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UPSC 2020 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

Cauchy's method of characteristics · PDEs · asked 4× in 13 yrs · Read the full method →

Question

Find the solution of the partial differential equation $z=\frac12(p^2+q^2)+(p-x)(q-y)$, $p\equiv\frac{\partial z}{\partial x}$, $q\equiv\frac{\partial z}{\partial y}$, which passes through the $x$-axis.

Technique

Charpit’s method; the special structure gives $p-x=a,\ q-y=b$ directly; build complete integral, impose the strip/data conditions on the $x$-axis to get $b=-2a$, then take the envelope.

Solution

This is a nonlinear first-order PDE. Write

Step 1 — Charpit’s auxiliary equations

Compute the partials:

Step 2 — Two simple integrals

Look at $\dfrac{dp}{F_x+pF_z}$ and $\dfrac{dx}{-F_p}$:

Hence $\dfrac{dp}{-F_p}=\dfrac{dx}{-F_p}\Rightarrow dp=dx\Rightarrow$

Similarly $F_y+qF_z=(x-p)-q=-(p+q-x)=-F_q$, so $\dfrac{dq}{-F_q}=\dfrac{dy}{-F_q}\Rightarrow dq=dy\Rightarrow$

Step 3 — Complete integral

With $p=x+a,\ q=y+b$ substituted into the PDE:

(One checks $z_x=x+a=p$, $z_y=y+b=q$, so this is a genuine complete integral with parameters $a,b$.)

Step 4 — Impose the data: surface through the $x$-axis

The $x$-axis is $y=0,\ z=0$, parametrized by $x=t$. On a solution surface the strip condition $dz=p\,dx+q\,dy$ holds. Along the curve $dy=0,\ dz=0,\ dx=dt$, so $p=0$ there.

• $p=0$ on the curve: $p=x+a=t+a=0\Rightarrow a=-t$.
• $z=0$ on the curve: $\tfrac12(t+a)^2+\tfrac12(0+b)^2+ab=0$. With $a=-t$: $0+\tfrac12 b^2-tb=0\Rightarrow b\big(\tfrac{b}{2}-t\big)=0$.

Discard $b=0$ (gives the trivial plane $z=\tfrac12(x+a)^2$, not through the whole axis nontrivially); take $b=2t$. Then

Step 5 — Eliminate the parameter (envelope)

Impose $b=-2a$ in the complete integral and take the envelope ($\partial z/\partial a=0$):

Substitute $a=2y-x$:

Expanding: $\tfrac12\cdot4y^2+\tfrac12(4x^2-12xy+9y^2)-2(4y^2-4xy+x^2)$ $=2y^2+2x^2-6xy+\tfrac92y^2-8y^2+8xy-2x^2=2xy-\tfrac32y^2.$

Answer