← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Family of surfaces · PDEs · asked 7× in 14 yrs · Read the full method →

Question

Form a partial differential equation by eliminating the arbitrary functions f(x)f(x) and g(y)g(y) from z=y f(x)+x g(y)z=y\,f(x)+x\,g(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x>0, y>0x>0,\ y>0.

Technique

Eliminate two arbitrary one-variable functions by forming xp+yqxp+yq and identifying the mixed derivative ss; classify via discriminant B2−ACB^2-AC of the principal part.

Solution

We have two arbitrary functions, so we expect a second-order PDE.

Step 1 — First-order partial derivatives

z=y f(x)+x g(y).z=y\,f(x)+x\,g(y). p=∂z∂x=y f′(x)+g(y),q=∂z∂y=f(x)+x g′(y).p=\frac{\partial z}{\partial x}=y\,f'(x)+g(y),\qquad q=\frac{\partial z}{\partial y}=f(x)+x\,g'(y).

Step 2 — A mixed second derivative

s=∂2z∂x ∂y=f′(x)+g′(y).s=\frac{\partial^2 z}{\partial x\,\partial y}=f'(x)+g'(y).

Step 3 — Eliminate f,gf,g

Form the combination x p+y qx\,p+y\,q:

x p+y q=xy f′(x)+x g(y)+y f(x)+xy g′(y)=xy(f′(x)+g′(y))+(y f(x)+x g(y)).x\,p+y\,q=xy\,f'(x)+x\,g(y)+y\,f(x)+xy\,g'(y)=xy\big(f'(x)+g'(y)\big)+\big(y\,f(x)+x\,g(y)\big).

The last bracket is exactly zz, and f′(x)+g′(y)=sf'(x)+g'(y)=s. Hence

x p+y q=xy s+z.x\,p+y\,q=xy\,s+z.

Rearranging,

  xy ∂2z∂x ∂y=x ∂z∂x+y ∂z∂y−z  \boxed{\;xy\,\frac{\partial^2 z}{\partial x\,\partial y}=x\,\frac{\partial z}{\partial x}+y\,\frac{\partial z}{\partial y}-z\;}

This is the required PDE (all arbitrary functions eliminated).

Step 4 — Classification

Write the principal (second-order) part as

A zxx+2B zxy+C zyy+…A\,z_{xx}+2B\,z_{xy}+C\,z_{yy}+\dots

Here zxxz_{xx} and zyyz_{yy} are absent, so A=0, C=0A=0,\ C=0, and the coefficient of zxyz_{xy} is xyxy, giving 2B=xy2B=xy, i.e. B=xy2B=\tfrac{xy}{2}.

The discriminant is

B2−AC=(xy2)2−0=x2y24.B^2-AC=\left(\frac{xy}{2}\right)^2-0=\frac{x^2y^2}{4}.

In the region x>0, y>0x>0,\ y>0 we have x2y2/4>0x^2y^2/4>0, so B2−AC>0B^2-AC>0 everywhere there.

Answer

The equation is hyperbolic throughout the region x>0, y>0.\boxed{\text{The equation is \emph{hyperbolic} throughout the region } x>0,\ y>0.}
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