Home » Blog » Show a given equation satisfies a particular second or partial differential equation

Show a given equation satisfies a particular second or partial differential equation

by RoRi

Let

    \[ f(x,y) = \frac{xy}{(x^2+y^2)^2}, \qquad (x,y) \neq (0,0). \]

Show that

    \[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0. \]

Proof. First, we compute the first-order partial derivative with respect to x,

    \begin{align*} \frac{\partial f}{\partial x} &= \frac{y(x^2+y^2)^2 - (xy)(2)(2x)(x^2+y^2)}{(x^2+y^2)^4} \\ &= \frac{x^2y + y^3 - 4x^2y}{(x^2+y^2)^3} \\ &= \frac{y^3 - 3x^2y}{(x^2+y^2)^3}. \end{align*}

Then, we compute the first-order partial derivative with respect to y,

    \begin{align*} \frac{\partial f}{\partial y} &= \frac{x(x^2+y^2)^2 - (xy)(2)(2y)(x^2+y^2)}{(x^2+y^2)^4} \\ &= \frac{x^3 + xy^2 - 4xy^2}{(x^2+y^2)^3} \\ &= \frac{x^3 - 3xy^2}{(x^2+y^2)^3}. \end{align*}

Next, we compute the second-order partial derivatives in the statement we are trying to prove,

    \begin{align*} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{y^3 - 3x^2y}{(x^2+y^2)^3} \right) \\ &= \frac{-6xy (x^2+y^2)^3 - (y^3-3x^2y)(3)(2x)(x^2+y^2)^2}{(x^2+y^2)^6} \\ &= \frac{-6x^3y - 6xy^3 - 6xy^3 + 18x^3y}{(x^2+y^2)^4} \\ &= 12xy \frac{x^2 - y^2}{(x^2+y^2)^4}. \\ \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{x^3 - 3xy^2}{(x^2+y^2)^3} \right) \\ &= \frac{-6xy(x^2+y^2)^3 - (x^3 - 3xy^2)(3)(2y)(x^2+y^2)^2}{(x^2+y^2)^6} \\ &= \frac{-6x^3 y - 6xy^3 - 6x^3 y + 18xy^3}{(x^2+y^2)^4} \\ &= -12xy \frac{x^2 - y^2}{(x^2+y^2)}. \end{align*}

Finally, putting these together we have,

    \begin{align*} \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} &= 12xy \frac{x^2 - y^2}{(x^2+y^2)^4} - 12xy \frac{x^2 - y^2}{(x^2+y^2)^4} \\ &= 0. \qquad \blacksquare \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):