§62方程常点邻域内的解 三u1 G1∩G 试证明:1仍是方程(6.3)的解 证设 d +p(2)-+q(2)i1=g(2) g(2)在G2内解析.因为m1是方程(6.3)在区域G1内的解,故在子区域G1∩G2内,仍满足方程 p( )+g(2) dz 而在此子区域内,1(2)≡1(2),故 d2W1 +p(2) q(2)1=0, 即g(z)≡0,z∈G1∩G2,根据解析函数的唯一性,立即证得 g(2)≡0 亦即v1在G2内满足方程 +p(x) 例65设u1和ω2都是方程(63)的两个线性无关解,且均在区域G1内解析.若 1和2分别是t1和m2在区域G2内的解析延拓,即在z∈G1∩G2中 试证:如1和m2仍线性无关 证由例6.4知,m1和v2仍是方程(在G2内)的解.因为u1和u2线性无关, U1 2 ≠0,z∈G1 设 W1 7D g(2)在G2内解析.由于在z∈G1∩G2中 uI三1,2 故g(2)≠0,z∈G1∩G2.仍然根据解析函数的唯一性,就证得 g(2)≠0,z∈G 所以,1和2(在G2内)仍线性无关.口§6.2 ✡☛✞➦üý þ☞✏ ✒ 6 ✓ w1 ≡ we1, z ∈ G1 T G2, (6.4) ◗ ➽ ➾➍ we1 ❘▼❋● (6.3) ❍▲❑ ❙ ❖ d 2we1 dz 2 + p(z) dwe1 dz + q(z)we1 = g(z), g(z) ❹ G2 ➱▲❱❑➤ ❊ w1 ▼❋● (6.3) ❹●Ü G1 ➱❍▲❯❚ ❹❯●Ü G1 T G2 ➱❯❘❱❲❋● d 2w1 dz 2 + p(z) dw1 dz + q(z)w1 = 0. ✰❹ ➥ ❯●Ü ➱❯ w1(z) ≡ we1(z) ❯ ❚ d 2we1 dz 2 + p(z) dwe1 dz + q(z)we1 = 0, z ∈ G1 T G2, ➩ g(z) ≡ 0, z ∈ G1 T G2 ❑ ×Ø▲❱☞❏❍Ó➁❲❯❳➩➽ú g(z) ≡ 0, z ∈ G2, ❨ ➩ we1 ❹ G2 ➱❱❲❋● d 2we1 dz 2 + p(z) dwe1 dz + q(z)we1 = 0. ➅ 6.5 ❖ w1 ❈ w2 → ▼ ❋ ● (6.3) ❍ ➌ ➂ ✔ ❲ ➙ ☛ ▲ ❯Õ ò ❹ ● Ü G1 ➱▲ ❱ ❑P we1 ❈ we2 ❮❚▼ w1 ❈ w2 ❹●Ü G2 ➱❍▲❱▲▼❯➩❹ z ∈ G1 T G2 ❽ w1 ≡ we1, w2 ≡ we2. ◗ ➽➍ we1 ❈ we2 ❘✔❲ ➙ ☛❑ ❙ P✹ 6.4 ❩❯ we1 ❈ we2 ❘▼❋● (❹ G2 ➱) ❍▲❑➤ ❊ w1 ❈ w2 ✔❲ ➙ ☛❯ ∆[w1, w2] ≡      w1 w2 w 0 1 w 0 2      6= 0, z ∈ G1. ❖ ∆[we1, we2] ≡      we1 we2 we 0 1 we 0 2      = g(z), g(z) ❹ G2 ➱▲❱❑P✮❹ z ∈ G1 T G2 ❽❯ w1 ≡ we1, w2 ≡ we2, ❚ g(z) 6= 0, z ∈ G1 T G2 ❑❘ á×Ø▲❱☞❏❍Ó➁❲❯î➽ú g(z) 6= 0, z ∈ G2. ➎➏❯ we1 ❈ we2(❹ G2 ➱) ❘✔❲ ➙ ☛❑