Solve the system of partial differential equations (∂ u)/(∂…

Solve the system of partial differential equations \[ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial z} = 0, \qquad \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0, \] subject to the initial condition $u = z^2$ when $x = y = 0$.

Step 1 · General solution of the first equation

The characteristic direction of $u_x+u_z=0$ is $(1,0,1)$, with invariants $y$ and $\xi=x-z$. So every solution of the first equation alone has the form \[ u = f(y,\,x-z) \] for an arbitrary function $f$.

Step 2 · Impose the second equation

With $\xi=x-z$: $u_y=f_y$ and $u_z=-f_\xi$, so $u_y+u_z=0$ becomes \[ f_y - f_\xi = 0, \] a first-order linear PDE for $f(y,\xi)$ with characteristic direction $(1,-1)$, i.e. invariant $y+\xi$. Hence $f=g(y+\xi)=g(x+y-z)$ for an arbitrary function $g$, so \[ u = g(x+y-z), \] equivalently $u=\tilde g(z-x-y)$ after relabelling the arbitrary function.

Step 3 · Apply the initial condition

At $x=y=0$: $u=\tilde g(z)=z^2$, so $\tilde g(t)=t^2$. Hence \[ u = (z-x-y)^2. \]

Verification

With $s=z-x-y$: $u_x=-2s$, $u_y=-2s$, $u_z=2s$, so $u_x+u_z=0$ and $u_y+u_z=0$. $\checkmark$ At $x=y=0$: $u=z^2$. $\checkmark$