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Show that the Fredholm integral equation

$$\phi(x)=f(x)+ \lambda \int^\pi_0\cos(x-s)\phi(s)\,ds$$

If $\lambda \neq \pm \frac{2}{\pi}$

My solution:

$\phi(x) = f(x) + \lambda \sin x \int^\pi_0 \cos s \phi(s) ds + \lambda \cos x \int^\pi _0 \sin s \phi(s)\, ds \tag1$

Let $\int^\pi_0 \cos s\phi(s) ds = c_1$ and $\int^\pi_0 \sin s \phi(s)\, ds = c_2$

then

$\phi(x) = f(x) + c_1\lambda \sin x + c_2 \lambda \cos x \tag2$

sub $(2)$ into $(1)$ giving

$c_1\sin x+c_2\cos x = \sin x[f_1 + c_1\lambda \int^\pi_0 \cos s \sin s \space ds + \lambda c_2 \int^\pi_0 \cos^2s \space ds] $ $\qquad\qquad\qquad\qquad\quad + \cos x [f_2 + c_1\lambda \int^\pi_0 \sin^2s \space ds + c_2\lambda \int^\pi_0 \sin s \cos s \space ds]$

where $f_1 = \int^\pi_0 \cos sf(x) \space ds$ and $f_2 = \int^\pi_0 \sin sf(x) \space ds$

since {$\sin x, \cos x$} are linearly independent we can compare coefficients giving

$c_1 = f_1 + \pi \lambda c_2$

$c_2 = f_2 + \pi \lambda c_1$

$\begin{pmatrix}1 &\lambda \pi \\ \lambda \pi &1\end{pmatrix}$$\begin{pmatrix}c_1 \\ c_2 \end{pmatrix}=\begin{pmatrix}f_1 \\ f_2 \end{pmatrix}$

Looking at the determinant of the first matrix shown we have

$$1-\lambda^2\pi^2 = 0$$

Which doesn't correspond with the initial condition given, so I must've gone wrong somewhere.

TIA

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  • 1
    $\begingroup$ $\displaystyle \int_0^\pi \cos^2 s \, ds = \int_0^\pi \sin^2 s \, ds = \frac\pi2$, not $\pi$. $\endgroup$ – tilper Jan 2 at 15:39
  • $\begingroup$ I'm not sure I understand the first step of your solution also, or what the determinant of that matrix at the end is supposed to show. The standard way to show a solution is unique is to assume there are two solutions and then show that they're equal, usually by showing their difference is zero. Although I think for this particular thing the standard is to use contraction mappings. Is that allowed in your case? See here - math.stackexchange.com/questions/2045232/… $\endgroup$ – tilper Jan 2 at 15:41

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