← 2016 Paper 2
UPSC 2016 Maths Optional Paper 2 Q6a — Step-by-Step Solution 15 marks · Section B
Cauchy's method of characteristics · PDEs · asked 4× in 13 yrs · Read the full method →
Question
Determine the characteristics of the equation z = p 2 − q 2 z=p^2-q^2 z = p 2 − q 2 4 z + x 2 = 0 , y = 0 4z+x^2=0,\ y=0 4 z + x 2 = 0 , y = 0
Technique
Charpit / characteristic-strip equations for the nonlinear PDE F = p 2 − q 2 − z = 0 F=p^2-q^2-z=0 F = p 2 − q 2 − z = 0 p 0 , q 0 p_0,q_0 p 0 , q 0 d z 0 = p 0 d x 0 + q 0 d y 0 dz_0=p_0dx_0+q_0dy_0 d z 0 = p 0 d x 0 + q 0 d y 0
Solution
Setup. Write the nonlinear first-order PDE as
F ( x , y , z , p , q ) = p 2 − q 2 − z = 0 , p = z x , q = z y . F(x,y,z,p,q)=p^2-q^2-z=0,\qquad p=z_x,\ q=z_y. F ( x , y , z , p , q ) = p 2 − q 2 − z = 0 , p = z x , q = z y . Charpit’s characteristic (strip) equations are
x ˙ = F p , y ˙ = F q , z ˙ = p F p + q F q , p ˙ = − ( F x + p F z ) , q ˙ = − ( F y + q F z ) , \dot x=F_p,\quad \dot y=F_q,\quad \dot z=pF_p+qF_q,\quad \dot p=-(F_x+pF_z),\quad \dot q=-(F_y+qF_z), x ˙ = F p , y ˙ = F q , z ˙ = p F p + q F q , p ˙ = − ( F x + p F z ) , q ˙ = − ( F y + q F z ) , where ( ) ˙ = d / d s \dot{(\ )}=d/ds ( ) ˙ = d / d s
F p = 2 p , F q = − 2 q , F z = − 1 , F x = 0 , F y = 0. F_p=2p,\quad F_q=-2q,\quad F_z=-1,\quad F_x=0,\quad F_y=0. F p = 2 p , F q = − 2 q , F z = − 1 , F x = 0 , F y = 0. Step 1 — The characteristic equations
x ˙ = 2 p , y ˙ = − 2 q , z ˙ = 2 p 2 − 2 q 2 = 2 ( p 2 − q 2 ) = 2 z , \dot x=2p,\quad \dot y=-2q,\quad \dot z=2p^2-2q^2=2(p^2-q^2)=2z, x ˙ = 2 p , y ˙ = − 2 q , z ˙ = 2 p 2 − 2 q 2 = 2 ( p 2 − q 2 ) = 2 z , p ˙ = − ( 0 + p ( − 1 ) ) = p , q ˙ = − ( 0 + q ( − 1 ) ) = q . \dot p=-(0+p(-1))=p,\qquad \dot q=-(0+q(-1))=q. p ˙ = − ( 0 + p ( − 1 )) = p , q ˙ = − ( 0 + q ( − 1 )) = q . Step 2 — Solve the strip
From p ˙ = p , q ˙ = q \dot p=p,\ \dot q=q p ˙ = p , q ˙ = q p 0 , q 0 p_0,q_0 p 0 , q 0 s = 0 s=0 s = 0
p = p 0 e s , q = q 0 e s . p=p_0e^{s},\qquad q=q_0e^{s}. p = p 0 e s , q = q 0 e s . Then z ˙ = 2 z ⇒ z = z 0 e 2 s \dot z=2z\Rightarrow z=z_0e^{2s} z ˙ = 2 z ⇒ z = z 0 e 2 s z = p 2 − q 2 = ( p 0 2 − q 0 2 ) e 2 s = z 0 e 2 s z=p^2-q^2=(p_0^2-q_0^2)e^{2s}=z_0e^{2s} z = p 2 − q 2 = ( p 0 2 − q 0 2 ) e 2 s = z 0 e 2 s
x ˙ = 2 p 0 e s ⇒ x = x 0 + 2 p 0 ( e s − 1 ) , y ˙ = − 2 q 0 e s ⇒ y = y 0 − 2 q 0 ( e s − 1 ) . \dot x=2p_0e^{s}\Rightarrow x=x_0+2p_0(e^{s}-1),\qquad \dot y=-2q_0e^{s}\Rightarrow y=y_0-2q_0(e^{s}-1). x ˙ = 2 p 0 e s ⇒ x = x 0 + 2 p 0 ( e s − 1 ) , y ˙ = − 2 q 0 e s ⇒ y = y 0 − 2 q 0 ( e s − 1 ) . These are the characteristic strips ; the projected ( x , y ) (x,y) ( x , y ) ( p 0 , − q 0 ) (p_0,-q_0) ( p 0 , − q 0 )
Step 3 — Initial data on the parabola
The parabola is 4 z + x 2 = 0 , y = 0 4z+x^2=0,\ y=0 4 z + x 2 = 0 , y = 0 z = − x 2 / 4 z=-x^2/4 z = − x 2 /4 y = 0 y=0 y = 0 t t t
x 0 = t , y 0 = 0 , z 0 = − t 2 4 . x_0=t,\quad y_0=0,\quad z_0=-\tfrac{t^2}{4}. x 0 = t , y 0 = 0 , z 0 = − 4 t 2 . The initial p 0 , q 0 p_0,q_0 p 0 , q 0 strip condition d z 0 = p 0 d x 0 + q 0 d y 0 dz_0=p_0\,dx_0+q_0\,dy_0 d z 0 = p 0 d x 0 + q 0 d y 0 d y 0 = 0 dy_0=0 d y 0 = 0
d z 0 d t = p 0 d x 0 d t ⇒ − t 2 = p 0 ( 1 ) ⇒ p 0 = − t 2 . \frac{dz_0}{dt}=p_0\frac{dx_0}{dt}\ \Rightarrow\ -\frac{t}{2}=p_0(1)\ \Rightarrow\ p_0=-\frac{t}{2}. d t d z 0 = p 0 d t d x 0 ⇒ − 2 t = p 0 ( 1 ) ⇒ p 0 = − 2 t . The PDE p 0 2 − q 0 2 = z 0 p_0^2-q_0^2=z_0 p 0 2 − q 0 2 = z 0
q 0 2 = p 0 2 − z 0 = t 2 4 − ( − t 2 4 ) = t 2 2 ⇒ q 0 = t 2 . q_0^2=p_0^2-z_0=\frac{t^2}{4}-\Big(-\frac{t^2}{4}\Big)=\frac{t^2}{2}\ \Rightarrow\ q_0=\frac{t}{\sqrt2}. q 0 2 = p 0 2 − z 0 = 4 t 2 − ( − 4 t 2 ) = 2 t 2 ⇒ q 0 = 2 t . Step 4 — Build and eliminate
p = − t 2 e s , q = t 2 e s , z = − t 2 4 e 2 s , p=-\tfrac{t}{2}e^{s},\quad q=\tfrac{t}{\sqrt2}e^{s},\quad z=-\tfrac{t^2}{4}e^{2s}, p = − 2 t e s , q = 2 t e s , z = − 4 t 2 e 2 s , x = t + 2 ( − t 2 ) ( e s − 1 ) = t ( 2 − e s ) , y = 0 − 2 ( t 2 ) ( e s − 1 ) = 2 t ( 1 − e s ) . x=t+2\big(-\tfrac{t}{2}\big)(e^{s}-1)=t(2-e^{s}),\qquad y=0-2\big(\tfrac{t}{\sqrt2}\big)(e^{s}-1)=\sqrt2\,t(1-e^{s}). x = t + 2 ( − 2 t ) ( e s − 1 ) = t ( 2 − e s ) , y = 0 − 2 ( 2 t ) ( e s − 1 ) = 2 t ( 1 − e s ) . Eliminate t , s t,s t , s t = x − 1 2 y t=x-\tfrac{1}{\sqrt2}y t = x − 2 1 y t e s = 2 x − 2 y te^{s}=2x-\sqrt2\,y t e s = 2 x − 2 y
x − 2 y = t e s ⋅ ( − 1 ) + … ⇒ z = − t 2 e 2 s 4 = − 1 4 ( x − 2 y ) 2 . x-\sqrt2\,y=t e^{s}\cdot(-1)+\ldots\ \Rightarrow\ z=-\tfrac{t^2e^{2s}}{4}=-\tfrac14\big(x-\sqrt2\,y\big)^2. x − 2 y = t e s ⋅ ( − 1 ) + … ⇒ z = − 4 t 2 e 2 s = − 4 1 ( x − 2 y ) 2 . Step 5 — Integral surface
Answer
z = − 1 4 ( x − 2 y ) 2 , i.e. 4 z + ( x − 2 y ) 2 = 0. \boxed{\;z=-\tfrac14\big(x-\sqrt2\,y\big)^2,\qquad\text{i.e.}\quad 4z+\big(x-\sqrt2\,y\big)^2=0.\;} z = − 4 1 ( x − 2 y ) 2 , i.e. 4 z + ( x − 2 y ) 2 = 0.