Selected Solutions (IEL5-Quiz)

Question 2

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

  1. ϕ(0)=20exp(-1) – 8
  2. ϕ(0)=20e-8
  3. ϕ(0)=22 – 8e
  4. ϕ(0)=22 – 8exp(-1)

Solution 1

Let c =  \int_0^1 e^{-y}\phi(y)dy \implies \phi(x)=2x^2+2e^x c. Substituting in the equation, 2x^2+2e^x c = 2x^2+ 2e^x\int_0^1 e^{-y}(2y^2+2e^y c)dy\implies c=\int_0^1e^{-y}(2y^2+ce^y)dy\implies c=2[\frac5e - 2]

Hence, \phi(x)=2x^2+4e^x  [\frac5e - 2]

Concluding, ϕ(0)=20exp(-1) – 8.

Solution 2

For a Fredholm integral equation of 2nd Kind, using \Gamma(s,t;\lambda)=\sum_p=0^\infty \frac{(-\lambda)^p}{p!}C_p(s,t)/ \sum_p=0^\infty \frac{(-\lambda)^p}{p!}c_p, where

  1. c_0=1,C_0(s,t)=K(s,t)
  2. c_p=\int_0^1 C_{p-1}(s,s)ds
  3. C_p = c_pK(s,t)-p\int_0^1K(s,x)C_{p-1}(x,t)dx

Here, c_0=1,C_0(x,y)=e^{x-y},c_1=1,C_1(x,y)=0. Hence \Gamma(x,t;\lambda)=\frac{e^{x-y}}{1-2}=- e^{x-y}

Hence the solution is \phi(x)=2x^2+2\int_0^1\Gamma(x,y;\lambda)f(y)dy = 2x^2-4\int_0^1e^{x-y}y^2dy and you know how to finish the rest.

Solution 3

Using iterated kernels, \Gamma(s,t;\lambda)=\sum_{m=1}^\infty \lambda^{m-1}K_m(s,t)

where K_1(s,t) = K(s,t),K_m(s,t)=\int K(s,x)K_{m-1}(x,t)dx

We will get K_m(s,t) = e^{x-y}, \implies  \Gamma(s,t;\lambda)=\sum_{m=1}^\infty \lambda^{m-1} e^{x-y} =   e^{x-y}  \sum_{m=1}^\infty \lambda^{m-1} =  e^{x-y}  \times (1-\lambda)^{-1}

Here, \lambda = 2\implies \Gamma(s,t;\lambda)= -  e^{x-y}  and you know where to take it from here.

Observation

Use Leibniz Integral Rule (link to blog page quoting the result) – Wikipedia (In particular, a specific case called Fundamental Theorem of Calculus here, since limits are constant) to find derivative of the equation w.r to x. What do you observe? [Hint: you will get a differential equation, as \phi'(x)-\phi(x)=4x-2x^2 – try solving it!]

Question 4

Method

Solve the corresponding transposed homogenous integral equation, \phi(x)=\frac2\pi\int_0^\pi \cos(x+t)\phi(t)dt and find corresponding solution \psi(x) for \lambda= \frac2\pi.

Then, for each option, if \int_0^\pi f(x)\psi(x)dx. If it is zero, it has solutions, and if it is not zero, no solutions.

Here, solution, \psi(x) =c_1\cos x +c_2 \sin x and hence options 2,3 and 4. (\int_0^\pi (\cos x+\sin x)\cos(3x)dx=0 and so on).

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