Holomorfik funktsiyani topish uchun Milne-Tomson usuli - Milne-Thomson method for finding a holomorphic function

Matematikada Milne-Tomson usuli a ni topish usuli holomorfik funktsiya haqiqiy yoki xayoliy qismi berilgan.^[1] Uning nomi berilgan Lui Melvil Milne-Tomson.

Kirish

Ruxsat bering ${ displaystyle z = x + iy}$ va ${ displaystyle { bar {z}} = x-iy}$ qayerda ${ displaystyle x}$ va ${ displaystyle y}$ bor haqiqiy.

Ruxsat bering ${ displaystyle f (z) = u (x, y) + iv (x, y)}$ har qanday bo'ling holomorfik funktsiya.

1-misol: ${ displaystyle z ^ {4} = (x ^ {4} -6x ^ {2} y ^ {2} + y ^ {4}) + i (4x ^ {3} y-4xy ^ {3})}$

2-misol: ${ displaystyle exp (iz) = cos (x) exp (-y) + i sin (x) exp (-y)}$

Uning maqolasida^[1], Milne-Tomson topish muammosini ko'rib chiqadi ${ displaystyle f (z)}$ qachon 1. ${ displaystyle u (x, y)}$ va ${ displaystyle v (x, y)}$ berilgan, 2. ${ displaystyle u (x, y)}$ berilgan va ${ displaystyle f (z)}$ haqiqiy o'qda haqiqiy, 3. faqat ${ displaystyle u (x, y)}$ berilgan, 4. faqat ${ displaystyle v (x, y)}$ berilgan. U haqiqatan ham 3 va 4-sonli masalalar bilan qiziqadi, ammo 3-chi va 4-chi masalalarning javoblarini isbotlash uchun 1 va 2-sonli masalalarga javoblar kerak bo'ladi.

1^st muammo

Muammo: ${ displaystyle u (x, y)}$ va ${ displaystyle v (x, y)}$ ma'lum; nima bu ${ displaystyle f (z)}$?

Javob: ${ displaystyle f (z) = u (z, 0) + iv (z, 0)}$

So'z bilan aytganda: holomorfik funktsiya ${ displaystyle f (z)}$ qo'yish orqali olish mumkin ${ displaystyle x = z}$ va ${ displaystyle y = 0}$ yilda ${ displaystyle u (x, y) + iv (x, y)}$.

1-misol: bilan ${ displaystyle u (x, y) = x ^ {4} -6x ^ {2} y ^ {2} + y ^ {4}}$ va ${ displaystyle v (x, y) = 4x ^ {3} y-4xy ^ {3}}$ biz olamiz ${ displaystyle f (z) = z ^ {4}}$.

2-misol: bilan ${ displaystyle u (x, y) = cos (x) exp (-y)}$ va ${ displaystyle v (x, y) = sin (x) exp (-y)}$ biz olamiz ${ displaystyle f (z) = cos (z) + i sin (z) = exp (iz)}$.

Isbot:

Birinchi juft ta'riflardan ${ displaystyle x = { frac {z + { bar {z}}} {2}}}$ va ${ displaystyle y = { frac {z - { bar {z}}} {2i}}}$.

Shuning uchun ${ displaystyle f (z) = u chap ({ frac {z + { bar {z}}} {2}} , { frac {z - { bar {z}}} {2i}} o'ng) + iv chap ({ frac {z + { bar {z}}} {2}} , { frac {z - { bar {z}}} {2i}} o'ng)}$.

Bu qachon bo'lsa ham, bu shaxsiyat ${ displaystyle x}$ va ${ displaystyle y}$ haqiqiy emas, ya'ni. ikkita o'zgaruvchi ${ displaystyle z}$ va ${ displaystyle { bar {z}} }$ mustaqil deb hisoblanishi mumkin. Qo'yish ${ displaystyle { bar {z}} = z}$ biz olamiz ${ displaystyle f (z) = u (z, 0) + iv (z, 0)}$.

2^nd muammo

Muammo: ${ displaystyle u (x, y)}$ ma'lum, ${ displaystyle v (x, y)}$ noma'lum, ${ displaystyle f (x + i0)}$ haqiqiy; nima bu ${ displaystyle f (z)}$?

Javob: ${ displaystyle f (z) = u (z, 0)}$.

Bu erda faqat 1-misol amal qiladi: bilan ${ displaystyle u (x, y) = x ^ {4} -6x ^ {2} y ^ {2} + y ^ {4}}$ biz olamiz ${ displaystyle f (z) = z ^ {4}}$.

Isbot: "${ displaystyle f (x + i0)}$ haqiqiydir "degan ma'noni anglatadi ${ displaystyle v (x, 0) = 0}$. Bunday holda 1-savolga javob bo'ladi ${ displaystyle f (z) = u (z, 0)}$.

3^rd muammo

Muammo: ${ displaystyle u (x, y)}$ ma'lum, ${ displaystyle v (x, y)}$ noma'lum; nima bu ${ displaystyle f (z)}$?

Javob: ${ displaystyle f (z) = u (z, 0) -i int u_ {y} (z, 0) dz}$ (qayerda ${ displaystyle u_ {y} (x, y)}$ ning qisman hosilasi hisoblanadi ${ displaystyle u (x, y)}$ munosabat bilan ${ displaystyle y}$).

1-misol: bilan ${ displaystyle u (x, y) = x ^ {4} -6x ^ {2} y ^ {2} + y ^ {4}}$ va ${ displaystyle u_ {y} (x, y) = - 12x ^ {2} y + 4y ^ {3}}$ biz olamiz ${ displaystyle f (z) = z ^ {4} + iC}$ haqiqiy, ammo aniqlanmagan ${ displaystyle C}$.

2-misol: bilan ${ displaystyle u (x, y) = cos (x) exp (-y)}$ va ${ displaystyle u_ {y} (x, y) = - cos (x) exp (-y)}$ biz olamiz ${ displaystyle f (z) = cos (z) + i int cos (z) dz = cos (z) + i ( sin (z) + C) = exp (iz) + iC}$.

Isbot: Bu quyidagidan kelib chiqadi ${ displaystyle f (z) = u (z, 0) + i int v_ {x} (z, 0) dz}$ va 2^nd Koshi-Riman tenglamasi ${ displaystyle u_ {y} (x, y) = - v_ {x} (x, y)}$.

4^th muammo

Muammo: ${ displaystyle u (x, y)}$ noma'lum, ${ displaystyle v (x, y)}$ ma'lum; nima bu ${ displaystyle f (z)}$?

Javob: ${ displaystyle f (z) = int v_ {y} (z, 0) dz + iv (z, 0)}$.

1-misol: bilan ${ displaystyle v (x, y) = 4x ^ {3} y-4xy ^ {3}}$ va ${ displaystyle v_ {y} (x, y) = 4x ^ {3} -12xy ^ {2}}$ biz olamiz ${ displaystyle f (z) = int 4z ^ {3} dz + i0 = z ^ {4} + C}$ haqiqiy, ammo aniqlanmagan ${ displaystyle C}$.

2-misol: bilan ${ displaystyle v (x, y) = sin (x) exp (-y)}$ va ${ displaystyle v_ {y} (x, y) = - sin (x) exp (-y)}$ biz olamiz ${ displaystyle f (z) = - int sin (z) dz + i sin (z) = cos (z) + C + i sin (z) = exp (iz) + C}$.

Isbot: Bu quyidagidan kelib chiqadi ${ displaystyle f (z) = int u_ {x} (z, 0) dz + iv (z, 0)}$ va 1^st Koshi-Riman tenglamasi ${ displaystyle u_ {x} (x, y) = v_ {y} (x, y)}$.

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