Fredholm Equations

Question

I have the following problem to solve

$$\phi (x)=\lambda \int_{o}^{\pi }\cos(2x+y)\phi (y) dy+ \sin x$$

following the instructions from the following link early to conclude that:

$$\phi (x)=\lambda \int_{o}^{\pi }\cos(2x+y)\phi (y) dy+\sin x$$ $$=\lambda \int_{o}^{\pi }\cos^2(x)\cos(y)-\sin^2(x)\cos(y)-2\sin(x)\cos(x)\sin(y)\phi (y) dy+ \sin x$$

then start to calculate $a_{11}, a_{12},a_{13},a_{21},a_{22},a_{23},a_{31},a_{32},a_{33}$, such that (these will be the elements of the matrix A):

$$a_{1}=\cos^2(x);a_{2}=-\sin^2(x) ;a _{3}=-2\sin(x)\cos(x)$$

$$b_{1}=\cos(y);b_{2}=\cos(y) ;b_{3}=\sin(y)$$

thus calculated:

$$a_{11}=\int_{o}^{\pi }b_{1}(y)a_{1}(y)dy$$

$$a_{12}=\int_{o}^{\pi }b_{1}(y)a_{2}(y)dy$$

$$a_{13}=\int_{o}^{\pi }b_{1}(y)a_{3}(y)dy$$

calculating and so on until all $a_{nm}$ with $n =1,2,3$ and $m=1,2,3$.

After calculations we obtain the matrix

$$A=\begin{bmatrix} a_{11} &a_{12} & a_{13}\\ a _{21}&a_{22} & a_{23}\\ a_{31} &a_{32} & a_{33} \end{bmatrix}$$

making: $\;(I-\lambda A)$,

and calculating the determinant originated by $\;(I-\lambda A)$, get all values ​​of $\lambda$.

The solutions they have there is only one $\lambda=3/2\sqrt{2}$ and $\lambda=-3/2\sqrt{2}$

But never get to that result. I appreciate any and all help. I do not even know if I was wrong in whole or in calculating the determinant of the matrix.

Answer 1

Related problem. First note this, the solution is

$$ \phi \left( x \right) =\sin \left( x \right) +12\,{\frac {{\lambda}^{2 }\pi \, \cos^2\left( x \right) }{8\,{\lambda}^{2}-9 }}-6\,{\frac {{\lambda}^{2}\pi }{8\,{\lambda}^{2}-9}}+9\,{\frac {\pi \,\lambda\,\sin \left( x \right) \cos \left( x \right) }{8\,{\lambda}^ {2}-9}}.$$

Now, you can see the singular values or the wanted $\lambda$ is given by

$$ 8\,\lambda^{2}-9 = 0 \implies \lambda = -\frac{3}{2\sqrt{2}},\,\frac{3}{2\sqrt{2}}.$$

Added: Here is a start

$$ \phi (x)=\lambda \int_{o}^{\pi }\cos(2x+y)\phi (y) dy+\sin x $$

$$ \phi(x) = \sin(x) +\lambda \cos(2x)\int_{0}^{\pi }\cos(y)\phi (y) dy - \sin(2x) \int_{0}^{\pi }\sin(y)\phi (y) dy $$

$$ \phi(x) = \sin(x) +\lambda c_1\cos(2x) - \lambda c_2 \sin(2x). $$

Now, just follow the technique in the link and you should not have a problem.