212HomeWork3sol - Physics 212A Homework#3 Name 1 If f[z...

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Physics 212A Homework #3 Name: 1. If f[z] =u[x,y]+ v[x,y] is analytic, and u[x,y]= - 3x 2 + x 4 + y - 9x 2 y + 3y 2 - 6x 2 y 2 + 3y 3 + y 4 , what is v[x,y] ? We know u : u1 = - 3x 2 + x 4 + y - 9x 2 y + 3y 2 - 6x 2 y 2 + 3y 3 + y 4 - 3x 2 + x 4 + y - 9x 2 y + 3y 2 - 6x 2 y 2 + 3y 3 + y 4 make sure it is harmonic : D @ u1, 8 x,2 <D + D @ u1, 8 y,2 <D 0 The gradient of v is ( using the CR relations ) : gradv = 8 - D @ u1,y D ,D @ u1,x D< 9 - 1 + 9x 2 - 6y + 12x 2 y - 9y 2 - 4y 3 , - 6x + 4x 3 - 18xy - 12xy 2 = int1 = gradv. 8 dx,dy < dy I - 6x + 4x 3 - 18xy - 12xy 2 M + dx I - 1 + 9x 2 - 6y + 12x 2 y - 9y 2 - 4y 3 M Start from {0, 0} and integrate gradv up y axis I1 = Integrate @ int1 . 8 x fi 0,dx fi 0,dy fi 1 < , 8 y,0,y <D 0 Now integrate along x : I2 = Integrate @ int1 . 8 dx fi 1,dy fi 0 < , 8 x,0,x <D - x + 3x 3 - 6xy + 4x 3 y - 9xy 2 - 4xy 3
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v1 = I1 + I2 - x + 3x 3 - 6xy + 4x 3 y - 9xy 2 - 4xy 3 verfy that it is harmonic D @ v1, 8 x,2 <D + D @ v1, 8 y,2 <D 0 Verfy that the CR conditions are OK D @ u1,x D D @ v1,y D True D @ u1,y D - D @ v1,x D True The analytic function f[z] with real part u1 and imaginary part v1 can be found as follows : f1 = u1 + Iv1 . 8 x fi H z + Conjugate @ z DL 2,y fi H z - Conjugate @ z DL H 2I L< Simplify z H + z L 3 2 212HomeWork3sol.nb
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ComplexExpand @ Re @ f1 .z fi x + Iy DD - 3x 2 + x 4 + y - 9x 2 y + 3y 2 - 6x 2 y 2 + 3y 3 + y 4 2. Calculate the principal value of 0 ¥ Cos @ m x D x 2 - b 2 x with m and b real using several techniques a) Evaluate the appropriate contour integral and utilize the rule that poles on the contour contribute Π *residue Use ª mz z 2 - b 2 ; the real part is the integral we are after. The integrand is symmetric in x , so we can extend the range of integration and take half the result. The integral around C1 vanishes for m>0. The poles are at x= b , x= -b. They are on the real axis, so we take Π times the sum of the residues C1 C2 b -b r1 = Residue B ª mz z 2 - b 2 , 8 z, b <F ª bm 2b 212HomeWork3sol.nb 3
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r2 = Residue B ª mz z 2 - b 2 , 8 z, - b <F - ª - bm 2b The value of ª mz z 2 - b 2 dz is cint = Π H r1 + r2 L - ª - bm 2b + ª bm 2b Π we want 1/2 the real part, which is rp2 = ComplexExpand @ Re @ cint DD 2 - Π Sin @ bm D 2b b) Use contour integration to calculate the integral for b->b0+ Ε and then take the limit as Ε ->0. ª mz z 2 - b 2 . b fi b0 + Ε ª mz z 2 - H b0 + Ε L 2 The location of the poles is z2 = z .Solve A z 2 - H b0 + Ε L 2 0,z E 8 - b0 - Ε ,b0 + Ε < The first one is outside the contour, so does not contribute. Including the contribution from the second one gives cint2 = Residue B ª mz z 2 - H b0 + Ε L 2 , 8 z, b0 + Ε <F H 2 Π L ª b0m - m Ε Π b0 + Ε Take the limit : Limit @ % , Ε fi 0 D ª b0m Π b0 4 212HomeWork3sol.nb
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again, we want 1/2 the real part ComplexExpand @ Re @ % DD 2 - Π Sin @ b0m D 2b0 , same as before. In this method, we pick up the full contribution of one pole and none of the other, instead of half of each. c) Use Integrate, but avoid the singularity by omitting a symmetric region 2 Ε wide centered on the singularity . The integrals can be expressed in terms of special functions. Take the limit as Εfi 0 I1 = Integrate B Cos @ mx D x 2 - b 2 , 8 x,0, b - ep < , Assumptions fi 8 m > 0, b > 0, b > ep > 0 <F 1 2b H Cos @ bm D H CosIntegral @ epm D - CosIntegral @ 2bm - epm DL + Sin @ bm D H SinIntegral @ epm D - SinIntegral @ 2bm - epm DLL I2 = Integrate B Cos @ mx D x 2 - b 2 , 8 x, b + ep, ¥ < , Assumptions fi 8 m > 0, b > 0, b > ep > 0 <F 1 2b H Cos @
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