Question #135768

Show that the function f(x,y)=arctan(y/x) satisfies Laplace's equation:



Expert's answer

Δf=∂2f∂x2+∂2f∂y2=0\Delta f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0

So let's find second derivatives:

∂f∂x=(arctan(yx))x′=−yx2+y2∂2f∂x2=(−yx2+y2)x′=2xy(x2+y2)2∂f∂y=(arctan(yx))y′=xx2+y2∂2f∂y2=(xx2+y2)y′=−2xy(x2+y2)2\frac{\partial f}{\partial x} = (arctan(\frac{y}{x}))^\prime_x = -\frac{y}{x^2 + y^2} \\ \frac{\partial^2 f}{\partial x^2} = ( -\frac{y}{x^2 + y^2})^{\prime}_x = \frac{2xy}{(x^2 + y^2)^2}\\ \frac{\partial f}{\partial y} = (arctan(\frac{y}{x}))^\prime_y = \frac{x}{x^2 + y^2} \\ \frac{\partial^2 f}{\partial y^2} = ( \frac{x}{x^2 + y^2} )^{\prime}_y =-\frac{2xy}{(x^2 + y^2)^2}


And now substitute in the formula:

Δf=2xy(x2+y2)2−2xy(x2+y2)2=0\Delta f = \frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} = 0


So f(x,y)=arctan(yx)f(x,y) = arctan(\frac{y}{x}) satisfies Laplace's equation


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