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UPSC 2026 Maths Optional Paper 2 Q6a — Step-by-Step Solution

20 marks · Section B

Classification and reduction to canonical form · PDEs · asked 9× in 14 yrs · Read the full method →

Question

Reduce the partial differential equation

to canonical form and hence solve it.

Technique

A second-order linear PDE with variable coefficients: the first move is always the discriminant $B^2-4AC$, which here is the perfect square $(x-y)^2$ — so the equation is hyperbolic everywhere except on the line $y=x$, where it is parabolic. Off that line the two characteristic families $y-x=c_1$ and $y^2-x^2=c_2$ are the new coordinates; in them the equation collapses to $\partial_\xi(\xi z_\eta)=0$, which integrates twice by inspection and delivers the two arbitrary functions. The trap is the sign of the middle term in the characteristic ODE ($A\,dy^2 - B\,dx\,dy + C\,dx^2=0$, minus $B$), and the trap after that is forgetting to say where the reduction is legitimate.

Solution

Throughout write

so that the equation is $A z_{xx}+B z_{xy}+C z_{yy}=0$, with $z=z(x,y)$ assumed $C^2$.

Step 1 — Classify: compute the discriminant.

Hence

• $B^2-4AC>0$ whenever $y\neq x$ — the equation is hyperbolic on each of the two open half-planes $\{y>x\}$ and $\{y<x\}$;
• $B^2-4AC=0$ on the line $y=x$ — the equation is parabolic there;
• at the single point $(0,0)$ all three of $A,B,C$ vanish and the equation degenerates to the vacuous statement $0=0$.

So the type is not constant on the plane, and a hyperbolic canonical form can only be sought on a region that does not meet $y=x$. Fix once and for all such a region, say

Step 2 — The characteristic equation (mind the sign).

If $\varphi(x,y)=\text{const}$ is a characteristic curve, then the coefficient of $z_{\varphi\varphi}$ after the change of variable must vanish, i.e.

Along $\varphi=\text{const}$ we have $\varphi_x\,dx+\varphi_y\,dy=0$, so $\varphi_x=-\varphi_y\,\dfrac{dy}{dx}$. Substituting into (1) and cancelling $\varphi_y^{\,2}\neq0$:

The middle sign is a minus — this is exactly where careless work loses the question, because with a $+B$ one gets $y-x$ and $y^2-x^2$ replaced by nonsense.

Here (2) reads

which factorises without any division (so no special pleading is needed when $y=0$):

Step 3 — Integrate the two characteristic families.

Take as new independent variables

Figure to draw (small, one-third of a page, worth its space because it makes the classification visible): draw the two axes; draw three or four parallel lines of slope $1$ (the family $y-x=c_1$) and, on top of them, two branches each of the hyperbolas $y^2-x^2=c_2$ for one positive and one negative $c_2$; draw the line $y=x$ as a bold dashed line and label it “parabolic locus $B^2-4AC=0$: the two families are tangent here”; label the line family ”$\xi=$ const” and the hyperbola family ”$\eta=$ const”; shade the region $y>x$ and label it $\Omega$ (hyperbolic).

The geometry the figure exposes is worth one sentence in the answer: on $y=x$ the hyperbola of the second family degenerates to $y^2-x^2=0$, i.e. to the pair of lines $y=\pm x$, one branch of which is the line characteristic $y-x=0$. The two families therefore coincide exactly on the parabolic locus — which is the geometric reason the reduction fails precisely there.

Step 4 — Check the change of variable is admissible.

So $J\neq0$ exactly when $y\neq x$: the map $(x,y)\mapsto(\xi,\eta)$ is a genuine change of coordinates on $\Omega$ and nowhere else. (Explicitly it inverts as $x=\tfrac12\!\left(\dfrac{\eta}{\xi}-\xi\right)$, $y=\tfrac12\!\left(\dfrac{\eta}{\xi}+\xi\right)$ for $\xi>0$, so $\Omega$ maps onto the half-plane $\xi>0$.) The vanishing of $J$ on $y=x$ is the same degeneracy as the vanishing of the discriminant — a consistency check worth stating.

Step 5 — Transform the equation.

Under $z(x,y)=Z(\xi,\eta)$ the standard transformation formula gives

with

Compute each:

(Both vanish — as they must, since $\xi$ and $\eta$ were built to be characteristic. This is the check that the Step 2 sign was right.)

Hence the equation becomes

On $\Omega$ we have $x-y\neq0$, so divide by $-2(x-y)^2$:

Finally $x-y=-\xi$, giving the canonical (normal) form

This is the first canonical form of a hyperbolic equation: the second-order part is the single mixed derivative $Z_{\xi\eta}$, and only a first-order term survives.

Step 6 — Integrate the canonical form.

Write (3) as $\dfrac{\partial}{\partial\xi}\bigl(\xi Z_\eta\bigr)=0$. Integrating with respect to $\xi$ (at fixed $\eta$), the bracket is a function of $\eta$ alone; call it $F'(\eta)$, with $F$ an arbitrary $C^2$ function:

Now integrate with respect to $\eta$ at fixed $\xi$; the “constant” of integration is an arbitrary $C^2$ function $G$ of $\xi$:

Step 7 — Return to $(x,y)$.

with $F,G$ arbitrary twice-differentiable functions. Two arbitrary functions for a second-order PDE in two variables — the count is right. $\qquad\blacksquare$

Step 8 — Two checks and one remark that earn marks.

(i) Direct verification of the $G$-part. For $z=G(y-x)$: $z_{xx}=G''$, $z_{xy}=-G''$, $z_{yy}=G''$, so

(ii) Direct verification of a non-trivial $F$-part. Take $F(\eta)=\eta^{2}$, so $z=\dfrac{(y^2-x^2)^2}{y-x}=(y-x)(y+x)^2$. Put $u=y-x$, $v=y+x$; then $z=uv^{2}$ and

so

(iii) The degenerate line, again. Since $\eta=y^2-x^2=(y-x)(y+x)=\xi\,(x+y)$, the first term can be rewritten

So a solution of this family extends continuously across the parabolic line $y=x$ only if $F(0)=0$, in which case, writing $H(\eta)=F(\eta)/\eta$, the general solution takes the equivalent and manifestly regular form

(For example $H\equiv1$ gives $z=x+y$, which indeed solves the equation trivially.) Saying this explicitly is what turns a correct answer into a complete one: the printed instruction “reduce to canonical form” is only meaningful off $y=x$, and the general solution itself remembers that.

Answer