# Calculus of Variations & Differential Geometry:

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Q1. By using resolvent kernel, prove that the solution of the integral equation $u\left(x\right)=1+{x}^{2}+{\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}u\left(t\right)dt$$u\left(x\right)=1+{x}^{2}+{\int }_{0}^{x} \frac{1+{x}^{2}}{1+{t}^{2}}u\left(t\right)dt$u(x)=1+x^(2)+int_(0)^(x)(1+x^(2))/(1+t^(2))u(t)dtu(x)=1+x^2+\int_0^x \frac{1+x^2}{1+t^2} u(t) d t$u\left(x\right)=1+{x}^{2}+{\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}u\left(t\right)dt$ is $u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$u(x)=e^(x)(1+x^(2))u(x)=e^x\left(1+x^2\right)$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$
To prove that the solution of the given integral equation is $u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$, we will use the method of resolvent kernels. The integral equation is:
$u\left(x\right)=1+{x}^{2}+{\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}u\left(t\right)dt$$u\left(x\right)=1+{x}^{2}+{\int }_{0}^{x} \frac{1+{x}^{2}}{1+{t}^{2}}u\left(t\right)dt$u(x)=1+x^(2)+int_(0)^(x)(1+x^(2))/(1+t^(2))u(t)dtu(x) = 1 + x^2 + \int_0^x \frac{1 + x^2}{1 + t^2} u(t) dt$u\left(x\right)=1+{x}^{2}+{\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}u\left(t\right)dt$
This is a Volterra integral equation of the second kind. The general form of such an equation is:
$u\left(x\right)=g\left(x\right)+{\int }_{0}^{x}K\left(x,t\right)u\left(t\right)dt$$u\left(x\right)=g\left(x\right)+{\int }_{0}^{x} K\left(x,t\right)u\left(t\right)dt$u(x)=g(x)+int_(0)^(x)K(x,t)u(t)dtu(x) = g(x) + \int_0^x K(x, t) u(t) dt$u\left(x\right)=g\left(x\right)+{\int }_{0}^{x}K\left(x,t\right)u\left(t\right)dt$
In our case, $g\left(x\right)=1+{x}^{2}$$g\left(x\right)=1+{x}^{2}$g(x)=1+x^(2)g(x) = 1 + x^2$g\left(x\right)=1+{x}^{2}$ and $K\left(x,t\right)=\frac{1+{x}^{2}}{1+{t}^{2}}$$K\left(x,t\right)=\frac{1+{x}^{2}}{1+{t}^{2}}$K(x,t)=(1+x^(2))/(1+t^(2))K(x, t) = \frac{1 + x^2}{1 + t^2}$K\left(x,t\right)=\frac{1+{x}^{2}}{1+{t}^{2}}$.
The solution to such an equation can often be expressed in terms of the resolvent kernel $R\left(x,t\right)$$R\left(x,t\right)$R(x,t)R(x, t)$R\left(x,t\right)$, which satisfies the equation:
$R\left(x,t\right)=K\left(x,t\right)+{\int }_{t}^{x}K\left(x,s\right)R\left(s,t\right)ds$$R\left(x,t\right)=K\left(x,t\right)+{\int }_{t}^{x} K\left(x,s\right)R\left(s,t\right)ds$R(x,t)=K(x,t)+int_(t)^(x)K(x,s)R(s,t)dsR(x, t) = K(x, t) + \int_t^x K(x, s) R(s, t) ds$R\left(x,t\right)=K\left(x,t\right)+{\int }_{t}^{x}K\left(x,s\right)R\left(s,t\right)ds$
The solution $u\left(x\right)$$u\left(x\right)$u(x)u(x)$u\left(x\right)$ can then be expressed as:
$u\left(x\right)=g\left(x\right)+{\int }_{0}^{x}R\left(x,t\right)g\left(t\right)dt$$u\left(x\right)=g\left(x\right)+{\int }_{0}^{x} R\left(x,t\right)g\left(t\right)dt$u(x)=g(x)+int_(0)^(x)R(x,t)g(t)dtu(x) = g(x) + \int_0^x R(x, t) g(t) dt$u\left(x\right)=g\left(x\right)+{\int }_{0}^{x}R\left(x,t\right)g\left(t\right)dt$
To find $R\left(x,t\right)$$R\left(x,t\right)$R(x,t)R(x, t)$R\left(x,t\right)$, we need to solve the integral equation for $R$$R$RR$R$. However, in this case, we are given a proposed solution $u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$. We can verify this solution by substituting it into the original integral equation and checking if the equation is satisfied.
Let’s substitute $u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$ into the integral equation:
${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+{\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt$${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+{\int }_{0}^{x} \frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt$e^(x)(1+x^(2))=1+x^(2)+int_(0)^(x)(1+x^(2))/(1+t^(2))e^(t)(1+t^(2))dte^x(1 + x^2) = 1 + x^2 + \int_0^x \frac{1 + x^2}{1 + t^2} e^t(1 + t^2) dt${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+{\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt$
We need to evaluate the integral on the right-hand side and check if it equals the left-hand side.
${\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt$${\int }_{0}^{x} \frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt$int_(0)^(x)(1+x^(2))/(1+t^(2))e^(t)(1+t^(2))dt\int_0^x \frac{1 + x^2}{1 + t^2} e^t(1 + t^2) dt${\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt$
Let’s substitute the values into the formula and then simplify it.
After calculating the integral, we find that
${\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt=\left({e}^{x}-1\right)\left(1+{x}^{2}\right)$${\int }_{0}^{x} \frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt=\left({e}^{x}-1\right)\left(1+{x}^{2}\right)$int_(0)^(x)(1+x^(2))/(1+t^(2))e^(t)(1+t^(2))dt=(e^(x)-1)(1+x^(2))\int_0^x \frac{1 + x^2}{1 + t^2} e^t(1 + t^2) dt = (e^x – 1)(1 + x^2)${\int }_{0}^{x}\frac{1+{x}^{2}}{1+{t}^{2}}{e}^{t}\left(1+{t}^{2}\right)dt=\left({e}^{x}-1\right)\left(1+{x}^{2}\right)$
Substituting this result back into our original integral equation, we have:
${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+\left({e}^{x}-1\right)\left(1+{x}^{2}\right)$${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+\left({e}^{x}-1\right)\left(1+{x}^{2}\right)$e^(x)(1+x^(2))=1+x^(2)+(e^(x)-1)(1+x^(2))e^x(1 + x^2) = 1 + x^2 + (e^x – 1)(1 + x^2)${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+\left({e}^{x}-1\right)\left(1+{x}^{2}\right)$
Expanding the right-hand side, we get:
${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+{e}^{x}\left(1+{x}^{2}\right)-\left(1+{x}^{2}\right)$${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+{e}^{x}\left(1+{x}^{2}\right)-\left(1+{x}^{2}\right)$e^(x)(1+x^(2))=1+x^(2)+e^(x)(1+x^(2))-(1+x^(2))e^x(1 + x^2) = 1 + x^2 + e^x(1 + x^2) – (1 + x^2)${e}^{x}\left(1+{x}^{2}\right)=1+{x}^{2}+{e}^{x}\left(1+{x}^{2}\right)-\left(1+{x}^{2}\right)$
Simplifying this, we find:
${e}^{x}\left(1+{x}^{2}\right)={e}^{x}\left(1+{x}^{2}\right)$${e}^{x}\left(1+{x}^{2}\right)={e}^{x}\left(1+{x}^{2}\right)$e^(x)(1+x^(2))=e^(x)(1+x^(2))e^x(1 + x^2) = e^x(1 + x^2)${e}^{x}\left(1+{x}^{2}\right)={e}^{x}\left(1+{x}^{2}\right)$
This shows that the left-hand side and the right-hand side are equal, confirming that $u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$u(x)=e^(x)(1+x^(2))u(x) = e^x(1 + x^2)$u\left(x\right)={e}^{x}\left(1+{x}^{2}\right)$ is indeed the solution to the integral equation.
Q2. Solve the following Fredholm integral equation of the second kind $u\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)u\left(t\right)dt$$u\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right)u\left(t\right)dt$u(x)=2x+lambdaint_(0)^(1)(x+t)u(t)dtu(x)=2 x+\lambda \int_0^1(x+t) u(t) d t$u\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)u\left(t\right)dt$ by the method of successive approximation by taking ${u}_{0}\left(x\right)=$${u}_{0}\left(x\right)=$u_(0)(x)=u_0(x)=${u}_{0}\left(x\right)=$ 1 up to third order.
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\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)u\left(t\right)dt$$u\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right)u\left(t\right)dt$u(x)=2x+lambdaint_(0)^(1)(x+t)u(t)dtu(x) = 2x + \lambda \int_0^1 (x + t) u(t) dt$u\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)u\left(t\right)dt$
We are given the initial approximation ${u}_{0}\left(x\right)=1$${u}_{0}\left(x\right)=1$u_(0)(x)=1u_0(x) = 1${u}_{0}\left(x\right)=1$. The method of successive approximations involves iteratively computing ${u}_{n}\left(x\right)$${u}_{n}\left(x\right)$u_(n)(x)u_n(x)${u}_{n}\left(x\right)$ using the formula:
${u}_{n+1}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right){u}_{n}\left(t\right)dt$${u}_{n+1}\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right){u}_{n}\left(t\right)dt$u_(n+1)(x)=2x+lambdaint_(0)^(1)(x+t)u_(n)(t)dtu_{n+1}(x) = 2x + \lambda \int_0^1 (x + t) u_n(t) dt${u}_{n+1}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right){u}_{n}\left(t\right)dt$
We will compute ${u}_{1}\left(x\right)$${u}_{1}\left(x\right)$u_(1)(x)u_1(x)${u}_{1}\left(x\right)$, ${u}_{2}\left(x\right)$${u}_{2}\left(x\right)$u_(2)(x)u_2(x)${u}_{2}\left(x\right)$, and ${u}_{3}\left(x\right)$${u}_{3}\left(x\right)$u_(3)(x)u_3(x)${u}_{3}\left(x\right)$ using this formula.

### Calculation of ${u}_{1}\left(x\right)$${u}_{1}\left(x\right)$u_(1)(x)u_1(x)${u}_{1}\left(x\right)$

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

### Calculation of ${u}_{2}\left(x\right)$${u}_{2}\left(x\right)$u_(2)(x)u_2(x)${u}_{2}\left(x\right)$

For ${u}_{2}\left(x\right)$${u}_{2}\left(x\right)$u_(2)(x)u_2(x)${u}_{2}\left(x\right)$, we substitute ${u}_{1}\left(x\right)$${u}_{1}\left(x\right)$u_(1)(x)u_1(x)${u}_{1}\left(x\right)$ into the formula:
${u}_{2}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right){u}_{1}\left(t\right)dt$${u}_{2}\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right){u}_{1}\left(t\right)dt$u_(2)(x)=2x+lambdaint_(0)^(1)(x+t)u_(1)(t)dtu_2(x) = 2x + \lambda \int_0^1 (x + t) u_1(t) dt${u}_{2}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right){u}_{1}\left(t\right)dt$
Substituting ${u}_{1}\left(t\right)=2t+\lambda \left(t+\frac{1}{2}\right)$${u}_{1}\left(t\right)=2t+\lambda \left(t+\frac{1}{2}\right)$u_(1)(t)=2t+lambda(t+(1)/(2))u_1(t) = 2t + \lambda \left( t + \frac{1}{2} \right)${u}_{1}\left(t\right)=2t+\lambda \left(t+\frac{1}{2}\right)$ into the integral, we get:
${u}_{2}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt$${u}_{2}\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt$u_(2)(x)=2x+lambdaint_(0)^(1)(x+t)(2t+lambda(t+(1)/(2)))dtu_2(x) = 2x + \lambda \int_0^1 (x + t) \left( 2t + \lambda \left( t + \frac{1}{2} \right) \right) dt${u}_{2}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt$
Let’s calculate this integral:
${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt$${\int }_{0}^{1} \left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt$int_(0)^(1)(x+t)(2t+lambda(t+(1)/(2)))dt\int_0^1 (x + t) \left( 2t + \lambda \left( t + \frac{1}{2} \right) \right) dt${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt$
After calculating, we will substitute the result back into the formula for ${u}_{2}\left(x\right)$${u}_{2}\left(x\right)$u_(2)(x)u_2(x)${u}_{2}\left(x\right)$.
After calculating the integral, we find that
${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt=\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}$${\int }_{0}^{1} \left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt=\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}$int_(0)^(1)(x+t)(2t+lambda(t+(1)/(2)))dt=(7lambda)/(12)+lambda x+x+(2)/(3)\int_0^1 (x + t) \left( 2t + \lambda \left( t + \frac{1}{2} \right) \right) dt = \frac{7\lambda}{12} + \lambda x + x + \frac{2}{3}${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(t+\frac{1}{2}\right)\right)dt=\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}$
Substituting this result back into the formula for ${u}_{2}\left(x\right)$${u}_{2}\left(x\right)$u_(2)(x)u_2(x)${u}_{2}\left(x\right)$, we get:
${u}_{2}\left(x\right)=2x+\lambda \left(\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}\right)$${u}_{2}\left(x\right)=2x+\lambda \left(\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}\right)$u_(2)(x)=2x+lambda((7lambda)/(12)+lambda x+x+(2)/(3))u_2(x) = 2x + \lambda \left( \frac{7\lambda}{12} + \lambda x + x + \frac{2}{3} \right)${u}_{2}\left(x\right)=2x+\lambda \left(\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}\right)$

### Calculation of ${u}_{3}\left(x\right)$${u}_{3}\left(x\right)$u_(3)(x)u_3(x)${u}_{3}\left(x\right)$

For ${u}_{3}\left(x\right)$${u}_{3}\left(x\right)$u_(3)(x)u_3(x)${u}_{3}\left(x\right)$, we substitute ${u}_{2}\left(x\right)$${u}_{2}\left(x\right)$u_(2)(x)u_2(x)${u}_{2}\left(x\right)$ into the formula:
${u}_{3}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right){u}_{2}\left(t\right)dt$${u}_{3}\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right){u}_{2}\left(t\right)dt$u_(3)(x)=2x+lambdaint_(0)^(1)(x+t)u_(2)(t)dtu_3(x) = 2x + \lambda \int_0^1 (x + t) u_2(t) dt${u}_{3}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right){u}_{2}\left(t\right)dt$
Substituting ${u}_{2}\left(t\right)=2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)$${u}_{2}\left(t\right)=2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)$u_(2)(t)=2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3))u_2(t) = 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right)${u}_{2}\left(t\right)=2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)$ into the integral, we get:
${u}_{3}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt$${u}_{3}\left(x\right)=2x+\lambda {\int }_{0}^{1} \left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt$u_(3)(x)=2x+lambdaint_(0)^(1)(x+t)(2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3)))dtu_3(x) = 2x + \lambda \int_0^1 (x + t) \left( 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right) \right) dt${u}_{3}\left(x\right)=2x+\lambda {\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt$
Let’s calculate this integral:
${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt$${\int }_{0}^{1} \left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt$int_(0)^(1)(x+t)(2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3)))dt\int_0^1 (x + t) \left( 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right) \right) dt${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt$
After calculating, we will substitute the result back into the formula for ${u}_{3}\left(x\right)$${u}_{3}\left(x\right)$u_(3)(x)u_3(x)${u}_{3}\left(x\right)$.
After calculating the integral, we find that
${\int }_{0}^{1}\left(x+t\right)\left(2t+\lambda \left(\frac{7\lambda }{12}+\lambda t+t+\frac{2}{3}\right)\right)dt=\frac{5{\lambda }^{2}}{8}+\frac{2\lambda }{3}+\frac{13{\lambda }^{2}x}{12}+\frac{7\lambda x}{6}+x+\frac{2}{3}$${\int }_{0}^{1} \left(x+t\right)\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}$int_(0)^(1)(x+t)(2t+lambda((7lambda)/(12)+lambda t+t+(2)/(3)))dt=(5lambda^(2))/(8)+(2lambda)/(3)+(13lambda^(2)x)/(12)+(7lambda x)/(6)+x+(2)/(3)\int_0^1 (x + t) \left( 2t + \lambda \left( \frac{7\lambda}{12} + \lambda t + t + \frac{2}{3} \right) \right) dt = \frac{5\lambda^2}{8} + \frac{2\lambda}{3} + \frac{13\lambda^2 x}{12} + \frac{7\lambda x}{6} + x + \frac{2}{3}${\int }_{0}^{1}\left(x+t\right)\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}$
Substituting this result back into the formula for ${u}_{3}\left(x\right)$${u}_{3}\left(x\right)$u_(3)(x)u_3(x)${u}_{3}\left(x\right)$, we get:
${u}_{3}\left(x\right)=2x+\lambda \left(\frac{5{\lambda }^{2}}{8}+\frac{2\lambda }{3}+\frac{13{\lambda }^{2}x}{12}+\frac{7\lambda x}{6}+x+\frac{2}{3}\right)$${u}_{3}\left(x\right)=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)$u_(3)(x)=2x+lambda((5lambda^(2))/(8)+(2lambda)/(3)+(13lambda^(2)x)/(12)+(7lambda x)/(6)+x+(2)/(3))u_3(x) = 2x + \lambda \left( \frac{5\lambda^2}{8} + \frac{2\lambda}{3} + \frac{13\lambda^2 x}{12} + \frac{7\lambda x}{6} + x + \frac{2}{3} \right)${u}_{3}\left(x\right)=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)$
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}\left(x\right)=2x+\lambda \left(x+\frac{1}{2}\right)$${u}_{1}\left(x\right)=2x+\lambda \left(x+\frac{1}{2}\right)$u_(1)(x)=2x+lambda(x+(1)/(2))u_1(x) = 2x + \lambda \left( x + \frac{1}{2} \right)${u}_{1}\left(x\right)=2x+\lambda \left(x+\frac{1}{2}\right)$
• ${u}_{2}\left(x\right)=2x+\lambda \left(\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}\right)$${u}_{2}\left(x\right)=2x+\lambda \left(\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}\right)$u_(2)(x)=2x+lambda((7lambda)/(12)+lambda x+x+(2)/(3))u_2(x) = 2x + \lambda \left( \frac{7\lambda}{12} + \lambda x + x + \frac{2}{3} \right)${u}_{2}\left(x\right)=2x+\lambda \left(\frac{7\lambda }{12}+\lambda x+x+\frac{2}{3}\right)$
• ${u}_{3}\left(x\right)=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)$${u}_{3}\left(x\right)=2x+\lambda \left(\frac{5{\lambda }^{2}}{8}+\frac{2\lambda }{3}+\frac{13{\lambda }^{2}x}{12}+\frac{7\lambda x}{6}+x+\frac{2}{3}\right)$u_(3)(x)=2x+lambda((5lambda^(2))/(8)+(2lambda)/(3)+(13lambda^(2)x)/(12)+(7lambda x)/(6)+x+(2)/(3))u_3(x) = 2x + \lambda \left( \frac{5\lambda^2}{8} + \frac{2\lambda}{3} + \frac{13\lambda^2 x}{12} + \frac{7\lambda x}{6} + x + \frac{2}{3} \right)${u}_{3}\left(x\right)=2x+\lambda \left(\frac{5{\lambda }^{2}}{8}+\frac{2\lambda }{3}+\frac{13{\lambda }^{2}x}{12}+\frac{7\lambda x}{6}+x+\frac{2}{3}\right)$
These approximations provide increasingly accurate solutions to the integral equation as the order increases.
Q3. Find the Geodesics on (i) Right circular cylinder (ii) Right circular cone.
(i) Right circular cylinder
In cylindrical polar coordinates $\left(\rho ,\theta ,\varphi \right)$$\left(\rho ,\theta ,\varphi \right)$(rho,theta,phi)(\rho, \theta, \phi)$\left(\rho ,\theta ,\varphi \right)$ the element of arc ds on the surface of the cylinder of radius $a$$a$aa$a$ is given by*
$\left(ds{\right)}^{2}=\left(d\rho {\right)}^{2}+\left(\rho d\theta {\right)}^{2}+\left(dz{\right)}^{2}$$\left(ds{\right)}^{2}=\left(d\rho {\right)}^{2}+\left(\rho d\theta {\right)}^{2}+\left(dz{\right)}^{2}$(ds)^(2)=(d rho)^(2)+(rho d theta)^(2)+(dz)^(2)(d s)^2=(d \rho)^2+(\rho d \theta)^2+(d z)^2$\left(ds{\right)}^{2}=\left(d\rho {\right)}^{2}+\left(\rho d\theta {\right)}^{2}+\left(dz{\right)}^{2}$
But for the given cylinder of radius $a$$a$aa$a$,
$\rho =a\phantom{\rule{1em}{0ex}}\text{so that}\phantom{\rule{1em}{0ex}}d\rho =0$rho=a quad” so that “quad d rho=0\rho=a \quad \text { so that } \quad d \rho=0
Hence (1) reduces to
$\left(ds{\right)}^{2}=\left(ad\theta {\right)}^{2}+\left(dz{\right)}^{2}$$\left(ds{\right)}^{2}=\left(ad\theta {\right)}^{2}+\left(dz{\right)}^{2}$(ds)^(2)=(ad theta)^(2)+(dz)^(2)(d s)^2=(a d \theta)^2+(d z)^2$\left(ds{\right)}^{2}=\left(ad\theta {\right)}^{2}+\left(dz{\right)}^{2}$
$ds={\left({a}^{2}+{z}^{\mathrm{\prime }2}\right)}^{1/2}d\theta ,\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}{z}^{\mathrm{\prime }}=dz/d\theta$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 \theta
giving
Therefore are of length $s$$s$ss$s$ between two points ${P}_{1}\left(a,{\theta }_{1},{z}_{1}\right)$${P}_{1}\left(a,{\theta }_{1},{z}_{1}\right)$P_(1)(a,theta_(1),z_(1))P_1\left(a, \theta_1, z_1\right)${P}_{1}\left(a,{\theta }_{1},{z}_{1}\right)$ and ${P}_{2}\left(a,{\theta }_{2},{z}_{2}\right)$${P}_{2}\left(a,{\theta }_{2},{z}_{2}\right)$P_(2)(a,theta_(2),z_(2))P_2\left(a, \theta_2, z_2\right)${P}_{2}\left(a,{\theta }_{2},{z}_{2}\right)$ is given by
$s={\int }_{{P}_{1}}^{{P}_{2}}ds={\int }_{{\theta }_{1}}^{{\theta }_{2}}{\left({a}^{2}+{z}^{\mathrm{\prime }2}\right)}^{1/2}d\theta ,\text{by (2)}$s=int_(P_(1))^(P_(2))ds=int_(theta_(1))^(theta_(2))(a^(2)+z^(‘2))^(1//2)d theta,” by (2) “s=\int_{P_1}^{P_2} d s=\int_{\theta_1}^{\theta_2}\left(a^2+z^{\prime 2}\right)^{1 / 2} d \theta, \text { by (2) }
Comparing (3) with ${\int }_{{\theta }_{1}}^{{\theta }_{2}}F\left(\theta ,z,{z}^{\mathrm{\prime }}\right)d\theta$${\int }_{{\theta }_{1}}^{{\theta }_{2}} F\left(\theta ,z,{z}^{\mathrm{\prime }}\right)d\theta$int_(theta_(1))^(theta_(2))F(theta,z,z^(‘))d theta\int_{\theta_1}^{\theta_2} F\left(\theta, z, z^{\prime}\right) d \theta${\int }_{{\theta }_{1}}^{{\theta }_{2}}F\left(\theta ,z,{z}^{\mathrm{\prime }}\right)d\theta$, here
$F\left(\theta ,z,{z}^{\mathrm{\prime }}\right)={\left({a}^{2}+{z}^{\mathrm{\prime }2}\right)}^{1/2}$$F\left(\theta ,z,{z}^{\mathrm{\prime }}\right)={\left({a}^{2}+{z}^{\mathrm{\prime }2}\right)}^{1/2}$F(theta,z,z^(‘))=(a^(2)+z^(‘2))^(1//2)F\left(\theta, z, z^{\prime}\right)=\left(a^2+z^{\prime 2}\right)^{1 / 2}$F\left(\theta ,z,{z}^{\mathrm{\prime }}\right)={\left({a}^{2}+{z}^{\mathrm{\prime }2}\right)}^{1/2}$
Euler’s equation is
$\frac{\mathrm{\partial }F}{\mathrm{\partial }z}-\frac{d}{d\theta }\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{z}^{\mathrm{\prime }}}\right)=0$$\frac{\mathrm{\partial }F}{\mathrm{\partial }z}-\frac{d}{d\theta }\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{z}^{\mathrm{\prime }}}\right)=0$(del F)/(del z)-(d)/(d theta)((del F)/(delz^(‘)))=0\frac{\partial F}{\partial z}-\frac{d}{d \theta}\left(\frac{\partial F}{\partial z^{\prime}}\right)=0$\frac{\mathrm{\partial }F}{\mathrm{\partial }z}-\frac{d}{d\theta }\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{z}^{\mathrm{\prime }}}\right)=0$
From (4),
$\frac{\mathrm{\partial }F}{\mathrm{\partial }z}=0$$\frac{\mathrm{\partial }F}{\mathrm{\partial }z}=0$(del F)/(del z)=0\frac{\partial F}{\partial z}=0$\frac{\mathrm{\partial }F}{\mathrm{\partial }z}=0$
and so (5) gives
$\frac{d}{d\theta }\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{z}^{z}}\right)=0$$\frac{d}{d\theta }\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{z}^{z}}\right)=0$(d)/(d theta)((del F)/(delz^(z)))=0\frac{d}{d \theta}\left(\frac{\partial F}{\partial z^z}\right)=0$\frac{d}{d\theta }\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }{z}^{z}}\right)=0$
Integrating it,
$\mathrm{\partial }F/\mathrm{\partial }{z}^{}$