Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ z = f(x+z) + g(x+y), \] where $f$ and $g$ are arbitrary differentiable functions.
Explanation of Method
The solution contains two arbitrary functions, so we expect a second-order PDE. Let $\alpha = x + z$, $\beta = x + y$, so that $z = f(\alpha) + g(\beta)$. Standard notation: $p = \dfrac{\partial z}{\partial x}$, $q = \dfrac{\partial z}{\partial y}$, $r = \dfrac{\partial^2 z}{\partial x^2}$, $s = \dfrac{\partial^2 z}{\partial x\,\partial y}$, $t = \dfrac{\partial^2 z}{\partial y^2}$.
Step 1 · First-order partial derivatives
Differentiate $z = f(\alpha)+g(\beta)$ with respect to $x$, noting $\alpha_x = 1+p$ and $\beta_x = 1$: \[ p = f'(\alpha)(1+p) + g'(\beta). \tag{1} \] Differentiate with respect to $y$, noting $\alpha_y = q$ and $\beta_y = 1$: \[ q = f'(\alpha)\,q + g'(\beta). \tag{2} \]
Step 2 · Solve for $f'(\alpha)$ and $g'(\beta)$
Subtract (2) from (1): \[ p - q = f'(\alpha)(1+p) - f'(\alpha)\,q = f'(\alpha)(1 + p - q). \] \[ f'(\alpha) = \frac{p-q}{1+p-q}. \tag{3} \] Substitute (3) into (2): \[ g'(\beta) = q - f'(\alpha)\,q = q\!\left(1 - \frac{p-q}{1+p-q}\right) = \frac{q}{1+p-q}. \tag{4} \]
Step 3 · Second-order partial derivatives
Differentiate (1) with respect to $x$: \[ r = f''(\alpha)(1+p)^2 + f'(\alpha)\,r + g''(\beta). \tag{5} \] Differentiate (1) with respect to $y$: \[ s = f''(\alpha)(1+p)\,q + f'(\alpha)\,s + g''(\beta). \tag{6} \] Differentiate (2) with respect to $y$: \[ t = f''(\alpha)\,q^2 + f'(\alpha)\,t + g''(\beta). \tag{7} \]
Step 4 · Eliminate $f''(\alpha)$ and $g''(\beta)$
Subtract (7) from (5): \[ r - t = f''(\alpha)\!\left[(1+p)^2 - q^2\right] + f'(\alpha)(r-t). \tag{8} \] Subtract (7) from (6): \[ s - t = f''(\alpha)\,q\!\left[(1+p) - q\right] + f'(\alpha)(s-t). \tag{9} \] From (8): $f''(\alpha) = \dfrac{(r-t)(1-f'(\alpha))}{(1+p-q)(1+p+q)}$. \tag{8'} From (9): $f''(\alpha) = \dfrac{(s-t)(1-f'(\alpha))}{q(1+p-q)}$. \tag{9'} Equate (8') and (9'), and cancel the common factor $(1-f'(\alpha))$ (nonzero in general) and $(1+p-q)$: \[ \frac{r-t}{1+p+q} = \frac{s-t}{q}. \] \[ q(r-t) = (s-t)(1+p+q). \] Expand and simplify: \[ qr - qt = s + ps + qs - t - pt - qt. \] \[ qr - (1+p+q)s + (1+p)t = 0. \]
Verification of structure
The equation is of the form $Ar + Bs + Ct = 0$ with $A = q$, $B = -(1+p+q)$, $C = 1+p$. The discriminant is: \[ B^2 - 4AC = (1+p+q)^2 - 4q(1+p) = 1 + p^2 + q^2 + 2p + 2q + 2pq - 4q - 4pq = (1+p-q)^2, \] which is a perfect square, confirming the PDE is parabolic (consistent with having two arbitrary functions of linear combinations involving $z$). $\checkmark$
Result
Eliminating $f$ and $g$ (together with their derivatives) yields the second-order linear PDE: