In this section we find the function

y_{0}∈ D={y∈ C^{1}[0,1]:y(0)=1,y(1)=2/3} |

for which the area of the corresponding surface of revolution is minimal. The result is not necessarily intuitive.

The area of the surface of revolution is measured by the functional

F(y)=2 π | ∫ |
| y(x) | √ |
| dx. |

We set f(y,y′,x)=y(x) √1+y′(x)^{2} and compute the associated Euler-Lagrange equation :

dy:=diff(y(x),x):; f:=y*sqrt(1+dy

`^`

2):; eq:=euler_lagrange(f)
Output :

[-(y(x))/(sqrt(diff(y(x),x)

`^`

2+1))=K_0, diff(y(x),x,2)=((diff(y(x),x)`^`

2+1)/(y(x)))]
We obtain the stationary function by finding the general solution of the first equation. Input :

sol:=collect(simplify(dsolve(eq[0],x,y)))

Output :

[-K_0,K_0*(-exp((x-c_1)/K_0)

`^`

2-1)/(2*exp((x-c_1)/K_0))]
Obviously the constant solution −K_{0} is not in D, so we set y_{0} to be the second element of the above list. That function, which can be written as

y_{0}(x)=−K_{0} cosh | ⎛ ⎜ ⎜ ⎝ |
| ⎞ ⎟ ⎟ ⎠ | , |

is called a *catenary*. Input :

y0:=sol[1]:; p:=[K_0,c_1]:;

To find the values of K_{0} and c_{1} from the boundary conditions, we first plot the curves y_{0}(0)=1 and y_{0}(1)=2/3 for K_{0}∈[−1,1] and c_{1}∈[−1,2] to see where they intersect each other. Input :

eq1:=subs(y0,x=0)=1:; eq2:=subs(y0,x=1)=2/3:; implicitplot([eq1,eq2],K_0=-1..1,c_1=-1..2)

Output :

We observe that there are exactly two catenaries satisfying the Euler-Lagrange necessary conditions and the given boundary conditions : the first with K_{0}≈ −0.5 and c_{1}≈ 0.6 resp. the second with K_{0}≈ −0.3 and c_{1}≈ 0.5. We obtain the values of these constants more precisely by using fsolve. Input :

p1:=fsolve([eq1,eq2],p,[-0.5,0.6]);
p2:=fsolve([eq1,eq2],p,[-0.3,0.5])

Output :

[-0.56237423894,0.662588703113], [-0.30613431407,0.567138261119]

We check, for each catenary, whether the strong Legendre condition

f_{y′ y′}(x,y_{k},y_{k}′)>0 |

holds for k=1,2. Input :

y1:=subs(y0,p,p1):; y2:=subs(y0,p,p2):;
D2f:=diff(f,diff(y(x),x),2):;
solve([eval(subs(D2f,y=y1,y(x)=y1))<=0,x>=0,x<=1],x);
solve([eval(subs(D2f,y=y2,y(x)=y2))<=0,x>=0,x<=1],x)

Output :

[],[]

We conclude that the strong Legendre condition is satisfied in both cases, so we proceed by attempting to find the points conjugate to 0 for each catenary. The function y_{0} depends on two parameters, so we use conjugate_equation to find these points easily. Input :

fsolve(conjugate_equation(y0,p,p1,x,0)=0,x=0..1)
fsolve(conjugate_equation(y0,p,p2,x,0)=0,x=0..1)

Output :

[0.0], [0.0,0.799514772606]

We conclude that there are no points conjugate to 0 in (0,1] for the catenary y_{1}, so it minimizes the functional F. However, for the other catenary there is a conjugate point in the relevant interval, therefore y_{2} is not a minimizer.

We can verify the above conclusions by computing the surface area for catenaries y_{1} and y_{2} and comparing them. Input :

int(y1*sqrt(1+diff(y1,x)

`^`

2),x=0..1); int(y2*sqrt(1+diff(y2,x)`^`

2),x=0..1)
Output :

0.81396915825,0.826468466845

We see that the surface formed by rotating the curve y_{1} is indeed smaller than the area of the surface formed by rotating the curve y_{2}. Finally, we visualize both surfaces for convenience. Input :

plot3d([y1*cos(t),y1*sin(t),x],x=0..1,t=0..2*pi, display=yellow+filled)

Output :

Input :

plot3d([y2*cos(t),y2*sin(t),x],x=0..1,t=0..2*pi, display=yellow+filled)

Output :