Shortest distance between $\frac{x^2}{25}+\frac{y^2}{16}+\frac{z^2}9=1$ and $x^2+y^2+z^2=4$ by the calculus of variations

Question

Find the shortest distance between the surfaces $$\frac{x^2}{25}+\frac{y^2}{16}+\frac{z^2}{9}=1 \qquad\text{and}\qquad x^2+y^2+z^2=4.$$

I tried to solve this problem as follows:

The problem reduces to testing for an extremum of the function

$$v[y(x),z(x)]=\int_{x_0}^{x_1} \sqrt{1+y'^2+z'^2}\;dx.....\tag{1}$$ Let $F=\sqrt{1+y'^2+z'^2}$.

Let $S_1$ be the surface $\displaystyle{\frac{x^2}{25}+\frac{y^2}{16}+\frac{z^2}{9}=1}$ and $S_2$ be the surface $x^2+y^2+z^2=4$. Consider the point of contact of the extremals of the functional on the surface $S_1$ as $A(x_0,y_0,z_0)$ and the point of contact of the extremals on $S_2$ as $B(x_1,y_1,z_1)$.

The extremals of the functional in Equation (1) are

$y=C_1x+C_2 \qquad...\tag{2}$ and $z=C_3x+C_4 \qquad...\tag{3}$

Since the ends of these extremals move on the surfaces $S_1$ and $S_2$, we get the following equations

$C_1x_0+C_2=y_0 \qquad .... \tag{4}$

$C_3x_0+C_4=z_0 \qquad .... \tag{5}$

$C_1x_1+C_2=y_1 \qquad .... \tag{6}$

$C_3x_1+C_4=z_1 \qquad .... \tag{7}$

The transversality conditions are

$F-y'F_{y'}+(\phi_x -z')F_{z'} \Bigg \vert_{x=x_0}=0 \qquad ....\tag{9}$

$F_{y'}+\phi_y F_{z'} \Bigg \vert_{x=x_0}=0 \qquad ....\tag{10}$

$F-y'F_{y'}+(\psi_x -z')F_{z'} \Bigg \vert_{x=x_1}=0 \qquad ....\tag{11}$

$F_{y'}+\psi_y F_{z'} \Bigg \vert_{x=x_1}=0 \qquad ....\tag{12}$

where $\phi(x,y)$ is the surface defining $S_1$ and $\psi(x,y)$ is the surface defining $S_2$.

Solving $S_1$ and $S_2$ for $z$, we get

$$\phi(x,y)=3\sqrt{1-\frac{x^2}{25}-\frac{y^2}{16}}$$ and $$\psi(x,y)=\sqrt{4-x^2-y^2}$$. Differentiating $\phi$ and $\psi$ partially with respect to $x$ and $y$, we get

$$\phi_x=-\frac{3x}{25\displaystyle{\sqrt{1-\frac{x^2}{25}-\frac{y^2}{16}}}}=-\frac{9x}{25z},$$

$$\phi_y=-\frac{3y}{16\displaystyle{\sqrt{1-\frac{x^2}{25}-\frac{y^2}{16}}}}=-\frac{9y}{16z},$$

$$\psi_x=-\frac{x}{\sqrt{4-x^2-y^2}}=-\frac{x}{z},$$

$$\psi_y=-\frac{y}{\sqrt{4-x^2-y^2}}=-\frac{y}{z},$$

Now substituting for $F$, $\phi_x$, $\phi_y$, $\psi_x$, $\psi_y$, $y'=C_1$ and $z'=C_3$ in Equations (9) - (12), we get the following equations.

$25z_0-9x_0C_3=0 \qquad .... \tag{13}$

$16z_0C_1-9y_0C_3=0 \qquad .... \tag{14}$

$z_1-C_3x_1=0 \qquad .... \tag{15}$

$C_1z_1-C_3y_1=0 \qquad .... \tag{16}$

Now we have to solve Equations (4) - (7) and Equations (13) - (16) for the eight unknowns $x_0$, $y_0$, $x_1$, $y_1$, $C_1$, $C_2$, $C_3$ and $C_4$, which I could not.

Hence, I request to verify whether the procedure adopted in solving this problem is correct and if there is any mistake, please point out. Thank you in advance.

Answer 1

Of course, the geometrical consideration of the two surfaces ( sphere inside ellipsoid) and symmetries obviously show that the couples of points $(0,0,2); (0,0,3)$ and $(0,0,-2) ; (0,0-3)$ are both at shorter distance $D=1$.

But we have to proceed with calculus of variation :

Let $(x,y,z)$ a point on the surface $\frac{x^2}{25}+\frac{y^2}{16}+\frac{z^2}{9}=1 \qquad (1)$

and $(X,Y,Z)$ a point on the surface $X^2+Y^2+Z^2=4 \qquad (2)$

Let $D$ be the distance between them : $D^2=(X-x)^2+(Y-y)^2+(Z-z)^2 \qquad (3)$

The extremum of $D$ are so that the differentials of Eqs.(1), (2), (3) be $=0$ : $$\begin{cases} \quad\frac{x\:dx}{25}+\frac{y\:dy}{16}+\frac{z\:dz}{9}=0 \qquad (4)\\ \quad X\:dX+Y\:dY+Z\:dZ=0 \qquad (5)\\ \quad (X-x)(dX-dx)+(Y-y)(dY-dy)+(Z-z)(dZ-dz)=0 \qquad (6)\\ \end{cases}$$ From (4) : $dz=-\frac{9x\:dx}{25}-\frac{9y\:dy}{16} \qquad (7)$

From (5) : $dZ=-\frac{X\:dX+Y\:dY}{Z} \qquad (8)$

Bringing (7) and (8) into (6) leads to : $$\left(x-X+\frac{9y}{25z}(Z-z)\right)dx+\left(y-Y+\frac{9y}{16z}(Z-z)\right)dy+\frac{Xz-xZ}{Z}dX+\frac{Yz-yZ}{Z}dY=0$$

The extremum correspond to any $dx$, $dy$, $dX$, $dY$ which imply : $$\begin{cases} x-X+\frac{9y}{25z}(Z-z)=0 \qquad (10)\\ y-Y+\frac{9y}{16z}(Z-z)=0\qquad (11)\\ \frac{Xz-xZ}{Z}=0\qquad (12)\\ \frac{Yz-yZ}{Z}=0\qquad (13) \end{cases}$$ From (12) and (13) : $X=\frac{Z}{z}x$ and $Y=\frac{Z}{z}y$. Putting them into (10) and (11) leads to :

$$(z-Z)\frac{x}{z}=0\quad \text{and} \quad(z-Z)\frac{y}{z}=0$$

The consideration of the case $z-Z=0$ lead to maximum values of $D$, which is not what we want.

The case $x=y=0$ leads to : $\begin{cases}\frac{z^2}{9}=1\\ Z^2=4 \end{cases}\quad$ from Eqs.(1) and (2).

$z=\pm 3$ and $Z=\pm 2$.

From Eq.(3) : $D^2=(Z-z)^2$. As a consequence, the cases for a minimum of $D^2$ are : $(z=3\:,\:Z=2)$ and $(z=-3\:,\:Z=-2)$