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To demonstrate that the function [tex]\( v = \log(x^2 + y^2) \)[/tex] is harmonic and to find a function [tex]\( u \)[/tex] such that [tex]\( u + iv \)[/tex] is analytic, we'll go through the following steps:
1. Compute the first-order partial derivatives of [tex]\( v \)[/tex]:
[tex]\[ v_x = \frac{\partial v}{\partial x} = \frac{2x}{x^2 + y^2} \][/tex]
[tex]\[ v_y = \frac{\partial v}{\partial y} = \frac{2y}{x^2 + y^2} \][/tex]
2. Compute the second-order partial derivatives of [tex]\( v \)[/tex]:
[tex]\[ v_{xx} = \frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) = \frac{(x^2 + y^2) \cdot 2 - 2x \cdot 2x}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2) - 4x^2}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4x^2}{(x^2 + y^2)^2} = \frac{-2x^2 + 2y^2}{(x^2 + y^2)^2} \][/tex]
Thus:
[tex]\[ v_{xx} = -\frac{4x^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \][/tex]
Now, compute [tex]\( v_{yy} \)[/tex]:
[tex]\[ v_{yy} = \frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) = \frac{(x^2 + y^2) \cdot 2 - 2y \cdot 2y}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2) - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \][/tex]
Thus:
[tex]\[ v_{yy} = -\frac{4y^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \][/tex]
3. Check if [tex]\( v \)[/tex] is harmonic by summing the second partial derivatives:
[tex]\[ v_{xx} + v_{yy} = \left( -\frac{4x^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \right) + \left( -\frac{4y^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \right) \][/tex]
[tex]\[ v_{xx} + v_{yy} = -\frac{4x^2}{(x^2 + y^2)^2} - \frac{4y^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} + \frac{2}{x^2 + y^2} \][/tex]
[tex]\[ = -\frac{4(x^2 + y^2)}{(x^2 + y^2)^2} + \frac{4}{x^2 + y^2} \][/tex]
[tex]\[ = -\frac{4}{x^2 + y^2} + \frac{4}{x^2 + y^2} = 0 \][/tex]
Since [tex]\( v_{xx} + v_{yy} = 0 \)[/tex], the function [tex]\( v \)[/tex] is indeed harmonic.
4. Find a function [tex]\( u \)[/tex] such that [tex]\( u + iv \)[/tex] is analytic.
According to the Cauchy-Riemann equations:
[tex]\[ u_x = -v_y \quad \text{and} \quad u_y = v_x \][/tex]
Given:
[tex]\[ v_y = \frac{2y}{x^2 + y^2} \Rightarrow u_x = -\frac{2y}{x^2 + y^2} \][/tex]
[tex]\[ v_x = \frac{2x}{x^2 + y^2} \Rightarrow u_y = \frac{2x}{x^2 + y^2} \][/tex]
Therefore, the function [tex]\( v = \log(x^2 + y^2) \)[/tex] is harmonic, and the function [tex]\( u \)[/tex] obtained by using the Cauchy-Riemann equations satisfies the conditions for [tex]\( u + iv \)[/tex] to be analytic.
1. Compute the first-order partial derivatives of [tex]\( v \)[/tex]:
[tex]\[ v_x = \frac{\partial v}{\partial x} = \frac{2x}{x^2 + y^2} \][/tex]
[tex]\[ v_y = \frac{\partial v}{\partial y} = \frac{2y}{x^2 + y^2} \][/tex]
2. Compute the second-order partial derivatives of [tex]\( v \)[/tex]:
[tex]\[ v_{xx} = \frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) = \frac{(x^2 + y^2) \cdot 2 - 2x \cdot 2x}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2) - 4x^2}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4x^2}{(x^2 + y^2)^2} = \frac{-2x^2 + 2y^2}{(x^2 + y^2)^2} \][/tex]
Thus:
[tex]\[ v_{xx} = -\frac{4x^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \][/tex]
Now, compute [tex]\( v_{yy} \)[/tex]:
[tex]\[ v_{yy} = \frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) \][/tex]
[tex]\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) = \frac{(x^2 + y^2) \cdot 2 - 2y \cdot 2y}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2) - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 + 2y^2 - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \][/tex]
Thus:
[tex]\[ v_{yy} = -\frac{4y^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \][/tex]
3. Check if [tex]\( v \)[/tex] is harmonic by summing the second partial derivatives:
[tex]\[ v_{xx} + v_{yy} = \left( -\frac{4x^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \right) + \left( -\frac{4y^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} \right) \][/tex]
[tex]\[ v_{xx} + v_{yy} = -\frac{4x^2}{(x^2 + y^2)^2} - \frac{4y^2}{(x^2 + y^2)^2} + \frac{2}{x^2 + y^2} + \frac{2}{x^2 + y^2} \][/tex]
[tex]\[ = -\frac{4(x^2 + y^2)}{(x^2 + y^2)^2} + \frac{4}{x^2 + y^2} \][/tex]
[tex]\[ = -\frac{4}{x^2 + y^2} + \frac{4}{x^2 + y^2} = 0 \][/tex]
Since [tex]\( v_{xx} + v_{yy} = 0 \)[/tex], the function [tex]\( v \)[/tex] is indeed harmonic.
4. Find a function [tex]\( u \)[/tex] such that [tex]\( u + iv \)[/tex] is analytic.
According to the Cauchy-Riemann equations:
[tex]\[ u_x = -v_y \quad \text{and} \quad u_y = v_x \][/tex]
Given:
[tex]\[ v_y = \frac{2y}{x^2 + y^2} \Rightarrow u_x = -\frac{2y}{x^2 + y^2} \][/tex]
[tex]\[ v_x = \frac{2x}{x^2 + y^2} \Rightarrow u_y = \frac{2x}{x^2 + y^2} \][/tex]
Therefore, the function [tex]\( v = \log(x^2 + y^2) \)[/tex] is harmonic, and the function [tex]\( u \)[/tex] obtained by using the Cauchy-Riemann equations satisfies the conditions for [tex]\( u + iv \)[/tex] to be analytic.
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