Selected Solutions (IEL6 Quiz)

Question 1

Question: Let $\phi$ be the solution of the integral equation $\frac12\phi(x)-\int_0^1e^{x-y}\phi(y)dy=x^2; (0\le x\le1$ . Then

$\phi(0)=20e^{-1}-8$
$\phi(0)=20e-8$
$\phi(1)=22-8e$
$\phi(1)=22-8 e^{-1}$

A couple of solutions to this problem are discussed in Unit Quiz 1 (Question 2, only difference is that options 3 and 4 are about value of phi at x=1), even using methods mentioned in Unit 2.

Question 2

For the linear IE $\phi(x)=x+\int_0^\frac12 \phi(\eta)d\eta$ , $R(x,\eta;1)$ is

Solution 1: Using Method 2

Method description can be found here: Unit 2 Summary

$K( x,\eta ) = 1 = K_1( x,\eta )$ .

$K_2( x,\eta ) = \int_0^\frac12 K(x,s)K_1(s,\eta)dx = \int_0^\frac12 dx = \frac12$

Similarly, $K_3(x,\eta) = \int_0^\frac12 K(x,s)K_2(s,\eta)dx = \frac1{2^2}$

Thus, $K_m(s,t) = \frac1{2^{m-1}}$

and hence, $R(x,\eta;1)=\sum_{m=1}^\infty 1^{m-1}K_m(x,\eta) = \sum_{m=1}^\infty \frac1{2^{m-1}} = {1\over{1-\frac12}}=2$

Solution 2: Using Method 4

Method description can be found here: Unit 2 Summary.

$R(x,\eta;\lambda)={D(x,\eta;\lambda)\over D(\lambda)}$

$c_0=1,C_0(x,\eta)=1$

$c_1=\int_0^\frac12 C_0(x,x)dx= \int_0^\frac12 dx =\frac12$

$C_1(x,\eta)=\frac12\times 1)-1\int_0^\frac12 dx=\frac12 - \frac12 =0$

and $c_p=0,C_p(x,\eta)=0,\forall p>1$ .

Hence, $R(x,\eta;\lambda)=\frac1{1-\lambda\frac12}\implies R(x,\eta;1) = 2$

Solution 3: Taking constant c

Given $\phi(x)=x+\int_0^\frac12 \phi(\eta)d\eta$ . We need to find $R(x,\eta;1)$ ; where resolvent kernel is nothing but a function such that $\phi(x)=x+\int_0^\frac12 R(x,\eta;1) f(\eta)d\eta = x+\int_0^\frac12 R(x,\eta;1) \eta d\eta$

All given answers are constants, hence, substituting $R(x,\eta;1)=c$ in above equation, we get $\phi(x)= x+\int_0^\frac12 c\eta d\eta = x+\frac c8$ . Now substituting this expression of $\phi(x)$ in the original equation and simplifying, we get $c=2$ .

Question 3

The integral equation $\phi(x) = f(x)+\int_0^1 K(x,y)\phi(y)dy$ for $K(x,y)=xy^2$ has the solution

$\phi(x)=f(x)$
$\phi(x)=K(x,x)$
$\phi(x)=x^3$
$\phi(x)=f(x)+\frac43x\int_0^1x^2f(x)dx$

Solution 1: Using Method 1

Method description can be found here: Unit 1 Summary.

Put $c= \int_0^1 y^2\phi(y)dy$ . We get $\phi(x)=f(x)+xc$ . Substituting it back in original equation,

$f(x)+xc = f(x)+\int_0^1 xy^2(f(y)+yc)dy$ Simplifying further, we get $c= \frac43 \int_0^1 y^2 f(y)dy$ . Hence, $\phi(x)=f(x)+ \frac43 x\int_0^1 y^2 f(y)dy$ , option 4.

Solution 2: Using Method 2

Method description can be found here: Unit 2 Summary.

$K_1(x,y)=xy^2$

$K_2(x,y)=\int_0^1 xs^2\times sy^2 ds = xy^2\times (\frac {s^4}4)_0^1={xy^2\over4}$

Continuing, $K_m(x,y)= {xy^2\over4^{n-1}}$ . Hence, $R(x,t;\lambda)=\sum_{n=1}^\infty \lambda^{m-1}K_m(x,y)= \sum_{n=1}^\infty \lambda^{m-1} {xy^2\over4^{m-1}}$

$R(x,t;1)= \sum_{n=1}^\infty {xy^2\over4^{m-1}}=\frac1{ 1-1/4}=\frac43 xy^2$ .

Hence, the solution, $\phi(x)=f(x)+ \frac43 x\int_0^1 y^2 f(y)dy$ , option 4.

Solution 3: Using Method 4

Method description can be found here: Unit 2 Summary. I leave it to reader to try solve using this method, ample enough samples are given above.

Solution 4: Elimination method

Substitute each option, see which one satisfies the equation.

$\phi(x)=f(x)\implies f(x)=f(x)+\int_0^1 xy^2f(y)dy$ , and it leads to $\int_0^1 xy^2f(y)dy=0$ , which is absurd.
Note that options 2 and 3 are the same – $K(x,y)=xy^2\implies K(x,x)=x^3$ . Substituting in the equation, $x^3=f(x)+\int_0^1 xy^2\times y^3dy=f(x)+xc$ where c is a constant – which again, doesn’t satisfy the equation.