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

STEP 1

1. We are given two functions u(x,y) u(x, y) and need to show they are harmonic.
2. We need to find the harmonic conjugate v(x,y) v(x, y) for each function.
3. The Cauchy-Riemann conditions and harmonic conditions are provided.

STEP 2

1. Verify that u(x,y)=2x(1y) u(x, y) = 2x(1-y) is harmonic.
2. Find the harmonic conjugate v(x,y) v(x, y) for u(x,y)=2x(1y) u(x, y) = 2x(1-y) .
3. Verify that u(x,y)=sinh(x)sin(y) u(x, y) = \sinh(x) \sin(y) is harmonic.
4. Find the harmonic conjugate v(x,y) v(x, y) for u(x,y)=sinh(x)sin(y) u(x, y) = \sinh(x) \sin(y) .

STEP 3

Calculate the second partial derivatives of u(x,y)=2x(1y) u(x, y) = 2x(1-y) :
ux=2(1y) \frac{\partial u}{\partial x} = 2(1-y) 2ux2=0 \frac{\partial^2 u}{\partial x^2} = 0
uy=2x \frac{\partial u}{\partial y} = -2x 2uy2=0 \frac{\partial^2 u}{\partial y^2} = 0
Verify harmonic condition:
2ux2+2uy2=0+0=0 \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 + 0 = 0

STEP 4

Use Cauchy-Riemann conditions to find v(x,y) v(x, y) :
ux=vy2(1y)=vy \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \Rightarrow 2(1-y) = \frac{\partial v}{\partial y}
Integrate with respect to y y :
v(x,y)=2yy2+g(x) v(x, y) = 2y - y^2 + g(x)
Use the second Cauchy-Riemann condition:
uy=vx2x=vx \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \Rightarrow -2x = -\frac{\partial v}{\partial x}
Integrate with respect to x x :
vx=2xv(x,y)=x2+h(y) \frac{\partial v}{\partial x} = 2x \Rightarrow v(x, y) = x^2 + h(y)
Combine results:
v(x,y)=2yy2+x2 v(x, y) = 2y - y^2 + x^2

STEP 5

Calculate the second partial derivatives of u(x,y)=sinh(x)sin(y) u(x, y) = \sinh(x) \sin(y) :
ux=cosh(x)sin(y) \frac{\partial u}{\partial x} = \cosh(x) \sin(y) 2ux2=sinh(x)sin(y) \frac{\partial^2 u}{\partial x^2} = \sinh(x) \sin(y)
uy=sinh(x)cos(y) \frac{\partial u}{\partial y} = \sinh(x) \cos(y) 2uy2=sinh(x)sin(y) \frac{\partial^2 u}{\partial y^2} = -\sinh(x) \sin(y)
Verify harmonic condition:
2ux2+2uy2=sinh(x)sin(y)sinh(x)sin(y)=0 \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \sinh(x) \sin(y) - \sinh(x) \sin(y) = 0

STEP 6

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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