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please solve the initial value problem $dy/dx=2y^{1/2}$, y(0)=a. I wanted to know that this problem admits infinitely many solutions for a=0 or admits infinitely many solutions for $a\geq 0$

pradeep shende
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If $a>0$ then you can just separate variables, at least for solutions going forward in time. You can also just separate variables backward in time up until the point where $y$ hits zero. Once $y$ hits zero, separation of variables (or the corresponding technique using the chain rule and no explicit differentials) is no longer guaranteed to work. Instead, solutions to your DE can "stay at zero" as long as you want. It can then "change back" to the separation of variables solution. Once it does so you recover uniqueness again.

This is one of the standard "counterexamples" to the Picard-Lindelöf existence and uniqueness theorem. It is not really a counterexample, but rather demonstrates what can happen if one of the hypotheses of the theorem (in this case that the right hand side is Lipschitz continuous in $y$) is violated. $y'=y^{2/3}$ is a similar example.

Daniel Fischer
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Ian
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$$\frac{dy}{2y^{1/2}} = dx$$

By integrating from two sides : $$y^{1/2} = x + c$$

$$y = (x+c)^2$$

And since $y(0) = 0$, we deduce $0 = (0+c)^2$, hence $c=0$

So , there is only one answer for $y$.

The answer is $y = x^2$

For $y(0) = a$ with $a\ge0$

$a = c^2$, hence $c=\pm\sqrt a$

two answers for $c$, so two answers for $y$

BusyAnt
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mahdokht
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  • This is entirely false: a constant solution exists, and "switching" solutions also exist. The whole problem is with the division by $y^{1/2}$, which is not admissible if $y=0$. – Ian Aug 03 '15 at 11:01
  • @Ian then pls tell me how many solution when a is greater than or equal to 0 – avijit kundu Sep 20 '15 at 05:03
  • @Ian when a=0 then there are two solutions. y=0 and y=x^2/4. Am i right? – avijit kundu Sep 20 '15 at 05:09
  • @avijitkundu As I said in my answer, there are infinitely many solutions. Given any constants $c_1 \leq 0 \leq c_2$, $y(x)=-(x-c_1)^2/4$ for $x \leq c_1$, $0$ for $c_1 \leq x \leq c_2$, and $(x-c_2)^2/4$ for $x \geq c_2$, is a solution. This looks complicated, but it just means that the system can stay at zero as long as it wants going both forward and backward in time, before returning to the ordinary behavior. – Ian Sep 20 '15 at 12:26