- #1
sarrah1
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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 $X(0)=I$, $A,B\in{R}^{n\times n}$ , $t\in[a,b]$
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 $[a,b]$ for sufficiently small $\epsilon$
Proof: $||{P}_{1}(t)||\le \int_{0}^{t} \,e^{||A||(t-\tau)}||B|| {e}^{||A||\tau} d\tau$
by Bonnet 2nd mean value theorem $=||B||{e}^{||A||t}\int_{\zeta}^{t} \,{e}^{||A||(t-\tau)}d\tau$
$\le||B||{e}^{||A||t}\int_{0}^{t} \,{e}^{||A||(t-\tau)}d\tau\le||B||{e}^{||A||t}t{e}^{||A||t}$
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 $[a,b]$
very grateful
Sarrah
NB: the first result is trivial, it's the 2nd result which i care for, i.e. the uniform convergence of the series.
thanks
$X(t)=e^{At}+\int_{0}^{t} \,e^{A(t-\tau)}\epsilon BX(\tau) d\tau$ (1) ,
of matrix solution $X(t)$ and $X(0)=I$, $A,B\in{R}^{n\times n}$ , $t\in[a,b]$
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 $[a,b]$ for sufficiently small $\epsilon$
Proof: $||{P}_{1}(t)||\le \int_{0}^{t} \,e^{||A||(t-\tau)}||B|| {e}^{||A||\tau} d\tau$
by Bonnet 2nd mean value theorem $=||B||{e}^{||A||t}\int_{\zeta}^{t} \,{e}^{||A||(t-\tau)}d\tau$
$\le||B||{e}^{||A||t}\int_{0}^{t} \,{e}^{||A||(t-\tau)}d\tau\le||B||{e}^{||A||t}t{e}^{||A||t}$
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 $[a,b]$
very grateful
Sarrah
NB: the first result is trivial, it's the 2nd result which i care for, i.e. the uniform convergence of the series.
thanks
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