samples/texts/2487562/page_6.md

Table 2

Exact and approximate solutions of integral equation given in example-2(iv).

x	0	π/4	π/2	3π/4	π
φ(x) (exact sol.)	-1.3067	-0.4507	0.2016	0.2681	-0.2901
φ1(x) (least-squares sol.)	-0.7838	-0.4721	-0.1603	0.1515	0.4633
φ2(x) (sol. by artificial way)	-0.7942	-0.4795	-0.1648	0.1499	0.4646
|φ - φ1|	0.5228	0.0214	0.3619	0.1166	0.7534
|φ - φ2|	0.5125	0.0288	0.3664	0.1182	0.7548

The eigenvalues of the integral equation are

$$\begin{array}{cccc}& \lambda =\frac{0.5\{-12(\alpha +1)\pm 12\sqrt{\alpha +2\sqrt{\alpha +0.5}}\}}{\alpha },(\alpha \mathrm{\ne }0),& & \text{(3.43)}\end{array}\backslash lambda\; =\; \backslash frac\{0.5\backslash \{-12(\backslash alpha\; +\; 1)\; \backslash pm\; 12\backslash sqrt\{\backslash alpha\; +\; 2\backslash sqrt\{\backslash alpha\; +\; 0.5\}\}\backslash \}\}\{\backslash alpha\},\; \backslash quad\; (\backslash alpha\; \backslash neq\; 0),\; \backslash tag\{3.43\}$$

and hence, for any non-eigenvalue $\mu \neq \lambda$, the integral Eq. (3.42) has a unique solution.

Now, let $\phi(x) = a_0 + a_1 x$ be an approximate solution to the Eq. (3.42). Substituting this approximate solution in the Eq. (3.42), we obtain

$$\begin{array}{cccc}& a_0-\mu (\frac{2}{3}+\alpha \sqrt{x})a_0-\frac{\mu }{10}(4+5\alpha \sqrt{x})a_1+x{a}_1=f(x),0\le x\le 1.& & \text{(3.44)}\end{array}a\_0\; -\; \backslash mu\; \backslash left(\; \backslash frac\{2\}\{3\}\; +\; \backslash alpha\; \backslash sqrt\{x\}\; \backslash right)\; a\_0\; -\; \backslash frac\{\backslash mu\}\{10\}\; \backslash left(\; 4\; +\; 5\backslash alpha\backslash sqrt\{x\}\; \backslash right)\; a\_1\; +\; x\; a\_1\; =\; f(x),\; \backslash quad\; 0\; \backslash le\; x\; \backslash le\; 1.\; \backslash tag\{3.44\}$$

Multiplying the Eq. (3.44) by $1$ and $x$ and then integrating from $x=0$ to $x=1$, we get

$$\begin{array}{cccc}& -\frac{1}{3}\{-3+2(1+\alpha )\mu \}{a}_0-\frac{1}{30}\{-15+2(6+5\alpha )\mu \}{a}_1=f_1& & \text{(3.45)}\end{array}-\backslash frac\{1\}\{3\}\backslash \{-3+2(1+\backslash alpha)\backslash mu\backslash \}a\_0\; -\; \backslash frac\{1\}\{30\}\backslash \{-15+2(6+5\backslash alpha)\backslash mu\backslash \}a\_1\; =\; f\_1\; \backslash tag\{3.45\}$$

and

$$\begin{array}{cccc}& \frac{1}{30}\{15-2(5+6\alpha )\mu \}{a}_0+\frac{1}{15}\{5-3(1+\alpha )\mu \}{a}_1=f_2,& & \text{(3.46)}\end{array}\backslash frac\{1\}\{30\}\backslash \{15-2(5+6\backslash alpha)\backslash mu\backslash \}a\_0\; +\; \backslash frac\{1\}\{15\}\backslash \{5-3(1+\backslash alpha)\backslash mu\backslash \}a\_1\; =\; f\_2,\; \backslash tag\{3.46\}$$

where

$$\begin{array}{cccc}& f_1={\int }_0^{1}f(x)dx,f_2={\int }_0^{1}xf(x)dx\mathrm{.}& & \text{(3.47)}\end{array}f\_1\; =\; \backslash int\_0^1\; f(x)\; dx,\; \backslash quad\; f\_2\; =\; \backslash int\_0^1\; x\; f(x)\; dx.\; \backslash tag\{3.47\}$$

Eqs. (3.45) and (3.46) are solvable if and only if the determinant of their coefficients is non-zero which leads to

$$\begin{array}{cccc}& \mu \mathrm{\ne }\frac{0.125\{-50(1+\alpha )\pm 50\sqrt{\alpha +0.5068\sqrt{\alpha +1.9732}}\}}{\alpha },(\alpha \mathrm{\ne }0),& & \text{(3.48)}\end{array}\backslash mu\; \backslash neq\; \backslash frac\{0.125\backslash \{-50(1+\backslash alpha)\; \backslash pm\; 50\backslash sqrt\{\backslash alpha\; +\; 0.5068\backslash sqrt\{\backslash alpha\; +\; 1.9732\}\}\backslash \}\}\{\backslash alpha\},\; \backslash quad\; (\backslash alpha\; \backslash neq\; 0),\; \backslash tag\{3.48\}$$

showing that there exists a value of $\mu \neq \lambda$ for which the matrix of the system of Eqs. (3.45) and (3.46) becomes singular.

Example-II:

$$\begin{array}{cccc}& \varphi (x)-\lambda {\int }_0^{\pi }(\alpha \sin x+\cos t)\varphi (t)dt=f(x),0\le x\le \pi \mathrm{.}& & \text{(3.49)}\end{array}\backslash phi(x)\; -\; \backslash lambda\; \backslash int\_\{0\}^\{\backslash pi\}\; (\backslash alpha\; \backslash sin\; x\; +\; \backslash cos\; t)\; \backslash phi(t)\; dt\; =\; f(x),\; \backslash quad\; 0\; \backslash le\; x\; \backslash le\; \backslash pi.\; \backslash tag\{3.49\}$$

The eigenvalues of the integral Eq. (3.49) are

$$\begin{array}{cccc}& \lambda =\frac{1}{2\alpha },(\alpha \mathrm{\ne }0)\mathrm{.}& & \text{(3.50)}\end{array}\backslash lambda\; =\; \backslash frac\{1\}\{2\backslash alpha\},\; (\backslash alpha\; \backslash neq\; 0).\; \backslash tag\{3.50\}$$

If $\phi(x) = a_0 + a_1 x$ is an approximate solution of the Eq. (3.49), we obtain

$$\begin{array}{cccc}& (1-\mu \alpha \pi \sin x)a_0+(x+2\mu -\frac{{\pi }^2}{2}\mu \alpha \sin x)a_1=f(x),0\le x\le 1.& & \text{(3.51)}\end{array}(1\; -\; \backslash mu\backslash alpha\backslash pi\backslash sin\; x)a\_0\; +\; \backslash left(x\; +\; 2\backslash mu\; -\; \backslash frac\{\backslash pi^2\}\{2\}\backslash mu\backslash alpha\backslash sin\; x\backslash right)a\_1\; =\; f(x),\; \backslash quad\; 0\; \backslash le\; x\; \backslash le\; 1.\; \backslash tag\{3.51\}$$

Then, multiplying the Eq. (3.51) by $x$ and $x^2$ and integrating from $x=0$ to $x=\pi$, we get

$$\begin{array}{cccc}& (\frac{{\pi }^2}{2}-\alpha {\pi }^2\mu )a_0+(-\frac{{\pi }^3}{2}\alpha \mu +\frac{1}{3}{\pi }^2(3\mu +\pi ))a_1=f_3& & \text{(3.52)}\end{array}\backslash left(\backslash frac\{\backslash pi^2\}\{2\}\; -\; \backslash alpha\backslash pi^2\backslash mu\backslash right)a\_0\; +\; \backslash left(-\backslash frac\{\backslash pi^3\}\{2\}\backslash alpha\backslash mu\; +\; \backslash frac\{1\}\{3\}\backslash pi^2(3\backslash mu\; +\; \backslash pi)\backslash right)a\_1\; =\; f\_3\; \backslash tag\{3.52\}$$

and

$$\begin{array}{cccc}& (\frac{{\pi }^3}{3}+4\alpha \pi \mu -\alpha {\pi }^3\mu )a_0+(\frac{{\pi }^2}{4}+2\alpha {\pi }^2\mu +\frac{2}{3}{\pi }^3\mu -\frac{1}{2}\alpha {\pi }^4\mu )a_1=f_4,& & \text{(3.53)}\end{array}\backslash left(\backslash frac\{\backslash pi^3\}\{3\}\; +\; 4\backslash alpha\backslash pi\backslash mu\; -\; \backslash alpha\backslash pi^3\backslash mu\backslash right)a\_0\; +\; \backslash left(\backslash frac\{\backslash pi^2\}\{4\}\; +\; 2\backslash alpha\backslash pi^2\backslash mu\; +\; \backslash frac\{2\}\{3\}\backslash pi^3\backslash mu\; -\; \backslash frac\{1\}\{2\}\backslash alpha\backslash pi^4\backslash mu\backslash right)a\_1\; =\; f\_4,\; \backslash tag\{3.53\}$$

where

$$\begin{array}{cccc}& f_3={\int }_0^{\pi }xf(x)dx,f_4={\int }_0^{\pi }{x}^{2}f(x)dx\mathrm{.}& & \text{(3.54)}\end{array}f\_3\; =\; \backslash int\_0^\backslash pi\; x\; f(x)\; dx,\; \backslash quad\; f\_4\; =\; \backslash int\_0^\backslash pi\; x^2\; f(x)\; dx.\; \backslash tag\{3.54\}$$