所以在有界区间上K(t)是有界的.再来看它是否满足更新方 程.由式（1)，有 K(t)=H(t)+M*H(t) =H(t)+(F+F*M)*H(t) =H(t)+F*(H+M*H)(t) =H(t)+F*K(t) 17/53 最后要证惟一性，只需证明更新方程的任何解都有（1)式 的形式.设K(t)是更新方程的解，并且满足有界性条件，则 由 衣(t)=H+F*区(t) 连续代换区(t),有 GoBack FullScreen Close Quit
17/53 kJ Ik J I GoBack FullScreen Close Quit §±3k.´m˛K(t)¥k.. 25wߥƒ˜vç#ê ß. d™£1§ßk K(t) = H(t) + M ∗ H(t) = H(t) + (F + F ∗ M) ∗ H(t) = H(t) + F ∗ (H + M ∗ H)(t) = H(t) + F ∗ K(t) Åáyéò5ßêIy²ç#êß?¤)—k£1§™ /™. K˜ (t)¥ç#êß)ßøÖ˜vk.5^áßK d K˜ (t) = H + F ∗ K˜ (t) ÎYìÜK˜ (t), k
花 衣=H(t)+F*(H+F*K)(t) =H(t)+F*H(t)+F*(F*K)(t) =H(t)+F*H(t)+F2*K(t) =H(t)+F*H(t)+F*(H+F*)(t) =H(t)+F*H(t)+F2*H(t)+F3*K(t) 18/53 =H+ A)H+RR0 n-1 k=1 注意到对任何t, IE*K()=IR(t-2)dFa() ≤{sup|K(t-x)}·Fn(t) GoBack 0≤x≤t FullScreen Close Quit
18/53 kJ Ik J I GoBack FullScreen Close Quit K˜ = H(t) + F ∗ (H + F ∗ K˜ )(t) = H(t) + F ∗ H(t) + F ∗ (F ∗ K˜ )(t) = H(t) + F ∗ H(t) + F2 ∗ K˜ (t) = H(t) + F ∗ H(t) + F2 ∗ (H + F ∗ K˜ )(t) = H(t) + F ∗ H(t) + F2 ∗ H(t) + F3 ∗ K˜ (t) = · · · · · · = H + X n−1 k=1 Fk ! ∗ H + Fn ∗ K˜ (t). 5øÈ?¤t, |Fn ∗ K˜ (t)| = | Z t 0 K˜ (t − x)dFn(x)| ≤ { sup 0≤x≤t |K˜ (t − x)|} · Fn(t)
由假定supo≤z≤t|K(t-x)川<o,并且M(t)=∑1Fn(t)< o知， l1imFn(t)=0,∀t 从而 lim|Fn*(t)川=0. 而 19/53 lim n→0∞ (空=（区小m =M*H(t) 于是推出 m-1 交(t)=H(t)+lim n→oo *H0+R.+K国 k=1 GoBack =H(t)+M*H(t), FullScreen Close Quit
19/53 kJ Ik J I GoBack FullScreen Close Quit db½sup0≤x≤t |K˜ (t−x)| < ∞,øÖM(t) = P∞ n=1 Fn(t) < ∞, lim n→∞ Fn(t) = 0, ∀t l lim n→∞ |Fn ∗ K˜ (t)| = 0. lim n→∞ X n−1 k=1 Fk ∗ H(t) ! = X ∞ k=1 Fk ! ∗ H(t) = M ∗ H(t) u¥Ì— K˜ (t) = H(t) + lim n→∞ "X n−1 k=1 Fk ∗ H(t) + Fn ∗ K˜ (t) # = H(t) + M ∗ H(t)