Classification and reduction to canonical form

At a Glance

• Frequency: 8 sub-parts across 8 of 13 years (2013, 2014, 2015, 2017, 2019, 2022, 2023, 2024)
• Priority tier: T1
• Marks (count): 10 (1), 15 (6), 20 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 6, hard 2
• Section: B | Dominant type: derivation

Why This Chapter Matters

PDE classification appears in 8 of the last 13 years, almost always as a 15-mark question. The procedure is algorithmic and rewards preparation: compute a discriminant, solve the characteristic ODE, change variables, substitute — and the messy original equation becomes something integrable. Every question follows this skeleton; the only variation is whether the type is hyperbolic or parabolic, and whether the coefficient of $u_{xx}$ is zero.

Minimum Theory

Standard form. A second-order PDE in the form $Au_{xx}+2Bu_{xy}+Cu_{yy}+\ldots=0$ (where $A,B,C$ may depend on $x,y$) is classified by its discriminant $\Delta=B^2-AC$:

• $\Delta>0$: hyperbolic — two real characteristic families; canonical form $u_{\xi\eta}=$ lower-order terms.
• $\Delta=0$: parabolic — one repeated characteristic family; canonical form $u_{\eta\eta}=$ (function of $\xi,\eta,u_\xi,u_\eta,u$).
• $\Delta<0$: elliptic — no real characteristics (not tested in this atom).

Characteristic ODE. The characteristics satisfy $A\left(\frac{dy}{dx}\right)^2-2B\frac{dy}{dx}+C=0,$ or in $(dx,dy)$ form: $A\,dy^2-2B\,dx\,dy+C\,dx^2=0$. Each real solution gives a family of curves $\phi(x,y)=$ const.

Change of variables. Take $\xi=\phi_1(x,y)$, $\eta=\phi_2(x,y)$ (the two characteristic integrals for hyperbolic; one characteristic + any independent function for parabolic). Apply the chain rule to transform $u_{xx},u_{xy},u_{yy}$. The second-order mixed term $u_{\xi\eta}$ (hyperbolic) or pure term $u_{\eta\eta}$ (parabolic) is the only surviving second-order piece in the canonical form.

Key rule for $A=0$. If $A=0$, the characteristic ODE $A\,dy^2-2B\,dx\,dy+C\,dx^2=0$ factors as $dx(C\,dx-2B\,dy)=0$. One characteristic is always $dx=0$ (i.e. $\xi=x$); the other comes from the linear factor.

Question Archetypes

One procedure covers all questions. The sub-cases depend on the type.

canonical-reduction (8 question(s); 2013, 2014, 2015, 2017, 2019, 2022, 2023, 2024)

Recognition Cues

• “Reduce to canonical form” — the entire question.
• Coefficients may be functions of $x,y$ (non-constant coefficients are common and change the characteristic curves to curves, not straight lines).
• Sometimes asks to “solve” after reducing: integrate the canonical ODE in the second new variable.

Solution Template

1. Identify $A,2B,C$. Compute $\Delta=B^2-AC$; state the type.
2. Characteristic ODE. Write $A\,dy^2-2B\,dx\,dy+C\,dx^2=0$; factor or use the quadratic formula.
3. Integrate each factor to get $\xi=\phi_1(x,y)=$ const and $\eta=\phi_2(x,y)=$ const.
4. Compute the chain-rule derivatives $u_x,u_y,u_{xx},u_{xy},u_{yy}$ in terms of $u_\xi,u_\eta,u_{\xi\xi},u_{\xi\eta},u_{\eta\eta}$.
5. Substitute. The $u_{\xi\xi}$ and $u_{\eta\eta}$ (hyperbolic) or the $u_{\xi\xi}$ and $u_{\xi\eta}$ (parabolic) coefficients vanish by construction. Collect remaining terms.
6. Express any surviving $x,y$ in $(\xi,\eta)$. Use the characteristic integrals to re-express.
7. Solve (if asked): the canonical form is an ODE in $\eta$ (with $\xi$ as parameter); integrate once or twice.

Worked Example(s)

2019 Paper 2, 2019-P2-Q7c (20 marks) — Parabolic, solve

Reduce $u_{xx}-2xu_{xy}+x^2u_{yy}=u_y+12x$ to canonical form; find the general solution.

Type. $A=1$, $B=-x$, $C=x^2$; $\Delta=x^2-x^2=0$. Parabolic.

Characteristic. $(dy+x\,dx)^2=0\Rightarrow y+x^2/2=$ const. Set $\xi=y+x^2/2$; choose $\eta=x$.

Derivatives. $\xi_x=x,\xi_y=1,\eta_x=1,\eta_y=0$. So $u_y=u_\xi$. Also $u_{xx}=x^2u_{\xi\xi}+2xu_{\xi\eta}+u_{\eta\eta}+u_\xi$ (the $u_\xi$ comes from $\xi_{xx}=1$).

Substitute into LHS $u_{xx}-2xu_{xy}+x^2u_{yy}$: the $u_{\xi\xi}$ and $u_{\xi\eta}$ terms cancel; remaining: $u_{\eta\eta}+u_\xi$. RHS $=u_y+12x=u_\xi+12\eta$.

$u_{\eta\eta}=12\eta.\quad\text{(Canonical form)}$

Solve. Integrate twice: $u=2\eta^3+\eta f(\xi)+g(\xi)$.

Back-substitute $\eta=x$, $\xi=y+x^2/2$:

$\boxed{\;u=2x^3+x\,f\!\Bigl(y+\tfrac{x^2}{2}\Bigr)+g\!\Bigl(y+\tfrac{x^2}{2}\Bigr).\;}$

2015 Paper 2, 2015-P2-Q8a (15 marks) — Parabolic, solve

Reduce $x^2u_{xx}-2xyu_{xy}+y^2u_{yy}+xu_x+yu_y=0$; find general solution.

Type. $A=x^2$, $B=-xy$, $C=y^2$; $\Delta=x^2y^2-x^2y^2=0$. Parabolic.

Characteristic. $(x\,dy+y\,dx)^2=0\Rightarrow d(xy)=0\Rightarrow xy=$ const. Set $\xi=xy$; choose $\eta=x$.

Reduce. After chain-rule substitution (and using $xu_x+yu_y$), the PDE reduces to: $\eta u_{\eta\eta}+u_\eta=0\quad(\text{canonical form, i.e. }\partial_\eta(\eta u_\eta)=0).$

Solve. $\eta u_\eta=f(\xi)\Rightarrow u=f(\xi)\ln|\eta|+g(\xi)$.

$\boxed{\;u=f(xy)\ln|x|+g(xy).\;}$

2017 Paper 2, 2017-P2-Q7a (15 marks) — Parabolic, solve

Reduce $y^2z_{xx}-2xyz_{xy}+x^2z_{yy}=(y^2/x)z_x+(x^2/y)z_y$; solve.

Type. $A=y^2$, $B=-xy$, $C=x^2$; $\Delta=0$. Parabolic.

Characteristic. $(y\,dy+x\,dx)^2=0\Rightarrow x^2+y^2=$ const. Set $\xi=x^2+y^2$; choose $\eta=y^2$.

Canonical form. After substitution (the first-order RHS cancels all lower-order terms): $Z_{\eta\eta}=0.$

Solve. $Z=\eta f(\xi)+g(\xi)$.

$\boxed{\;z=y^2f(x^2+y^2)+g(x^2+y^2).\;}$

2014 Paper 2, 2014-P2-Q6a (15 marks) — Hyperbolic, non-constant

Reduce $z_{xx}=x^2z_{yy}$ to canonical form.

Type. $A=1$, $B=0$, $C=-x^2$; $\Delta=x^2>0$ (for $x\ne0$). Hyperbolic.

Characteristics. $(dy-x\,dx)(dy+x\,dx)=0$. Two families: $y\pm x^2/2=$ const. Set $\xi=y-x^2/2$, $\eta=y+x^2/2$.

Substitute. After chain rule: $z_{xx}-x^2z_{yy}=(z_\eta-z_\xi)-4x^2z_{\xi\eta}=0$. Since $x^2=\eta-\xi$:

$\boxed{\;z_{\xi\eta}=\frac{z_\eta-z_\xi}{4(\eta-\xi)}.\;}$

2013/2022 — Hyperbolic, $yz_{xx}+(x+y)z_{xy}+xz_{yy}=0$

(2013: 10 marks, 2022: 15 marks including solve)

Sources:,

Type. $A=y$, $B=(x+y)/2$, $C=x$; $\Delta=(x-y)^2/4>0$ for $x\ne y$. Hyperbolic.

Characteristics. Quadratic factors: $(dy/dx-x/y)(dy/dx-1)=0$. Two families: $y^2-x^2=$ const ($\xi=y^2-x^2$) and $y-x=$ const ($\eta=y-x$).

Canonical form (after substitution, using $x-y=-\eta$): $\eta u_{\xi\eta}+u_\xi=0\quad(\text{i.e.}\;\partial_\xi(\eta u_\xi)=0).$

Solve (2022): $\eta u_\xi=F(\eta)\Rightarrow u_\xi=F(\eta)/\eta\Rightarrow u=G(\eta)/\eta+H(\eta)$ (rename):

$\boxed{\;z=\frac{F(y^2-x^2)}{y-x}+H(y-x).\;}$

2023 Paper 2, 2023-P2-Q8a (15 marks) — Hyperbolic, $A=0$

Reduce $z_{yy}-z_{xy}+z_x-z_y(1+1/x)+z/x=0$ to canonical form.

Type. $A=0$, $2B=-1$, $C=1$; $\Delta=B^2=1/4>0$. Hyperbolic.

Characteristics. $A=0\Rightarrow dx(dy+dx)=0$. Two families: $\xi=x$ and $\eta=x+y$.

Canonical form (chain rule with $z_x=z_\xi+z_\eta$, $z_y=z_\eta$): $z_{\xi\eta}-z_\xi+\frac{z_\eta-z}{\xi}=0.$

2024 Paper 2, 2024-P2-Q8a (15 marks) — Substitution approach

Solve $\partial_y(\phi_x+\phi)+2x^2y(\phi_x+\phi)=0$.

Key insight. The expression $\psi=\phi_x+\phi$ appears twice. Substitute directly: $\psi_y+2x^2y\psi=0.$

This is a separable first-order ODE in $y$: $\ln|\psi|=-x^2y^2+h(x)\Rightarrow\psi=h(x)e^{-x^2y^2}$.

Unwind from $\phi_x+\phi=\psi$: integrating factor $e^x$: $\phi=e^{-x}\!\left[k(y)+\int h(x)\,e^{x-x^2y^2}\,dx\right].$

$\boxed{\;\phi=e^{-x}\!\left[k(y)+\int h(x)\,e^{x-x^2y^2}\,dx\right].\;}$

Common Traps

• The sign of $B$. The standard form is $Au_{xx}+2Bu_{xy}+Cu_{yy}$. If the given PDE has coefficient $-2xy$ on $u_{xy}$, then $2B=-2xy$ so $B=-xy$. Writing $B$ with the wrong sign gives the wrong discriminant and wrong characteristics.
• For $A=0$: the characteristic ODE $A\,dy^2-2B\,dx\,dy+C\,dx^2=0$ has $dx$ as an automatic factor. Don’t apply the quadratic formula; factor first.
• The $u_\xi$ term from parabolic reduction. When $\xi_{xx}\ne0$ (e.g., $\xi=y+x^2/2$, $\xi_{xx}=1$), differentiating $u_x$ w.r.t. $x$ introduces a $u_\xi$ term from the $\xi_{xx}$ derivative of $u_\xi$. Missing it leaves a spurious first-order term that won’t cancel the RHS.
• The perfect-square structure. For parabolic equations, the characteristic ODE is always a perfect square — recognising this saves time (e.g., $(x\,dy+y\,dx)^2=0$; $(y\,dy+x\,dx)^2=0$; $(dy+x\,dx)^2=0$).
• Express $x,y$ in $(\xi,\eta)$ after substitution. Any surviving $x$ or $y$ in the canonical form must be re-expressed using the characteristic integrals (e.g., $x^2=\eta-\xi$ for the 2014 problem).

Marks-Aware Writing

10-mark questions (2013): State type (one line), write characteristic ODE and find one family (two lines), write the second family (one line), state canonical form. No need to solve.

15-mark questions (2014, 2015, 2017, 2022, 2023, 2024): Full canonical reduction: type, both characteristics, chain-rule derivatives (show at least the key ones), substitute and verify cancellations, state canonical form. If solve is asked: integrate once or twice in $\eta$, back-substitute.

20-mark questions (2019): All of the above plus: show the $u_\xi$ cancellation explicitly (since $\xi_{xx}=1$ generates it), integrate twice, write out the two-arbitrary-function general solution in $(x,y)$ variables.

Practice Set