ODE - Qualitative study of solutions

[gugo82]

Here's an exercise for Analisi II or Sistemi Dinamici students.

***

Exercise:

Consider the Cauchy's problem:

(CP) [tex]$\begin{cases} [y(x)-x^2]\ y^\prime (x)=1 \\ y(0)=a\end{cases}$[/tex].

1. Find all the values of the parameter [tex]$a$[/tex] for which (CP) has unique local solution in a neighbourhood of [tex]$0$[/tex].

2. Is there any value of [tex]$a$[/tex] for which (CP) has no solution at all?

3. Do a qualitative study of the maximal solutions to (CP).

[gugo82]

"gugo82":Exercise:

Consider the Cauchy's problem:

(CP) [tex]$\begin{cases} [y(x)-x^2]\ y^\prime (x)=1 \\ y(0)=a\end{cases}$[/tex].

1. Find all the values of the parameter [tex]$a$[/tex] for which (CP) has unique local solution in a neighbourhood of [tex]$0$[/tex].

2. Is there any value of [tex]$a$[/tex] for which (CP) has no solution at all?

The equation in (CP) is a first order implicit ODE, i.e. it comes in the form [tex]$F(x,y(x),y^\prime (x))=0$[/tex] with [tex]$F(x,y,v):=(y-x^2)\ v-1$[/tex]; moreover the function [tex]$F$[/tex] is of class [tex]$C^\infty (\mathbb{R}^3)$[/tex] with gradient [tex]$\text{D}F(x,y,v):=(-2xv,yv,y-x^2)$[/tex].

To put the ODE in normal form, i.e. to write the equation in the form [tex]$y^\prime (x)=f(x,y(x))$[/tex] (at least locally), we have to "stay away" from the singular set [tex]$\mathcal{S}:=\{(x,y)\in \mathbb{R}^2: y-x^2=0\}$[/tex] (by Dini's theorem), which is a parabola in the plane [tex]$Oxy$[/tex] (with vertex in [tex]$o=(0,0)$[/tex] and symmetric w.r.t. the positive [tex]$y$[/tex] semiaxis; see figure below).

[asvg]xmin=-3;xmax=3; ymin=-2;ymax=4;
axes("","");
stroke="red";
plot("x^2",-3,3);[/asvg]

If we choose [tex]$a\neq 0$[/tex], then Dini's theorem applies to the equation [tex]$F(x,y,v)=0$[/tex], which can be solved w.r.t. [tex]$v$[/tex]: hence we can put the ODE in normal form in a neighbourhood of the point [tex]$(x_0,y_0)=(0,a)$[/tex] and write (CP) in the following form:

(CP') [tex]$\begin{cases} y^\prime (x)=\frac{1}{y(x)-x^2}\\
y(0)=a\end{cases}$[/tex]

The r.h.s. of the equation in (CP'), [tex]$f(x,y):=\frac{1}{y-x^2}$[/tex], is a continuous (w.r.t. [tex]$(x,y)$[/tex]), locally Lipschitz (w.r.t. [tex]$y$[/tex]) function in the set [tex]$\mathbb{R}^2\setminus \mathcal{S}$[/tex], hence classical results applies: we have existence and uniqueness for the maximal solution of (CP').

On the other hand, if we choose [tex]$a=0$[/tex], the ODE becomes meaningless for [tex]$x=0$[/tex]: in fact there is no real number [tex]$y^\prime (0)$[/tex] s.t. [tex]$0\cdot y^\prime (0)=[y(0)-0]\ y^\prime (0)=1$[/tex]. Hence for [tex]$a=0$[/tex] problem (CP) has no solutions.

"gugo82":3. Do a qualitative study of the maximal solutions to (CP).

In what follows we fix [tex]$a\neq 0$[/tex].

The function [tex]$f(x,y):=\frac{1}{y-x^2}$[/tex] is positive in [tex]$P:=\{ (x,y)\in \mathbb{R}^2:\ y>x^2\}$[/tex] and negative in [tex]$N:=\{ (x,y)\in \mathbb{R}^2:\ y
[asvg]xmin=-3;xmax=3; ymin=-2;ymax=4;
axes("","");
text([0.5,2.5],"P"); text([0.5,-1.5],"N");
marker="arrow"; line([-1.5,3],[-1,3.5]); line([1,3],[1.5,3.5]); line([-0.25,1.5],[0.25,2]);
line([-0.25,-0.5],[0.25,-1]); line([-2.5,-1.5],[-2,-2]); line([2,-1.5],[2.5,-2]); line([-2.5,1.5],[-2,1]); line([2,1.5],[2.5,1]);
stroke="red";
plot("x^2",-3,3);[/asvg]
Let [tex]$y(x)$[/tex] be a maximal solution of (CP) with [tex]$a>0$[/tex] and let [tex]$(X_-,X_+) \subseteq \mathbb{R}$[/tex] (with [tex]$X_- In a neighbourhood of [tex]$0$[/tex] the solution [tex]$y(x)$[/tex] increases stricly, hence [tex]$y(x)>0$[/tex] around [tex]$0$[/tex]; as a consequence for [tex]$x<0$[/tex]:

[tex]$y^\prime (x)=\frac{1}{y(x)-x^2} \geq \frac{1}{y(x)}$[/tex]
[tex]$\Rightarrow \quad 2\ y(x)\ y^\prime (x) \geq 2$[/tex]
[tex]$\Rightarrow \quad a^2-y^2(x)=\int_x^0 2\ y(x)\ y^\prime (x)\ \text{d}x \geq \int_x^02\ \text{d}x =-2x$[/tex]
[tex]$\Rightarrow \quad y(x)\leq \sqrt{2x+a^2}$[/tex].

The function [tex]$\sqrt{2x+a^2}-x^2$[/tex] is positive in [tex]$0$[/tex] and negative in [tex]$-\tfrac{a^2}{2}$[/tex] and continuous between them, therefore the graph of [tex]$\sqrt{2x+a^2}$[/tex] meets the singular set [tex]$\mathcal{S}$[/tex] in a point [tex]$(\xi,\eta)$[/tex] with abscissa [tex]$\xi \in ]-\tfrac{a^2}{2},0[$[/tex]. It follows that [tex]$X_-$[/tex] is finite and [tex]$X_- \in ]\xi , 0[$[/tex].

This is not the end, but I'm going to leave the exercise open.
My reasons are really simple: 1. Calculus is boring and 2. who's studying can try to solve the remaining part as an exercise.
Good luck.

[j18eos]

No bad for first time

[Benny24]

"dissonance":those corrections are correct!

Is that a tongue-twister?

[dissonance]

"j18eos":the vector field is not definite for

This one sounds weird, I'm afraid it's an error. "The vector field is not defined" might be better.

[tex]y(x)=x^2[/tex], for conseguence if [tex]a=0[/tex] this problem is impossible!

Again, weird. Consequently or hence seem to me to be better choices.

EDIT: Is correct it english?

Here I'm definitely sure you're wrong: is my English correct? should be the one.

PLEASE NOTE: I'm no English teacher. Also, my English dictionary is away from my desk at the moment (where did I put it last time...? ). So there is no guarantee those corrections are correct!

[gugo82]

Your considerations are true: I made a mistake in writing the second question, for the condition [tex]$a=0$[/tex] cannot be satisfied by any integral of the ODE.
Hence, I have to correct the text of the problem.
Thank you for pointing out my error.


P.S.: The first period sounds somewhat strange... I'd have preferred "The ODE can be written in the form..." to your incipit.

[j18eos]

This Cauchy's problem is writeable as [tex]\dot y(x)=\frac{1}{y(x)-x^2}[/tex] and the the vector field is not defined for [tex]y(x)=x^2[/tex], consequently if [tex]a=0[/tex] this problem is impossible!