Subtopic: Homogeneous function – Euler's theorem on homogeneous functions
Question Statement:
If $u(x, y) = \sin^{-1}\left( \frac{x + 2y}{\sqrt{x} + \sqrt{y}} \right)$, the value of the differential expression:
$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2}$
when $u = \frac{\pi}{6}$ is ________ (round off to three decimal places).
Answer Key:
-0.096
Step-by-Step Solution & Explanation
Let $z = \sin(u) = \frac{x + 2y}{\sqrt{x} + \sqrt{y}}$.
The degree of homogeneity of $z$ is $n = 1 - \frac{1}{2} = \frac{1}{2}$.
By Euler's theorem for homogeneous functions:
$x u_x + y u_y = n \frac{F(u)}{F'(u)} = \frac{1}{2} \frac{\sin u}{\cos u} = \frac{1}{2} \tan u = g(u)$
Using the second-order Euler differential identity:
$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = g(u)[g'(u) - 1]$
Here $g(u) = \frac{1}{2}\tan u$, so $g'(u) = \frac{1}{2}\sec^2 u$.
$g(u)[g'(u) - 1] = \frac{1}{2}\tan u \left[ \frac{1}{2}\sec^2 u - 1 \right]$
At $u = \frac{\pi}{6}$:
$\tan(\pi/6) = \frac{1}{\sqrt{3}}, \quad \sec^2(\pi/6) = \frac{4}{3}$
$g(\pi/6) = \frac{1}{2\sqrt{3}}$
$g'(\pi/6) - 1 = \frac{1}{2}\left(\frac{4}{3}\right) - 1 = \frac{2}{3} - 1 = -\frac{1}{3}$
$x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = \frac{1}{2\sqrt{3}} \left( -\frac{1}{3} \right) = -\frac{1}{6\sqrt{3}} \approx -0.0962$