Fredholm Equations - Seperable Kernel Technique

Calculus

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

Part A

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

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

$$C_y={\int }_0^{1}K(x,y)\Psi (y)dy$$ $$C_y={\int }_0^{1}K(x,y)(x+C_y)dy$$ $$C_y=(x+C_y){\int }_0^{1}K(x,y)dy$$ $$C_y=(x+C_y)(1+\frac{x}{2})$$ $$-\frac{C_{y}x}{2}=x+\frac{x^2}{2}$$ $$C_y=-2-x$$

Therefore:

$$\Psi (x)=-2$$

Part B

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

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

$$C_y=\lambda {\int }_0^1(C_y+y{S}_y)dy$$ $$C_y=\lambda (C_y+\frac{1}{2}{S}_y)$$ $$C_y=\frac{\lambda S_y}{2(1-\lambda )}$$

Equation 2: $S_y=\lambda {\int }_0^{1}y(C_y+y{S}_y)dy$ $S_y=\lambda (\frac{1}{2}{C}_y+\frac{1}{3}{S}_y)$

$$S_y=\frac{\frac{\lambda }{2}{C}_y}{1-\frac{\lambda }{3}}$$

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

Notes

• 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