[EX] On ODEs

[gugo82]

Exercise:

1. Draw a qualitative graphic of the maximal solutions of the ODE:
\[
u^\prime (x) = \frac{1}{x\ \left( 1+\ln u(x) \right)}\; .
\]

2. Prove that the maximal solution \(U\) corresponding to the initial data \((x_0,u_0)=(1,1)\) satisfies the identity:
\[
\left( U(x)\right)^{U(x)} = x
\]
for any \(x\) in its domain.

[gugo82]

"Camillo":1 )Just to start, from the ODE it appears that :

$x ne 0 $ ; $ 1+ ln u ne 0 rarr ln u ne -1 rarr une 1/e $; $ u > 0 $

Study of the sign of $y' $

*$ u ' >0 rarr u $ increasing if :
$x>0 ; u >1/e $
or
$x <0 ; u<1/e $

* $ u ' <0 rarr u $ decreasing if :
$x <0 ; u >1/e $
or
$ x>0 ; u <1/e $
The qualitative graph .... ?

Qualitative means approximate, but not too much...

To be more precise, you proved that maximal solutions increase somewhere and decrease somewhere else, but what about other significative properties of solutions or of their graphics? What can be said about the domain of the maximal solutions and their regularity? Or, what about convexity/concavity of their graphics? And what about the existence of asymptotes?
All these informations (which are called qualitative, because they do not involve estimates[nota]As opposite to quantitative informations, which do involve esimates.
For example, Stirling's formula:
\[
n!\sim \sqrt{2\pi\ n}\ n^n\ e^{-n} \qquad \text{as } n\to \infty
\]
recasts into:
\[
n! = \sqrt{2\pi\ n}\ n^n\ e^{-n}\ \big( 1 + \text{o}(1)\big) \qquad \text{as } n\to \infty\;
\]
and it gives the information \(\delta (n) \to 0\) (i.e., \(\delta (n) = \text{o}(1)\)) about the relative error:
\[
\delta (n) = \frac{n! - \sqrt{2\pi\ n}\ n^n\ e^{-n}}{\sqrt{2\pi\ n}\ n^n\ e^{-n}}\; ;
\]
this information is purely qualitative, because it does not quantify the speed \(\delta(n)\) tends to \(0\). On the other hand, Robbins estimates:
\[
\sqrt{2\pi\ n}\ n^n\ e^{-n}\ \left( 1 + e^{\frac{1}{12n+1}}\right)\leq n! \leq \sqrt{2\pi\ n}\ n^n\ e^{-n}\ \left( 1 + e^{\frac{1}{12n}}\right)
\]
yield:
\[
\lim_n 12n\ \delta (n) = 1\; ,
\]
i.e. \(\delta (n) \sim \frac{1}{12n}\) as \(n\to \infty\), which is a quantitative information because it does measure the speed \(\delta (n)\) tends to \(0\).[/nota]) can be recovered using only the ODE... And this is great, because one has not to esplicitly solve the ODE to draw a (sufficiently approximate) picture of the graphic of the maximal solutions.
Such a picture is then called qualitative graphic.

"Camillo":2) The equation is of the separable variable type.
Integrating we get :
$u ln u = lnx +C $
the condition $u(1)= 1 $ implies $C=0 $ , consequently the solution is

$u*lnu = ln x rarr u^u=x $

Yep!

Another way to find 2 (which is substantially the same as yours) involves integral functions: in fact, from the equality:
\[
\big( 1+\ln U(x)\big)\ U^\prime (x) = \frac{1}{x}\; ,
\]
which holds in the whole interval of definition of \(U\), one infers equality between integral functions:
\[
\int_1^x \big( 1+\ln U(t)\big)\ U^\prime (t)\ \text{d} t = \int_1^x \frac{1}{t}\ \text{d} t\; ,
\]
i.e.:
\[
\left. U(t)\ \ln U(t)\right|_1^x = \left. \ln t \right|_1^x \qquad \stackrel{U(1)=1}{\Rightarrow} \qquad U(x)\ \ln U(x) = \ln x
\]
which is the claim.

[Camillo]

1 )Just to start, from the ODE it appears that :

$x ne 0 $ ; $ 1+ ln u ne 0 rarr ln u ne -1 rarr une 1/e $; $ u > 0 $

Study of the sign of $y' $

*$ u ' >0 rarr u $ increasing if :
$x>0 ; u >1/e $
or
$x <0 ; u<1/e $

* $ u ' <0 rarr u $ decreasing if :
$x <0 ; u >1/e $
or
$ x>0 ; u <1/e $
The qualitative graph .... ?

2) The equation is of the separable variable type.
Integrating we get :
$u ln u = lnx +C $
the condition $u(1)= 1 $ implies $C=0 $ , consequently the solution is

$u*lnu = ln x rarr u^u=x $