Q.NO.1: Show that the function z=ln(x^2 + y^2)+2tan^-1(y/x) satisfies Laplace’s equation.
(∂^2z/∂x^2)+(∂^2z/∂y^2)=0
I've done a couple of these for you now. Where do you get stuck? What do you have so far? Doing the partials can be tedious, but not difficult.
You can get some help here, for example, to verify your work:
http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2+ln(x%5E2+%2B+y%5E2)%2B2tan%5E-1(y%2Fx)
To show that the function z = ln(x^2 + y^2) + 2tan^(-1)(y/x) satisfies Laplace's equation, we need to calculate the second partial derivatives (∂^2z/∂x^2) and (∂^2z/∂y^2) and verify that their sum is zero.
Let's start by finding the first partial derivative (∂z/∂x).
∂z/∂x = 2x/(x^2 + y^2) + 2(y/x^2) / (1 + (y/x)^2)
Next, we can find the second partial derivative (∂^2z/∂x^2) by differentiating (∂z/∂x) with respect to x:
∂^2z/∂x^2 = (2(x^2 + y^2) - 4x^2) / (x^2 + y^2)^2 + (-2(y/x^2) * (2x^2 - 2y^2)) / (1 + (y/x)^2)^2
Simplifying the above expression, we have:
∂^2z/∂x^2 = (2y^2 - 2x^2) / (x^2 + y^2)^2 + (4xy^2)/(x^2 + y^2)^2
Similarly, we can find the second partial derivative (∂^2z/∂y^2) by differentiating (∂z/∂y) with respect to y:
∂^2z/∂y^2 = (2(x^2 + y^2) - 4y^2) / (x^2 + y^2)^2 + (2(y/x^2) * (2x^2 - 2y^2)) / (1 + (y/x)^2)^2
Simplifying the above expression, we have:
∂^2z/∂y^2 = (2x^2 - 2y^2) / (x^2 + y^2)^2 - (4xy^2)/(x^2 + y^2)^2
Now, let's add (∂^2z/∂x^2) and (∂^2z/∂y^2) to verify if it equals to zero:
(∂^2z/∂x^2) + (∂^2z/∂y^2) = (2y^2 - 2x^2) / (x^2 + y^2)^2 + (4xy^2)/(x^2 + y^2)^2 + (2x^2 - 2y^2) / (x^2 + y^2)^2 - (4xy^2)/(x^2 + y^2)^2
After simplification, we get:
(∂^2z/∂x^2) + (∂^2z/∂y^2) = 0
Therefore, we have shown that the function z = ln(x^2 + y^2) + 2tan^(-1)(y/x) satisfies Laplace's equation (∂^2z/∂x^2) + (∂^2z/∂y^2) = 0.
To show that the given function satisfies Laplace's equation (∂^2z/∂x^2) + (∂^2z/∂y^2) = 0, we need to calculate the second derivatives of z with respect to x and y, and then verify that their sum is zero.
Given function: z = ln(x^2 + y^2) + 2tan^-1(y/x)
Step 1: Calculate the first-order partial derivatives.
∂z/∂x = 2x/(x^2 + y^2) + 2y/(x^2 + y^2)
∂z/∂y = 2y/(x^2 + y^2) - 2x/(x^2 + y^2)
Step 2: Calculate the second-order partial derivatives.
∂^2z/∂x^2 = [(2(x^2 + y^2) - 4x^2)/(x^2 + y^2)^2]
= (2y^2 - 2x^2)/(x^2 + y^2)^2
∂^2z/∂y^2 = [(2(x^2 + y^2) - 4y^2)/(x^2 + y^2)^2]
= (2x^2 - 2y^2)/(x^2 + y^2)^2
Step 3: Add the second derivatives.
(∂^2z/∂x^2) + (∂^2z/∂y^2) = (2y^2 - 2x^2)/(x^2 + y^2)^2 + (2x^2 - 2y^2)/(x^2 + y^2)^2
= (2y^2 - 2x^2 + 2x^2 - 2y^2)/(x^2 + y^2)^2
= 0/(x^2 + y^2)^2
= 0
Therefore, (∂^2z/∂x^2) + (∂^2z/∂y^2) = 0. The function z = ln(x^2 + y^2) + 2tan^-1(y/x) satisfies Laplace's equation, and it is a solution to Laplace's equation.