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i am curious why following initial-value problem

$$ \frac{dy}{dt}=\frac{1}{y},\quad y(0)=0$$ has no solution

if we solve it by method of seperation of variables, we get that

$$y(t)=\pm\sqrt{2t\ {}} $$

we have assumption that our function has form $f(t,y)$; book from which i have taken this example,says that ,it has not solution because of it does not contain $t$ variable (or at book language,does not include $t$ axis) i need to understand it well, as if i met such type of problem, i won't to mixes and say that,it has solution, thanks a lot of,as a additional fact, in book there is written,if change $y(0)=1$, then $y(t)=\sqrt{2t+1}$, it is defined on this interval $(-1/2,\infty)$, does it have solution here?if yes than, $2t$ would be defined on $[0,\infty]$ right? thanks

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$dy/dt=1/t$, $dy=dt/t$, $\int\,dy=\int\,dt/t$, where do you get that square root? –  Gerry Myerson May 17 '12 at 13:42
    
@ Gerry Myerson,see please edited –  dato datuashvili May 17 '12 at 13:47
    
This equation is meaningless. Substitute $t=0$ –  userNaN May 17 '12 at 13:48
    
then $y(0)=0$ so no problem with solution right? –  dato datuashvili May 17 '12 at 13:51
    
@dato: With the DE $\frac{dy}{dt}=\frac{1}{y}$, you would end up with square roots of the type you mention. Can you check carefully which DE is under discussion? There is a big difference between $\frac{dy}{dt}=\frac{1}{t}$ and $\frac{dy}{dt}=\frac{1}{y}$. –  André Nicolas May 17 '12 at 13:51

2 Answers 2

up vote 4 down vote accepted

Let us first solve the problem with initial condition $y(0)=1$. Rewrite our equation in the usual style as $y\,dy=dt$. Integrate. We get $\frac{1}{2}y^2=t+C$, or equivalently $y^2=2t+C'$. From the condition $y(0)=1$ we obtain $C'=1$. We have arrived at the implicit function $y^2=2t+1$. If we want an explicit expression for $y$, we get the two solutions $y=\sqrt{2t+1}$ and $y=-\sqrt{2t+1}$.

As to where these solutions, implicit or explicit, are defined, note that there is no problem if $t>-1/2$, since then $2t+1 \gt 0$. And, (for real solutions) there is a fatal problem if $t<-1/2$. At $t=-1/2$, the derivative of $\sqrt{2t+1}$ is not defined, so technically neither $\sqrt{2t+1}$ nor $-\sqrt{2t+1}$ satisfies the DE at $t=-1/2$. We conclude that there are two solutions, both valid only for $t>-1/2$.

Now let us turn to the initial condition $y(0)=0$. The procedure we used above gives $y^2=2t$. But note that the derivative of $\sqrt{2t}$ is not defined at $t=0$, since there does not exist an open interval about $0$ in which $\sqrt{2t}$ is defined.

We could, by stretching things a little, accept $y=\sqrt{2t}$ and $y=-\sqrt{2t}$ as solutions for $t \gt 0$. We would need to reinterpret the condition $y(0)=0$ as meaning that $\lim_{t\to 0+} y(t)=0$, and to interpret $\frac{dy}{dt}=\frac{1}{y}$ at $t=0$ as meaning that $\lim_{t\to 0+}y\frac{dy}{dt}=1$. That seems to be an interpretation your book does not wish to make.

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thanks too much,i have understood everything now –  dato datuashvili May 17 '12 at 14:54

Before attemting to solve an IVP by analytically or numerically of a first-order ODEs in the form

$y\prime= f(x, y),\ \ y(x_0)= y_0$

we should check whether it has a solution or not. If it has a solution, we ask whether it is unique. These questions are answered by the basic existence-uniquness theorem which briefly states that

If $f$ and $\frac{\partial f}{\partial y}$ are continuous in a rectangle containing the point $(x_0, y_0)$, then the IVP has a unique solution. There is also a weak condition on $f$ (instead of $\frac{\partial f}{\partial y}$) which is the Lipschitz continuity. In your case $(x_0, y_0)=(0, 0)$ and $f(x, y)= 1/y$ where $f$ is even not defined at $y=0$. Examples of such problems including your question you can look at the notes http://www.math.ust.hk/~mamu/courses/303/Notes.pdf

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