If a function u(x,y)=arctan(y/x)
deduce the laplace equation of the function.
plz show working i don't no where to start from...plz
thanks
How about starting with the definition of the equation?
∇^2(u) = 0
∂u/∂x = 1/(1+y^2/x^2) (-y/x^2)
= x^2/(x^2+y^2) (-y/x^2)
= -y/(x^2+y^2)
∂^u/∂x^2 = 2xy/(x^2+y^2)^2
Now do ∂u/∂y and then form the equation.
To deduce the Laplace equation for the function u(x, y) = arctan(y/x), we will need to calculate its second partial derivatives with respect to both x and y.
Let's start by finding the first partial derivatives of u with respect to x and y:
∂u/∂x = ∂(arctan(y/x))/∂x
To differentiate arctan(y/x) with respect to x, we can apply the chain rule:
= -(1/(1 + (y/x)^2)) * (y/x^2)
= (-y)/(x^2 + y^2)
∂u/∂y = ∂(arctan(y/x))/∂y
To differentiate arctan(y/x) with respect to y, again, we apply the chain rule:
= (1/(1 + (y/x)^2)) * (1/x)
= (1)/(x^2/y^2 + 1) * (1/x)
= (x)/(x^2 + y^2)
Now, let's calculate the second partial derivatives:
∂^2u/∂x^2 = ∂/∂x (∂u/∂x)
= ∂(-y)/(x^2 + y^2)/∂x
= 2xy/(x^2 + y^2)^2
∂^2u/∂y^2 = ∂/∂y (∂u/∂y)
= ∂(x)/(x^2 + y^2)/∂y
= -2xy/(x^2 + y^2)^2
Finally, we can now deduce the Laplace equation for u(x, y) by summing the second partial derivatives:
∂^2u/∂x^2 + ∂^2u/∂y^2 = 2xy/(x^2 + y^2)^2 - 2xy/(x^2 + y^2)^2
= 0
Therefore, the Laplace equation for the function u(x, y) = arctan(y/x) is simply 0.