UPSC 2013 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

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

Question

Reduce the equation

y\frac{\partial^{2}z}{\partial x^{2}}+(x+y)\frac{\partial^{2}z}{\partial x\partial y}+x\frac{\partial^{2}z}{\partial y^{2}}=0

to its canonical form when $x\ne y$ .

Discriminant → characteristic ODEs → coordinate change $(x,y)\to(\xi,\eta)$ → chain-rule substitution.

Strategy. Compute discriminant; find characteristic equations; introduce characteristic coordinates $(\xi,\eta)$ ; transform.

Compare with $Az_{xx}+2Bz_{xy}+Cz_{yy}=\ldots$ : $A=y$ , $2B=x+y$ (so $B=(x+y)/2$ ), $C=x$ .

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

For $x\ne y$ : discriminant $>0\Rightarrow$ hyperbolic.

$A(dy/dx)^{2}-2B(dy/dx)+C=0$ :

y\,m^{2}-(x+y)m+x=0,\qquad m=dy/dx.

Roots:

m=\dfrac{(x+y)\pm\sqrt{(x+y)^{2}-4xy}}{2y}=\dfrac{(x+y)\pm|x-y|}{2y}.

For $x>y$ (the case $x<y$ is symmetric):

Two families:

$dy/dx=1\Rightarrow y-x=$ const. Coordinate $\xi=y-x$ .
$dy/dx=x/y\Rightarrow y\,dy=x\,dx\Rightarrow y^{2}-x^{2}=$ const. Coordinate $\eta=y^{2}-x^{2}$ .

With $\xi=y-x,\;\eta=y^{2}-x^{2}$ : $\xi_x=-1,\;\xi_y=1,\;\eta_x=-2x,\;\eta_y=2y$ .

By the chain rule:

z_x=-z_\xi-2x z_\eta,\quad z_y=z_\xi+2y z_\eta.

Second-order partials (after applying chain rule twice):

z_{xx}=z_{\xi\xi}+4xz_{\xi\eta}+4x^{2}z_{\eta\eta}-2z_\eta,

z_{yy}=z_{\xi\xi}+4yz_{\xi\eta}+4y^{2}z_{\eta\eta}+2z_\eta,

z_{xy}=-z_{\xi\xi}-2(x+y)z_{\xi\eta}-4xy\,z_{\eta\eta}.

y\,z_{xx}+(x+y)z_{xy}+x\,z_{yy}=0.

Collect coefficients of each $(\xi,\eta)$ -second-derivative:

$z_{\xi\xi}$ : $y-(x+y)+x=0$ ✓.

$z_{\eta\eta}$ : $4x^{2}y-4xy(x+y)+4xy^{2}=4xy[x-(x+y)+y]=0$ ✓.

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

$z_\eta$ : $-2y+0+2x=2(x-y)$ .

$z_\xi$ : $0$ .

So the PDE reduces to

-2(x-y)^{2}\,z_{\xi\eta}+2(x-y)\,z_\eta=0.

Divide by $-2(x-y)$ (allowed since $x\ne y$ ):

(x-y)\,z_{\xi\eta}-z_\eta=0.

Now use $\xi=y-x$ , so $x-y=-\xi$ :

-\xi\,z_{\xi\eta}-z_\eta=0,\;\text{i.e.}\;\xi\,z_{\xi\eta}+z_\eta=0.

\boxed{\;z_{\xi\eta}+\dfrac{1}{\xi}z_\eta=0,\quad\xi=y-x,\;\eta=y^{2}-x^{2}.\;}