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

  1. \phi(0)=20e^{-1}-8
  2. \phi(0)=20e-8
  3. \phi(1)=22-8e
  4. \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

  1. 1/2
  2. 2
  3. 3/2
  4. 4

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

  1. \phi(x)=f(x)
  2. \phi(x)=K(x,x)
  3. \phi(x)=x^3
  4. \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.

  1. \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.
  2. 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.

Hence we have eliminated three options – since at least one option must be true, we are left with option 4.

Questions 4,5

These questions are simply about finding the integral and rearranging it.

Question 7

For the integral equation y(x)=1+x^3+\int_0^xK9x,t)y(t)dt where K(x,t)=2^{x-t}, the iterated kernel K_3(x,t) is

  1. 2^{x-t} (x-t)^2
  2. 2^{x-t} (x-t)^3
  3. 2^{x-t-1} (x-t)^2
  4. 2^{x-t-1} (x-t)^3

This question should be solved using Method 2. Method description can be found here: Unit 2 Summary. I leave it to the reader to find the answer.

Question 9

Let \phi(x) be the solution of the integral equation \int_0^x e^{x-t}\phi(t)dt=x,x>0 . Then \phi(1) equals,

  1. -1
  2. 0
  3. 1
  4. 2

Attempt to solve using other methods are left to the reader. Note that this is a Volterra IE.

Solution by Leibniz Integral Rule

Leibniz Integral Rule (link) talks about differentiation of integral. Differentiating this equation using the above rule w.r. to x, we get

\int_0^x e^{x-t}\phi(t)dt + \phi(x) = 1, using the original equation, this reduces to x+\phi(x)=1\implies \phi(x)=1-x. Trivially, \phi(1)=0.

If you are interested, please explore solution by Laplace transforms and we can further expand this article; useful for many others around the globe.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.