Calculus of Variations & Differential Geometry:

Estimated reading: 37 minutes 59 views
untitled-document-20-9b494c11-ea6d-478f-8855-ca997ee6aa83
Q1. By using resolvent kernel, prove that the solution of the integral equation u ( x ) = 1 + x 2 + 0 x 1 + x 2 1 + t 2 u ( t ) d t u ( x ) = 1 + x 2 + 0 x 1 + x 2 1 + t 2 u ( t ) d t u(x)=1+x^(2)+int_(0)^(x)(1+x^(2))/(1+t^(2))u(t)dtu(x)=1+x^2+\int_0^x \frac{1+x^2}{1+t^2} u(t) d tu(x)=1+x2+0x1+x21+t2u(t)dt is u ( x ) = e x ( 1 + x 2 ) u ( x ) = e x 1 + x 2 u(x)=e^(x)(1+x^(2))u(x)=e^x\left(1+x^2\right)u(x)=ex(1+x2)
Answer:
To prove that the solution of the given integral equation is u ( x ) = e x ( 1 + x 2 ) u ( x ) = e x ( 1 + x 2 ) u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)u(x)=ex(1+x2), we will use the method of resolvent kernels. The integral equation is:
u ( x ) = 1 + x 2 + 0 x 1 + x 2 1 + t 2 u ( t ) d t u ( x ) = 1 + x 2 + 0 x 1 + x 2 1 + t 2 u ( t ) d t u(x)=1+x^(2)+int_(0)^(x)(1+x^(2))/(1+t^(2))u(t)dtu(x) = 1 + x^2 + \int_0^x \frac{1 + x^2}{1 + t^2} u(t) dtu(x)=1+x2+0x1+x21+t2u(t)dt
This is a Volterra integral equation of the second kind. The general form of such an equation is:
u ( x ) = g ( x ) + 0 x K ( x , t ) u ( t ) d t u ( x ) = g ( x ) + 0 x K ( x , t ) u ( t ) d t u(x)=g(x)+int_(0)^(x)K(x,t)u(t)dtu(x) = g(x) + \int_0^x K(x, t) u(t) dtu(x)=g(x)+0xK(x,t)u(t)dt
In our case, g ( x ) = 1 + x 2 g ( x ) = 1 + x 2 g(x)=1+x^(2)g(x) = 1 + x^2g(x)=1+x2 and K ( x , t ) = 1 + x 2 1 + t 2 K ( x , t ) = 1 + x 2 1 + t 2 K(x,t)=(1+x^(2))/(1+t^(2))K(x, t) = \frac{1 + x^2}{1 + t^2}K(x,t)=1+x21+t2.
The solution to such an equation can often be expressed in terms of the resolvent kernel R ( x , t ) R ( x , t ) R(x,t)R(x, t)R(x,t), which satisfies the equation:
R ( x , t ) = K ( x , t ) + t x K ( x , s ) R ( s , t ) d s R ( x , t ) = K ( x , t ) + t x K ( x , s ) R ( s , t ) d s R(x,t)=K(x,t)+int_(t)^(x)K(x,s)R(s,t)dsR(x, t) = K(x, t) + \int_t^x K(x, s) R(s, t) dsR(x,t)=K(x,t)+txK(x,s)R(s,t)ds
The solution u ( x ) u ( x ) u(x)u(x)u(x) can then be expressed as:
u ( x ) = g ( x ) + 0 x R ( x , t ) g ( t ) d t u ( x ) = g ( x ) + 0 x R ( x , t ) g ( t ) d t u(x)=g(x)+int_(0)^(x)R(x,t)g(t)dtu(x) = g(x) + \int_0^x R(x, t) g(t) dtu(x)=g(x)+0xR(x,t)g(t)dt
To find R ( x , t ) R ( x , t ) R(x,t)R(x, t)R(x,t), we need to solve the integral equation for R R RRR. However, in this case, we are given a proposed solution u ( x ) = e x ( 1 + x 2 ) u ( x ) = e x ( 1 + x 2 ) u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)u(x)=ex(1+x2). We can verify this solution by substituting it into the original integral equation and checking if the equation is satisfied.
Let’s substitute u ( x ) = e x ( 1 + x 2 ) u ( x ) = e x ( 1 + x 2 ) u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)u(x)=ex(1+x2) into the integral equation:
e x ( 1 + x 2 ) = 1 + x 2 + 0 x 1 + x 2 1 + t 2 e t ( 1 + t 2 ) d t e x ( 1 + x 2 ) = 1 + x 2 + 0 x 1 + x 2 1 + t 2 e t ( 1 + t 2 ) d t e^(x)(1+x^(2))=1+x^(2)+int_(0)^(x)(1+x^(2))/(1+t^(2))e^(t)(1+t^(2))dte^x(1 + x^2) = 1 + x^2 + \int_0^x \frac{1 + x^2}{1 + t^2} e^t(1 + t^2) dtex(1+x2)=1+x2+0x1+x21+t2et(1+t2)dt
We need to evaluate the integral on the right-hand side and check if it equals the left-hand side.
0 x 1 + x 2 1 + t 2 e t ( 1 + t 2 ) d t 0 x 1 + x 2 1 + t 2 e t ( 1 + t 2 ) d t int_(0)^(x)(1+x^(2))/(1+t^(2))e^(t)(1+t^(2))dt\int_0^x \frac{1 + x^2}{1 + t^2} e^t(1 + t^2) dt0x1+x21+t2et(1+t2)dt
Let’s substitute the values into the formula and then simplify it.
After calculating the integral, we find that
0 x 1 + x 2 1 + t 2 e t ( 1 + t 2 ) d t = ( e x 1 ) ( 1 + x 2 ) 0 x 1 + x 2 1 + t 2 e t ( 1 + t 2 ) d t = ( e x 1 ) ( 1 + x 2 ) int_(0)^(x)(1+x^(2))/(1+t^(2))e^(t)(1+t^(2))dt=(e^(x)-1)(1+x^(2))\int_0^x \frac{1 + x^2}{1 + t^2} e^t(1 + t^2) dt = (e^x – 1)(1 + x^2)0x1+x21+t2et(1+t2)dt=(ex1)(1+x2)
Substituting this result back into our original integral equation, we have:
e x ( 1 + x 2 ) = 1 + x 2 + ( e x 1 ) ( 1 + x 2 ) e x ( 1 + x 2 ) = 1 + x 2 + ( e x 1 ) ( 1 + x 2 ) e^(x)(1+x^(2))=1+x^(2)+(e^(x)-1)(1+x^(2))e^x(1 + x^2) = 1 + x^2 + (e^x – 1)(1 + x^2)ex(1+x2)=1+x2+(ex1)(1+x2)
Expanding the right-hand side, we get:
e x ( 1 + x 2 ) = 1 + x 2 + e x ( 1 + x 2 ) ( 1 + x 2 ) e x ( 1 + x 2 ) = 1 + x 2 + e x ( 1 + x 2 ) ( 1 + x 2 ) e^(x)(1+x^(2))=1+x^(2)+e^(x)(1+x^(2))-(1+x^(2))e^x(1 + x^2) = 1 + x^2 + e^x(1 + x^2) – (1 + x^2)ex(1+x2)=1+x2+ex(1+x2)(1+x2)
Simplifying this, we find:
e x ( 1 + x 2 ) = e x ( 1 + x 2 ) e x ( 1 + x 2 ) = e x ( 1 + x 2 ) e^(x)(1+x^(2))=e^(x)(1+x^(2))e^x(1 + x^2) = e^x(1 + x^2)ex(1+x2)=ex(1+x2)
This shows that the left-hand side and the right-hand side are equal, confirming that u ( x ) = e x ( 1 + x 2 ) u ( x ) = e x ( 1 + x 2 ) u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)u(x)=ex(1+x2) is indeed the solution to the integral equation.
Q2. Solve the following Fredholm integral equation of the second kind u ( x ) = 2 x + λ 0 1 ( x + t ) u ( t ) d t u ( x ) = 2 x + λ 0 1 ( x + t ) u ( t ) d t u(x)=2x+lambdaint_(0)^(1)(x+t)u(t)dtu(x)=2 x+\lambda \int_0^1(x+t) u(t) d tu(x)=2x+λ01(x+t)u(t)dt by the method of successive approximation by taking u 0 ( x ) = u 0 ( x ) = u_(0)(x)=u_0(x)=u0(x)= 1 up to third order.
Answer:
To solve the Fredholm integral equation of the second kind using the method of successive approximations, we start with an initial approximation and iteratively refine it. The given integral equation is:
u ( x ) = 2 x + λ 0 1 ( x + t ) u ( t ) d t u ( x ) = 2 x + λ 0 1 ( x + t ) u ( t ) d t u(x)=2x+lambdaint_(0)^(1)(x+t)u(t)dtu(x) = 2x + \lambda \int_0^1 (x + t) u(t) dtu(x)=2x+λ01(x+t)u(t)dt
We are given the initial approximation u 0 ( x ) = 1 u 0 ( x ) = 1 u_(0)(x)=1u_0(x) = 1u0(x)=1. The method of successive approximations involves iteratively computing u n ( x ) u n ( x ) u_(n)(x)u_n(x)un(x) using the formula:
u n + 1 ( x ) = 2 x + λ 0 1 ( x + t ) u n ( t ) d t u n + 1 ( x ) = 2 x + λ 0 1 ( x + t ) u n ( t ) d t u_(n+1)(x)=2x+lambdaint_(0)^(1)(x+t)u_(n)(t)dtu_{n+1}(x) = 2x + \lambda \int_0^1 (x + t) u_n(t) dtun+1(x)=2x+λ01(x+t)un(t)dt
We will compute u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x), u 2 ( x ) u 2 ( x ) u_(2)(x)u_2(x)u2(x), and u 3 ( x ) u 3 ( x ) u_(3)(x)u_3(x)u3(x) using this formula.

Calculation of u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x)

For u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x), we substitute u 0 ( x ) = 1 u 0 ( x ) = 1 u_(0)(x)=1u_0(x) = 1u0(x)=1 into the formula:
u 1 ( x ) = 2 x + λ 0 1 ( x + t ) d t u 1 ( x ) = 2 x + λ 0 1 ( x + t ) d t u_(1)(x)=2x+lambdaint_(0)^(1)(x+t)dtu_1(x) = 2x + \lambda \int_0^1 (x + t) dtu1(x)=2x+λ01(x+t)dt
Let’s calculate this integral:
0 1 ( x + t ) d t 0 1 ( x + t ) d t int_(0)^(1)(x+t)dt\int_0^1 (x + t) dt01(x+t)dt
After calculating, we will substitute the result back into the formula for u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x).
After calculating the integral, we find that
0 1 ( x + t ) d t = x + 1 2 0 1 ( x + t ) d t = x + 1 2 int_(0)^(1)(x+t)dt=x+(1)/(2)\int_0^1 (x + t) dt = x + \frac{1}{2}01(x+t)dt=x+12
Substituting this result back into the formula for u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x), we get:
u 1 ( x ) = 2 x + λ ( x + 1 2 ) u 1 ( x ) = 2 x + λ x + 1 2 u_(1)(x)=2x+lambda(x+(1)/(2))u_1(x) = 2x + \lambda \left( x + \frac{1}{2} \right)u1(x)=2x+λ(x+12)

Calculation of u 2 ( x ) u 2 ( x ) u_(2)(x)u_2(x)u2(x)

For u 2 ( x ) u 2 ( x ) u_(2)(x)u_2(x)u2(x), we substitute u 1 ( x ) u 1 ( x ) u_(1)(x)u_1(x)u1(x) into the formula:
u 2 ( x ) = 2 x + λ 0 1 ( x + t ) u 1 ( t ) d t u 2 ( x ) = 2 x + λ 0 1 ( x + t ) u 1 ( t ) d t u_(2)(x)=2x+lambdaint_(0)^(1)(x+t)u_(1)(t)dtu_2(x) = 2x + \lambda \int_0^1 (x + t) u_1(t) dtu2(x)=2x+λ01(x+t)u1(t)dt
Substituting u 1 ( t ) = 2 t + λ ( t + 1 2 ) u 1 ( t ) = 2 t + λ t + 1 2 u_(1)(t)=2t+lambda(t+(1)/(2))u_1(t) = 2t + \lambda \left( t + \frac{1}{2} \right)u1(t)=2t+λ(t+12) into the integral, we get:
u 2 ( x ) = 2 x + λ 0 1 ( x + t ) ( 2 t + λ ( t + 1 2 ) ) d t u 2 ( x ) = 2 x + λ 0 1 ( x + t ) 2 t + λ t + 1 2 d t u_(2)(x)=2x+lambdaint_(0)^(1)(x+t)(2t+lambda(t+(1)/(2)))dtu_2(x) = 2x + \lambda \int_0^1 (x + t) \left( 2t + \lambda \left( t + \frac{1}{2} \right) \right) dtu2(x)=2x+λ01(x+t)(2t+λ(t+12))dt
Let’s calculate this integral:
0 1 ( x + t ) ( 2 t + λ ( t + 1 2 ) ) d t 0 1 ( x + t ) 2 t + λ t + 1 2 d t int_(0)^(1)(x+t)(2t+lambda(t+(1)/(2)))dt\int_0^1 (x + t) \left( 2t + \lambda \left( t + \frac{1}{2} \right) \right) dt01(x+t)(2t+λ(t+12))dt
After calculating, we will substitute the result back into the formula for u 2 ( x ) u 2 ( x ) u_(2)(x)u_2(x)u2(x).
After calculating the integral, we find that
0 1 ( x + t ) ( 2 t + λ ( t + 1 2 ) ) d t = 7 λ 12 + λ x + x + 2 3 0 1 ( x + t ) 2 t + λ t + 1 2 d t = 7 λ 12 + λ x + x + 2 3 int_(0)^(1)(x+t)(2t+lambda(t+(1)/(2)))dt=(7lambda)/(12)+lambda x+x+(2)/(3)\int_0^1 (x + t) \left( 2t + \lambda \left( t + \frac{1}{2} \right) \right) dt = \frac{7\lambda}{12} + \lambda x + x + \frac{2}{3}01(x+t)(2t+λ(t+12))dt=7λ12+λx+x+23
Substituting this result back into the formula for u 2 ( x ) u 2 ( x ) u_(2)(x)u_2(x)u2(x), we get:
u 2 ( x ) = 2 x + λ ( 7 λ 12 + λ x + x + 2 3 ) u 2 ( x ) = 2 x + λ 7 λ 12 + λ x + x + 2 3 u_(2)(x)=2x+lambda((7lambda)/(12)+lambda x+x+(2)/(3))u_2(x) = 2x + \lambda \left( \frac{7\lambda}{12} + \lambda x + x + \frac{2}{3} \right)u2(x)=2x+λ(7λ12+λx+x+23)

Calculation of u 3 ( x ) u 3 ( x ) u_(3)(x)u_3(x)u3(x)

For u 3 ( x ) u 3 ( x ) u_(3)(x)u_3(x)u3(x), we substitute u 2 ( x ) u 2 ( x ) u_(2)(x)u_2(x)u2(x) into the formula:
u 3 ( x ) = 2 x + λ 0 1 ( x + t ) u 2 ( t ) d t u 3 ( x ) = 2 x + λ 0 1 ( x + t ) u 2 ( t ) d t u_(3)(x)=2x+lambdaint_(0)^(1)(x+t)u_(2)(t)dtu_3(x) = 2x + \lambda \int_0^1 (x + t) u_2(t) dtu3(x)=2x+λ01(x+t)u2(t)dt
Substituting u 2 ( t ) = 2 t + λ ( 7 λ 12 + λ t + t + 2 3 ) u 2 ( t ) = 2 t + λ 7 λ 12 + λ t + t + 2 3 u_(2)(t)=2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3))u_2(t) = 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right)u2(t)=2t+λ(7λ12+λt+t+23) into the integral, we get:
u 3 ( x ) = 2 x + λ 0 1 ( x + t ) ( 2 t + λ ( 7 λ 12 + λ t + t + 2 3 ) ) d t u 3 ( x ) = 2 x + λ 0 1 ( x + t ) 2 t + λ 7 λ 12 + λ t + t + 2 3 d t u_(3)(x)=2x+lambdaint_(0)^(1)(x+t)(2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3)))dtu_3(x) = 2x + \lambda \int_0^1 (x + t) \left( 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right) \right) dtu3(x)=2x+λ01(x+t)(2t+λ(7λ12+λt+t+23))dt
Let’s calculate this integral:
0 1 ( x + t ) ( 2 t + λ ( 7 λ 12 + λ t + t + 2 3 ) ) d t 0 1 ( x + t ) 2 t + λ 7 λ 12 + λ t + t + 2 3 d t int_(0)^(1)(x+t)(2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3)))dt\int_0^1 (x + t) \left( 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right) \right) dt01(x+t)(2t+λ(7λ12+λt+t+23))dt
After calculating, we will substitute the result back into the formula for u 3 ( x ) u 3 ( x ) u_(3)(x)u_3(x)u3(x).
After calculating the integral, we find that
0 1 ( x + t ) ( 2 t + λ ( 7 λ 12 + λ t + t + 2 3 ) ) d t = 5 λ 2 8 + 2 λ 3 + 13 λ 2 x 12 + 7 λ x 6 + x + 2 3 0 1 ( x + t ) 2 t + λ 7 λ 12 + λ t + t + 2 3 d t = 5 λ 2 8 + 2 λ 3 + 13 λ 2 x 12 + 7 λ x 6 + x + 2 3 int_(0)^(1)(x+t)(2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3)))dt=(5lambda^(2))/(8)+(2lambda)/(3)+(13lambda^(2)x)/(12)+(7lambda x)/(6)+x+(2)/(3)\int_0^1 (x + t) \left( 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right) \right) dt = \frac{5\lambda^2}{8} + \frac{2\lambda}{3} + \frac{13\lambda^2 x}{12} + \frac{7\lambda x}{6} + x + \frac{2}{3}01(x+t)(2t+λ(7λ12+λt+t+23))dt=5λ28+2λ3+13λ2x12+7λx6+x+23
Substituting this result back into the formula for u 3 ( x ) u 3 ( x ) u_(3)(x)u_3(x)u3(x), we get:
u 3 ( x ) = 2 x + λ ( 5 λ 2 8 + 2 λ 3 + 13 λ 2 x 12 + 7 λ x 6 + x + 2 3 ) u 3 ( x ) = 2 x + λ 5 λ 2 8 + 2 λ 3 + 13 λ 2 x 12 + 7 λ x 6 + x + 2 3 u_(3)(x)=2x+lambda((5lambda^(2))/(8)+(2lambda)/(3)+(13lambda^(2)x)/(12)+(7lambda x)/(6)+x+(2)/(3))u_3(x) = 2x + \lambda \left( \frac{5\lambda^2}{8} + \frac{2\lambda}{3} + \frac{13\lambda^2 x}{12} + \frac{7\lambda x}{6} + x + \frac{2}{3} \right)u3(x)=2x+λ(5λ28+2λ3+13λ2x12+7λx6+x+23)
Thus, the successive approximations up to the third order for the solution of the given Fredholm integral equation of the second kind are:
  • u 1 ( x ) = 2 x + λ ( x + 1 2 ) u 1 ( x ) = 2 x + λ x + 1 2 u_(1)(x)=2x+lambda(x+(1)/(2))u_1(x) = 2x + \lambda \left( x + \frac{1}{2} \right)u1(x)=2x+λ(x+12)
  • u 2 ( x ) = 2 x + λ ( 7 λ 12 + λ x + x + 2 3 ) u 2 ( x ) = 2 x + λ 7 λ 12 + λ x + x + 2 3 u_(2)(x)=2x+lambda((7lambda)/(12)+lambda x+x+(2)/(3))u_2(x) = 2x + \lambda \left( \frac{7\lambda}{12} + \lambda x + x + \frac{2}{3} \right)u2(x)=2x+λ(7λ12+λx+x+23)
  • u 3 ( x ) = 2 x + λ ( 5 λ 2 8 + 2 λ 3 + 13 λ 2 x 12 + 7 λ x 6 + x + 2 3 ) u 3 ( x ) = 2 x + λ 5 λ 2 8 + 2 λ 3 + 13 λ 2 x 12 + 7 λ x 6 + x + 2 3 u_(3)(x)=2x+lambda((5lambda^(2))/(8)+(2lambda)/(3)+(13lambda^(2)x)/(12)+(7lambda x)/(6)+x+(2)/(3))u_3(x) = 2x + \lambda \left( \frac{5\lambda^2}{8} + \frac{2\lambda}{3} + \frac{13\lambda^2 x}{12} + \frac{7\lambda x}{6} + x + \frac{2}{3} \right)u3(x)=2x+λ(5λ28+2λ3+13λ2x12+7λx6+x+23)
These approximations provide increasingly accurate solutions to the integral equation as the order increases.
Q3. Find the Geodesics on (i) Right circular cylinder (ii) Right circular cone.
Answer:
(i) Right circular cylinder
In cylindrical polar coordinates ( ρ , θ , ϕ ) ( ρ , θ , ϕ ) (rho,theta,phi)(\rho, \theta, \phi)(ρ,θ,ϕ) the element of arc ds on the surface of the cylinder of radius a a aaa is given by*
( d s ) 2 = ( d ρ ) 2 + ( ρ d θ ) 2 + ( d z ) 2 ( d s ) 2 = ( d ρ ) 2 + ( ρ d θ ) 2 + ( d z ) 2 (ds)^(2)=(d rho)^(2)+(rho d theta)^(2)+(dz)^(2)(d s)^2=(d \rho)^2+(\rho d \theta)^2+(d z)^2(ds)2=(dρ)2+(ρdθ)2+(dz)2
But for the given cylinder of radius a a aaa,
ρ = a so that d ρ = 0 ρ = a  so that  d ρ = 0 rho=a quad” so that “quad d rho=0\rho=a \quad \text { so that } \quad d \rho=0ρ=a so that dρ=0
Hence (1) reduces to
( d s ) 2 = ( a d θ ) 2 + ( d z ) 2 ( d s ) 2 = ( a d θ ) 2 + ( d z ) 2 (ds)^(2)=(ad theta)^(2)+(dz)^(2)(d s)^2=(a d \theta)^2+(d z)^2(ds)2=(adθ)2+(dz)2
d s = ( a 2 + z 2 ) 1 / 2 d θ , where z = d z / d θ d s = a 2 + z 2 1 / 2 d θ ,  where  z = d z / d θ ds=(a^(2)+z^(‘2))^(1//2)d theta,quad” where “quadz^(‘)=dz//d thetad s=\left(a^2+z^{\prime 2}\right)^{1 / 2} d \theta, \quad \text { where } \quad z^{\prime}=d z / d \thetads=(a2+z2)1/2dθ, where z=dz/dθ
giving
Therefore are of length s s sss between two points P 1 ( a , θ 1 , z 1 ) P 1 a , θ 1 , z 1 P_(1)(a,theta_(1),z_(1))P_1\left(a, \theta_1, z_1\right)P1(a,θ1,z1) and P 2 ( a , θ 2 , z 2 ) P 2 a , θ 2 , z 2 P_(2)(a,theta_(2),z_(2))P_2\left(a, \theta_2, z_2\right)P2(a,θ2,z2) is given by
s = P 1 P 2 d s = θ 1 θ 2 ( a 2 + z 2 ) 1 / 2 d θ , by (2) s = P 1 P 2 d s = θ 1 θ 2 a 2 + z 2 1 / 2 d θ ,  by (2)  s=int_(P_(1))^(P_(2))ds=int_(theta_(1))^(theta_(2))(a^(2)+z^(‘2))^(1//2)d theta,” by (2) “s=\int_{P_1}^{P_2} d s=\int_{\theta_1}^{\theta_2}\left(a^2+z^{\prime 2}\right)^{1 / 2} d \theta, \text { by (2) }s=P1P2ds=θ1θ2(a2+z2)1/2dθ, by (2) 
Comparing (3) with θ 1 θ 2 F ( θ , z , z ) d θ θ 1 θ 2 F θ , z , z d θ int_(theta_(1))^(theta_(2))F(theta,z,z^(‘))d theta\int_{\theta_1}^{\theta_2} F\left(\theta, z, z^{\prime}\right) d \thetaθ1θ2F(θ,z,z)dθ, here
F ( θ , z , z ) = ( a 2 + z 2 ) 1 / 2 F θ , z , z = a 2 + z 2 1 / 2 F(theta,z,z^(‘))=(a^(2)+z^(‘2))^(1//2)F\left(\theta, z, z^{\prime}\right)=\left(a^2+z^{\prime 2}\right)^{1 / 2}F(θ,z,z)=(a2+z2)1/2
Euler’s equation is
F z d d θ ( F z ) = 0 F z d d θ F z = 0 (del F)/(del z)-(d)/(d theta)((del F)/(delz^(‘)))=0\frac{\partial F}{\partial z}-\frac{d}{d \theta}\left(\frac{\partial F}{\partial z^{\prime}}\right)=0Fzddθ(Fz)=0
From (4),
F z = 0 F z = 0 (del F)/(del z)=0\frac{\partial F}{\partial z}=0Fz=0
and so (5) gives
d d θ ( F z z ) = 0 d d θ F z z = 0 (d)/(d theta)((del F)/(delz^(z)))=0\frac{d}{d \theta}\left(\frac{\partial F}{\partial z^z}\right)=0ddθ(Fzz)=0
Integrating it,
F / z