Q.NO.3: Show that the function z=tan^-1(2xy/x^2-y^2)satisfies Laplace’s equation; then make the substitution
x=r cosθ, y= r sinθ
and show that the resulting function of satisfies the polar form of laplace’s equation
(∂^2z/∂r^2)+(1/r^2)(∂^2z/∂θ^2)+(1/r)(∂z/∂r)=0
well, you have
tanz = r^2 tan2?
z = arctan(r^2 tan2?)
?z/?r = (2r tan2?)/((r^2 tan2?)^2+1)
?z/?? = (2r^2 sec^2(2?))/((r^2 tan2?)^2+1)
the second partials get a bit messy, but you can get them here:
http://www.wolframalpha.com/input/?i=d%5E2%2Fdr%5E2+arctan(r%5E2+tan2%CE%B8)
http://www.wolframalpha.com/input/?i=d%5E2%2Fdy%5E2+arctan(r%5E2+tan(2y))
(where I used y instead of ? because it choked on ?
The rest is just algebra!
To show that the function z = tan^-1(2xy / x^2 - y^2) satisfies Laplace's equation, we need to compute the partial derivatives and substitute them into the Laplace's equation.
Starting with the function z = tan^-1(2xy / x^2 - y^2), let's find the partial derivatives:
∂z/∂x = ∂/∂x(tan^-1(2xy / x^2 - y^2))
To compute this derivative, we can use the chain rule. Let u = 2xy / (x^2 - y^2).
Now, we can rewrite z = tan^-1(u) and differentiate with respect to u:
dz/du = 1/(1 + u^2) (derivative of tan^-1(u) with respect to u)
Next, we compute the derivative of u with respect to x:
du/dx = (2y(x^2 - y^2) - 2xy(2x)) / (x^2 - y^2)^2
= -4xy^3 / (x^2 - y^2)^2
Finally, we apply the chain rule to the partial derivative:
∂z/∂x = dz/du * du/dx
= (-4xy^3 / (x^2 - y^2)^2) * (1/(1 + (2xy / x^2 - y^2)^2))
= (-4xy^3 / (x^2 - y^2)^2) * (1/(1 + 4x^2y^2 / (x^2 - y^2)^2))
= -4xy^3 / (x^2 - y^2)^2 + 4x^3y / (x^2 - y^2)^2
= 4xy(x^2 - y^2 + y^2 - x^2) / (x^2 - y^2)^2
= 0
Similarly, we can compute ∂z/∂y and show that it is also equal to 0. You can verify this as an exercise.
Now, we take the given substitution x = r cosθ and y = r sinθ and find the partial derivatives of z with respect to r and θ using the chain rule.
∂z/∂r = ∂z/∂x * ∂x/∂r + ∂z/∂y * ∂y/∂r
∂z/∂r = 0 * cosθ + 0 * sinθ = 0
∂z/∂θ = ∂z/∂x * ∂x/∂θ + ∂z/∂y * ∂y/∂θ
∂z/∂θ = 0 * (-r sinθ) + 0 * r cosθ = 0
Now, let's substitute these derivatives back into the polar form of Laplace's equation:
(∂^2z/∂r^2) + (1/r^2)(∂^2z/∂θ^2) + (1/r)(∂z/∂r)
= (0) + (1/r^2)(0) + (1/r)(0)
= 0
Hence, the resulting function of z satisfies the polar form of Laplace's equation.