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Show that the first order differential equation $y'(x)=\sqrt{|y(x)|}$ with intial value $y(1/2)= 1/16$ has infinite solutions on the interval [−1, 1].

My thought were to show that this equation has two solutions (just by ansatz, or just looking at the interval $]0, 1]$ to get $y'(x) = \sqrt{|y(x)|}$) and deduce from that that there must be infinitely many solutions since the solution set forms a vector space. I've got a nagging feeling this isn't the way to go though. Can anybody give me a hint where to look?

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Check if I entered the correct TeX syntax. There were symbols after $1/16$ that my browser could ont display. –  Siminore Sep 19 '12 at 12:18
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Better to say: "...has infinitely many solutions..." When you say it "has infinite solutions" we may think there are solutions with the value infinity or something. –  GEdgar Sep 19 '12 at 12:56
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1 Answer 1

up vote 2 down vote accepted

This is a classical example of non uniqueness due to the fact that the right hand side of the equation, $\sqrt{|y|}$, is not Lipschitz at $y=0$. You show this by direct computation. The equation is in separeted variables: $$ \frac{dy}{\sqrt{|y|}}=dt,\quad y(1/2)=1/16. $$ Integrating, and taking into account the absolute value, we see that $$ y(t)=\frac{t\,|t|}{4}=\begin{cases} t^2/4 & t\ge0,\\-t^2/4 & t<0, \end{cases} $$ is a solution. But for any $\tau\in(-1,0)$, the function $$ y_\tau(t)=\begin{cases} t^2/4 & t\ge0,\\0 & \tau<t<0\\-(t-\tau)^2/4 & t<\tau, \end{cases} $$ is also a solution. Below is the graph of $y_{-1/2}$.

Graph of $y_{-1/2}$

By the way, the set of solutions is not a vector space.

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I'm having some trouble seeing how $y_\tau(t)=0$ is a solution. Does this function satisfy $y(1/2) = 1/16$? –  user1437901 Sep 19 '12 at 18:26
    
$y_\tau(t)=0$ only for $\tau<t<0$. For $0\le t\le1$, $y_\tau(t)=t^2/4$, so that $y\tau(1/2)=1/16$, as required. I have added th graph of $y_{-1/2}$. –  Julián Aguirre Sep 19 '12 at 18:51
    
This is crystal clear to me now, thank you. –  user1437901 Sep 19 '12 at 19:25
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