Problem


\Psi(x) = f(x) + \lambda \int _0^1 (1 + xy)\Psi(y)dy

A. Given f(x) = x, \lambda = 1, find \Psi B. Given f(x) = 0, find eigenvalues \lambda and corresponding eigenfunctions \Psi

Solutions


\Psi(x) = x + \int _0^1 (1 + xy)\Psi(y)dy

Set the kernel as K(x, y) = 1 + xy, C_y = \int _0^1 K(x, y)\Psi(y)dy:


C_y = \int _0^1 K(x, y)\Psi(y)dy


C_y = \int _0^1 K(x, y)(x + C_y)dy


C_y = (x + C_y) \int _0^1 K(x, y) dy


C_y = (x + C_y)(1 + \frac x 2)


-\frac {C_yx} 2 = x + \frac {x^2} 2


C_y = - 2 - x

Therefore:


\Psi(x) = -2


\Psi(x) = \lambda \int _0^1 (1 + xy)\Psi(y)dy


\Psi(x) = \lambda(\int _0^1 \Psi(y) dy + x\int _0^1 y\Psi(y)dy)

Let C_y = \int _0^1 \Psi (y) dy, S_y = \int _0^1 y \Psi(y) dy: Equation 1:


C_y = \lambda \int _0^1 (C_y + yS_y) dy


C_y = \lambda (C_y + \frac 1 2 S_y)


C_y = \frac {\lambda S_y} {2(1 - \lambda)}

Equation 2:

S_y = \lambda \int _0^1 y(C_y + yS_y) dy S_y = \lambda (\frac 1 2 C_y + \frac 1 3 S_y)


S_y = \frac {\frac \lambda 2 C_y} {1 - \frac \lambda 3}

S_y = \frac {3\lambda C_y} {2(3 - \lambda)}

Substituting:


C_y = \frac {\lambda} {2(1 - \lambda)} \cdot \frac {3\lambda C_y} {6 - 2\lambda}


C_y = \frac {3\lambda^2} {4(1 - \lambda)(3 - \lambda)} C_y


1 = \frac {3\lambda^2} {4(1 - \lambda)(3 - \lambda)}

4(1 - \lambda)(3 - \lambda) = 3\lambda^2


\lambda^2 - 16\lambda + 12 = 0


\lambda = 8 \pm 2\sqrt{13}

Substitute in for eigenfunctions!

Reference used to figure out Fredholm equations and kernel integration
This was part of a really old final from SJSU! Thank you Mr. A, this was a very cool problem :D