UPSC 2017 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

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

Question

Reduce the equation

y^2\frac{\partial^2 z}{\partial x^2}-2xy\frac{\partial^2 z}{\partial x\partial y}+x^2\frac{\partial^2 z}{\partial y^2}=\frac{y^2}{x}\frac{\partial z}{\partial x}+\frac{x^2}{y}\frac{\partial z}{\partial y}

to canonical form and hence solve it.

Technique

Discriminant $H^2-AB=0$ ⇒ parabolic; one repeated characteristic $x^2+y^2=$ const taken as $\xi$ , independent $\eta=y^2$ ; transform reduces to $Z_{\eta\eta}=0$ ; integrate twice.

Solution

Setup. Compare with $A z_{xx}+2H z_{xy}+B z_{yy}+\cdots=0$ :

A=y^2,\qquad 2H=-2xy\ (H=-xy),\qquad B=x^2.

Step 1 — Classify

Discriminant:

H^2-AB=(-xy)^2-(y^2)(x^2)=x^2y^2-x^2y^2=0.

The equation is parabolic everywhere (away from the axes).

Step 2 — Characteristic curves

The characteristics satisfy $A\,dy^2-2H\,dx\,dy+B\,dx^2=0$ :

y^2\,dy^2+2xy\,dx\,dy+x^2\,dx^2=0\ \Longrightarrow\ (y\,dy+x\,dx)^2=0.

A single (repeated) family:

y\,dy+x\,dx=0\ \Longrightarrow\ x^2+y^2=\text{const}.

Step 3 — Choose canonical coordinates

Take the characteristic as one coordinate and any independent function as the second:

\xi=x^2+y^2,\qquad \eta=y^2.

(The choice $\eta=y^2$ is convenient; any $\eta$ functionally independent of $\xi$ works.)

Step 4 — Transform the equation

Writing $z=Z(\xi,\eta)$ and substituting the chain-rule derivatives, the second-order part collapses (as it must for a parabolic equation) to a single pure second derivative, and the given first-order right-hand side exactly cancels the first-order terms generated by the transformation. The equation reduces to the canonical form

\boxed{\;\frac{\partial^2 Z}{\partial\eta^2}=0.\;}

Step 5 — Solve the canonical equation

Integrating $Z_{\eta\eta}=0$ twice in $\eta$ (treating $\xi$ as a parameter):

Z=\eta\,f(\xi)+g(\xi),

where $f,g$ are arbitrary functions. Returning to $x,y$ with $\eta=y^2,\ \xi=x^2+y^2$ :

Answer

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