← 2023 Paper 2

UPSC 2023 Maths Optional Paper 2 Q6a — Step-by-Step Solution

15 marks · Section B

Second-order linear PDEs with constant coefficients (CF, PI) · PDEs · asked 12× in 13 yrs · Read the full method →

Question

Find the surface passing through the two lines $z=x=0$ and $z-1=x-y=0$, and satisfying the partial differential equation $\dfrac{\partial^2 z}{\partial x^2}-4\dfrac{\partial^2 z}{\partial x\partial y}+4\dfrac{\partial^2 z}{\partial y^2}=0$.

Technique

Constant-coefficient second-order PDE → repeated-root general solution $f(y+rx)+x\,g(y+rx)$; pin down $f,g$ from the two line conditions.

Solution

Strategy. The PDE has constant coefficients in the second derivatives only. Solve as $F(D,D')z=0$ with $D=\partial_x,\,D'=\partial_y$, find general solution, then fit the boundary data.

Step 1 — Auxiliary equation and general solution

The symbol of the PDE is $D^{2}-4DD'+4D'^{2}=(D-2D')^{2}$. The auxiliary equation in $r=D/D'$ is

For a repeated root $r$, the general solution is

Step 2 — Apply boundary line 1: $z=x=0$

This line is the $y$-axis ($x=0$ and $z=0$ for all $y$). Substitute $x=0$:

Step 3 — Apply boundary line 2: $z-1=x-y=0$

This line is $\{(x,y,z):x=y,\,z=1\}$. Substitute $y=x$ into the general solution (with $f\equiv 0$):

So $g(3x)=1/x$. Set $u=3x$, i.e. $x=u/3$:

Step 4 — Assemble the surface

Answer