Understanding a solution to the differential equation $\frac{dx}{dt} = - \sqrt{x}$

Question

I am trying to understand a solution to the differential equation $$ \frac{dx}{dt} = - \sqrt{x}, \qquad x(0)=1, $$ where $x$ is a function of $t$ for $t \geq 0$. The problem statement claims that there is a unique, non-negative, continuously differentiable solution.

The solution I am reading leads off with what it claims is the unique solution: $$ x(t) = \begin{cases} \frac{(t-2)^2}{4} & \text{ $t \leq 2$} \\ 0 & \text{$t \geq 2$} \end{cases} $$ The first case I believe I understand. But I am unsure on why the solution tends to $0$ as $t \geq 2$.

I tried to solve this using separation of variables. In order to do that, I need to divide by $\sqrt{x}$, so I can't do that when $x = 0$. So I noted that $x = 0$ is an equilibrium solution with $\frac{dx}{dt} < 0$ for every $x > 0$. (I believe this makes it "asymptotically stable," which I think may explain the end behavior of the system.) For $x > 0$, I separate variables and integrate to get $$ x^{1/2} = \frac{-t + C}{2}. $$ I'm not sure at what point I should apply the initial condition or try to restrict $t$. The first thing I notice is that square roots have to be non-negative (in this case positive since I ruled out $x = 0$ above) so I need to have $-t + C > 0$ and therefore $t<C$. By plugging in $x(0) = 1$, I get $C = 2$. I can then square both sides to get the first solution $$ \frac{(-t+2)^2}{4} = \frac{(t-2)^2}{4} $$ for $t < 2$. But at this point, I am unsure of what to do when $t \geq 2$. I can check that the above solution works, but I don't understand how I would come up with this without either guessing it or being told beforehand. I believe it has something to do with the "asymptotically stable" equilibrium solution deduction above, but how would I know for which values of $t$ this applies to?

Answer 1

How to find the solution. Let us try to find solutions of $$\dot{x} = - \sqrt{x}. \tag{1}$$ This implies $$(\dot{x})^2 = x.$$ Differentiating, we obtain $$2 \dot{x}\ddot{x} = \dot{x},$$ so $$\dot{x}\left( 2 \ddot{x} - 1 \right) = 0.$$ for every $t$. It is natural (but it does not exhaust all possibilities) to consider two cases:

Case I. $\dot{x} = 0$ for every $t$. That is, $x(t) = c$ for some constant $c$.
By $(1)$ we conclude that $c = 0$. Thus $x = 0$ is a solution, defined for every $t$.

Case II. $2 \ddot{x} - 1 = 0$ for every $t$. So $x(t) = t^2/4 + bt + c$.

We have the initial condition $x(0) = 1$, and by $(1)$ we get $\dot{x}(0) = -1$.
The solutions of Case II satisfy these initial conditions only when $c = 1$ and $b = -1$, so

$$x(t) = \frac{(t-2)^2}{4}.$$

However, such $x$ satisfies $(1)$ only for $t \leq 2$, since for $t > 2$ we have

$$\frac{t-2}{2} = \dot{x} \neq -\sqrt{\frac{(t-2)^2}{4}} = -\frac{|t-2|}{2} = \frac{2-t}{2}.$$

The solution $x=0$ in Case I does not satisfy the initial conditions, but it coincides with the solution $x(t) = (t-2)^2/4$ at $t=2$.
Thus, if you want a global solution (defined for all $t$), you can glue together these two solutions:

$$ x(t) = \begin{cases} \frac{(t-2)^2}{4}, & t \leq 2, \\ 0, & t \geq 2. \end{cases} $$

Why this solution is unique. Note that local solutions are unique for initial conditions $x(t_0) = c$ with $c > 0$, since $-\sqrt{x}$ is locally Lipschitz near $c$. Indeed this implies $\dot{x}(t_0)=-\sqrt{c}$ and an analysis of Case II show that the unique solution of $(1)$ for this Cauchy problem is

$$x(t) = \frac{(t-t_0-2\sqrt{c})^2}{4},$$ that is a solution for $t< t_0+2\sqrt{c}.$

Thus $x(t) = (t-2)^2/4$ is really the only choice as long as $t < 2$.

To obtain a solution defined for every $t$, we must find another solution of $(1)$ valid for $t \geq 2$ that coincides with $(t-2)^2/4$ at $t=2$, that is, $x(2) = 0$.

We claim that such a solution must be $x=0$.

Added. Eric pointed out in a comment a much faster way to handle this claim, which I reproduce here: if $x(t)$ is a solution, then by $(1)$ we have $\dot{x}(t) \leq 0$ for every $t$. Thus $t\mapsto x(t)$ is a non-increasing function. It follows that $x(t) \leq x(2) = 0$ for $t \geq 2$. On the other hand, $(1)$ implies that $x(t)$ must belong to the domain of the square root, so $x(t) \geq 0$. We conclude that $x(t) = 0$ for $t \geq 2$.

Answer 2

To expand on my comment: yes, the first part of the solution is obtained by separation of variables. Note that since $x(0)=1\neq 0$ you are allowed to divide and integrate, no problem here. The slight problem arises when $t$ approaches $2$ since it is better to keep the square root in your solution and see that at the point $t=2$ your separable solution may lead to the case when $\sqrt{x}<0$, which is nonsense.

At this point you note that $x(t)=0$ solves your equation (just plug it in). Moreover, if you take the left derivative of the first part of your solution at the point $t=2$, then it is zero. Hence, if you glue these two solutions together, you get a unique continuously differentiable solution.