Unique solutions with given inital value problem for ODE

Question

For the equation

$$y' = \sqrt{y}, \quad y(0) = a$$

I got the solution $y = \frac{1}{4} (t+a)^2$. How can I determine the values of $a$ for which this IVP has a solution and for what values the solution is unique? $\sqrt{y}$ is continuous on the interval $[0,\infty]$, so are there unique solutions on this interval? Thank you.

Answer 1

For uniqueness of solutions, it is not enough that the vector field is continuous. For a differential equation $y'=f(y)$, if $f$ is continuous on an open set $U$, then for any $a \in U$ the IVP $y'=f(y), \ y(0)=a$ has at least one solution over a sufficiently small time interval $[0,\varepsilon)$. But if $f$ is Lipschitz on an open set $U \ni a$, then the IVP $y'=f(y), \ y(0)=a$ has exactly one solution over a sufficiently small time interval $[0,\varepsilon)$.

(By the way, all the above statements hold for backward-time solutions as well as for forward-time solutions.)

In your case, $\sqrt{y}$ is continuous on $[0,\infty)$, so at least one solution always exists (at least for a small finite time - although in this particular case it turns out to exist for all time); but $\sqrt{y}$ is not Lipschitz on any neighbourhood of $0$.

The full answer to your question is:

For $a>0$: The solution $y(t)=\frac{1}{4}(t+2\sqrt{a})^2$ is a solution; and it must be the only solution, since $\sqrt{y}$ is locally Lipschitz on $(0,\infty)$ and it is clear that $y(t)$ cannot fall to $0$. ...

[By the way, I think that when you calculated your solution, you incorrectly assumed that the constant of integration is equal to the initial condition $a$. You cannot assume this: to compute the constant of integration $c$ in terms of $a$, you need to use your expression for $y(t)$ in terms of $c$, set $t=0$, and then equate the result to $a$.]

... But for $a=0$: The solution $y(t)=\frac{1}{4}t^2$ is a solution, but it is not the only solution; for, indeed, an even more obvious solution is $y(t)=0$. In fact, for every $r \geq 0$, the function $$ y_r(t) \ = \ \left\{ \begin{array}{c l} 0 & 0 \leq t \leq r \\ \tfrac{1}{4}(t-r)^2 & t \geq r \end{array} \right. $$ is a solution.

If you want a fun exercise, see if you can prove that for $a=0$, the solutions mentioned above are all the solutions.