How to solve $\frac{dy}{dx}=xy^{1/2}, y(0)=0$?

Question

On page 48 of [Zill2022], there is an initial-value problem (IVP): $$ \frac{dy}{dx}=xy^{1/2}, y(0)=0. \tag{1} $$ I try to solve this IVP by myself. My solution is as follows:

Solution: First, it is easily seen that $y = 0$ is a solution to (1). Second, separating variables gives $$ y^{-1/2} dy = x dx. \tag{2} $$ Integrating (2), we have $$ \int y^{-1/2} dy = \int x dx, \tag{3} $$ then $$ 2 y^{1/2} = \frac{x^2}{2} + c. \tag{4} $$ Applying the initial condition $y(0) = 0$, we obtain $c = 0$. Then $$ 2 y^{1/2} = \frac{x^2}{2} \tag{5} $$ which implies $$ y = \frac{x^4}{16}. \tag{6} $$ Therefore, IVP (1) has two solutions: $y=0$ and $y = \frac{x^4}{16}$. $\blacksquare$

I thought I had found all the solutions. But [Zill2022] says:

The initial-value problem (1) actually possesses many more solutions, since for any choice of the parameter $a \geq 0$ the piecewise-defined function $$ y = \begin{cases} 0, & x < a \\ \frac{1}{16} (x^2 - a^2)^2, & x \geq a \end{cases} \tag{7}$$ satisfies both the differential equation and the initial condition.

Thus, my question is:

Question: How to solve (1) to get the solution in (7)?

Any comments and answers are welcome. Thanks in advance.

Reference

[Zill2022] D. G. Zill, Advanced Engineering Mathematics, Seventh Edition, International Student Edition, Jones & Bartlett Learning, 2022.

Answer 1

A simpler question that illustrates the same concept is $y'=y^{1/2},y(0)=0$. This has a "trivial solution" $y \equiv 0$ and a "separable solution" $y=\begin{cases} x^2/4 & x>0 \\ 0 & \text{otherwise} \end{cases}$. But this problem is autonomous, so there is nothing special about it starting to increase at $0$ instead of somewhere else. You can just run the trivial solution up to some $a \geq 0$ and then view that $a$ like how the separable solution views $0$, resulting in $y=\begin{cases} (x-a)^2/4 & x>a \\ 0 & \text{otherwise} \end{cases}$ for any $a \geq 0$.

Your problem is basically the same, except that it isn't autonomous, so the various separable solutions that start increasing at various different $a$'s are not simply translates of one another. Nevertheless the idea is the same: you can run the trivial solution up to whatever $a \geq 0$ you like and then follow the separable solution starting at that $a$.

Getting to avoid doing this kind of analysis is why we like having an existence/uniqueness theorem to lean on. Picard-Lindelof is the usual one, but it requires the equation to be of the form $y'=f(x,y)$ where $f$ is locally Lipschitz in $y$. Your equation fails this condition ($f$ becomes too sensitive to changes in $y$ near $y=0$), which means Picard-Lindelof makes no guarantees of uniqueness.

Answer 2

By integration assuming either $y=0$ or $y\ne0$, the general solution is

$$y=c_0\text{ or }\sqrt y=\frac{x^2}4+c_1.$$

The second part requires $\dfrac{x^2}4+c_1\ge0$, or $x^2\ge-4c_1$.

So even if $y(0)=0$, a nonzero solution remains possible in the positive $x$, with some negative $c_1$.

Due to the singular nature of the equation, the nonzero solution does not depend on the initial condition.

Answer 3

If $f(x,y)$ is Lipschitz with respect to $y$ in the interior of a domain $D\subset\mathbb{R}^{n+1}$, the solution of the ODE $$ y'=f(x,y),y(x_0)=y_0$$ for $(x_0,y_0)\in D$ is unique (see https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem). In your case, $f(x,y)=xy^\frac12$ is not Lipschitz near $(0,0)$ and hence it is not guaranteed that it has a unique solution.

Update: You already give two solutions $y\equiv0$ and $y=\frac1{16}x^2$. To seek other solutions, first solving the following IVP $$ y'=xy^\frac12, y(a)=0$$ for $a>0$ gives $y=\frac1{16}(x^2-a^2)^2$. Define the piecewise function $$y=\begin{cases} \frac1{16}(x^2-a^2)^2 & x>a, \\ 0 & \text{otherwise}. \end{cases}$$ It is easy to see that $y(0)=0$ and $$ \lim_{x\to a^-}y'(x)=\lim_{x\to a^+}y'(x)=0$$ which means that the piecewise function is differentiable. So the IVP has infinite many solutions.

Answer 4

Square equation (4) to get:

$$4y = \frac{1}{4}x^4 + cx^2 + c^2$$ $$y = \frac{1}{16}\left(x^4 + 4cx^2 + 4c^2\right)$$ $$y = \frac{1}{16}\left(x^2 + 2c\right)^2$$

But since $c$ is an arbitrary constant, we can replace it with a different constant expression, so let $c = -\frac{a^2}{2}$.

$$y = \frac{1}{16}\left(x^2 - a^2\right)^2 \tag{1}$$

It can easily be shown that:

$$\frac{dy}{dx} = \frac{1}{4}x(x^2 - a^2)$$ $$xy^{1/2} = \frac{1}{4}x(x^2 - a^2)$$

And thus the differential equation $\frac{dy}{dx} = xy^{1/2}$ is satisfied. But, we haven't done anything with the initial value $y(0) = 0$. The “natural” thing to do is to just plug $x = 0$ and $y = 0$ into equation (1), giving: $0 = \frac{1}{16}\left(0-a^2\right)^2$, from which $a = 0$. This gives the solution $y = \frac{1}{16}x^4$, as you found.

But what if we “cheat” by just defining $y = 0$ on some interval near $x = 0$? Pick some $b > 0$ and let:

$$y(x) = \begin{cases} 0 & x < b \\ \frac{1}{16}(x^2 - a^2)^2 & x > b \end{cases}$$

Then this trivially satisfies the condition $y(0) = 0$. And both cases satisfy the differential equation (without the initial condition). All we need to do is ensure that the equation is still satisfied at $x = b$.

In order for $y$ to be differentiable at $x = b$, it must be continuous, so

$$0 = \frac{1}{16}(b^2 - a^2)^2$$ $$0 = \frac{1}{16}(b + a)^2(b - a)^2$$ $$b = \pm a$$

So, just choose $b = a$, giving:

$$y(x) = \begin{cases} 0 & x \le a \\ \frac{1}{16}(x^2 - a^2)^2 & x \ge a \end{cases}$$

And then $y'(a) = 0$, thus satisfying $\frac{dy}{dx}=xy^{1/2}$ with $0 = a\sqrt{0}$ at the point $(a, 0)$.

So that's how you get the solution in Zill's equation (7).

Yeah, it's contrary to the “normal” approach, but that's kind of the point. There's technically no constraint in the problem that $y$ can't be piecewise-defined, and $xy^{1/2}$ breaks the separable equation assumption of being Lipschitz near $y = 0$. Hence this “extra” class of solutions that you didn't consider.

Answer 5

Apparently $y\equiv 0$ is a solution to the IVP. If $y(x_0)\not=0$ (hence $y(x_0)>0$ due to the presence of $y^{1/2}$ in the ODE) for some $x_0>0$, then by differentiability hence continuity of $y$, $y$ is nonzero around $x_0$, then the derivation (2)-(4) are completely justified. And note that since $y'=xy^{1/2}\ge0$ for $x\ge 0$, $y$ is increasing, hence $y(x)>0$ for all $x\ge x_0$. This means that once (2)-(4) kicks in, they will always be valid.

This proves that the solution over the interval $[0, \infty)$ can be separated into two pieces, one is just constantly $0$ and the other is of the form (4). But their connection has to be continuous and this will give us (7), with an arbitrary separation point $a$. $^{[1]}$

The same argument works for $(-\infty, 0]$, so the more general solution is $$y=\begin{cases} \frac{1}{16}(x^2-b^2)^2 & x<-b \\ 0 & -b\le x \le a \\ \frac{1}{16}(x^2-a^2)^2 & x>a \end{cases}$$

where $a,b$ are non-negative and allowed to be $\infty$.

That said, these are examples that are designed to show IVPs doesn't necessarily have unique solutions without further constraints. They are not very relevant in applications (though a more knowledgeable person would almost certainly disagree). In reality, the zero solution is apparently very unstable hence coudln't last for long, i.e. ubiquitious random noise can easily push the system to evolve (diverge to infinity).

---

$^{[1]}$ To add a bit more details, multiply $2$ on both sides of (4), replace $2c$ by $c$, we have $4y^{1/2} = x^2+c$, and $y=\frac{1}{16}(x^2+c)^2$. Assume $a$ is the separation point between the zero solution and the above solution, then $$y(a)=\lim_{x\rightarrow a^-} y(x)=0$$ $$y(a)=\lim_{x\rightarrow a^+}y(x)=\lim_{x\rightarrow a^+}\frac{1}{16}(x^2+c)^2=\frac{1}{16}(a^2+c)^2$$ hence $$c=-a^2$$

Similar analysis can be done for $x\le 0$.

Answer 6

For intuition's sake, consider $x$ as a "time" variable and let's look at how the function changes as time increases from $0$. There are essentially two cases to consider:

1. When $y=0$, then $\frac{dy}{dx}=0$. Hence, the function stays flat whenever it is on the $x$-axis.
2. When $y>0$, then $\frac{dy}{dx}>0$ (when $x>0$), so the function must strictly increase.

We can now think about the problem as follows: at time $x=0$, the function is on the $x$-axis, so it stays flat. But it's also possible at some point in time the function curves up away from $0$ (while maintaining a derivative of $0$ at its last point on the $x$-axis, which works as long as the derivative is continuous). Once this happens, the derivative $\frac{dy}{dx}>0$, and the function continually increases.

When $y(x)\not\equiv 0$, we can set $x=a$ as our "last point" that lies on the $x$-axis (or technically speaking, the supremum of the $x$-values of all points that lie on the $x$-axis). Once the function goes past $a$, the differential equation becomes "nice"—that is, $xy^{1/2}$ comes Lipschitz continuous, so by Picard-Lindelöf, there is a unique solution now. Your work in steps (2)-(4) apply, so we can conclude that when the differential equation is bounded by $x>a$, there is a unique solution of the form $2y^{1/2}=\frac{x^2}{2}+c$ for some constant $c$. Since $y$ is positive, this can be rewritten as $y=\left(\frac{x^2}{4}+\frac{c}{2}\right)^2$.

But we also know that $y(x)$ must be continuous, so the right limit as $x$ approaches $a$ must be $0$. Hence, $\frac{a^2}{4}+\frac{c}{2}=0\implies c=-\frac{a^2}{2}$. So for $x\geq a$, we have $y=\frac{1}{16}(x^2-a^2)^2$. And in order for $y(a)=0$, all points before $a$ must also lie on the $x$-axis. Hence, this gives

$$y=\begin{cases}0 & \text{if } x<a\\ \frac{1}{16}(x^2-a^2)^2&\text{if }x\geq a\end{cases},$$

as one of the possible families of solutions to this differential equation, which Zill describes. Note that this solution assumes the differential equation is on the interval $[0,\infty)$. We can apply the same logic to the negative case to completely characterize all solutions.