### Author Topic: Can there exists infinite number of solutions given initial conditions.  (Read 969 times)

#### Xinyu Jiao

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##### Can there exists infinite number of solutions given initial conditions.
« 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.
« Last Edit: September 25, 2018, 09:18:16 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Can there exists infinite number of solutions given initial conditions.
« Reply #1 on: September 25, 2018, 09:27:59 PM »
For condition see Section 2.8 of the textbook or this Lecture Note

Yes, this equation $y'=3 y^{2/3}$ (I modified it for simplicity) has a general solution $y=(x-c)^{3}$ 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,\\ &(x-c)^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 $y=0$.

#### Kathryn Bucci

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##### Re: Can there exists infinite number of solutions given initial conditions.
« Reply #2 on: October 06, 2018, 10:59:07 AM »
If 𝑦′=𝑦𝛼 with 0 < 𝛼 < 1 e.g. 𝑦′=3𝑦2/3= 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.

#### Victor Ivrii

Continuity of $\frac{\partial f}{\partial y}$ is not required, but "Hölder  property" $|f(x,y)-f(x,z)|\le M$ is.