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$${\partial z \over \partial x}f(x,y,z)+{\partial z \over \partial y}g(x,y,z)=h(x,y,z) $$ $${\partial z \over \partial x}F(x,y,z)+{\partial z \over \partial y}G(x,y,z)=H(x,y,z) $$

$$f(x,y,z)={x-x_1 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}-{x-x_2 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }} $$

$$g(x,y,z)={y-y_1 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}-{y-y_2 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }} $$

$$h(x,y,z)={z-z_1 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }}-{z-z_2 \over {\sqrt {(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} }} $$

$$F(x,y,z)=(y-y_1)(z-z_2)-(y-y_2)(z-z_1) $$ $$G(x,y,z)=(x-x_2)(z-z_1)-(x-x_1)(z-z_2) $$ $$ H(x,y,z)=(x-x_1)(y-y_2)-(x-x_2)(y-y_1) $$

$ x_1, y_1, z_1, x_2, y_2, z_2 $ Are constants

What is the standard approach to solving these equations? I've gone through my textbook, but the author has not discussed this case. Moreover, I'm not being able to word my problem correctly, as a consequence, the Google Searches on the Internet are resulting in utter failure. Please help

Thank You.

bbgodfrey
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  • Perhaps you should ask this on [math.se] instead. This is a site for the computing software Mathematica. –  Jan 07 '16 at 04:40
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    Is this really a PDE? what is the unknown? $z$ seems to be a function $z(x,y)$, is $z(x,y)$ the unknown? if that's the case, then $F(x,y,z)$ are actually $F(x,y,z(x,y))$ – lurscher Jan 07 '16 at 05:08
  • Solve the two equations for D[z[x,y],x] and D[z[x,y],y], and integrate the first with respect to x and the second with respect to y. The solutions will contain arbitrary functions of y and x respectively. Since the two solutions must be the same to be valid, choose the arbitrary functions to make the solutions equal. Note, however, that not all such systems of PDEs have solutions. – bbgodfrey Jan 07 '16 at 07:51
  • See, for instance, 1601657. – bbgodfrey Jan 07 '16 at 08:09

1 Answers1

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This approach is not standard but this set of PDE can be solved using geometric arguments.

Let $\vec{p} = (x,y,z)$ be any generic point in $\mathbb{R}^3$.

Let $\vec{p}_i = (x_i,y_i,z_i)$, $\vec{r}_i = \vec{p} - \vec{p}_i$ and $r_i = |\vec{r}_i|$ for $i = 1,2$.

Given any function $f(x,y)$, the vector $$\vec{N_f(x,y)} \stackrel{def}{=} \left(\frac{\partial f(x,y)}{\partial x}, \frac{\partial f(x,y)}{\partial y}, -1\right)$$ is proportional to the normal vector for the surface $z = f(x,y)$ at point $(x,y, f(x,y))$.

The set of given PDE can be rewritten as $$\begin{align} \vec{N}_f(x,y) \cdot \left[ \nabla (r_1 - r_2) \right]_{(x,y,f(x,y))} &= 0\\ \vec{N}_f(x,y) \cdot \left[ \vec{r}_1 \times \vec{r}_2 \right]_{(x,y,f(x,y))} &= 0 \end{align} $$ Geometrically, solving the PDE is equivalent to finding a surface, in the form $z = f(x,y)$, such that for any point on it, the normal vector $N_f$ is perpendicular to two vector fields:

  1. The vector field $\nabla (r_1 - r_2)$ which is proportional to the normal vectors on the families of two-sheet hyperboloid $r_1 - r_2 = const.$ which have foci at $\vec{p}_1$ and $\vec{p}_2$.

  2. The vector field $\vec{r}_1 \times \vec{r}_2$ which is proportional to the normal vectors on the families of planes passing through the line joining $\vec{p}_1$ and $\vec{p}_2$.

It is well known on the plane, the families of ellipses and hyperbolas sharing the same foci are orthogonal to each other. Lift this phenomenon to $\mathbb{R}^3$, we immediately find

The set of ellipsoid $r_1 + r_2 = const.$ with foci at $\vec{p}_1$ and $\vec{p}_2$ is a family of surfaces that "solves" the given PDE.

I'm lazy, I will leave the task to unwind the expression $r_1 + r_2 = const.$ to the explicit form $z = f(x,y)$ to you.

achille hui
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