus problem $y'=3 y^{2/3}$, $y(0)=0$ has an infinite number of solutions. Restricting ourselves by $x>0$ we get solutions $y=\left\{\begin{aligned} &0 &&0<x<c,\\ &(xc)^3 && x\ge c\end{aligned}\right.$ with any $c\ge 0$ and similarly for $x< 0$.
This happens because this Lipschitz condition is violated at each point of the solution $y=0$.

If 𝑦′=𝑦^{𝛼} with 0 < 𝛼 < 1 e.g. 𝑦′=3𝑦^{2/3}= f(t,y), then ∂f/∂y=2y^{1/3} is not continuous at (0,0).
According to theorem 2.4.2 (existence and uniqueness for 1st order nonlinear equations), both f and ∂f/∂y have to be continuous on an interval containing the initial point (0,0)  ∂f/∂y is not continuous there so you can't infer that there is a unique solution.

Continuity of $\frac{\partial f}{\partial y}$ is not required, but "Hölder property" $f(x,y)f(x,z)\le M$ is.
There is a notion of the singular solution