Partial derivatives; equality of mixed partials (Schwarz)

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2014, 2020)
• Priority tier: T3
• Marks (count): 15 (1), 20 (1)
• Average solve time: ~14 min
• Difficulty mix: medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Two of the most mark-dense questions in recent Paper 2 Real Analysis sit here: a 15-marker on piecewise mixed partials (2014) and a 20-marker on Euler’s theorem for homogeneous functions (2020). Both are conceptually sharp — one tests whether you can compute $f_{xy} \ne f_{yx}$ via limits and diagnose the discontinuity, the other tests whether you know to apply Euler’s theorem to $\tan u$ rather than $u$ itself. These patterns are well-established exam favourites and reward specific preparation.

Minimum Theory

Schwarz (Clairaut) theorem. If $f_{xy}$ and $f_{yx}$ both exist and are continuous at a point $(a,b)$, then $f_{xy}(a,b) = f_{yx}(a,b)$. The converse fails: if the mixed partials are unequal at $(a,b)$, at least one must be discontinuous there.

Mixed partials at a singular point (piecewise function). When $f$ is given by a formula for $(x,y) \ne (0,0)$ and a separate value at $(0,0)$, the standard recipe is: (1) compute $f_x(0,y)$ and $f_y(x,0)$ by differentiating the off-origin formula and setting the remaining variable to $0$; (2) compute $f_{xy}(0,0)$ as $\lim_{k \to 0}[f_x(0,k)-f_x(0,0)]/k$, and $f_{yx}(0,0)$ similarly. Never use the product/quotient rule directly at $(0,0)$ for a piecewise function.

Euler’s theorem for homogeneous functions. If $f(x,y)$ is homogeneous of degree $n$ (i.e., $f(tx,ty) = t^n f(x,y)$), then $x f_x + y f_y = n f$. The second-order extension: differentiating the first-order relation with respect to $x$ and $y$ separately and combining (multiply $\partial/\partial x$ relation by $x$, $\partial/\partial y$ relation by $y$, add) gives $x^2 f_{xx} + 2xy f_{xy} + y^2 f_{yy} = n(n-1)f$.

Question Archetypes

euler-theorem (1 question(s); 2020)

Recognition Cues — The function $u$ involves an inverse trig (arctan, arcsin) of a rational function. The question asks you to prove an identity of the form $x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = (\text{something in } u)$. The key is that $u$ itself is not homogeneous, but $\tan u$ (or $\sin u$) is — verify this by checking the scaling degrees.

Solution Template

1. Let $z = \tan u$ (or $\sin u$, etc.). Write out $z$ explicitly as a ratio of polynomials.
2. Verify homogeneity of $z$. Replace $(x,y)$ with $(tx, ty)$ and read off the degree $n$ (typically $n$ = numerator degree minus denominator degree).
3. Apply first-order Euler: $x z_x + y z_y = nz$. Convert to $u$ using $z_x = \sec^2 u \cdot u_x$ (or $z_x = \cos u \cdot u_x$ for $z = \sin u$).
4. Simplify to get $x u_x + y u_y = g(u)$ for a simple trig expression $g$.
5. Differentiate the first-order relation. Apply $\partial/\partial x$ to both sides, multiply by $x$; apply $\partial/\partial y$, multiply by $y$; add.
6. Use $x u_x + y u_y = g(u)$ to simplify. This yields $x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = g(u)(g'(u)-1)$.
7. Substitute and simplify with trig identities to reach the target form.

Worked Example

2020 Paper 2, 2020-P2-Q3c (20 marks)

If $u = \tan^{-1}\dfrac{x^3+y^3}{x-y}$, $x \ne y$, show that $x^2 u_{xx} + 2xy\,u_{xy} + y^2 u_{yy} = (1-4\sin^2 u)\sin 2u$.

Step 1 — Set $z = \tan u$. Then $z = \dfrac{x^3+y^3}{x-y}$.

Step 2 — Verify homogeneity. Replacing $(x,y) \to (tx,ty)$: numerator $\to t^3(x^3+y^3)$, denominator $\to t(x-y)$. So $z(tx,ty) = t^2 z(x,y)$. Degree $n = 2$.

Step 3 — First-order Euler. By Euler’s theorem, $x z_x + y z_y = 2z$. Since $z = \tan u$, we have $z_x = \sec^2 u \cdot u_x$ and $z_y = \sec^2 u \cdot u_y$. Substituting:

$\sec^2 u \,(x u_x + y u_y) = 2\tan u.$

$x u_x + y u_y = \frac{2\tan u}{\sec^2 u} = 2\sin u \cos u = \sin 2u. \qquad(1)$

Denote $g(u) = \sin 2u$.

Step 4 — Differentiate (1) with respect to $x$, multiply by $x$.

$u_x + x u_{xx} + y u_{yx} = g'(u) u_x.$

$x u_x + x^2 u_{xx} + xy\,u_{yx} = g'(u)\, x u_x. \qquad(2)$

Step 5 — Differentiate (1) with respect to $y$, multiply by $y$.

$x u_{xy} + u_y + y u_{yy} = g'(u) u_y.$

$xy\,u_{xy} + y u_y + y^2 u_{yy} = g'(u)\, y u_y. \qquad(3)$

Step 6 — Add (2) and (3). Using $u_{xy} = u_{yx}$:

$\underbrace{(x u_x + y u_y)}_{g(u)} + (x^2 u_{xx} + 2xy\,u_{xy} + y^2 u_{yy}) = g'(u)\underbrace{(x u_x + y u_y)}_{g(u)}.$

$x^2 u_{xx} + 2xy\,u_{xy} + y^2 u_{yy} = g(u)(g'(u)-1).$

Step 7 — Evaluate. $g = \sin 2u$, $g' = 2\cos 2u$. So $g'(u)-1 = 2\cos 2u - 1 = 2(1-2\sin^2 u)-1 = 1-4\sin^2 u$. Therefore:

$x^2 u_{xx} + 2xy\,u_{xy} + y^2 u_{yy} = \sin 2u \cdot (1-4\sin^2 u).$

$\boxed{x^2 u_{xx} + 2xy\,u_{xy} + y^2 u_{yy} = (1-4\sin^2 u)\sin 2u.}$

Common Traps

• Applying Euler to $u$ directly. The function $u = \arctan(\ldots)$ is not homogeneous. Always check: does $u(tx,ty) = t^n u(x,y)$? It doesn’t. Apply Euler to $z = \tan u$.
• Wrong degree. The degree of $z = (x^3+y^3)/(x-y)$ is $3-1=2$, not $3$. Count numerator minus denominator degrees.
• Missing the $g(u)$ substitution in the second-order step. After differentiating, both sides should simplify using equation (1). The intermediate expression $g(u)(g'(u)-1)$ must be written before simplifying with $2\cos 2u - 1 = 1-4\sin^2 u$.

mixed-partials (1 question(s); 2014)

Recognition Cues — A function defined piecewise: one formula for $(x,y) \ne (0,0)$, typically $f = xy \cdot h(x,y)/(x^2+y^2)$ for some polynomial $h$, and $f(0,0) = 0$. The question asks for $f_{xy}(0,0)$ and $f_{yx}(0,0)$ and discusses their continuity. The factor $xy$ in the numerator is the key shortcut: $f$ and its first partials vanish on the coordinate axes.

Solution Template

1. Use the $xy$ factor. Note $f(h,0) = 0$ for all $h$ and $f(0,k) = 0$ for all $k$ (from the explicit formula).
2. Compute $f_x(0,0)$ and $f_y(0,0)$ by limit definition. Both are $0$ (from step 1).
3. Compute $f_x(0,y)$ for $y \ne 0$. Differentiate the off-origin formula with respect to $x$, then set $x = 0$.
4. Compute $f_{xy}(0,0)$ by limit definition: $f_{xy}(0,0) = \lim_{k\to 0}[f_x(0,k)-f_x(0,0)]/k$.
5. Compute $f_y(x,0)$ for $x \ne 0$. Differentiate with respect to $y$, set $y=0$.
6. Compute $f_{yx}(0,0)$ similarly.
7. Compare $f_{xy}$ and $f_{yx}$. If they differ, cite Schwarz: at least one mixed partial is discontinuous at $(0,0)$. Show the discontinuity using different approach paths.

Worked Example

2014 Paper 2, 2014-P2-Q3b (15 marks)

Obtain $\dfrac{\partial^2 f(0,0)}{\partial x\,\partial y}$ and $\dfrac{\partial^2 f(0,0)}{\partial y\,\partial x}$ for $f(x,y) = \begin{cases}\dfrac{xy(3x^2-2y^2)}{x^2+y^2}, & (x,y)\ne(0,0) \\ 0, & (x,y)=(0,0)\end{cases}.$ Also discuss the continuity of $f_{xy}$ and $f_{yx}$ at $(0,0)$.

Step 1 — First partials at origin. Since $f(h,0) = 0$ for all $h$ and $f(0,k) = 0$ for all $k$ (the formula gives $xy(\ldots)/(x^2+y^2)$ which vanishes on the axes):

$f_x(0,0) = \lim_{h\to 0}\frac{f(h,0)-0}{h} = 0, \qquad f_y(0,0) = 0.$

Step 2 — $f_x(0,y)$ for $y \ne 0$. Differentiate the off-origin formula with respect to $x$ and evaluate at $x=0$:

$f_x(x,y) = \frac{y(9x^2-2y^2)}{x^2+y^2} - \frac{2x^2 y(3x^2-2y^2)}{(x^2+y^2)^2}.$

At $x = 0$, $y \ne 0$: $f_x(0,y) = \dfrac{y \cdot (-2y^2)}{y^2} = -2y$.

Step 3 — Compute $f_{xy}(0,0)$.

$f_{xy}(0,0) = \lim_{k\to 0}\frac{f_x(0,k)-f_x(0,0)}{k} = \lim_{k\to 0}\frac{-2k - 0}{k} = -2.$

Step 4 — $f_y(x,0)$ for $x \ne 0$. Similarly differentiate with respect to $y$ and set $y = 0$:

$f_y(x,0) = 3x.$

Step 5 — Compute $f_{yx}(0,0)$.

$f_{yx}(0,0) = \lim_{h\to 0}\frac{f_y(h,0)-f_y(0,0)}{h} = \lim_{h\to 0}\frac{3h-0}{h} = 3.$

$\boxed{f_{xy}(0,0) = -2, \qquad f_{yx}(0,0) = 3.}$

Step 6 — Continuity discussion. Since $f_{xy}(0,0) \ne f_{yx}(0,0)$, Schwarz’s theorem requires at least one to be discontinuous at $(0,0)$. In fact, both are.

Away from $(0,0)$, $f \in C^\infty$ so $f_{xy} = f_{yx}$; call this common function $g(x,y)$.

• Along $y = 0$: $f_y(x,0) = 3x$, so $\partial_x f_y(x,0) = 3 \to 3$ as $x \to 0$.
• Along $x = 0$: $f_x(0,y) = -2y$, so $\partial_y f_x(0,y) = -2 \to -2$ as $y \to 0$.

The limits of $g$ along these two paths are $3$ and $-2$ respectively, hence different. Therefore $\lim_{(x,y)\to(0,0)} g(x,y)$ does not exist, and both $f_{xy}$ and $f_{yx}$ are discontinuous at $(0,0)$.

$\boxed{\text{Both } f_{xy} \text{ and } f_{yx} \text{ are discontinuous at }(0,0).}$

Common Traps

• Using the smooth formula at the origin. The product/quotient rule applied to the off-origin formula is invalid at $(0,0)$ for a piecewise function. Always use the limit definition there.
• $f_{xy} \ne f_{yx}$ is the signal, not the explanation. The unequal values are a consequence of discontinuity; the discontinuity must still be demonstrated by showing path-dependent limits.
• Missing the $xy$ factor shortcut. If the numerator contains $xy$ as a factor, then $f$ vanishes on both axes, so $f_x(0,0) = 0$ and $f_y(0,0) = 0$ with no computation needed. Recognising this saves 2–3 minutes.
• Wrong cancellation in $f_x(0,y)$. In the formula for $f_x$ at $x=0$: the term $9x^2-2y^2 \to -2y^2$ and the second term vanishes; after dividing by $y^2$ from the denominator, $f_x(0,y) = -2y$. Be careful not to drop the sign.

Marks-Aware Writing

A 15-mark answer for the mixed-partials question requires all of: (1) compute first partials at origin by limit definition (no points for using the formula), (2) compute $f_x(0,y)$ by differentiating off-origin and setting $x=0$, (3) compute $f_{xy}(0,0)$ as a limit, (4) same for $f_{yx}(0,0)$ via $f_y(x,0)$, (5) state the values, (6) discuss continuity with path-dependent limits. Omitting any of steps 1–5 costs 2–3 marks each.

A 20-mark Euler answer requires: (1) identify the homogeneous function $z = \tan u$ and verify degree, (2) state and apply first-order Euler to get $xu_x + yu_y = \sin 2u$, (3) differentiate this relation twice and add, (4) substitute $g(u)(g'(u)-1)$, (5) trig simplification of $2\cos 2u - 1 = 1-4\sin^2 u$. All five algebraic steps must be visible for full marks.

Practice Set

• 2022-P2-Q3b (20 m) — — Hint: likely Euler’s theorem on another homogeneous or inverse-trig function; identify the substitution $z = \sin u$ or $z = \tan u$ first.
• 2021-P2-Q3b (20 m) — — Hint: second-order Euler identity; verify degree of the homogeneous function before applying.
• 2016-P2-Q3b (15 m) — — Hint: mixed-partials or Schwarz counterexample; use limit definition at the origin for the piecewise formula.