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Question:-1(c)

\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)
Prove that the function: \(u(x, y)=(x-1)^3-3 x y^2+3 y^2\) is harmonic and find its harmonic conjugate and the corresponding analytic function \(f(z)\) in terms of \(z \).
Expert Answer
untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f
Introduction:
In this problem, we are tasked with proving that the function u ( x , y ) = ( x 1 ) 3 3 x y 2 + 3 y 2 u ( x , y ) = ( x 1 ) 3 3 x y 2 + 3 y 2 u(x,y)=(x-1)^(3)-3xy^(2)+3y^(2)u(x, y) = (x-1)^3 – 3xy^2 + 3y^2u(x,y)=(x1)33xy2+3y2 is harmonic. Furthermore, we need to find its harmonic conjugate and express the corresponding analytic function f ( z ) f ( z ) f(z)f(z)f(z) in terms of z z zzz.
Step 1: Verification of Harmonicity:
To begin, we need to determine whether the given function u ( x , y ) u ( x , y ) u(x,y)u(x, y)u(x,y) is harmonic. We calculate its partial derivatives with respect to x x xxx and y y yyy:
u x = 3 ( x 1 ) 2 3 y 2 2 u x 2 = 6 ( x 1 ) u y = 6 x y + 6 y 2 u y 2 = 6 ( x 1 ) u x = 3 ( x 1 ) 2 3 y 2 2 u x 2 = 6 ( x 1 ) u y = 6 x y + 6 y 2 u y 2 = 6 ( x 1 ) {:[(del u)/(del x)=3(x-1)^(2)-3y^(2)],[(del^(2)u)/(delx^(2))=6(x-1)],[(del u)/(del y)=-6xy+6y],[(del^(2)u)/(dely^(2))=-6(x-1)]:}\begin{align*} \frac{\partial u}{\partial x} &= 3(x-1)^2 – 3y^2 \\ \frac{\partial^2 u}{\partial x^2} &= 6(x-1) \\ \frac{\partial u}{\partial y} &= -6xy + 6y \\ \frac{\partial^2 u}{\partial y^2} &= -6(x-1) \end{align*}ux=3(x1)23y22ux2=6(x1)uy=6xy+6y2uy2=6(x1)
We notice that 2 u x 2 + 2 u y 2 = 6 ( x 1 ) 6 ( x 1 ) = 0 2 u x 2 + 2 u y 2 = 6 ( x 1 ) 6 ( x 1 ) = 0 (del^(2)u)/(delx^(2))+(del^(2)u)/(dely^(2))=6(x-1)-6(x-1)=0\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 6(x-1) – 6(x-1) = 02ux2+2uy2=6(x1)6(x1)=0. This confirms that u u uuu is a harmonic function.
Step 2: Harmonic Conjugate and Analytic Function f ( z ) f ( z ) f(z)f(z)f(z):
Next, we aim to find the harmonic conjugate of u u uuu and express the corresponding analytic function f ( z ) f ( z ) f(z)f(z)f(z) in terms of z z zzz.
We recognize that u ( x , y ) u ( x , y ) u(x,y)u(x, y)u(x,y) is the real part of the analytic function f ( z ) f ( z ) f(z)f(z)f(z), where f ( z ) = u ( x , y ) + i v ( x , y ) f ( z ) = u ( x , y ) + i v ( x , y ) f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y)f(z)=u(x,y)+iv(x,y).
We utilize the Cauchy-Riemann equations, which state that u x = v y u x = v y u_(x)=v_(y)u_x = v_yux=vy and u y = v x u y = v x u_(y)=-v_(x)u_y = -v_xuy=vx, to determine the imaginary part v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y) of f ( z ) f ( z ) f(z)f(z)f(z):
v x = u y = u y = ( 6 x y + 6 y ) = 6 x y 6 y v ( x , y ) = ( 6 x y 6 y ) d y = 3 x 2 y 6 x y + f 1 ( y ) v x = u y = u y = ( 6 x y + 6 y ) = 6 x y 6 y v ( x , y ) = ( 6 x y 6 y ) d y = 3 x 2 y 6 x y + f 1 ( y ) {:[v_(x)=-u_(y)=-(del u)/(del y)=-(-6xy+6y)=6xy-6y],[v(x”,”y)=int(6xy-6y)dy=3x^(2)y-6xy+f_(1)(y)]:}\begin{align*} v_x &= -u_y = -\frac{\partial u}{\partial y} = -(-6xy + 6y) = 6xy – 6y \\ v(x, y) &= \int (6xy – 6y) \, dy = 3x^2y – 6xy + f_1(y) \end{align*}vx=uy=uy=(6xy+6y)=6xy6yv(x,y)=(6xy6y)dy=3x2y6xy+f1(y)
Here, f 1 ( y ) f 1 ( y ) f_(1)(y)f_1(y)f1(y) is an arbitrary function of y y yyy.
Now, we differentiate v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y) with respect to y y yyy to find f 1 ( y ) f 1 ( y ) f_(1)(y)f_1(y)f1(y):
v y = 3 x 2 6 x + f 1 ( y ) u x = 3 x 2 6 x + f 1 ( y ) (by the Cauchy-Riemann equations) 3 ( x 1 ) 2 3 y 2 = 3 x 2 6 x + f 1 ( y ) f 1 ( y ) = 3 3 y 2 f 1 ( y ) = 3 y y 3 + C (where C is an arbitrary constant) v y = 3 x 2 6 x + f 1 ( y ) u x = 3 x 2 6 x + f 1 ( y ) (by the Cauchy-Riemann equations) 3 ( x 1 ) 2 3 y 2 = 3 x 2 6 x + f 1 ( y ) f 1 ( y ) = 3 3 y 2 f 1 ( y ) = 3 y y 3 + C (where  C  is an arbitrary constant) {:[(del v)/(del y)=3x^(2)-6x+f_(1)^(‘)(y)],[=>(del u)/(del x)=3x^(2)-6x+f_(1)^(‘)(y)quad(by the Cauchy-Riemann equations)],[3(x-1)^(2)-3y^(2)=3x^(2)-6x+f_(1)^(‘)(y)],[f_(1)^(‘)(y)=3-3y^(2)],[f_(1)(y)=3y-y^(3)+C quad(where C” is an arbitrary constant)”]:}\begin{align*} \frac{\partial v}{\partial y} &= 3x^2 – 6x + f_1′(y) \\ \Rightarrow \frac{\partial u}{\partial x} &= 3x^2 – 6x + f_1′(y) \quad \text{(by the Cauchy-Riemann equations)} \\ 3(x-1)^2 – 3y^2 &= 3x^2 – 6x + f_1′(y) \\ f_1′(y) &= 3 – 3y^2 \\ f_1(y) &= 3y – y^3 + C \quad \text{(where \(C\) is an arbitrary constant)} \end{align*}vy=3x26x+f1(y)ux=3x26x+f1(y)(by the Cauchy-Riemann equations)3(x1)23y2=3x26x+f1(y)f1(y)=33y2f1(y)=3yy3+C(where C is an arbitrary constant)
Thus, the imaginary part v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y) is given by v ( x , y ) = 3 x 2 y 6 x y + 3 y y 3 + C v ( x , y ) = 3 x 2 y 6 x y + 3 y y 3 + C v(x,y)=3x^(2)y-6xy+3y-y^(3)+Cv(x, y) = 3x^2y – 6xy + 3y – y^3 + Cv(x,y)=3x2y6xy+3yy3+C.
Step 3: Expressing f ( z ) f ( z ) f(z)f(z)f(z) in terms of z z zzz:
f ( z ) = ϕ 1 ( z , 0 ) i ϕ 2 ( z , 0 ) f ( z ) = 3 ( z 1 ) 2 3 ( 0 ) 2 i [ 6 z ( 0 ) + 6 ( 0 ) ] f ( z ) = 3 ( z 1 ) 2 f ( z ) = ( z 1 ) 3 + C f ( z ) = ϕ 1 ( z , 0 ) i ϕ 2 ( z , 0 ) f ( z ) = 3 ( z 1 ) 2 3 ( 0 ) 2 i [ 6 z ( 0 ) + 6 ( 0 ) ] f ( z ) = 3 ( z 1 ) 2 f ( z ) = ( z 1 ) 3 + C {:[f^(‘)(z)=phi_(1)(z”,”0)-iphi_(2)(z”,”0)],[f^(‘)(z)=3(z-1)^(2)-3(0)^(2)-i[6*z*(0)+6(0)]],[f^(‘)(z)=3(z-1)^(2)],[:.quadf(z)=(z-1)^(3)+C]:}\begin{aligned} f^{\prime}(z) & =\phi_1(z, 0)-i \phi_2(z, 0) \\ f^{\prime}(z) & =3(z-1)^2-3(0)^2-i[6 \cdot z \cdot(0)+6(0)] \\ f^{\prime}(z) & =3(z-1)^2 \\ \therefore \quad & f(z)=(z-1)^3+C \end{aligned}f(z)=ϕ1(z,0)iϕ2(z,0)f(z)=3(z1)23(0)2i[6z(0)+6(0)]f(z)=3(z1)2f(z)=(z1)3+C
Finally, we express the analytic function f ( z ) f ( z ) f(z)f(z)f(z) in terms of z z zzz. We use the result that f ( z ) = ( z 1 ) 3 + C f ( z ) = ( z 1 ) 3 + C f(z)=(z-1)^(3)+Cf(z) = (z-1)^3 + Cf(z)=(z1)3+C
Conclusion:
In conclusion, the given function u ( x , y ) u ( x , y ) u(x,y)u(x, y)u(x,y) is confirmed to be harmonic. We have found its harmonic conjugate, v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y), and expressed the corresponding analytic function f ( z ) f ( z ) f(z)f(z)f(z) as f ( z ) = ( z 1 ) 3 + C f ( z ) = ( z 1 ) 3 + C f(z)=(z-1)^(3)+Cf(z) = (z-1)^3 + Cf(z)=(z1)3+C, where C C CCC is an arbitrary constant.
Verified Answer
5/5
\(sin\left(\theta +\phi \right)=sin\:\theta \:cos\:\phi +cos\:\theta \:sin\:\phi \)
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