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Find $A^{-1}$ if $Af(x)=f(x)+\int\limits_{0}^{1}\cos(x+y)f(y)dy$.

Any ideas? I am not sure how to approach it...

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3 Answers 3

Hints:

  1. Define notation $c:=\cos$ and $s:=\sin$.

  2. Define inner product $\langle f, g \rangle := \int_0^1 \overline{f(x)} g(x) dx $.

  3. Then $(Af)(x)~=~f(x) + c(x) \langle c, f \rangle - s(x)\langle s, f \rangle .$

  4. Guess that the inverse is of the form $$(A^{-1}f)(x)~=~f(x) + [c(x)~s(x)] M \left[\begin{array}{c} \langle c, f \rangle \cr \langle s, f \rangle \end{array}\right] ,$$ where $M$ is a constant $2\times 2$ matrix independent of the function $f$ and the argument $x$.

  5. Determine $M$ by checking either $A^{-1}A=id$ or $AA^{-1}=id$.

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Actually, your method, i.e. determine $M$, helped me solve a similar exercise I had to deal with. Would you have any textbook references to get more familiar with these kind of tricks? –  johnny Jul 5 '12 at 14:44
    
No, unfortunately I don't have a reference. I just guessed the Ansatz. –  Qmechanic Jul 5 '12 at 15:04

Thank you all, I solved it myself last night. Here is my solution: $g(x)=f(x)-\int\limits_{0}^{1}\cos(x+y)f(y)dy=f(x)-\cos{x}\int\limits_{0}^{1}\cos{y}f(y)dy+\sin{x}\int\limits_{0}^{1}\sin{y}f(y)dy=\\$ $f(x)-\cos{x}c_f+\sin{x}d_f$, where $c_f=\int\limits_{0}^{1}\cos{y}f(y)dy$ and $d_f=\int\limits_{0}^{1}\sin{y}f(y)dy$. Now: $f(x)=g(x)+\cos{x}c_f-\sin{x}d_f$ and if we substitute $f(x)$ in expressions for $c_f$ and $d_f$ we get: $c_f=\int\limits_{0}^{1}\cos{x}f(x)dx=\int\limits_{0}^{1}\cos{x}(g(x)+\cos{x}c_f-\sin{x}d_f)=c_g+c_f\int\limits_{0}^{1}\cos^2{x}-d_f\int\limits_{0}^{1}\sin{x}\cos{x}dx.$ We can easily calculate these integrals and obtain first equation (we get another equation by doing the same thing, only starting with $d_f$). At the end we get a 2x2 system, where $c_g$ and $d_g$ are "constants" and $c_f$ and $d_f$ are unknown. By doing this we get $f(x)=g(x)+$something that depends only of $g(x)$ and some trigonometric functions and constants.

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$\def\l{\lambda}$ We assume the integral is over $[0,\pi]$ and not $[0,1]$. If $[0,1]$ is the intended interval the method below will work but the eigenfunctions and eigevalues will be different. We use the standard shorthand $(K f)(x) = \int_0^\pi dy\, K(x,y) f(y)$ and $\langle f,g\rangle = \int_0^\pi dx f(x)^* g(x)$. The norm of $f$ is $\sqrt{\langle f,f\rangle}$.

Let $A f = g$. We wish to solve $g = f + K f$, for $f$, that is, to solve the inhomogeneous integral equation $$\begin{equation*} f = g + \l K f,\tag{1} \end{equation*}$$ where $\l = -1$. The kernel $K(x,y) = \cos(x+y)$ is that of a degenerate Hilbert-Schmidt integral operator. We expect there to be a finite number of orthogonal eigenfunctions.

A standard technique involves first examining the homogeneous eigenvalue equation $$u_i = \l_i K u_i.$$ This is straightforward to solve. In detail $$\begin{eqnarray*} u_i(x) &=& \l_i \int_0^\pi dy\, \cos(x+y) u_i(y) \\ &=& \l_i \left( \cos x \int_0^\pi dy\, \cos(y) u_i(y) -\sin x \int_0^\pi dy\, \sin(y) u_i(y)\right). \end{eqnarray*}$$ The eigenfunctions must be of the form $A \cos x + B \sin x$. Plugging this into the equation above allows us to find the eigenfunctions (up to an overall constant) and eigenvalues. We find the normalized eigenfunctions are $\sqrt{\frac{2}{\pi}}\cos x$ and $\sqrt{\frac{2}{\pi}}\sin x$, with eigenvalues $2/\pi$ and $-2/\pi$, respectively.

Having the eigenfunctions, one can show that the solution to (1) is $$f = g + \l \sum_i \frac{u_i}{\l_i-\l} \langle u_i,g\rangle,$$ that is, $$\begin{equation*} A^{-1}f = f + \l \sum_i \frac{u_i}{\l_i-\l} \langle u_i,f\rangle.\tag{2} \end{equation*}$$

Addendum: For the interval $[0,1]$, the eigenfunctions are $$\begin{eqnarray*} u_1(x) &=& c_1(\cos x - \alpha \sin x) \\ u_2(x) &=& c_2(\sin x - \alpha \cos x), \end{eqnarray*}$$ where $\alpha = \frac{1}{2} \left(2-\sqrt{6-2\cos2} \, \cos1\right)\csc^2 1$. It it straightforward to verify the $u_i$ are orthogonal. The eigenvalues are $$\begin{eqnarray*} \lambda_1 &=& (\sqrt{6-2\cos 2} - 2\sin 1)\sec 1\\ \lambda_2 &=& -(\sqrt{6-2\cos 2} + 2\sin 1)\sec 1. \end{eqnarray*}$$ The $c_i$ are got by imposing $\langle u_i,u_i\rangle = 1$, $$\begin{eqnarray*} c_1 &=& \frac{2}{\sqrt{(2-\sin 2) \alpha^2-4 \alpha\sin ^2 1 +\sin 2+2}} \\ c_2 &=& \frac{2}{\sqrt{(2+\sin 2) \alpha^2-4 \alpha\sin ^2 1 -\sin 2+2}}. \end{eqnarray*}$$

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