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

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Unformatted text preview: Physics 212A Homework #3 Name: 1. If f[z] =u[x,y]+ v[x,y] is analytic, and u[x,y]=- 3 x 2 + x 4 + y- 9 x 2 y + 3 y 2- 6 x 2 y 2 + 3 y 3 + y 4 , what is v[x,y] ? We know u : u1 = - 3 x 2 + x 4 + y- 9 x 2 y + 3 y 2- 6 x 2 y 2 + 3 y 3 + y 4- 3 x 2 + x 4 + y- 9 x 2 y + 3 y 2- 6 x 2 y 2 + 3 y 3 + y 4 make sure it is harmonic : D @ u1, 8 x, 2 <D + D @ u1, 8 y, 2 <D The gradient of v is ( using the CR relations ) : gradv = 8- D @ u1, y D , D @ u1, x D< 9- 1 + 9 x 2- 6 y + 12 x 2 y- 9 y 2- 4 y 3 ,- 6 x + 4 x 3- 18 x y- 12 x y 2 = int1 = gradv. 8 dx, dy < dy I- 6 x + 4 x 3- 18 x y- 12 x y 2 M + dx I- 1 + 9 x 2- 6 y + 12 x 2 y- 9 y 2- 4 y 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 Now integrate along x : I2 = Integrate @ int1 . 8 dx fi 1, dy fi < , 8 x, 0, x <D- x + 3 x 3- 6 x y + 4 x 3 y- 9 x y 2- 4 x y 3 v1 = I1 + I2- x + 3 x 3- 6 x y + 4 x 3 y- 9 x y 2- 4 x y 3 verfy that it is harmonic D @ v1, 8 x, 2 <D + D @ v1, 8 y, 2 <D 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 + I v1 . 8 x fi H z + Conjugate @ z DL 2, y fi H z- Conjugate @ z DL H 2 I L< Simplify z H + z L 3 2 212HomeWork3sol.nb ComplexExpand @ Re @ f1 . z fi x + I y DD- 3 x 2 + x 4 + y- 9 x 2 y + 3 y 2- 6 x 2 y 2 + 3 y 3 + y 4 2. Calculate the principal value of 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 *residue Use m z 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 times the sum of the residues C1 C2 b-b r1 = Residue B m z z 2- b 2 , 8 z, b <F b m 2 b 212HomeWork3sol.nb 3 r2 = Residue B m z z 2- b 2 , 8 z,- b <F- - b m 2 b The value of m z z 2- b 2 dz is cint = H r1 + r2 L- - b m 2 b + b m 2 b we want 1/2 the real part, which is rp2 = ComplexExpand @ Re @ cint DD 2- Sin @ b m D 2 b b) Use contour integration to calculate the integral for b->b0+ and then take the limit as ->0. m z z 2- b 2 . b fi b0 + m z z 2- H b0 + L 2 The location of the poles is z2 = z . Solve A z 2- H b0 + L 2 0, z E 8- b0- , b0 + < The first one is outside the contour, so does not contribute. Including the contribution from the second one gives cint2 = Residue B m z z 2- H b0 + L 2 , 8 z, b0 + <F H 2 L b0 m- m b0 + Take the limit : Limit @ % , fi D b0 m b0 4 212HomeWork3sol.nb again, we want 1/2 the real part ComplexExpand @ Re @ % DD 2- Sin @ b0 m D 2 b0 , same as before. In this method, we pick up the full contribution of one pole and none of the other, instead , same as before....
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212HomeWork3sol - Physics 212A Homework #3 Name: 1. If f[z]...

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