← 2022 Paper 2

UPSC 2022 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 $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$ to canonical form and solve.

Technique

Compute discriminant (hyperbolic); solve characteristic ODEs $A(y')^2-2By'+C=0$ to find $\xi,\eta$; transform all derivatives; canonical form reduces to a first-order ODE in $u_\xi$.

Solution

Identify type. $A=y$, $2B=x+y$ (so $B=(x+y)/2$), $C=x$.

Discriminant: $B^2-AC=\dfrac{(x+y)^2}{4}-xy=\dfrac{x^2+2xy+y^2-4xy}{4}=\dfrac{(x-y)^2}{4}$.

For $x\ne y$: $B^2-AC>0$, hyperbolic.

For $x=y$: $B^2-AC=0$, parabolic line.

Step 1 — Characteristic equations (hyperbolic case)

$A(dy/dx)^2-2B(dy/dx)+C=0$: $y(dy/dx)^2-(x+y)(dy/dx)+x=0$.

Treat as quadratic in $dy/dx$: $dy/dx=\dfrac{(x+y)\pm\sqrt{(x+y)^2-4xy}}{2y}=\dfrac{(x+y)\pm|x-y|}{2y}$.

For $x>y$: $|x-y|=x-y$:

• $(dy/dx)_+=\dfrac{(x+y)+(x-y)}{2y}=\dfrac{2x}{2y}=x/y$.
• $(dy/dx)_-=\dfrac{(x+y)-(x-y)}{2y}=\dfrac{2y}{2y}=1$.

First family: $dy/dx=x/y\Rightarrow y\,dy=x\,dx\Rightarrow y^2-x^2=$ const. So $\xi=y^2-x^2$.

Second family: $dy/dx=1\Rightarrow y-x=$ const. So $\eta=y-x$.

Step 2 — Transform derivatives

$\xi=y^2-x^2$, $\eta=y-x$.

$\xi_x=-2x$, $\xi_y=2y$, $\eta_x=-1$, $\eta_y=1$.

$u_x=u_\xi\xi_x+u_\eta\eta_x=-2x u_\xi-u_\eta$. $u_y=u_\xi\xi_y+u_\eta\eta_y=2y u_\xi+u_\eta$.

$u_{xx}=\partial_x(-2x u_\xi-u_\eta)=-2u_\xi-2x(u_{\xi\xi}\xi_x+u_{\xi\eta}\eta_x)-(u_{\eta\xi}\xi_x+u_{\eta\eta}\eta_x)$ $=-2u_\xi-2x(-2x u_{\xi\xi}-u_{\xi\eta})-(-2x u_{\xi\eta}-u_{\eta\eta})$ $=-2u_\xi+4x^2 u_{\xi\xi}+2x u_{\xi\eta}+2x u_{\xi\eta}+u_{\eta\eta}$ $=4x^2 u_{\xi\xi}+4x u_{\xi\eta}+u_{\eta\eta}-2u_\xi$.

$u_{yy}=\partial_y(2y u_\xi+u_\eta)=2u_\xi+2y(u_{\xi\xi}\xi_y+u_{\xi\eta}\eta_y)+(u_{\eta\xi}\xi_y+u_{\eta\eta}\eta_y)$ $=2u_\xi+2y(2y u_{\xi\xi}+u_{\xi\eta})+(2y u_{\xi\eta}+u_{\eta\eta})$ $=4y^2 u_{\xi\xi}+4y u_{\xi\eta}+u_{\eta\eta}+2u_\xi$.

$u_{xy}=\partial_y(-2x u_\xi-u_\eta)=-2x(u_{\xi\xi}\cdot 2y+u_{\xi\eta}\cdot 1)-(u_{\eta\xi}\cdot 2y+u_{\eta\eta}\cdot 1)$ $=-4xy u_{\xi\xi}-2x u_{\xi\eta}-2y u_{\xi\eta}-u_{\eta\eta}$ $=-4xy u_{\xi\xi}-2(x+y)u_{\xi\eta}-u_{\eta\eta}$.

Step 3 — Substitute into PDE

$y u_{xx}+(x+y)u_{xy}+x u_{yy}$:

$y u_{xx}=y[4x^2 u_{\xi\xi}+4x u_{\xi\eta}+u_{\eta\eta}-2u_\xi]$ $=4x^2 y u_{\xi\xi}+4xy u_{\xi\eta}+y u_{\eta\eta}-2y u_\xi$.

$(x+y)u_{xy}=(x+y)[-4xy u_{\xi\xi}-2(x+y)u_{\xi\eta}-u_{\eta\eta}]$ $=-4xy(x+y)u_{\xi\xi}-2(x+y)^2 u_{\xi\eta}-(x+y)u_{\eta\eta}$.

$x u_{yy}=x[4y^2 u_{\xi\xi}+4y u_{\xi\eta}+u_{\eta\eta}+2u_\xi]$ $=4xy^2 u_{\xi\xi}+4xy u_{\xi\eta}+x u_{\eta\eta}+2x u_\xi$.

Sum coefficient of $u_{\xi\xi}$: $4x^2 y-4xy(x+y)+4xy^2=4xy[x-(x+y)+y]=4xy\cdot 0=0$ ✓.

Sum coefficient of $u_{\xi\eta}$: $4xy-2(x+y)^2+4xy=8xy-2(x^2+2xy+y^2)=8xy-2x^2-4xy-2y^2=-2x^2+4xy-2y^2=-2(x-y)^2$.

Sum coefficient of $u_{\eta\eta}$: $y-(x+y)+x=0$ ✓.

Sum coefficient of $u_\xi$: $-2y+0+2x=2(x-y)$.

Sum coefficient of $u_\eta$: $0$ (no $u_\eta$ terms in expansions).

So PDE becomes: $-2(x-y)^2 u_{\xi\eta}+2(x-y)u_\xi=0$,

$(x-y)^2 u_{\xi\eta}=(x-y)u_\xi$,

$(x-y)u_{\xi\eta}=u_\xi$ (assuming $x\ne y$).

Express $x-y$ in terms of $\xi,\eta$: $\eta=y-x=-(x-y)$, so $x-y=-\eta$.

$-\eta\cdot u_{\xi\eta}=u_\xi$, i.e.

Step 4 — Canonical form and solve

Let $v=u_\xi$. Then $\eta v_\eta+v=0$, i.e., $\partial_\eta(v\eta)=0$.

So $v\eta=g(\xi)$ for some function $g$. I.e., $u_\xi=g(\xi)/\eta$.

Integrate w.r.t. $\xi$: $u=G(\xi)/\eta+h(\eta)$ where $G(\xi)=\int g(\xi)d\xi$.

Renaming arbitrary functions: $G(\xi)$ is arbitrary, so:

Answer