Solving a Cauchy Problem for $f(x,y)$, Defined on $(-b,b)$

[evinda]

Hello! (Wave)

We consider the following Cauchy problem

$$y'=f(x,y), y(0)=\frac{1}{b}$$

Find an example of a function $f(x,y)$ such that the solution exists only on the interval $(-b,b)$ for $b>0$.

I thought to pick $y(x)=\frac{1}{\sqrt{b^2-x^2}}$ since this is defined on $(-b,b)$.

Then differentiating we get $y'(x)=\frac{x}{(b^2-x^2)^{\frac{3}{2}}}$.

So I thought to pick $f(x,y)=\frac{x}{(b^2-x^2)^{\frac{3}{2}}}$.

Is my idea right? (Thinking)

In my notes there is the same example but for the interval $(-\sqrt{b}, \sqrt{b})$.

There the following is done:$y'=g(x) y^2 , y(0)=a$ where $a=\frac{1}{b}$.

$\frac{dy}{y^2}=g(x) dx$

$\frac{1}{y}=\frac{1}{a}-\int_0^x g(\xi) d \xi=\frac{1-a \int_0^x g(\xi) d \xi}{a}$

$y(x)=\frac{a}{1-a \int_0^x g(\xi) d \xi}$

$g(x)=2x$

$y(x)=\frac{1}{b-x^2}$If we would also want to do it as in my notes, could we again look for a solution in the form $y'=g(x) y^2$ ? (Thinking)

Because I tried it and I got $y=\frac{1}{b-\int_0^x g(\xi) d \xi}$.

But can we get from this a solution that is defined on $(-b,b)$ ?

 

[jvicens]

Yes, your idea is correct. Your chosen function $f(x,y)=\frac{x}{(b^2-x^2)^{\frac{3}{2}}}$ satisfies the given Cauchy problem and its solution is indeed only defined on the interval $(-b,b)$.

To show this, we can use the existence and uniqueness theorem for first-order differential equations. Since $f$ is continuous and satisfies the Lipschitz condition with respect to $y$, there exists a unique solution to the Cauchy problem on some interval containing $x=0$. In this case, it is clear that the solution must be defined on $(-b,b)$ since this is the interval where the initial condition $y(0)=\frac{1}{b}$ is valid.

Regarding your second question, it is possible to also solve the given Cauchy problem using the form $y'=g(x)y^2$. However, it is important to note that the solution obtained in this way will not necessarily be the same as the one obtained using your chosen function $f(x,y)$. In fact, the solution obtained using $y'=g(x)y^2$ will be different for different choices of $g(x)$. Therefore, it is not always possible to obtain a solution that is defined on $(-b,b)$ using this method.