← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q3b — Step-by-Step Solution

15 marks · Section A

Partial derivatives; equality of mixed partials (Schwarz) · Real Analysis · asked 2× in 14 yrs · Read the full method →

Question

Obtain $\dfrac{\partial^{2}f(0,0)}{\partial x\,\partial y}$ and $\dfrac{\partial^{2}f(0,0)}{\partial y\,\partial x}$ for the function

Also, discuss the continuity of $\dfrac{\partial^{2}f}{\partial x\,\partial y}$ and $\dfrac{\partial^{2}f}{\partial y\,\partial x}$ at $(0,0)$.

Technique

Same as 2013 P1 Q3(b): compute partial derivatives off-origin by smooth formulas; use limit definition at origin; check Clairaut’s hypothesis (continuity of 2nd partials).

Solution

Strategy. Use the limit definition to compute partial derivatives at $(0,0)$; the mixed partials $f_{xy}(0,0)$ and $f_{yx}(0,0)$ can differ when $f_{xy}$ and $f_{yx}$ are discontinuous at the origin (Clairaut’s theorem fails).

Step 1 — First partials at $(0,0)$

$f(h,0)=0$ for any $h$ (since $f$ has factor $xy$ in numerator). So

Similarly $f_y(0,0)=0$.

Step 2 — $f_x(0,y)$ for $y\ne 0$

For $(x,y)\ne(0,0)$, the formula for $f$ is smooth. Differentiate w.r.t. $x$:

At $x=0$, $y\ne 0$:

Step 3 — Compute $f_{xy}(0,0)$

Step 4 — $f_y(x,0)$ for $x\ne 0$

Differentiating $f$ w.r.t. $y$ gives:

At $y=0$, $x\ne 0$:

Step 5 — Compute $f_{yx}(0,0)$

Summary

Since these are different, Clairaut’s theorem must fail — meaning at least one of $f_{xy}$ or $f_{yx}$ is discontinuous at $(0,0)$. In fact, both are.

Continuity discussion

Away from $(0,0)$, $f\in C^{\infty}$, so $f_{xy}=f_{yx}$ as functions (Clairaut). Call this common formula $g(x,y)$. From the calculations above:

• Along $y=0$: $f_y(x,0)=3x$, so $\lim_{x\to 0}f_y(x,0)=0$. Hence $g(x,0)\to ?$ via $\partial f_y/\partial x$ along $y=0$: $\frac{d}{dx}(3x)=3$. So $g(x,0)\to 3$ as $x\to 0$.
• Along $x=0$: $f_x(0,y)=-2y$. Similarly $g(0,y)\to -2$ as $y\to 0$.

Different limits along different paths through the origin ⇒ $g$ has no limit at $(0,0)$. Hence both $f_{xy}$ and $f_{yx}$ (which equal $g$ off the origin) are discontinuous at $(0,0)$ — the limit doesn’t exist, regardless of how we set the value at the origin.

Answer