Let u u x y exx cos y y sin y Show that u ux y is a harm

Let u = u (x, y) = e^x(x cos y - y sin y). Show that u = u(x, y) is a harmonic function. Find the harmonic conjugate of u = u(x, y) Write the analytic function f(z) = u(x, y) + v(x, y)i in the terms of z. we need to show that u_xx + u_yy = 0. So we have: u_x = partial differential u (x, y)/partial differential x = e^x (x cos y - y sin y) + e^x cos y implies u_xx = partial differential^2 u (x, y)/partial differential x^2 = e^x (x cos y - y sin y) + e^x cos y + e^x cos y u_y = partial differential u (x, y)/partial differential y = e^x (-x sin y - sin y - y cos y) implies u_yy = partial differential^2 u (x, y)/partial differential y^2 = e^x (-x cos y - cos y - cos y + y sin y) u_xx + u_yy = e^x (x cos y - y sin y) + e^x cos y + e^x cos y + e^x (-x cos y - cos y - cos y + y sin y) = e^x (x cos y - y sin y + cos y + cos y - x cos y - cos y - cos y + y sin y) = 0

Solution

Solution :

Given,
u = e^x(xcosy - ysiny)

( a )

u = (e^x.x.cosy - e^x.y.siny) -------------( 1 )

Here du/dx means partial differentiation of u with respect to \' x \'

diff 1 w.r.t \' x \' we get,

du/dx = cosy(e^x.1 + x.e^x) - y.siny.(e^x)

du/dx = e^x.cosy + x.e^x.cosy - e^x.y.siny ---------------- ( 2 )

diff 2 again w.r.t x we get,

d²u/dx² = cosy.(e^x) + cosy(e^x.1 + x.e^x)...

d²u/dx² = e^x.cosy + e^x.cosy + x.e^x.cosy...

d²u/dx² = 2.e^x.cosy + x.e^x.cosy - y.e^x....

Now diff 1 again w.r.t y we get,

du/dy = e^x.x.(-siny) - e^x(y.cosy + siny.1)

du/dy = - x.e^x.siny - y.e^x.cosy - e^x.siny....

Now diff 4 w.r.t y again we get,

d²u/dy² = - x.e^x.(cosy) - e^x(y.( - siny) +...

d²u/dy² = - x.e^x.cosy + y.e^x.siny - e^x...

d²u/dy² = - x.e^x.cosy + y.e^x.siny - 2.e^x....

Now a function is said to be harmonic if ,

d²u/dx² + d²u/dy² = 0

Now adding we get,

(2.e^x.cosy + x.e^x.cosy - y.e^x.siny) + ( - 2...

since both cancel out each other thereby giving zero thus u = u(x, y) is a harmonic function.

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( b )

let us consider cauchy - riemann equation which is given by,

du/dx = dv/dy and dv/dx = - du/dy

since we are having the real part let us find the imaginary part,

dv/dy = e^x(cosy + xcosy - ysiny)

dv = e^x(cosy + xcosy - ysiny).dy

Now integrating we get,

INT(dv) = e^x[ INT(cosy)dy + x.INT(cosy)dy - INT(ysiny)dy ] + f(x)   here INT = INTEGRATION

v = e^x.(siny + xsiny + ycosy - siny) + f(x)

v = x.e^x.siny + e^x.y.cosy + f(x)

Now,

dv/dx = e^x(x.siny + y.cosy + siny)

dv = e^x(x.siny + y.cosy + siny).dx

Integrating we get,

INT(dv) = siny.INT(x.e^x)dx + y.cosy.INT(e^x)dx + siny...

v = x.e^x.siny + e^x.y.cosy + g(y)----------...

Now Comparing the value of f(x)and g(y) are,

f(x) = 0 and g(x) = 0,

So therefore,

v = x.e^x.siny + e^x.y.cosy

v = e^x.(x.siny + y.cosy)

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( c )

Let us consider,

f(z) = u + iv

f(z) = e^x.(x.cosy - y.siny) + i.e^x.(x.siny + y.cosy...

Putting x = z and y = 0 in above equation.

f(z) = e^z.(z.cos0 - 0.sin0) + i.e^z(zsin0 + 0...

Now cos0 = 1 and sin0 = 0 so therefore,

f(z) = z.e^z