# Show that the Fredholm integral equation has a unique solution

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

• $\displaystyle \int_0^\pi \cos^2 s \, ds = \int_0^\pi \sin^2 s \, ds = \frac\pi2$, not $\pi$. – tilper Jan 2 at 15:39
• 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/… – tilper Jan 2 at 15:41