← 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 13 yrs · Read the full method →
Question
Form a partial differential equation by eliminating the arbitrary functions f(x) and g(y) from z=yf(x)+xg(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x>0, y>0.
Technique
Eliminate two arbitrary one-variable functions by forming xp+yq and identifying the mixed derivative s; classify via discriminant B2−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=yf(x)+xg(y).
p=∂x∂z=yf′(x)+g(y),q=∂y∂z=f(x)+xg′(y).
Step 2 — A mixed second derivative
s=∂x∂y∂2z=f′(x)+g′(y).
Step 3 — Eliminate f,g
Form the combination xp+yq:
xp+yq=xyf′(x)+xg(y)+yf(x)+xyg′(y)=xy(f′(x)+g′(y))+(yf(x)+xg(y)).
The last bracket is exactly z, and f′(x)+g′(y)=s. Hence
xp+yq=xys+z.
Rearranging,
xy∂x∂y∂2z=x∂x∂z+y∂y∂z−z
This is the required PDE (all arbitrary functions eliminated).
Step 4 — Classification
Write the principal (second-order) part as
Azxx+2Bzxy+Czyy+…
Here zxx and zyy are absent, so A=0, C=0, and the coefficient of zxy is xy, giving 2B=xy, i.e. B=2xy.
The discriminant is
B2−AC=(2xy)2−0=4x2y2.
In the region x>0, y>0 we have x2y2/4>0, so B2−AC>0 everywhere there.
Answer
The equation is hyperbolic throughout the region x>0, y>0.