Answer :
Let's solve the problem step by step.
### Determine if [tex]\( v(x, y) = \log(x^2 + y^2) \)[/tex] is Harmonic
To determine if [tex]\( v = \log(x^2 + y^2) \)[/tex] is harmonic, we need to compute the Laplacian of [tex]\( v \)[/tex]:
[tex]\[ \Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} \][/tex]
First, we find the partial derivatives of [tex]\( v \)[/tex].
#### First Partial Derivatives:
[tex]\[ v = \log(x^2 + y^2) \][/tex]
[tex]\[ \frac{\partial v}{\partial x} = \frac{2x}{x^2 + y^2} \][/tex]
[tex]\[ \frac{\partial v}{\partial y} = \frac{2y}{x^2 + y^2} \][/tex]
#### Second Partial Derivatives:
[tex]\[ \frac{\partial^2 v}{\partial x^2} \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) \][/tex]
Using the quotient rule:
[tex]\[ \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) = \frac{(2(x^2 + y^2) - 2x \cdot 2x)}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2 - 2x^2)}{(x^2 + y^2)^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} \][/tex]
Similarly for [tex]\( \frac{\partial^2 v}{\partial y^2} \)[/tex]:
[tex]\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) \][/tex]
Using the quotient rule:
[tex]\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) = \frac{(2(x^2 + y^2) - 2y \cdot 2y)}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2 - 2y^2)}{(x^2 + y^2)^2} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} \][/tex]
Now, summing these up for the Laplacian:
[tex]\[ \Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} + \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} = 0 \][/tex]
Since [tex]\( \Delta v = 0 \)[/tex], [tex]\( v(x, y) = \log(x^2 + y^2) \)[/tex] is harmonic.
### Finding a Function [tex]\( u \)[/tex] such that [tex]\( u + iv \)[/tex] is Analytic
We need to find [tex]\( u(x, y) \)[/tex] such that [tex]\( u + iv \)[/tex] is an analytic function. For this, the Cauchy-Riemann equations must be satisfied:
[tex]\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \][/tex]
[tex]\[ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \][/tex]
We already found:
[tex]\[ \frac{\partial v}{\partial x} = \frac{2x}{x^2 + y^2} \][/tex]
[tex]\[ \frac{\partial v}{\partial y} = \frac{2y}{x^2 + y^2} \][/tex]
Using the Cauchy-Riemann equations:
[tex]\[ \frac{\partial u}{\partial x} = \frac{2y}{x^2 + y^2} \][/tex]
[tex]\[ \frac{\partial u}{\partial y} = -\frac{2x}{x^2 + y^2} \][/tex]
We can integrate these equations to find [tex]\( u \)[/tex]:
To find [tex]\( u \)[/tex]:
[tex]\[ u = \int \frac{\partial u}{\partial x} \, dx \][/tex]
[tex]\[ u = \int \frac{2y}{x^2 + y^2} \, dx \][/tex]
Here, [tex]\( y \)[/tex] is treated as a constant:
[tex]\[ u = 2y \int \frac{1}{x^2 + y^2} \, dx \][/tex]
Recognizing the standard integral:
[tex]\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) \][/tex]
Thus:
[tex]\[ u = 2y \cdot \frac{1}{y} \arctan \left( \frac{x}{y} \right) = 2 \arctan \left( \frac{x}{y} \right) \][/tex]
But we also need to integrate with respect to [tex]\( y \)[/tex]:
[tex]\[ \int -\frac{2x}{x^2 + y^2} \, dy \][/tex]
This is solved similarly:
[tex]\[ u = -2x \arctan \left( \frac{y}{x} \right) \][/tex]
Combining these, we arrive at the form:
[tex]\[ u - i \arctan \left( \frac{x}{y} \right) + i \arctan \left( \frac{y}{x} \right) \][/tex]
This leads to a logarithmic form:
[tex]\[ u + iv = i \log \left( \frac{x - iy}{x + iy}, \right) \][/tex]
Thus:
[tex]\[ u + iv = -I \log(x - iy) + I \log(x + iy) + I \log(-I x + y) - I \log(I x + y) \][/tex]
This is the analytic function combining [tex]\( u \)[/tex] and [tex]\( iv \)[/tex].
### Determine if [tex]\( v(x, y) = \log(x^2 + y^2) \)[/tex] is Harmonic
To determine if [tex]\( v = \log(x^2 + y^2) \)[/tex] is harmonic, we need to compute the Laplacian of [tex]\( v \)[/tex]:
[tex]\[ \Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} \][/tex]
First, we find the partial derivatives of [tex]\( v \)[/tex].
#### First Partial Derivatives:
[tex]\[ v = \log(x^2 + y^2) \][/tex]
[tex]\[ \frac{\partial v}{\partial x} = \frac{2x}{x^2 + y^2} \][/tex]
[tex]\[ \frac{\partial v}{\partial y} = \frac{2y}{x^2 + y^2} \][/tex]
#### Second Partial Derivatives:
[tex]\[ \frac{\partial^2 v}{\partial x^2} \][/tex]
[tex]\[ \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) \][/tex]
Using the quotient rule:
[tex]\[ \frac{\partial}{\partial x} \left( \frac{2x}{x^2 + y^2} \right) = \frac{(2(x^2 + y^2) - 2x \cdot 2x)}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2 - 2x^2)}{(x^2 + y^2)^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} \][/tex]
Similarly for [tex]\( \frac{\partial^2 v}{\partial y^2} \)[/tex]:
[tex]\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) \][/tex]
Using the quotient rule:
[tex]\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^2 + y^2} \right) = \frac{(2(x^2 + y^2) - 2y \cdot 2y)}{(x^2 + y^2)^2} = \frac{2(x^2 + y^2 - 2y^2)}{(x^2 + y^2)^2} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} \][/tex]
Now, summing these up for the Laplacian:
[tex]\[ \Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} + \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} = 0 \][/tex]
Since [tex]\( \Delta v = 0 \)[/tex], [tex]\( v(x, y) = \log(x^2 + y^2) \)[/tex] is harmonic.
### Finding a Function [tex]\( u \)[/tex] such that [tex]\( u + iv \)[/tex] is Analytic
We need to find [tex]\( u(x, y) \)[/tex] such that [tex]\( u + iv \)[/tex] is an analytic function. For this, the Cauchy-Riemann equations must be satisfied:
[tex]\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \][/tex]
[tex]\[ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \][/tex]
We already found:
[tex]\[ \frac{\partial v}{\partial x} = \frac{2x}{x^2 + y^2} \][/tex]
[tex]\[ \frac{\partial v}{\partial y} = \frac{2y}{x^2 + y^2} \][/tex]
Using the Cauchy-Riemann equations:
[tex]\[ \frac{\partial u}{\partial x} = \frac{2y}{x^2 + y^2} \][/tex]
[tex]\[ \frac{\partial u}{\partial y} = -\frac{2x}{x^2 + y^2} \][/tex]
We can integrate these equations to find [tex]\( u \)[/tex]:
To find [tex]\( u \)[/tex]:
[tex]\[ u = \int \frac{\partial u}{\partial x} \, dx \][/tex]
[tex]\[ u = \int \frac{2y}{x^2 + y^2} \, dx \][/tex]
Here, [tex]\( y \)[/tex] is treated as a constant:
[tex]\[ u = 2y \int \frac{1}{x^2 + y^2} \, dx \][/tex]
Recognizing the standard integral:
[tex]\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) \][/tex]
Thus:
[tex]\[ u = 2y \cdot \frac{1}{y} \arctan \left( \frac{x}{y} \right) = 2 \arctan \left( \frac{x}{y} \right) \][/tex]
But we also need to integrate with respect to [tex]\( y \)[/tex]:
[tex]\[ \int -\frac{2x}{x^2 + y^2} \, dy \][/tex]
This is solved similarly:
[tex]\[ u = -2x \arctan \left( \frac{y}{x} \right) \][/tex]
Combining these, we arrive at the form:
[tex]\[ u - i \arctan \left( \frac{x}{y} \right) + i \arctan \left( \frac{y}{x} \right) \][/tex]
This leads to a logarithmic form:
[tex]\[ u + iv = i \log \left( \frac{x - iy}{x + iy}, \right) \][/tex]
Thus:
[tex]\[ u + iv = -I \log(x - iy) + I \log(x + iy) + I \log(-I x + y) - I \log(I x + y) \][/tex]
This is the analytic function combining [tex]\( u \)[/tex] and [tex]\( iv \)[/tex].
