UPSC Maths 2025 Paper 2 Q8a — Solution

15 marks · Section B

Question

Find the characteristics of the partial differential equation $p^2 + q^2 = 2$ ; $p \equiv \dfrac{\partial z}{\partial x}$ , $q \equiv \dfrac{\partial z}{\partial y}$ and determine the integral surface which passes through $x = 0$ , $z = y$ .

Technique

Use the method of characteristics (the Lagrange–Charpit strip equations) for a nonlinear first-order PDE $F(p,q)=0$ . Since $F$ depends only on $p,q$ , the characteristics are straight lines and $p,q$ are constant along them; the initial data curve fixes the constants.

Solution

Let $F(x,y,z,p,q) = p^2 + q^2 - 2 = 0.$

Characteristic (strip) equations

$\frac{dx}{dt} = F_p = 2p,\quad \frac{dy}{dt} = F_q = 2q,$ $\frac{dz}{dt} = pF_p + qF_q = 2(p^2 + q^2) = 4,$ $\frac{dp}{dt} = -(F_x + pF_z) = 0,\quad \frac{dq}{dt} = -(F_y + qF_z) = 0.$

So along each characteristic $p =$ const and $q =$ const (since $F_x=F_y=F_z=0$ ). Integrating with $p,q$ constant: $x = 2p\,t + x_0,\qquad y = 2q\,t + y_0,\qquad z = 4t + z_0.$

Characteristics: straight lines in space with direction $(2p, 2q, 4)$ , projecting onto the $xy$ -plane as straight lines of slope $q/p$ , with $p^2+q^2=2$ fixed.

Apply the initial curve $x = 0$ , $z = y$

Parametrise the initial curve by $s$ : $x_0 = 0$ , $y_0 = s$ , $z_0 = s$ (since $z = y = s$ there), at $t = 0$ .

The strip condition on the initial curve requires $dz_0 = p\,dx_0 + q\,dy_0$ : $\frac{dz_0}{ds} = p\frac{dx_0}{ds} + q\frac{dy_0}{ds}\implies 1 = p\cdot 0 + q\cdot 1 = q.$ So $q = 1$ . From the PDE $p^2 + q^2 = 2 \Rightarrow p^2 = 1 \Rightarrow p = \pm 1$ .

Build the integral surface

Take $p = 1$ , $q = 1$ . Then $x = 2t,\qquad y = s + 2t,\qquad z = s + 4t.$

Eliminate $s,t$ : from $x = 2t$ we get $t = x/2$ . Then $y = s + x \Rightarrow s = y - x$ . Hence $z = s + 4t = (y - x) + 2x = x + y.$

$\boxed{z = x + y.}$

Check: $z = x+y \Rightarrow p = 1,\,q = 1 \Rightarrow p^2+q^2 = 2$ ✓; at $x=0$ , $z = y$ ✓.

The other branch

Taking $p = -1$ , $q = 1$ gives, analogously, the surface $z = y - x,$ which also passes through $x=0,\,z=y$ and satisfies $p^2+q^2=(-1)^2+1^2=2$ . Both planes are integral surfaces through the given curve; the principal answer is $z = x + y$ .

Answer

Characteristics: straight lines $x = 2pt + x_0$ , $y = 2qt + y_0$ , $z = 4t + z_0$ with $p,q$ constant and $p^2+q^2=2$ .

Integral surface through $x=0,\,z=y$ : $\;z = x + y\;$ (equivalently $z = y - x$ for the other root $p=-1$ ).