ODE -Space of solutions

[Camillo]

Let's consider the solutions $y in RR $ of the following ODE :

$y'''-y''+y'-y = 0 $ .

a) Determine the dimension of the space of the solutions defined in $RR$ .
b) Determine the dimension of the space of the limited solutions defined in $RR $.
c) Determine the dimension of the space of the solutions defined in $RR$ and such that : $lim_( x rarr -oo) y(x)= 0 $.

[Leonardo891]

Don't give up Gugo82!
I speak for me but i think others people will agree with me.
Many problems in this sections are too much advanced. I'm studyng now analysis in more variables (or also in normed vector space) and differential equations: I don't now yet the thery to theory to the base of many problems you posted! I don't study yet Lagrange multipliers!
Shall we make an agreement? If you post a problem of analysis in one variable that a student of the second year can (or shoul be able to) solve and I will try to solve it.
However, I know how much English is important and if someone wants to correct my mistakes he's welcome.

[Camillo]

I do agree with Gugo considerations

quote
As usual, in these last months, nobody tried to solve our simple exercise.
This one and many other easy problems in this section are posted in order to give young students a chance of writing in English.
English is the international language of scientists since the 50s, hence a mathematician has to write in a decent English. Since there is no other way to learn something than to practice it as much as possible, we're offering to young students a great opportunity.
Nevertheless no one seems to appreciate our efforts, for no one (or just few guys) attempts to answer here.
unquote

and I am very disappointed that quite few people try to solve the problems shown in this section and this is not due to the difficulty of the problems, but to the laziness and shyness in practicing English.
Good Knowledge of English language is today fundamental in each scientific /technical field .
So make an effort today and you will be rewarded tomorrow in your job .
No English ----> No Job
Take adavantage of this opportunity to practice .

@ Gugo : do not give up, your contributions, support and enthusiasm are essential for this section !

[dissonance]

"Gugo82":Therefore this and the solution to the exercise therein will be my last elementary post in this section; I'm just too disappointed to go on with this kind of stuff.

I beg you to reconsider your choice, which is really bad news for the whole forum. It's "this kind of stuff" that gives life to this section; without your work and that of a few other contributors it would be dead and gone. They say you start appreciating what you have the moment you're losing it: well, they're right. I believe we've learnt a lesson here.
To start with, I ask you not to reveal the solution to the other open problem for another day. I'm working on it.

[gugo82]

As usual, in these last months, nobody tried to solve our simple exercise.

This one and many other easy problems in this section are posted in order to give young students a chance of writing in English.
English is the international language of scientists since the 50s, hence a mathematician has to write in a decent English. Since there is no other way to learn something than to practice it as much as possible, we're offering to young students a great opportunity.
Nevertheless no one seems to appreciate our efforts, for no one (or just few guys) attempts to answer here.

Therefore this and the solution to the exercise therein will be my last elementary post in this section; I'm just too disappointed to go on with this kind of stuff.

***

"Gugo82":d) Choose initial conditions:

$\{(y(0)=a),(y'(0)=b),(y''(0)=c):}$

with $a,b,c \in RR$ and transform the third order Cauchy problem

$\{(y'''-y''+y'-y=0),(y(0)=a),(y'(0)=b),(y''(0)=c):}$

into a linear integral equation (l.i.e.) in $y$; try to put this l.i.e. in the form:

$y(x)=f(x)+\int_(-oo)^(+oo) K(x,t) y(t)" d"t$

finding the appropriate kernel $K(x,t)$ defined in $RR^2$.

We put our ODE in normal form and integrate both l.h. and r.h. sides three times:

[tex]$\int_0^x \int_0^t \int_0^\tau y^{\prime \prime \prime}(\sigma) \text{ d} \sigma \text{ d} \tau \text{ d} t =\int_0^x \int_0^t \int_0^\tau \left\{ y^{\prime \prime} (\sigma ) -y^{\prime} (\sigma) +y(\sigma)\right\} \text{ d} \sigma \text{ d} \tau \text{ d} t$[/tex].

Keeping in mind the initial conditions, i.e. [tex]$y(0)=a,\ y^\prime(0)=b,\ y^{\prime \prime}(0)=c$[/tex], the l.h.s. gives:

[tex]$\int_0^x \int_0^t \int_0^\tau y^{\prime \prime \prime}(\sigma) \text{ d} \sigma \text{ d} \tau \text{ d} t= \int_0^x \int_0^t [y^{\prime \prime}(\tau) -c] \text{ d} \tau \text{ d} t$[/tex]
[tex]$=\int_0^x [y^{\prime}(t)-b-ct] \text{ d} t$[/tex]
[tex]$=y(x)-a-bx-\frac{c}{2} x^2$[/tex];

coming to the r.h.s. we have to integrate the three summands separately and use a little integration by parts: we have:

- 1st summand:
$[tex]\int_0^x \int_0^t \int_0^\tau y^{\prime \prime} (\sigma ) \text{ d} \sigma \text{ d} \tau \text{ d} t =\int_0^x \int_0^t [y^{\prime} (\tau)-b] \text{ d} \text{ d} \tau \text{ d} t$[/tex]
[tex]$=\int_0^x [y(t)-a-bt] \text{ d} t$[/tex]
[tex]$=-ax-\frac{b}{2} x^2 +\int_0^x y(t) \text{ d} t$[/tex];

- 2nd summand:
[tex]$\int_0^x \int_0^t \int_0^\tau -y^\prime (\sigma ) \text{ d} \sigma \text{ d} \tau \text{ d} t =-\int_0^x \int_0^t [y(\tau) -a] \text{ d} \tau \text{ d} t$[/tex]
[tex]$=\frac{a}{2} x^2 -\int_0^x \int_0^t y(\tau ) \text{ d} \tau \text{ d} t$[/tex]
(integration by parts w.r.t. [tex]$t$[/tex] with finite factor [tex]$\int_0^t y(\tau) \text{ d} \tau$[/tex] and differential factor [tex]$1$[/tex])
[tex]$=\frac{a}{2} x^2-\Big[ t \int_0^t y(\tau) \text{ d} \tau \Big]_0^x +\int_0^x t y(t) \text{ d} t$[/tex]
([tex]$x$[/tex] can pass under integral sign because it's constant w.r.t. the integration variable, which we rename [tex]$t$[/tex])
[tex]$=\frac{a}{2} x^2 +\int_0^x (t-x) y(t) \text{ d} t$[/tex];

- 3rd summand:
[tex]$\int_0^x \int_0^t \int_0^\tau y(\sigma ) \text{ d} \sigma \text{ d} \tau \text{ d} t $[/tex]
(integration by parts w.r.t. [tex]$t$[/tex] with finite factor [tex]$\int_0^t \int_0^\tau y(\sigma ) \text{ d} \sigma \text{ d} \tau$[/tex] and differential factor [tex]$1$[/tex])
[tex]$=\left[ t \int_0^t \int_0^\tau y(\sigma ) \text{ d} \sigma \text{ d} \tau \right]_0^x-\int_0^x t \int_0^t y(\tau ) \text{ d} \tau \text{ d} t$[/tex]
([tex]$x$[/tex] can pass under integral sign because it's constant w.r.t. the integration variable, which we rename [tex]$t$[/tex])
[tex]$=-\int_0^x (t-x) \int_0^t y(\tau) \text{ d} \tau \text{ d} \text{ d} t$[/tex]
(integration by parts w.r.t. [tex]$t$[/tex] with f.f. [tex]$\int_0^t y(\tau ) \text{ d} \tau$[/tex] and d.f. [tex]$t-x$[/tex])
[tex]$=\left[ -\frac{1}{2}(t-x)^2 \int_0^t y(\tau ) \text{ d} \tau \right]_0^x +\int_0^x \frac{1}{2} (t-x)^2 y(t) \text{ d} t$[/tex]
[tex]$=\int_0^x \frac{1}{2} (t-x)^2 y(t) \text{ d} t$[/tex].

Putting what we've found together yelds the l.i.e.:

[tex]$y(x)=\frac{c-b+a}{2} x^2 + (b-a) x+a+\int_0^x \left\{ \frac{1}{2} (t-x)^2 +(t-x)+1\right\} y(t) \text{ d} t$[/tex]

which is equivalent to (i.e. has the same solution) our Cauchy problem.

Finally, if we let:

[tex]$K(x,t)=\begin{cases} \text{sign}(x) \left[ \frac{1}{2} (t-x)^2 +(t-x)+1\right] & ,\text{ if $0\leq t\leq x$ or $x\leq t \leq 0$}\\ 0 & ,\text{ otherwise}\end{cases}$[/tex]

the previous l.i.e. can be rewritten as a Volterra equation with kernel [tex]$K(x,t)$[/tex], i.e.:

[tex]$y(x)=\frac{c-b+a}{2} x^2 + (b-a) x+a+\int_{-\infty}^{+\infty} K(x,t)\ y(t) \text{ d} t$[/tex].

[gugo82]

Since Camillo's a-b-c can be solved by a direct computation, I'd like to add another point which can be solved analogously:

d) Choose initial conditions:

$\{(y(0)=a),(y'(0)=b),(y''(0)=c):}$

with $a,b,c \in RR$ and transform the third order Cauchy problem

$\{(y'''-y''+y'-y=0),(y(0)=a),(y'(0)=b),(y''(0)=c):}$

into a linear integral equation (l.i.e.) in $y$; try to put this l.i.e. in the form:

$y(x)=f(x)+\int_(-oo)^(+oo) K(x,t) y(t)" d"t$

finding the appropriate kernel $K(x,t)$ defined in $RR^2$.