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Lagrange's method for a PDE

Short answer

Auxillary equations give x+y = C1 and (z²+(x+y)²) e^{-2x} = C2. General solution: φ(x+y, (z²+(x+y)²)e^{-2x})=0, i.e., z²+(x+y)² = e^{2x} f(x+y) with arbitrary f.

Solve the PDE z(p − q) = z² + (x + y)² using Lagrange's method, where p = ∂z/∂x and q = ∂z/∂y.

Lagrange's linear PDE is Pp + Qq = R. Its auxiliary (subsidiary) equations are dx/P = dy/Q = dz/R. Here P = z, Q = −z, R = z² + (x + y)².

Auxiliary equations

dx/z = dy/(−z) = dz/[z² + (x + y)²]

First integral. From the first two ratios, dx = −dy, so dx + dy = 0 and

x + y = a ... (1)

Second integral. Use dx/z = dz/(z² + a²) with (x + y) = a, so z dz/(z² + a²) = dx.

Integrating: ½ ln(z² + a²) = x + c, that is ln(z² + a²) − 2x = 2c, so

(z² + (x + y)²) e^(−2x) = b ... (2)

General solution. With an arbitrary function φ:

φ( x + y, (z² + (x + y)²) e^(−2x) ) = 0

Equivalently, z² + (x + y)² = e^(2x) f(x + y) for an arbitrary function f.

Check: take u = (z² + (x+y)²)e^(−2x). Then z·u_x − z·u_y + (z² + (x+y)²)u_z = 0, as required for a solution of the auxiliary system.

✓ Answer checked by Claude Opus 5.5 (high effort)

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