← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q6a — Step-by-Step Solution

15 marks · Section B

Quasilinear first-order PDEs (Lagrange's method) · PDEs · asked 9× in 13 yrs · Read the full method →

Question

Solve the first order quasilinear partial differential equation by the method of characteristics:

Technique

Lagrange’s method / method of characteristics. Two independent integrals $C_1=(u+y)/x$ (from combining $dy+du$) and $C_2=x^2(x+2y-u)$ (from a linear ODE in $y$ along the characteristic); the Cauchy data fixes the relation $C_1=1+2C_2$.

Solution

Setup. This is Lagrange’s equation $P u_x+Q u_y=R$ with

Characteristic (Lagrange auxiliary) system:

Step 1 — First integral $C_1$

Add the $dy$ and $du$ rows: $\dfrac{dy+du}{(u-x-y)+(x+2y)}=\dfrac{d(u+y)}{u+y}$. Pair with $\dfrac{dx}{x}$:

Step 2 — Second integral $C_2$

Along a characteristic put $x=e^{s}$ (so $dx/ds=x$), and $u+y=C_1 x$. Then

i.e. $\dfrac{dy}{ds}+2y=(C_1-1)e^{s}$, whose solution is

Hence $3A=3x^2y-(C_1-1)x^3$ is constant. Using $C_1=(u+y)/x$,

(Quick check: along the characteristics both $C_1$ and $C_2$ are constant.)

Step 3 — Apply the Cauchy data $u=1+y$ on $x=1$

On the data curve ($x=1$, $u=1+y$):

Eliminating the parameter $y$: $C_1=1+2C_2$. The general solution is $\Phi(C_1,C_2)=0$; the data fixes the function as $C_1=1+2C_2$, i.e.

Step 4 — Solve for $u$

Answer