UPSC 2014 Maths Optional Paper 2 Q6a — 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 equation $\dfrac{\partial^{2}z}{\partial x^{2}}=x^{2}\dfrac{\partial^{2}z}{\partial y^{2}}$ to canonical form.

Standard hyperbolic-PDE reduction: discriminant > 0 → characteristic-coordinate change → canonical form $z_{\xi\eta}=$ lower-order terms.

Strategy. Identify type via discriminant; find characteristic curves; transform to characteristic coordinates.

Rewrite: $z_{xx}-x^{2}z_{yy}=0$ . Compare with $Az_{xx}+2Bz_{xy}+Cz_{yy}+\ldots=0$ : $A=1$ , $B=0$ , $C=-x^{2}$ .

Discriminant: $B^{2}-AC=0-(1)(-x^{2})=x^{2}\ge 0$ .

For $x\ne 0$ : $x^{2}>0$ ⇒ hyperbolic. At $x=0$ : degenerate (parabolic). Assume $x\ne 0$ .

$A(dy)^{2}-2B\,dx\,dy+C(dx)^{2}=0$ becomes $(dy)^{2}-x^{2}(dx)^{2}=0$ , factoring as

(dy-x\,dx)(dy+x\,dx)=0.

Two families: $dy=\pm x\,dx$ . Integrate:

\xi=y-\dfrac{x^{2}}{2}=\text{const},\qquad\eta=y+\dfrac{x^{2}}{2}=\text{const}.

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

z_x=-x\,z_\xi+x\,z_\eta=x(z_\eta-z_\xi),\qquad z_y=z_\xi+z_\eta.

For second derivatives:

z_{xx}=(z_\eta-z_\xi)+x^{2}(z_{\eta\eta}-2z_{\xi\eta}+z_{\xi\xi}),

(differentiating $z_x=x(z_\eta-z_\xi)$ and applying chain rule again with $\xi_x=-x,\;\eta_x=x$ );

z_{yy}=z_{\xi\xi}+2z_{\xi\eta}+z_{\eta\eta}.

$z_{xx}-x^{2}z_{yy}=0$ :

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

Simplify the $x^{2}$ terms: the $z_{\xi\xi}$ and $z_{\eta\eta}$ pieces cancel, leaving $-4x^{2}z_{\xi\eta}$ :

(z_\eta-z_\xi)-4x^{2}z_{\xi\eta}=0.

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

So $x^{2}=\eta-\xi$ . Substituting:

(z_\eta-z_\xi)-4(\eta-\xi)z_{\xi\eta}=0,

\boxed{\;z_{\xi\eta}=\dfrac{z_\eta-z_\xi}{4(\eta-\xi)},\quad\xi=y-x^{2}/2,\;\eta=y+x^{2}/2.\;}