只须证明级数∑1（纵(）一张-1(》一致收敛即可。可归纳证明 因为∑=cm-1收敛，故由Weierters的M-判别法可得结论。 )唯一性 假设有二解=(),=(),既有 uo）=0+f,u(ed, a=0+金ah. 则有 )-el=le,)-,a训 ≤La(a)-(ld ≤KIr-xol。K=maxu(o)-(川 由归纳可得 令w()=lu(r)-(ed,则有 w'(z)<Lw(r),2 ro,w()20. 于是 e-r-olu(≤0→e-olu()≤m(ro)=0 →w()≤0 →m()=0x≥x0êLy²?Í P∞ k=1(yk(x) − yk−1(x))òó¬Ò=å"å8By² |yn(x) − yn−1(x)| ≤ M L (L|x − x0|) n n! ≤ M L (Lh) n n! . œè P (Lh) n n! = e Ln − 1¬ÒßdWeierterasM−O{å(ÿ" 4) çò5 bk)y = u(x), y = v(x),Qk u(x) = y0 + Z x x0 f(x, u(x))dx, v(x) = y0 + Z x x0 f(x, v(x))dx. Kk |u(x) − v(x)| = | Z x x0 [f(x, u(x)) − f(x, v(x))]dx| ≤ L Z x x0 |u(x) − v(x)|dx, ≤ KL|x − x0|, K = max I |u(x) − v(x)| d8Bå |u(x) − v(x)| ≤ K (L|x − x0|) n n! → 0, as n → ∞. ************** -w(x) = R x x0 |u(x) − v(x)|dxßKk w 0 (x) ≤ Lw(x), x ≥ x0, w(x) ≥ 0. u¥ [e −L(x−x0)w(x)]0 ≤ 0 =⇒ e −L(x−x0)w(x) ≤ w(x0) = 0 =⇒ w(x) ≤ 0 =⇒ w(x) = 0, x ≥ x0