← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Classification and reduction to canonical form · PDEs · asked 8× in 13 yrs · Read the full method →

Question

Reduce the second-order PDE

into canonical form. Hence, find its general solution.

Technique

Identify parabolic (discriminant 0); find single family of characteristics ($xy=$ const); change variables to $\xi=xy,\eta=x$; transform derivatives via chain rule; PDE reduces to $\eta u_{\eta\eta}+u_\eta=0$ — solve as ODE in $\eta$.

Solution

Step 1 — Identify type

Coefficients: $A=x^2$, $B=-2xy$ (so the $u_{xy}$ coefficient is $2B=-2xy$, hence $B=-xy$), $C=y^2$.

Discriminant: $B^2-AC=x^2 y^2-x^2 y^2=0$. Parabolic PDE.

Step 2 — Characteristic ODE

For parabolic, single family of characteristics:

i.e. $x^2 dy^2+2xy\,dx\,dy+y^2 dx^2=0$, which factors as $(x\,dy+y\,dx)^2=0$.

So $x\,dy+y\,dx=0\Rightarrow d(xy)=0\Rightarrow xy=\text{const}$.

Characteristic: $\xi=xy$.

Step 3 — Second variable

Choose $\eta$ independent of $\xi$. Standard choice: $\eta=y$ (or $\eta=x$). Use $\eta=x$.

So $\xi=xy$, $\eta=x$. Then $x=\eta$, $y=\xi/\eta$.

Step 4 — Transform derivatives

Chain rule:

Step 5 — Substitute into PDE

$x^2 u_{xx}=x^2(y^2 u_{\xi\xi}+2y u_{\xi\eta}+u_{\eta\eta})=x^2 y^2 u_{\xi\xi}+2x^2 y u_{\xi\eta}+x^2 u_{\eta\eta}$.

$-2xy u_{xy}=-2xy(u_\xi+xy u_{\xi\xi}+x u_{\xi\eta})=-2xy u_\xi-2x^2 y^2 u_{\xi\xi}-2x^2 y u_{\xi\eta}$.

$y^2 u_{yy}=y^2\cdot x^2 u_{\xi\xi}=x^2 y^2 u_{\xi\xi}$.

Sum of second-order parts: $x^2 y^2 u_{\xi\xi}\cdot[1-2+1]+2x^2 y u_{\xi\eta}[1-1]+x^2 u_{\eta\eta}\cdot 1=x^2 u_{\eta\eta}$. ✓ Pure $u_{\eta\eta}$ (parabolic canonical).

Plus the lower-order: $-2xy u_\xi$.

Plus the first-order original terms: $xu_x+yu_y=x(yu_\xi+u_\eta)+y(xu_\xi)=xy u_\xi+x u_\eta+xy u_\xi=2xy u_\xi+xu_\eta$.

Sum everything: $x^2 u_{\eta\eta}-2xy u_\xi+2xy u_\xi+x u_\eta=x^2 u_{\eta\eta}+x u_\eta=0$.

Divide by $x$ (assuming $x\ne 0$): $x u_{\eta\eta}+u_\eta=0$, i.e. $\eta u_{\eta\eta}+u_\eta=0$ (since $\eta=x$).

Step 6 — Canonical form

This is an ODE in $\eta$ (with $\xi$ as parameter).

Step 7 — Solve the ODE

$\eta u_{\eta\eta}+u_\eta=(\eta u_\eta)_\eta-0=\dfrac{\partial}{\partial\eta}(\eta u_\eta)=0$? Let’s check: $\dfrac{\partial}{\partial\eta}(\eta u_\eta)=u_\eta+\eta u_{\eta\eta}$ ✓.

So $\dfrac{\partial}{\partial\eta}(\eta u_\eta)=0\Rightarrow\eta u_\eta=f(\xi)$ for some arbitrary $f$.

Then $u_\eta=f(\xi)/\eta$, integrate w.r.t. $\eta$: $u=f(\xi)\ln|\eta|+g(\xi)$.

Step 8 — Back to $(x,y)$ variables

$\xi=xy$, $\eta=x$:

Answer