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$$u_{x} ^{2} +u_{y}^{2}+u_{z}^{2}=1, \qquad u=k \;\text{ on the plane }\; \alpha x+\beta y + z =0, \text{ } $$ 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!

EditPiAf
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Ben
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    This seems like a form of the Eikonal equation, which has a known solution, if I'm not mistaken – GSofer May 23 '20 at 20:06
  • Welcome to math.se! You might be interested by the methods in this post and related ones. Could you add some context (where this problem comes from, why it's interesting, etc.)? Also it would be nice to tell what you tried, see how-to-ask-a-good-question – EditPiAf May 24 '20 at 00:46
  • Just by inspection : Try $u(x,y,z)=ax+by+cz+d$ and determine the constants $a,b,c,d$ so that the PDE and the condition be satisfied. – JJacquelin May 24 '20 at 06:38

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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}

spetrevski
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