Math  /  Calculus

Question1. Let the function f(z)=u(x,y)+iv(x,y)f(z)=u(x, y)+i v(x, y) and it satisfies the Cauchy-Riemann conditions: u(x,y)x=v(x,y)yu(x,y)y=v(x,y)x\begin{array}{l} \frac{\partial u(x, y)}{\partial x}=\frac{\partial v(x, y)}{\partial y} \\ \frac{\partial u(x, y)}{\partial y}=-\frac{\partial v(x, y)}{\partial x} \end{array} then f(z)f(z) is said to be analytical and v(x,y)v(x, y) is said to be harmonic conjugate of u(x,y)u(x, y). It is said to be harmonic if 2u(x,y)x2+2u(x,y)y2=02v(x,y)x2+2v(x,y)y2=0\begin{array}{l} \frac{\partial^{2} u(x, y)}{\partial x^{2}}+\frac{\partial^{2} u(x, y)}{\partial y^{2}}=0 \\ \frac{\partial^{2} v(x, y)}{\partial x^{2}}+\frac{\partial^{2} v(x, y)}{\partial y^{2}}=0 \end{array}
Show that the following u(x,y)u(x, y) are harmonic and find its harmonic conjugate (a) u(x,y)=2x(1y)u(x, y)=2 x(1-y) (b) u(x,y)=sinh(x)sin(y)u(x, y)=\sinh (x) \sin (y)

Studdy Solution
Use Cauchy-Riemann conditions to find v(x,y) v(x, y) :
ux=vycosh(x)sin(y)=vy \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \Rightarrow \cosh(x) \sin(y) = \frac{\partial v}{\partial y}
Integrate with respect to y y :
v(x,y)=cosh(x)cos(y)+g(x) v(x, y) = -\cosh(x) \cos(y) + g(x)
Use the second Cauchy-Riemann condition:
uy=vxsinh(x)cos(y)=vx \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \Rightarrow \sinh(x) \cos(y) = -\frac{\partial v}{\partial x}
Integrate with respect to x x :
vx=sinh(x)cos(y)v(x,y)=cosh(x)cos(y)+h(y) \frac{\partial v}{\partial x} = -\sinh(x) \cos(y) \Rightarrow v(x, y) = -\cosh(x) \cos(y) + h(y)
Combine results:
v(x,y)=cosh(x)cos(y) v(x, y) = -\cosh(x) \cos(y)
The harmonic conjugates are: (a) v(x,y)=2yy2+x2 v(x, y) = 2y - y^2 + x^2 (b) v(x,y)=cosh(x)cos(y) v(x, y) = -\cosh(x) \cos(y)

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