Question

$\red{ \rm{If \: u=( {x}^{2} + {y}^{2} ) ^{ \frac{1}{3} } } \: then \: evaluate} \\ \color{green} \rm {x}^{2} \frac{ \partial {}^{2}u }{ \partial {x}^{2} } + 2xy \frac{ \partial {}^{2} u}{2x \partial y} + {y}^{2} \frac{ { \partial}^{2}u }{ \partial {y}^{2} }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: u=\bigg( {x}^{2} + {y}^{2} \bigg) ^{ \dfrac{1}{3} }$

can be rewritten as

$\rm \: u= {\bigg(x\bigg) }^{\dfrac{2}{3} } \bigg(1 + \frac{ {y}^{2} }{ {x}^{2} } \bigg) ^{ \dfrac{1}{3} }$

$\rm\implies \:u \: is \: a \: homogenous \: function \: of \: degree \: \dfrac{2}{3}$

So, By Euler's Theorem, we have

$\rm \: x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \dfrac{2}{3}u - - - (1)$

On differentiating both sides partially w. r. t. x, we get

$\rm \: x\dfrac{\partial {}^{2} u}{\partial {x}^{2} } +\dfrac{\partial u}{\partial x} + y\dfrac{\partial {}^{2} u}{\partial y \: \partial x} = \dfrac{2}{3}\dfrac{\partial u}{\partial x}$

$\rm \: x\dfrac{\partial {}^{2} u}{\partial {x}^{2} } + y\dfrac{\partial {}^{2} u}{\partial y \: \partial x} = \dfrac{2}{3}\dfrac{\partial u}{\partial x} - \dfrac{\partial u}{\partial x}$

$\rm \: x\dfrac{\partial {}^{2} u}{\partial {x}^{2} } + y\dfrac{\partial {}^{2} u}{\partial y \: \partial x} = - \: \dfrac{1}{3}\dfrac{\partial u}{\partial x}$

On multiply both sides by x, we get

$\rm \: {x}^{2} \dfrac{\partial {}^{2} u}{\partial {x}^{2} } + xy\dfrac{\partial {}^{2} u}{\partial y \: \partial x} = - \: \dfrac{1}{3}x\dfrac{\partial u}{\partial x} - - - (2)$

Now, on differentiating both sides partially w. r. t. y, equation (1), we get

$\rm \: x\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + y\dfrac{\partial {}^{2} u}{\partial {y}^{2} } + \dfrac{\partial u}{\partial y} = \dfrac{2}{3}\dfrac{\partial u}{\partial y}$

$\rm \: x\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + y\dfrac{\partial {}^{2} u}{\partial {y}^{2} } = \dfrac{2}{3}\dfrac{\partial u}{\partial y} - \dfrac{\partial u}{\partial y}$

$\rm \: x\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + y\dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \: \dfrac{1}{3}\dfrac{\partial u}{\partial y}$

On multiply both sides by y, we get

$\rm \: xy\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + {y}^{2} \dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \: \dfrac{1}{3}y\dfrac{\partial u}{\partial y} - - - (3)$

On adding equation (2) and (3), we get

$\rm \: {x}^{2} \dfrac{\partial {}^{2}u}{\partial {x}^{2} } + 2xy\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + {y}^{2} \dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \dfrac{1}{3}x \dfrac{\partial u}{\partial x} - \dfrac{1}{3}y\dfrac{\partial u}{\partial y}$

$\rm \: {x}^{2} \dfrac{\partial {}^{2}u}{\partial {x}^{2} } + 2xy\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + {y}^{2} \dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \dfrac{1}{3}\bigg(x \dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} \bigg)$

Now, using equation (1), we get

$\rm \: {x}^{2} \dfrac{\partial {}^{2}u}{\partial {x}^{2} } + 2xy\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + {y}^{2} \dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \dfrac{1}{3}\bigg(\dfrac{2}{3}u \bigg)$

$\rm \: {x}^{2} \dfrac{\partial {}^{2}u}{\partial {x}^{2} } + 2xy\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + {y}^{2} \dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \dfrac{2}{9}u$

Hence,

$\color{green}\rm\implies \boxed{ \rm{ \:{x}^{2} \dfrac{\partial {}^{2}u}{\partial {x}^{2} } + 2xy\dfrac{\partial {}^{2} u}{\partial x \: \partial y} + {y}^{2} \dfrac{\partial {}^{2} u}{\partial {y}^{2} } = - \dfrac{2}{9}u}}$