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How many solutions does the differential equation $\frac{dy}{dx}=-\sqrt{y}, x>0$ with the initial condition $y(0)=0$ have? It is clear that zero is one of its solutions. However, when we attempt to solve it, we arrive at $\sqrt{y}=\frac{-x}{2}$ which is not a possible solution. How can we determine the number of solutions for this initial value problem (IVP)? If it has a unique solution, why is that the case? If it has infinitely many solutions, could you please provide at least one non-zero solution? Thank you.

neelkanth
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2 Answers2

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You correctly reason that the only solution is the zero solution. Your question of "why" is not very clear. A priori there is no "reason" that an equation does or does not have a solution - it just does or doesn't.

In this case, though, there is a relatively satisfying answer. Equations of the form $y' = f(y)$ are guaranteed to have a unique solution by the Picard–Lindelöf theorem provided that $f$ is Lipschitz continuous. In the case $f(y) = -\sqrt{y}$, $f$ does not have this property (can you show why?).

Note that in the case $f(y) = \sqrt{y}$, the theorem still doesn't apply, but there are non-zero solutions. So just having the conditions fail does not imply there are no solutions. The reason there are no solutions is the argument you provide in your question.

preferred_anon
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As already mentioned, this ODE doesn't have a unique solution. The step that is needed to quantify the number of solutions, lies in the fact that the equation is autonomous, hence "time invariant". This means that the particular solution $$\begin{equation} y(x) = \begin{cases} \frac{x^2}{2} & \text{if $x>0$}\\ 0 & \text{if $x\le0$} \end{cases} \end{equation}$$

is valid in a more general setting: $$\begin{equation} y(x) = \begin{cases} \frac{(x-\chi)^2}{2} & \text{if $x>\chi$}\\ 0 & \text{if $x\le\chi$} \end{cases} \end{equation}$$

From the information provided in the problem, so only the initial value, we cannot determine the value of $\chi$. This means that the equation has infinitely many solutions!

To give you more insights about this equation, if we consider $x$ to be the time $t$, we can say that there could be more than one future since we can't predict the time at which "the function changes". Indeed, with the equation $y''=\sqrt{y}$ (note that it's now 2nd order) we can actually show that Newtonian mechanics is non-deterministic. If you're interested, read about Norton's dome.

Sandro
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  • is it satisfying given Differential equation? – neelkanth Jun 28 '23 at 17:02
  • Yes, it does. By the way, I put a typo: the solution should be over 4, not over 2. – Sandro Jun 29 '23 at 12:57
  • How can $\frac{x^2}{2}$ satisfies $y'=-\sqrt{y}?$ – neelkanth Jun 30 '23 at 04:05
  • It's a separable ODE, which we can solve with traditional methods. Namely $\frac{dy}{dx}=-\sqrt{y}\implies \int\frac{dy}{\sqrt{y}}=-\int dx\implies 2\sqrt{y}=-x+c\implies \sqrt{y}=\frac{-x+c}{2}\implies y(x)=\frac{(-x+c)^2}{4}$

    Plugging in the initial condition you find $c=0$, so the solution is $y(x)=\frac{x^2}{4}$

     – Sandro Jun 30 '23 at 10:30
  • if $c=0$, then $\sqrt{y}=\frac{-x}{2}$, which is not possible as square root function is non negative. – neelkanth Jun 30 '23 at 10:45
  • I understand your problem and I honestly believe that I cannot provide you with a satisfactory answer. What I can tell you is that if you assume that to be true, and if you consider the solution $y(x) = \frac{x^2}{4}$, then $y'=\frac{x}{2}$. But $\frac{x}{2} = -\left(-\frac{x}{2}\right)=-\sqrt{y}$.

    So $y'\equiv -\sqrt{y}$. This means that the function is a solution to the ODE.

     – Sandro Jun 30 '23 at 11:34