Why is the solution to the Volterra integral equation written as a power series?

[sarrah1]

I have this Volterra integral equation

$X(t)=e^{At}+\int_{0}^{t} \,e^{A(t-\tau)}\epsilon BX(\tau) d\tau$ (1) ,
of matrix solution $X(t)$ and $X(0)=I$, $A,B\in{R}^{n\times n}$ ,

By iteration:
Result 1 : the solution $X(t)$ of (1) can be written as a power series in $\epsilon$ in the form
$X(t)=e^{At}+\sum_{1}^{\infty}{\epsilon}^{n} {P}_{n}(t)$ (2)
where ${P}_{n}(t)$ satisfies the recurrence
${P}_{n}(t)= \int_{0}^{t} \,e^{A(t-\tau)}B {P}_{n-1}(\tau) d\tau$ , ${P}_{0}(t)={e}^{At}$ (3)
proof is by substituting (2) into (1) and equating equal powers of $\epsilon$

Result 2 : The series (2) is uniformly convergent in in some interval of $t$ for sufficiently small $\epsilon$
Proof: $||{P}_{1}(t)||\le \int_{0}^{t} \,e^{||A||(t-\tau)}||B|| {e}^{||A||\tau} d\tau$
$\le||B||{e}^{||A||t}\int_{0}^{t} \,{e}^{||A||(t-\tau)}d\tau$
$\le||B||{e}^{||A||t}t{e}^{||A||t}$
similarly
$||{P}_{2}(t)||\le{e}^{||A||t}{(t||B||{e}^{||A||t})}^{2}$ etc...
Thus the series in (2) is absolutely convergent if $\epsilon t||B||{e}^{||A||t}<1$.
According to Wieirstrass M-test, the series $\sum_{1}^{\infty}{\epsilon}^{n} {P}_{n}(t)$ converges uniformly on some interval
$[0,t:\epsilon t||B||{e}^{||A||t}<1]$.

very grateful

Sarrah

it is result 2 which I care for
Sarrah

 

[Euge]

Hi sarrah,

Why is

$$\|P_1(t)\| \le \int_0^t e^{\|A\|(t - \tau)}\|B\|e^{\|A\|\tau}\,d\tau?$$

It appears you used a Minkowski type inequality, but what norm are you using?

Given your bounds are correct, the last statements you made do not make sense. The interval you speak of should be independent of $t$.

 

[sarrah1]

Hi sarrah,

Why is

$$\|P_1(t)\| \le \int_0^t e^{\|A\|(t - \tau)}\|B\|e^{\|A\|\tau}\,d\tau?$$

It appears you used a Minkowski type inequality, but what norm are you using?

Given your bounds are correct, the last statements you made do not make sense. The interval you speak of should be independent of $t$.

thank you Euge you always are there for help.
concerning the inequality I am using the rule "Norm of integral of a function is less than the integral of its norm" I assume this applies to any norm,

Now the norm of $||{e}^{A(t-\tau)}B{e}^{A\tau}||$ is less than the product of their norms (you are right I am using a sub-multiplicative norm)and I am using Minkowski inequality for the series say $||{e}^{A\tau}||\le{e}^{||A||\tau}$ as I am adding up the norm of each term of its Taylor's series expansion.
Concerning your second criticism, you are right, I couldn't define the closed interval in which the series converges uniformly. So I showed absolute convergence based upon the condition $t\epsilon||B||{e}^{||A||t}<1$ from which an interval of $t$ can be obtained say $[0,{t}_{0}]$, where ${t}_{0}$ is the max $t$ satisfying the inequality. I didn't designate this interval as I will have to solve for $t$ the foregoing inequality being difficult involving logarithm. Another problem now even if I could solve for $t$ then uniform convergence is on $[0,t)$ , but one is accustomed to be on a closed interval
again very grateful
sarrah