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There are theorems of existence and uniqueness of differential equations.

I was wondering if it is possible that a differential equations has a solution but it is not unique.

mariosangiorgio
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3 Answers3

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Consider for example the equation $x' = 2\sqrt{|x|}$. For every $a$, the function $$ x_a(t) = \begin{cases} 0 & t < a \\ (t-a)^2 & t \ge a \end{cases} $$ is a solution. Note that for $a \ge 0$ all $x_a$ have $x_a(0) = 0$, so they are all solutions to the IVP $x' = 2\sqrt{|x|}, x(0) = 0$ and you usually discuss uniqueness for initial value problems, as otherwise uniqueness will almost never hold ($x' = 0$ has all constants as solutions).

ttq
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martini
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  • Thanks. You mentioned initial value problems. Do they have an unique solution or no solution at all? Or even in that case we may have no solution, an unique solution or multiple solutions? – mariosangiorgio Nov 17 '12 at 15:44
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    @mariosangiorgio In my above answer, I gave an example of an non-unique solvable IVP. – martini Nov 17 '12 at 16:38
  • The definition of the solution on piecewise form should depend on the value of t. – Jorge E. Cardona May 01 '16 at 21:57
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Let your ODE be $y'-x\sqrt{y}=0, \; y(0)=0$. It is not difficult finding its solution on $\mathbb R$. It has at least two solutions as $y=0$ and $y=\frac{x^4}{16}$ passing through the origin. Can you see why the ODE has no unique solution?

Mikasa
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  • Is it because the two solutions shares only the point (0,0) but they do not intersect? – mariosangiorgio Nov 17 '12 at 15:55
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    @mariosangiorgio: Actually no. If we consider the equation as $y'=f(x,y)$ wherein $f(x,y)=xy^{\frac{1}{2}}$, then $\frac{\partial f}{\partial y}=\frac{x}{2y^{\frac{1}{2}}}$ which are both defined just for $y>0$. In fact Picard's theorem doesn't let the equation to have a unique solution in a rectangular region around the origin. – Mikasa Nov 17 '12 at 16:05
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    Thank you for the explanation – mariosangiorgio Nov 17 '12 at 16:22
  • Nice explanation, indeed! ;-) – amWhy Apr 05 '13 at 01:33
  • In fact since f is continuous and the IVP has 2 solutions thus the IVP should have infinite solutions by Kneser's theorem.and why is it that $f(x,y)$ is not defined for $y=0$. It is even continuous at $0$ – Upstart Nov 24 '20 at 13:16
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Since you didn't specify an IVP, it is trivial to find an ODE with non unique solution, such as $y'(x)=0 \implies y(x)=C$, where $C\in\mathbb R$.

jinawee
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