How to obtain all solutions of separable differential equation where the existence and uniqueness theorem does not apply?

Question

From blackpenredpen's February 2017 video on the existence & uniqueness theorem, the separable differential equation $$\frac{{\rm d} y}{{\rm d}x}= x\sqrt{y-3} \tag{1}$$ with the initial condition $y(4)=3$ has the only solutions $$y=\left(\frac{x^2}{4}-4\right)^2+3 \tag{2}$$

$$y(x)=3 \tag{3}$$ where the valid interval can be the whole real line. The solution is obtained by dividing both sides of $(1)$ by $\sqrt{y-3}$ and then taking the integral of both sides.

For me, it seems that it has not been proven that there aren't any more solutions $y$ to $(1)$ such that there exists a $x$ for which $y(x)=3$. I'm thinking when integrating both sides we first assume that $y(x) \neq 3$ so we can divide both sides. When this happens, we can not say we have covered all solutions $y$ of $(1)$ such that there is a $x$ for which $y(x)=3$. This is evident from $(3)$. I'm assuming obtaining $(2)$ was just a happy accident.

In short the proof has 2 cases. First case assumes $y\neq3$ for all $x$ and covers all those solutions and the second case is the constant solution $y=3$. I'm asking where the justification is that we don't need a third case for the solutions $y$ such that only for some $x$ in the valid interval is $y(x)=3$.

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_{Background: Solving separable differential equations}

Answer 1

Let $x_0 := 4$. Consider the following continuous and differentiable solution

$$ y (x) = \begin{cases} 3 & \text{if } x_0 \leq x < x_1\\ 3 + \left( \dfrac{x^2 - x_1^2}{4} \right)^2 & \text{if } x \geq x_1\end{cases} $$

where $x_1 > x_0$. Since there are infinitely many choices for $x_1$, there are infinitely many solutions.

Answer 2

Here is a slightly more detailed answer; I hope it'll be satisfactory for Rodrigo de Azevedo also. I will skip most of the calculations as they are standard and algebraic and they are not that different from what has been discussed already.

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The main idea is to first consider the ODE by itself, make full use of the Existence and Uniqueness Theorem where it's available, and finally realize that the initial conditions provided are edge cases. Why uniqueness is broken is clear: at the initial condition the RHS is only Hölder continuous in the space variable $y$, and so the standard cutting and pasting argument of Rodrigo de Azevedo becomes available.

First put $V:\mathbb{R}\times[3,\infty[\to \mathbb{R},\,\,V(x,y)=x\sqrt{y-3}$ and note that $V$ is $C^0$ on its domain of definition and $C^1$ (in fact real analytic) in the interior $\mathbb{R}\times]3,\infty[$ . Thus we have the Existence and Uniqueness Theorem available in the interior. To state it more explicitly (see also The proof of the fact that $[v,w]=0\;\Rightarrow\;\exp(\varepsilon v) \exp(\theta w)x=\exp(\theta w)\exp(\varepsilon v)x $), define for any $(x_0,y_0)\in \mathbb{R}\times ]3,\infty[$,

$$ J(x_0,y_0)=\begin{cases}\mathbb{R}&\text{, if }x_0^2-4\sqrt{y_0-3}<0\\ \left]\sqrt{x_0^2-4\sqrt{y_0-3}},\infty\right[&\text{, if }x_0^2-4\sqrt{y_0-3}\geq0\text{ and } x_0>0\\ \left]-\infty,-\sqrt{x_0^2-4\sqrt{y_0-3}}\right[&\text{, if }x_0^2-4\sqrt{y_0-3}\geq0\text{ and } x_0<0 \end{cases}.$$

Then by the Existence and Uniqueness Theorem (and some algebra) for any $(x_0,y_0)\in \mathbb{R}\times ]3,\infty[$,

$$y(\bullet;x_0,y_0): J(x_0,y_0)\to\mathbb{R},\,\, x\mapsto \left(\dfrac{x^2}{4}+\dfrac{C(x_0,y_0)}{2}\right)^2+3,$$

where $C(x_0,y_0)=2\sqrt{y_0-3}-\dfrac{x_0^2}{2}$, is the unique (polynomial) solution with maximal time interval to the IVP

$$\dfrac{dy}{dx}=V(x,y),\,\, y(x_0)=y_0.$$

Further, whenever the interval of definition $J(x_0,y_0)$ is not the whole real line, $y'(x;x_0,y_0)=0$ for $x$ the appropriate boundary point.

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Here is a humble graph to go along with this statement: https://www.desmos.com/calculator/zbgizhnfta . It seems to me going forward the graph will be much more beneficial compared to algebra. We claim that Rodrigo de Azevedo's list (together with the solutions mentioned in the OP; note that one can concatenate two unique nonconstant solutions to either end of a constant solution, considering backwards time as well) is the complete list of the $C^1$ ($\dagger$) solutions to the IVP. Indeed, by the vanishing of the derivative at the boundary concatenating a unique solution starting in $\mathbb{R}\times ]3,\infty[$ and the constant solution will produce a $C^1$ solution to the IVP with $(x_0,y_0)=(4,3)$. If $y$ were another solution to the IVP with $(x_0,y_0)=(4,3)$ that takes a value more than $3$ for some time $x_1$ (note that this means that its graph has to be in the purple region in the graph), then in finite time it would have to take the value $3$, beyond which time until $4$ it would not be able to take any other value. (Thus we have a uniqueness statement after all: for any two pairs of points, one on each side of the purple region in the graph, the IVP has a unique solution that passes through those points.)

($\dagger$) Similarly looking into higher regularity one sees that the concatenated solutions are not $C^2$ (though for instance for the initial condition $(x_0,y_0)=(0,3)$ there are exactly two extra concatenated solutions, one for either side of the boundary component of the purple region). In this sense, the list in the OP is the comprehensive list of $C^2$ solutions for the IVP at hand.