Question Bank • QB_EM_ADV_030
MCQ • 1 Mark
Section 1: Engineering Mathematics
Calculus
Subtopic: Partial derivatives
Question Statement:
If $u(x, y) = \ln(x^2 + y^2)$, the value of $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ for all $(x, y) \neq (0, 0)$ is:
Options:
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(A) $0$
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(B) $\frac{2}{x^2 + y^2}$
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(C) $\frac{4}{(x^2 + y^2)^2}$
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(D) $-\frac{2}{x^2 + y^2}$
Step-by-Step Solution & Explanation
Differentiating with respect to $x$:
$\frac{\partial u}{\partial x} = \frac{2x}{x^2 + y^2}$
$\frac{\partial^2 u}{\partial x^2} = \frac{2(x^2 + y^2) - 2x(2x)}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2}$
By symmetry with respect to $y$:
$\frac{\partial^2 u}{\partial y^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2}$
Adding the two second partial derivatives:
$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{(2y^2 - 2x^2) + (2x^2 - 2y^2)}{(x^2 + y^2)^2} = 0$
Thus $u(x, y)$ is harmonic in $\mathbb{R}^2 \setminus \{(0, 0)\}$.
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