# Thread: tough DE theory problem

1. ## tough DE theory problem

Hey Gang:

I am going to ask a question this time. I got a hold of the book

Ordinary DE and Stability Theory by Sanchez I picked up for a song. It has some tough problems. I thought it would be a nice book, but the notation is killer.

I have been wanting to strengthen my DE skills.

Here is one if anyone has a good idea.

Let's take the equation $\displaystyle x'=f(x)$, where x=x(t) is an unknown

scalar function and f and df/dx are defined and continuous on the strip

$\displaystyle B=[(t,x)|-\infty<t<{\infty}, \;\ a<x<b]$.

We can also assume $\displaystyle f(x) \neq 0, \;\ a<x<b$,

and let $\displaystyle F(x)=\int_{x_{0}}^{x}\frac{1}{f(s)}ds$,

where $\displaystyle a<x_{0}<b$.

Show all solutions are described by the family

$\displaystyle x(t)={\phi}(t-c)$, where $\displaystyle {\phi}$ is the inverse

of F, and c is a constant determined by initial conditions.

I tried using the second fundamental theorem of calculus and thought

about $\displaystyle \frac{d}{dx}[f(x)]=\frac{d}{dt}\int_{x_{0}}^{x(t)}\frac{1}{f(s)}ds$

Just a little hung up.

2. Wingless sent me the following. He gets the credit for this post

We know these:

$\displaystyle \frac{dx}{dt} = x'(t) = f[x(t)]$

$\displaystyle F[x(t)] = \int_{x_0}^{x} \frac{1}{f(s)}~ds$

Now, on the RHS of the second line, substitute $\displaystyle s = x(t)$ to get,

$\displaystyle F[x(t)] = \int_{x^{-1}(x_0)}^{x^{-1}[x(t)]} \frac{x'(t)}{f[x(t])}~dt$

Using the first line, replace x'(t) to get,

$\displaystyle F[x(t)] = \int_{x^{-1}(x_0)}^{x^{-1}[x(t)]} \frac{f[x(t)]}{f[x(t])}~dt = \int_{x^{-1}(x_0)}^{x^{-1}[x(t)]}dt = \underbrace{x^{-1}[x(t)]}_{\text{this is t}}-\underbrace{x^{-1}(x_0)}_{\text{some constant}}+C = t- C$

$\displaystyle x(t) = \phi (t-C)$

3. Thank you much fellas. Since the wing isn't here, tell him thanks as well.

I started out right and then went down the primrose path

4. My solution is also similar to that by wingless's, however I would like to give it here for showing some technical details.
Since $\displaystyle x^{\prime}(t)=f(x(t))\neq0$ for all $\displaystyle x\in(a,b)$, we see that $\displaystyle x$ is either increasing or decreasing, i.e. it has a inverse function $\displaystyle x^{-1}$.
Now, consider the followings.
$\displaystyle F[x(t)]=\int\limits_{x_{0}}^{x(t)}\frac{1}{f(s)}ds=\int\l imits_{x_{0}}^{x(t)}\frac{1}{x^{\prime}(x^{-1}(s))}ds=\int\limits_{x^{-1}(x_{0})}^{t}dv=t-x^{-1}(x_{0}),\qquad(\ast)$
where
$\displaystyle v=x^{-1}(s)\text{ and so }dv=(x^{-1}(s))^{\prime}ds=\frac{1}{x^{\prime}(x^{-1}(s))}ds$ (see Inverse functions and differentiation - Wikipedia).
Applying $\displaystyle \phi$ on both sides of (*), we get the desired result.