PDE Formation Calculator

Form a partial differential equation by eliminating arbitrary constants from z = f(x, y, a, b), an arbitrary function from φ(u, v) = 0, or two arbitrary functions from z = f(ax + by) + g(cx + dy), with the answer checked by substitution.

Eliminate	Arbitrary constants a, b from z = f(x, y, a, b)
z =	(x - a)^2 + (y - b)^2

Eliminate

z =

Use a and b for the constants.

The PDE

z = (x − ((−(p − 2x)) / 2))^2 + (y − ((−(q − 2y)) / 2))^2

Eliminating a and b gives z = (x − ((−(p − 2x)) / 2))^2 + (y − ((−(q − 2y)) / 2))^2, where p = ∂z/∂x and q = ∂z/∂y.

Exact form. Not independently checked.

Method: Eliminating arbitrary constants
Worked out with: AllCalcHub expression engine
Check: Every step that can be checked was checked

p: 2(x − a)
q: 2(y − b)

About the PDE Formation Calculator

With two constants a and b, differentiate z in x and y to get p and q, then eliminate a and b from the three equations. For example z = (x − a)² + (y − b)² gives p = 2(x − a), q = 2(y − b), so 4z = p² + q².

Eliminating an arbitrary function from φ(u, v) = 0 always gives Lagrange's linear equation Pp + Qq = R with P = ∂(u, v)/∂(y, z), Q = ∂(u, v)/∂(z, x), R = ∂(u, v)/∂(x, y). Two arbitrary functions of linear arguments give a second-order equation.

The formula

φ(u, v) = 0 ⇒ Pp + Qq = R, P = ∂(u, v)/∂(y, z), Q = ∂(u, v)/∂(z, x), R = ∂(u, v)/∂(x, y)

Worked example

z = (x − a)² + (y − b)²

p = 2(x − a), q = 2(y − b)
z = p²/4 + q²/4, i.e. 4z = p² + q²

Things to keep in mind

Constants are eliminated when z, p or q is linear in them (most textbook families); a family like x² + y² + (z − c)² = a² isn't handled.

Questions people ask

Why does eliminating a function give a linear PDE?

φ_u and φ_v appear linearly after differentiating, and removing them leaves an equation of first degree in p and q.

What are r, s and t?

The second derivatives ∂²z/∂x², ∂²z/∂x∂y and ∂²z/∂y².