complex6_ harmonic function & conjugate

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Transcript of complex6_ harmonic function & conjugate

Section 25 Harmonic Function :A real valued function u(x, y) is said to be harmonic in a given domain D if

(i) u x , u xx , u y & u yy exist & they are continuous in D, (ii) u satisfies Laplace eqution 2 u = u xx + u yy = 0

Theorem 1 : If f(z) = u(x, y) + i v(x, y) is analytic in a domain D, then u & v are harmonic in D

Proof : f(z) = u(x, y) + i v(x, y) is analytic in a domain D u x = v y , u y = v x throughout in D u xx = v yx , u yx = v xx ....... (1) u xy = v yy , u yy = v xy ...........( 2)

Use the results: (i) f(z) is analytic at a point, then Re f(z) & Im f(z) have continuous partial derivatives of all orders at that point. (ii) Continuity of partial derivatives of u & v uxy = uyx, vxy = vyx uxx+ uyy= 0 & vxx +vyy= 0 ( from 1 and 2) Hence proved.

Definition : Harmonic Conjugate : Let u and v be two harmonic functions in a domain D and there first partial derivatives satisfy CR equations. u x = v y , u y = v x through out in D ........... (1) Then v is said be Harmonic Conjugate of u.

Re mark 1 : v is a harmonic conjugate of u does not imply u is a harmonic conjugate of v. For, if u is a harmonic conjugate of v, then v x = u y & v y = u x which is not same as (1)

Remark 2 : v is a harmonic conjugate of u u is a harmonic conjugate of - v as - v x = u y , v y = u x i.e. u x = v y & u y = v x which is same as (1)

Theorem 2 : If a function f(z) = u(x, y) + i v(x, y) is analytic in a domain D iff v is a harmonic conjugate of u.

Page 74, Q.7 Let f(z) be analytic in a domain D. Prove that f(z) must be constant in D if (a) f(z) is real valued z in D. (b) f(z) is analytic in D. (c) f(z) is constant in D.

Solution : Since f(z) is analytic in a domain D. u x = v y , u y = v x & f ( z ) = u x + i v x (1) (2)

(a) Given f(z) is a real valued function z D f(z) = u (x, y) + i v (x, y), v(x, y) = 0 (x, y) D.

v(x, y) = 0 v x = 0, v y = 0 u x = v y , u y = v x u x ( x , y ) = 0 = u y ( x , y ) ( x , y ) D (2) f ( z ) = 0 z D f ( z ) constant z D.

(b ) f(z) = u( x, y ) + i v ( x , y ) f(z) = u( x , y ) i v ( x , y ) f(z) is analytic in D u & - v satisfy C R equations , viz. u x = v y , u y = (v x ) = v x ....... (3)

(1) and (3) ux = vy, ux = -vy

ux = 0uy = -vx, uy = vx vx = 0

f (z) = ux + i vx = 0 z D f(z) constant z D

(c ) f(z) = constant in D Let f(z) = c If c = 0, then f ( z ) = 0 z D. f ( z ) = cons tan t z D.

Assume c 0. Then, f(z) = c 0. f(z) = c f ( z ). f ( z ) = c 2 c f ( z) = f ( z) f ( z ) is analytic in D f ( z ) is analytic in D by case (b) : f(z) is constant in D2 2 2

Q.10 Show that u is harmonic & find a harmonic conjugate v when (a) u ( x, y ) = 2 x (1 - y ) u x = 2(1 y ), u xx = 0 u y = 2 x, u yy = 0 u xx + u yy = 0 u is harmonic.

v is a harmonic conjugate of u CR Equations are satisfied i.e. u x = v y , u y = v x Now v y = u x = 2(1 y ) 2 v = 2 y y + ( x) v x = ( x) = u y = 2 x

( x ) = 2 x ( x) = x + c2

v = 2 y y + x + c2 2

(b)u ( x, y ) = sinh x sin y u x = cosh x sin y, u xx = sinh x sin y u y = sinh x. cos y, u yy = sinh x sin y u xx + u yy = 0

Let v be a harmonic conjugate of u u x = v y , u y = v x v y = cosh x sin y v = cosh . cos y + ( x) v x = sinh x cos y + ( x) u y = sinh x. cos y ( x) = 0 ( x) = c v = cosh x. cos y + c

Problem: Show that if v and V are harmonic conjugates of u in a domain D, Then v(x,y) and V(x,y) can differ at most by an additive constant.

Solution : v is a harmonic conjugate of u ux = vy , u y = v x (1) v is a harmonic conjugate of u u x = v y , u y = v x ( 2)

(1) & (2) v x = Vx , v y = Vy v = V + (y), v = V + (x) v y = Vy + (y), v x = Vx + (x) (y) = 0, (x) = 0 (y) = c1 , (x) = c 2 v V = constant