← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Cauchy's method of characteristics · PDEs · asked 4× in 14 yrs · Read the full method →

Question

Find a complete integral of $p=(z+qy)^2$ using Charpit’s method.

Technique

Charpit’s method; auxiliary equations; identify a clean first integral from the ratios; combine with $F=0$ to derive complete integral.

Solution

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

Charpit’s auxiliary equations: $\dfrac{dx}{F_p}=\dfrac{dy}{F_q}=\dfrac{dz}{pF_p+qF_q}=\dfrac{dp}{-(F_x+pF_z)}=\dfrac{dq}{-(F_y+qF_z)}$.

Step 1 — Compute partials of $F$

$F=p-(z+qy)^2$.

$F_p=1$. $F_q=-2y(z+qy)$. $F_x=0$. $F_y=-2q(z+qy)$. $F_z=-2(z+qy)$.

Step 2 — Charpit equations

$\dfrac{dx}{1}=\dfrac{dy}{-2y(z+qy)}=\dfrac{dz}{p\cdot 1+q\cdot(-2y(z+qy))}=\dfrac{dp}{-(0+p\cdot(-2(z+qy)))}=\dfrac{dq}{-(−2q(z+qy)+q\cdot(-2(z+qy)))}$.

Simplify denominators:

• $dz$-denom: $p-2qy(z+qy)$. From $F=0$: $p=(z+qy)^2$, so this becomes $(z+qy)^2-2qy(z+qy)=(z+qy)[(z+qy)-2qy]=(z+qy)(z-qy)$.
• $dp$-denom: $-(-2p(z+qy))=2p(z+qy)$.
• $dq$-denom: $-(-2q(z+qy)-2q(z+qy))=4q(z+qy)$.

Step 3 — Look for clean integral

$\dfrac{dp}{2p(z+qy)}=\dfrac{dq}{4q(z+qy)}$.

Cancel $(z+qy)$: $\dfrac{dp}{2p}=\dfrac{dq}{4q}$, i.e., $\dfrac{dp}{p}=\dfrac{dq}{2q}$.

Integrate: $\ln p=\dfrac{1}{2}\ln q+C\Rightarrow p=c\sqrt q$ where $c$ = constant.

Hmm — that has $\sqrt q$. Let me try a different ratio.

From $\dfrac{dq}{4q}$ and the equation $F=0$, $p=(z+qy)^2$. Let me parametrise by $q$:

If $p=A\sqrt q$ for some constant $A$, then $(z+qy)^2=A\sqrt q$, so $z+qy=\sqrt{A\sqrt q}=A^{1/2}q^{1/4}$.

This is getting weird. Let me try another approach.

Alternative — try assuming form for $q$

Charpit aims to find a first integral $f(x,y,z,p,q)=$ const compatible with $F=0$. Then solve.

Try: from $dx/1=dy/(-2y(z+qy))$, get $dy/dx=-2y(z+qy)$.

If $z+qy=$ constant along charpit (call it $k$): then $dy/dx=-2ky$, $y=y_0 e^{-2kx}$. And $z+qy=k$ gives $q=(k-z)/y$.

Substitute into $F=0$: $p=(z+qy)^2=k^2$.

So if $z+qy=k$ and $p=k^2$ — a candidate Charpit integral.

Now solve: $p=z_x=k^2$, $q=z_y=(k-z)/y$.

From $p=k^2$: $z=k^2 x+g(y)$.

From $q=(k-z)/y$: $g'(y)=q=(k-z)/y=(k-k^2 x-g(y))/y$.

So $y g'(y)+g(y)=k-k^2 x$.

But LHS depends only on $y$, RHS has $x$. For consistency, $k^2=0$, i.e., $k=0$. Trivial.

Hmm, contradiction. Let me reconsider.

Try: $q=$ constant = $a$

Suppose along Charpit, $q=a$ (constant). From $F=0$: $p=(z+ay)^2$.

Now solve $z_x=(z+ay)^2$ and $z_y=a$ simultaneously.

From $z_y=a$: $z=ay+h(x)$.

Substitute into $z_x=h'(x)=(z+ay)^2=(2ay+h(x))^2$. But LHS depends only on $x$, RHS on both $x$ and $y$. Consistency requires the $y$-dependence to vanish, i.e., $a=0$. Trivial.

Try: assume special form $z+qy=f(x)$ along characteristic

Let $\xi=z+qy$. Charpit: $\dfrac{d\xi}{dx}=z_x+(dq/dx)y+q(dy/dx)=p+y\dot q+q\dot y$.

From Charpit equations: $dx=F_p\,dt=dt$, so $t=x$. $\dot q=dq/dx=-(F_y+qF_z)=2q(z+qy)+2q(z+qy)=4q(z+qy)$. So $\dot q=4q\xi$.

$\dot y=F_q=-2y\xi$. So $q\dot y=-2qy\xi$.

$p=\xi^2$ from $F=0$.

$\dot\xi=p+y\dot q+q\dot y=\xi^2+y(4q\xi)+(-2qy\xi)=\xi^2+2qy\xi=\xi(\xi+2qy)$.

Hmm, $\xi+2qy=z+qy+2qy=z+3qy$. Not as clean.

Actually: $\xi^2+4qy\xi-2qy\xi=\xi^2+2qy\xi=\xi(\xi+2qy)$. And $\xi+2qy=z+3qy$. Continues messy.

Cleaner Charpit choice

Let me try the ratio $\dfrac{dx}{1}=\dfrac{dp}{2p\xi}$ (where $\xi=z+qy$).

$dp/dx=2p\xi$. And $p=\xi^2$, so $\dot\xi^2/dx$… Actually $p$ is determined by $\xi$ via $p=\xi^2$, so $dp/dx=2\xi(d\xi/dx)$.

Combined: $2\xi(d\xi/dx)=2p\xi=2\xi^3$, so $d\xi/dx=\xi^2$, separable: $-1/\xi=x+C$, $\xi=-1/(x+C)$.

So along char: $z+qy=-1/(x+C)$, with $C$ a constant.

This is one Charpit relation. Combined with $F=0$: $p=\xi^2=1/(x+C)^2$.

Step 4 — Construct complete integral

Let’s use the simpler observation: the equation $p=(z+qy)^2$ has the structure that if we let $\xi=z+qy$, then… hmm, perhaps a more direct method works.

Try $z=$ function of $\xi=y/x$ (similarity solution)? Or try ansatz $z+qy=$ something specific.

Looking back at the Charpit form for clean solution: $p=(z+qy)^2$ is symmetric in structure suggesting $z+qy=$ function of $(x,y)$ — perhaps $z+qy=1/(x+a)$ for some constant $a$ (from the char eq above) leads to:

$z=-qy+1/(x+a)$.

Differentiate w.r.t. $x$: $p=z_x=-1/(x+a)^2$. But $F=0$ requires $p=(z+qy)^2=1/(x+a)^2$. Sign mismatch.

Hmm. Try $z+qy=1/(b-x)$ instead: $z=-qy+1/(b-x)$, $p=z_x=1/(b-x)^2=(z+qy)^2$ ✓.

So one solution is $z+qy=1/(b-x)$, with $q=z_y=-y$… no wait, $z_y=-q$? But $q$ is just a parameter here.

Take a different approach. Assume $q=a$ constant along the Charpit characteristic. Then $(z+ay)^2=p=z_x$. So $z_x=(z+ay)^2$, separable for fixed $y$.

Let $w=z+ay$. Then $w_x=z_x=w^2$, so $-1/w=x+f(y)$ for some function $f$.

$w=-1/(x+f(y))$, i.e., $z=-ay-1/(x+f(y))$.

Now $z_y=q=a$ requires $-a-\partial_y[1/(x+f(y))]=a$. Compute $\partial_y[1/(x+f(y))]=-f'(y)/(x+f(y))^2$.

So $-a+f'(y)/(x+f(y))^2=a$, $f'(y)/(x+f(y))^2=2a$.

LHS depends on $x$ (through denom); RHS is constant. So unless $f'(y)=0$ and $a=0$, no go. If $f(y)=b$ constant: $f'=0$, contradicts unless $a=0$.

So the ansatz ”$q=a$ constant” doesn’t directly give a complete integral.

Complete integral by direct construction

A complete integral has two arbitrary constants. Try: $z=(\text{function of }x,y,a,b)$ where $a,b$ are arbitrary.

The PDE $p=(z+qy)^2$ has degree 2 in $q$. Let me try:

$z+ay=\dfrac{1}{b-x}$ for arbitrary $a,b$. Check: $q=z_y=-a$. So $z+qy=z-ay+2ay=z+ay$ ✗ (signs don’t match).

Hmm. Try $z=\dfrac{1}{b-x}-ay$, so $z+ay=1/(b-x)$, $p=z_x=1/(b-x)^2=(1/(b-x))^2=(z+ay)^2$.

Then $q=z_y=-a$. And the PDE: $p=(z+qy)^2=(z+(-a)y)^2=(z-ay)^2$.

But $z-ay=1/(b-x)-2ay$, which is not the same as $1/(b-x)$. So $p\ne(z+qy)^2$ unless $a=0$.

OK the structure is subtle. Let me try $z=ax+\dfrac{1}{b-y}+\text{stuff}$. Actually I think the complete integral requires a non-trivial form.

Try $z+qy=1/(x+\phi(y))$ — but $q$ depends on $y$.

Given the difficulty, let me just present the complete integral as derived from Charpit:

From Charpit: along char, $z+qy=-1/(x+a)$ (where $a$ is one constant).

The complete integral has two arbitrary constants. After full Charpit machinery:

Answer