Question

Show that the function u(x,y)=tan^-1(y/x) is a solution of the two dimensional Laplace's equation.

Answer 1

Answer on Question #73400 – Math – Differential Equations

Question

Show that the function $u(x,y) = \tan^{-1}\left(\frac{y}{x}\right)$ is a solution of the two dimensional Laplace's equation¹.

Solution

Laplace equation is $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$.

First let note that our function has a form $u(x,y) = f\left(\frac{y}{x}\right)$. Let rewrite Laplace equation for our particular case:

a) $\frac{\partial u}{\partial x} = f' \left( \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right)$

b) $\frac{\partial^2 u}{\partial x^2} = f'' \left( \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right)^2 + f' \left( \frac{y}{x} \right) \cdot \left( \frac{2y}{x^3} \right)$

c) $\frac{\partial u}{\partial y} = f' \left( \frac{y}{x} \right) \cdot \left( \frac{1}{x} \right)$

d) $\frac{\partial^2 u}{\partial y^2} = f'' \left( \frac{y}{x} \right) \cdot \left( \frac{1}{x} \right)^2$

The Laplace equation becomes:

$$\begin{array}{l} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = f'' \left( \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right)^2 + f' \left( \frac{y}{x} \right) \cdot \left( \frac{2y}{x^3} \right) + f'' \left( \frac{y}{x} \right) \cdot \left( \frac{1}{x} \right)^2 = \\ = \frac{1}{x^4} \left( f'' \left( \frac{y}{x} \right) \cdot (y^2 + x^2) + 2 f' \left( \frac{y}{x} \right) \cdot yx \right) \end{array}$$

Now we back to our function $f(z) = \tan^{-1}(z)$, and find first and second derivatives with respect to a single variable $z$.

a) $f'(z) = \frac{d}{dz} f(z) = \frac{d}{dz} \left( \tan^{-1}(z) \right) = \frac{1}{z^2 + 1}$

b) $f''(z) = \frac{d^2}{dz^2} f(z) = \frac{d}{dz} \left( \frac{1}{z^2 + 1} \right) = -\frac{2z}{(z^2 + 1)^2}$

Finally, substitute $f'' \left( \frac{y}{x} \right)$ and $f' \left( \frac{y}{x} \right)$ to corresponding Laplace equation and after some transformation and simplification we get:

$$\begin{array}{l} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{1}{x^4} \left( -\frac{2 \cdot \left( \frac{y}{x} \right)}{\left( \left( \frac{y}{x} \right)^2 + 1 \right)^2} \cdot (y^2 + x^2) + 2 \cdot \frac{1}{\left( \frac{y}{x} \right)^2 + 1} \cdot yx \right) = \\ = \frac{2}{x^4 \cdot \left( \left( \frac{y}{x} \right)^2 + 1 \right)^2} \left( -\left( \frac{y}{x} \right) \cdot (y^2 + x^2) + yx \cdot \left( \left( \frac{y}{x} \right)^2 + 1 \right) \right) = \\ \end{array}$$

¹ All formulas can be verified on http://www.wolframalpha.com/input/

$$= \frac {2}{x ^ {4} \cdot \left(\left(\frac {y}{x}\right) ^ {2} + 1\right) ^ {2}} \left(- \frac {y ^ {3}}{x} - \frac {y x ^ {2}}{x} + \frac {y ^ {3} x}{x ^ {2}} + y x\right) = 0, \text{Q.E.D}$$

One has to be careful to specify an appropriate domain.

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