An easy one, just after Christmas

[gugo82]

Problem:

Let $alpha The linear initial value problem of the first order:

(PC) $\quad \{ (y'=a*y+b),(y(x_0)=y_0):}$

has unique solution $phi\in C^1([alpha,beta])$ (by Cauchy's Theorem).

1) Show that there exists a constant $c>0$, which depends only on $alpha,beta$ and $||a||_oo$, s.t.:

$\quad ||phi||_oo+||phi'||_oo<= c*(|y_0|+||b||_oo) \quad$.

2) Let $b_1,b_2\in C([alpha,beta])$, $y_1,y_2\in RR$ and let $phi_1,phi_2\in C^1([alpha ,beta])$ the solutions of the two linear initial value problems:

$\{(y'=a*y+b_1),(y(x_0)=y_1):}$ and $\{(y'=a*y+b_2),(y(x_0)=y_2):}$.

Show that if $y_1,y_2$ and $b_1,b_2$ are chosen close enough (with respect to the natural norms of $RR$ and $C([alpha,beta])$), then $\phi_1,phi_2$ are close enough with respect to the $C^1$ norm $||u||_(C^1):=||u||_oo+||u'||_oo$; in other words, show that for all $epsilon >0$ there exists $delta>0$ s.t.:

$\quad ||b_1-b_2||_oo \quad ||phi_1-phi_2||_(C^1)

Hint: Use the representation formula for the solution $phi$, i.e.:

$\quad phi(x)=e^(\int_(x_0)^x a(t)" d"t)*(y_0+\int_(x_0)^x e^(-\int_(x_0)^t a(tau )" d"tau) b(t)" d"t) \quad$,

the monotonicity of the exponential and the usual relation between the absolute value of an integral and the integral of the absolute value (e.g. $|\int_alpha^beta f(x) " d"x|<=\int_alpha^beta |f(x)|" d"x$).

[gugo82]

"Gugo82":
2) Let $b_1,b_2\in C([alpha,beta])$, $y_1,y_2\in RR$ and let $phi_1,phi_2\in C^1([alpha ,beta])$ the solutions of the two linear initial value problems:

$\{(y'=a*y+b_1),(y(x_0)=y_1):}$ and $\{(y'=a*y+b_2),(y(x_0)=y_2):}$.

Show that if $y_1,y_2$ and $b_1,b_2$ are chosen close enough (with respect to the natural norms of $RR$ and $C([alpha,beta])$), then $\phi_1,phi_2$ are close enough with respect to the $C^1$ norm $||u||_(C^1):=||u||_oo+||u'||_oo$; in other words, show that for all $epsilon >0$ there exists $delta>0$ s.t.:

$\quad ||b_1-b_2||_oo \quad ||phi_1-phi_2||_(C^1)
It's a consequence of 1).
Infact by linearity the function $phi:=phi_1-phi_2$ solves the problem:

$\{(y'=a*y+(b_1-b_2)),(y(x_0)=y_1-y_2):}$

and by 1) we have:

$||phi_1-phi_2||_(C^1)<= c*(|y_1-y_2|+||b_1-b_2||_oo)$;

so we get $||phi_1-phi_2||_(C^1)

Cita 0 0

Segnala

[gugo82]

It's been more than a month since I wrote the first post but no one even tried to solve the problem.
Now I propose my solution to the first question, hoping someone will read it.

"Gugo82":Problem:

Let $alpha The linear initial value problem of the first order:

(PC) $\quad \{ (y'=a*y+b),(y(x_0)=y_0):}$

has unique solution $phi\in C^1([alpha,beta])$ (by Cauchy's Theorem).

1) Show that there exists a constant $c>0$, which depends only on $alpha,beta$ and $||a||_oo$, s.t.:

$\quad ||phi||_oo+||phi'||_oo<= c*(|y_0|+||b||_oo) \quad$.

Hint: Use the representation formula for the solution $phi$, i.e.:

$\quad phi(x)=e^(\int_(x_0)^x a(t)" d"t)*(y_0+\int_(x_0)^x e^(-\int_(x_0)^t a(tau )" d"tau) b(t)" d"t) \quad$,

the monotonicity of the exponential and the usual relation between the absolute value of an integral and the integral of the absolute value (e.g. $|\int_alpha^beta f(x) " d"x|<=\int_alpha^beta |f(x)|" d"x$).

As the hint says, we have the following representaion for $phi$:

$\quad phi(x)=e^(\int_(x_0)^x a(t)" d"t)*(y_0+\int_(x_0)^x e^(-\int_(x_0)^t a(tau )" d"tau) b(t)" d"t) \quad$.

Taking absolute value in both members and remembering that $|e^z|<=e^|z|$, we deduce:

$|phi(x)|=e^(\int_(x_0)^x a(t)" d"t)*|y_0+\int_(x_0)^x e^(-\int_(x_0)^t a(tau )" d"tau) b(t)" d"t| <=$
$\quad <= e^(|\int_(x_0)^x a(t)" d"t|)*(|y_0|+|\int_(x_0)^x e^(-\int_(x_0)^t a(tau )" d"tau) b(t)" d"t|) <=$
$\quad <= e^(|\int_(x_0)^x |a(t)|" d"t|)*(|y_0|+|\int_(x_0)^x |e^(-\int_(x_0)^t a(tau )" d"tau)|*|b(t)|" d"t|) <=$
$\quad <= e^(|\int_(x_0)^x |a(t)|" d"t|)*(|y_0|+|\int_(x_0)^x e^(|\int_t^(x_0) a(tau )" d"tau|)*|b(t)|" d"t|) <=$
$\quad <= e^(|\int_(x_0)^x |a(t)|" d"t|)*(|y_0|+|\int_(x_0)^x e^(|\int_t^(x_0) |a(tau )|" d"tau|)*|b(t)|" d"t|)\quad$.

Since, for known results, we have:

$|int_(x_0)^x|a(t)|" d"t|<=||a||_oo*|\int_(x_0)^x" d"t|<=||a||_oo*(beta-alpha) \quad => \quad e^(|\int_(x_0)^x|a(t)|" d"t|)<=e^(||a||_oo*(beta-alpha))$

the last member of the previous inequalities chain is less than or equal to $e^(||a||_oo*(b-a))*(|y_0|+|\int_(x_0)^x e^(||a||_oo(b-a))*|b(t)|" d"t|)$, so we can write:

$|phi(x)|<= e^(||a||_oo*(beta-alpha))*(|y_0|+e^(||a||_oo(beta-alpha))*|\int_(x_0)^x |b(t)|" d"t|)$
$\quad <= e^(||a||_oo*(beta-alpha))*|y_0|+e^(2||a||_oo(beta-alpha))*(beta-alpha)*||b||_oo$

and finally:

(*) $\quad ||phi||_oo <= e^(||a||_oo*(beta-alpha))*|y_0|+e^(2||a||_oo(beta-alpha))*(beta-alpha)*||b||_oo \quad$.

Now we turn to $phi'$: since $phi$ is the solution of (PC), we obviously have $||phi'||_oo<=||a||_oo*||phi||_oo+||b||_oo$ and also:

(**) $\quad ||phi'||_oo <= ||a||_oo*e^(||a||_oo*(beta-alpha))*|y_0|+(1+||a||_oo*e^(2||a||_oo*(beta-alpha))*(beta-alpha))*||b||_oo$

by (*).
Putting (*) and (**) together we find:

$||phi||_(C^1):=||phi||_oo+||phi'||_oo<=(1+||a||_oo)*e^(||a||_oo*(beta-alpha))*|y_0|+[1+(1+||a||_oo)*e^(2||a||_oo*(beta-alpha))]*(beta-alpha)*||b||_oo \quad$;

putting $c_(alpha,beta,||a||_oo):=max \{ (1+||a||_oo)*e^(||a||_oo*(beta-alpha)), [1+(1+||a||_oo)*e^(2||a||_oo*(beta-alpha))]*(beta-alpha)\}$ in the last inequality we get:

$||phi||_(C^1)<=c_(alpha,beta,||a||_oo)*(|y_0|+||b||_oo)$

which is our claim since $c_(alpha,beta,||a||_oo)>0$.


P.S.: Most of the formulae look quite ugly, since the MathML has only "|" for the absolute value (while AMS-LaTeX has also "\left|" and "\right|").

[gugo82]

Oh, c'mon! It's easier than it seems at first look... None of you wants even try a solution?