UPSC 2023 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Classification and reduction to canonical form · PDEs · asked 9× in 14 yrs · Read the full method →

Question

Reduce the partial differential equation

\dfrac{\partial^2 z}{\partial y^2}-\dfrac{\partial^2 z}{\partial x\partial y}+\dfrac{\partial z}{\partial x}-\dfrac{\partial z}{\partial y}\Bigl(1+\dfrac{1}{x}\Bigr)+\dfrac{z}{x}=0

to canonical form.

Technique

Discriminant classification → characteristic equations → coordinate change to $(\xi,\eta)$ → chain-rule substitution.

Solution

Step 1 — Classify

Write the principal (second-order) part as $A\,z_{xx}+2B\,z_{xy}+C\,z_{yy}=0$ with

A=0,\qquad 2B=-1\;\Longrightarrow\;B=-\tfrac{1}{2},\qquad C=1.

Discriminant $B^{2}-AC=\tfrac{1}{4}-0=\tfrac{1}{4}>0\;\Longrightarrow$ hyperbolic.

Step 2 — Characteristic equations

The characteristic ODE is $A(dy)^{2}-2B\,dx\,dy+C(dx)^{2}=0$ :

0\cdot(dy)^{2}+dx\,dy+(dx)^{2}=0\;\Longrightarrow\;dx\bigl(dy+dx\bigr)=0.

Two families:

$dx=0\Rightarrow x=\text{const}$ , so $\xi=x$ ;
$dy+dx=0\Rightarrow x+y=\text{const}$ , so $\eta=x+y$ .

The Jacobian $\det\!\begin{pmatrix}\xi_x&\xi_y\\\eta_x&\eta_y\end{pmatrix}=\det\!\begin{pmatrix}1&0\\1&1\end{pmatrix}=1\ne 0$ , so this is a valid coordinate change.

Step 3 — Chain-rule derivatives

With $\xi=x,\,\eta=x+y$ :

z_x=z_\xi+z_\eta,\qquad z_y=z_\eta.

z_{xx}=z_{\xi\xi}+2z_{\xi\eta}+z_{\eta\eta},\quad z_{xy}=z_{\xi\eta}+z_{\eta\eta},\quad z_{yy}=z_{\eta\eta}.

Step 4 — Substitute

z_{yy}-z_{xy}=z_{\eta\eta}-(z_{\xi\eta}+z_{\eta\eta})=-z_{\xi\eta}.

z_x-z_y\!\left(1+\tfrac{1}{x}\right)+\tfrac{z}{x}=(z_\xi+z_\eta)-z_\eta\!\left(1+\tfrac{1}{\xi}\right)+\tfrac{z}{\xi}=z_\xi-\tfrac{z_\eta}{\xi}+\tfrac{z}{\xi}.

Combining:

-z_{\xi\eta}+z_\xi-\tfrac{z_\eta}{\xi}+\tfrac{z}{\xi}=0.

Multiplying by $-1$ and arranging:

Answer

\boxed{\;z_{\xi\eta}-z_\xi+\frac{z_\eta-z}{\xi}=0\;\Longleftrightarrow\;\xi\,z_{\xi\eta}-\xi\,z_\xi+z_\eta-z=0.\;}