Suppose we want to solve the differential equation $y'=x \sqrt{y}$. Easy right? Because you can transform the equation into a separable one. However, I think that there are more than meets the eye. Where are we looking for solutions? $\mathbb{R}$? Or some subset of $\mathbb{R}$? I am relatively new to differential equations, and I think most of them convey a sense of ambiguity. What happens when there exist points in the domain of $y$ such that $y(x)=0$? Oftentimes, we disregard them and proceed to divide by $\sqrt{y}$ to achieve the separable differential equation $\frac{y'}{\sqrt{y}}=x$. This implies $y(x)= (\frac{x^{2}}{4}+c)^{2}$, for some constant $c$. And since we presupposed $y \neq 0$, we investigate if $y=0$ is a possible solution, which it is. My problem with that argument is that we divide $y$ into two cases, either $y$ is $0$ or not. But we never pay attention to functions $y$ that can attain $0$ at single points of $\mathbb{R}$. Furthermore, if we pay attention to the case when $y$ can be $0$ at single points in $\mathbb{R}$, we find a ton of other families of solutions to this equation, namely:
- $y(x)=0$ for all $x \in \mathbb{R}$,
- $y(x)=(\frac{x^{2}}{4}+c)^{2}$ for all $x \in \mathbb{R}$, where $c \geq 0$ is constant,
- $y(x)=0$ for all $x \leq 0$ and $y(x)=\frac{x^{4}}{16}$ for all $x>0$,
- $y(x) = \frac{x^{4}}{16}$ for all $x < 0$ and $y(x) = 0$ for all $x \geq 0$,
- $y(x) = (\frac{x^{2}}{4} +c )^{2}$ for all $|x|>2 \sqrt{-c}$ and $y(x)=0$ for all $|x| \leq 2 \sqrt{-c}$, where $c<0$ is constant,
- $y(x)=0$ for all $x \leq 2 \sqrt{-c}$ and $y(x)=(\frac{x^{2}}{4} +c )^{2}$ for all $x > 2 \sqrt{-c}$, where $c<0$ is constant, and finally,
- $y(x)=(\frac{x^{2}}{4} +c )^{2}$ for all $x < - 2\sqrt{-c}$ and $y(x)=0$ for all $x \geq - 2\sqrt{-c}$, where $c<0$ is constant.
And these are the only solutions, as discussed in my solution below:
Solution. An obvious solution is $y \equiv 0$ (that is $y$ is $0$ everywhere on $\mathbb{R}$). Suppose that $y \not\equiv 0$ is a solution to the differential equation (that being said, $y$ can be $0$ at some points in $\mathbb{R}$ but not everywhere in $\mathbb{R}$). Then $I = \{x \in \mathbb{R}:y(x) \neq 0 \} \neq \emptyset$, and since $y$ is continuous ($y$ is differentiable on $\mathbb{R}$), $I$ must be an open set. We proceed to investigate the behaviour of $y$ on $I$. We are now allowed to rearrange the equation into $\frac{y'}{\sqrt{y}} = x$, that is $\sqrt{y} = \frac{x^{2}}{4} + c$ for some $c \in \mathbb{R}$. This means that $y(x)=(\frac{x^{2}}{4} + c )^{2}$ on $I$. Left is to investigate $y$ on $I^{c}$. The best way to do this is through some casework:
Case 1. Suppose $c>0$. Then for all $x \in I$ we must have that $y(x) \geq c^{2}$. Since $I$ is open, its complement $I^{c}$ must be closed. Suppose for the sake of contradiction that $I^{c} \neq \emptyset$. By definition of closed sets and that $I \neq \emptyset$, we can find $x_{0} \in I^{c}$ such that every neighbourhood of $x_{0}$ contains points of $I$, that is $(x_{0}-\delta,x_{0}+\delta) \cap I \neq \emptyset$ for every $\delta > 0$. This is problematic because this translates into the following: For every $\delta>0$ we can find $x$ (more specifically $x \in I$) such that $|x-x_{0}|<\delta$ but $c^{2} \leq y(x) = |y(x)-y(x_{0})|$, since $y(x_{0})=0$ by definition. This is exactly the Weierstrass definition of discontinuity of $y$ at $x_{0}$. Hence $y$ is not continuous, contradiction. Thus $I^{c}= \emptyset$, implying $I=\mathbb{R}$. Thus $y(x) = (\frac{x^{2}}{4} + c )^{2}$ for all $x \in \mathbb{R}$.
Case 2. Suppose that $c = 0$. For the sake of contradiction, suppose that $y(0) \neq 0$. Then by definition of $I$, $0 \in I$. But then $y(0)=(\frac{0^{2}}{4})^{2}=0$, contradiction. Thus $y(0)=0$. Next we analyse the intervals $(- \infty,0)$ and $(0, \infty)$. Since $I \neq \emptyset$, $y$ can not be $0$ everywhere on both intervals. Wlog. assume that there exist $a \in (0, \infty)$ with $y(a) \neq 0$. If there exists $b \in (0, \infty)$ with $y(b)=0$, consider the interval $[a,b]$ (here we can assume $a < b$, the case when $a>b$ is the same). Then $I^{c} \cap [a,b]$ and $I \cap [a,b]$ are non-empty, and since $I^{c} \cap [a,b]$ is closed (the intersection of closed sets remains closed), there exists $x_{0} \in [a,b]$ such that for every $\delta > 0$, $(x_{0}-\delta, x_{0}+\delta) \cap I \neq \emptyset$. Furthermore, $y$ is bounded on $I \cap [a,b]$ by some positive constant $K$. Ultimately, we end up with the same contradiction we arrived at in Case 1. Thus $y(x) \neq 0$ for all $x>0$, namely $y(x) = (\frac{x^{2}}{4})^{2}=\frac{x^{4}}{16}$ for all $x>0$. In this case, we end up with a total of 3 other solutions: (i)$y(x) = \frac{x^{4}}{16}$ for all $x \in \mathbb{R}$, (ii)$y(x) = 0$ for all $x \leq 0$ and $y(x)=\frac{x^{4}}{16}$ for all $x >0$, and finally (iii)$y(x)=\frac{x^{4}}{16}$ for all $x < 0$ and $y(x) = 0$ for all $x \geq 0$. Note that these are indeed solutions to our original differential equation.
Case 3. Suppose $c<0$. We first study the interval $(- 2\sqrt{-c}, 2 \sqrt{-c})$. If there exist $x_{0}$, with $|x_{0}| < 2 \sqrt{-c}$, then by definition we have $x_{0} \in I$. But then $\frac{x^{2}}{4} + c < 0$, that is $0 \leq \sqrt{y(x_{0})} = \frac{x^{2}}{4} + c < 0$, contradiction! Hence $y$ is $0$ everywhere on $(- 2\sqrt{-c}, 2 \sqrt{-c})$. Since $I^{c}$ is closed we must have that $y$ is $0$ also at the endpoints $x = \pm 2 \sqrt{-c}$. We proceed to study the intervals $(- \infty, - 2 \sqrt{-c})$ and $(2 \sqrt{-c}, \infty)$. This is identical to the study we conducted in Case 2. Consequently, we find three more families of solutions: (i)$y(x)=(\frac{x^{2}}{4}+c)^{2}$ for all $|x| > 2 \sqrt{-c}$ and $y(x)=0$ for all $|x| \leq 2 \sqrt{-c}$, (ii)$y(x) = 0$ for all $x \leq 2 \sqrt{-c}$ and $y(x) = (\frac{x^{2}}{4} + c)^{2}$ for $x > 2 \sqrt{-c}$, and finally (iii)$y(x) = (\frac{x^{2}}{4} + c)^{2}$ for all $x < - 2 \sqrt{-c}$ and $y(x)=0$ for all $x \geq - 2 \sqrt{-c}$. Notice that these are indeed solutions to our original differential equation.
We are done. $\blacksquare$
This boils down to the following question: How do we rigorously attack such a differential equation that involves dividing by functions that could potentially be equal to $0$? Thank you in advance!