$$u_{x} ^{2} +u_{y}^{2}+u_{z}^{2}=1, \qquad u=k \;\text{ on the plane }\; \alpha x+\beta y + z =0, \text{ } $$ where $k$, $\alpha$ and $\beta$ are constants
I've been trying to solve this PDE problem but I can only get to solve for $u_{x}, u_{y}$, and $u_{z}$ in terms of $\alpha$ and $\beta$. Can anyone help me out? Thanks!
Inspection works, but you may solve this more systematically by generalising Charpit's method to more than two independent variables. Parametrise everything using $s_1, s_2, t$ where the initial data is given by: $(x(s_1,s_2,0),y(s_1,s_2,0),z(s_1,s_2,0)) = (s_1,s_2,-\alpha s_1 - \beta s_2)$. With the dot denoting differentiation by $t$, writing $F(p,q,r)=p^2+q^2+r^2-1 \equiv 0$ we have: \begin{equation} \dot{x}= F_p = 2p, \\ \dot{y} = F_q =2q, \\ \dot{z} = F_r =2r, \\ \dot{p} = \dot{q} = \dot{r} = 0, \\ \dot{u} = pF_p + qF_q + rF_r = 2(p^2+q^2+r^2) = 2. \end{equation} From the penultimate line, $p(s_1,s_2,t) = p_0(s_1,s_2)$ and analogously for $q$ and $r$. But the initial data now gives us three equations that determine $p_0,q_0, r_0$: one is from the PDE itself at the data curve, and two come from differentiating along it with respect to each of the two parameters. Namely: \begin{equation} p_{0}^{2} + q_{0}^{2} + r_{0}^{2} = 1, \\ \dfrac{\partial u_0}{\partial s_1} = \dfrac{\partial u_0}{\partial x_0}\dfrac{\partial x_0}{\partial s_1} + \dfrac{\partial u_0}{\partial y_0}\dfrac{\partial y_0}{\partial s_1} + \dfrac{\partial u_0}{\partial z_0}\dfrac{\partial z_0}{\partial s_1} = p_0 - \alpha r_0, \\ \dfrac{\partial u_0}{\partial s_2} = \dfrac{\partial u_0}{\partial x_0}\dfrac{\partial x_0}{\partial s_2} + \dfrac{\partial u_0}{\partial y_0}\dfrac{\partial y_0}{\partial s_2} + \dfrac{\partial u_0}{\partial z_0}\dfrac{\partial z_0}{\partial s_2} = q_0 - \beta r_0. \end{equation} Therefore, since $u$ is just a constant initially, $p_0 = \alpha r_0$, $q_0 = \beta r_0$, so from the sum of squares \begin{equation} r_0 = \pm \dfrac{1}{\sqrt{1+ \alpha^2 + \beta^2}}. \end{equation}
Now everything may be recovered into a parametric solution: \begin{equation} x = \pm \dfrac{2\alpha t}{\sqrt{1+ \alpha^2 + \beta^2}} + s_1, \\ y = \pm \dfrac{2\beta t}{\sqrt{1+ \alpha^2 + \beta^2}} + s_2, \\ z = \pm \dfrac{2 t}{\sqrt{1+ \alpha^2 + \beta^2}}-\alpha s_1 - \beta s_2, \\ u = 2 t + k. \end{equation} Now notice that: \begin{equation} \dfrac{\alpha x + \beta y + z}{\sqrt{1+ \alpha^2 + \beta^2}} = \pm 2 t = \pm (u-k). \end{equation}
This gives us two solutions: \begin{equation} u(x,y,z) = \dfrac{\alpha x + \beta y + z}{\sqrt{1+ \alpha^2 + \beta^2}} + k, \end{equation} or the alternative \begin{equation} u(x,y,z) = -\dfrac{\alpha x + \beta y + z}{\sqrt{1+ \alpha^2 + \beta^2}} + k. \end{equation}