Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question


  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 $AA^{-1}=id$.

share|cite|improve this answer
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.

share|cite|improve this answer

$\def\l{\lambda}$ 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 $\langle u_i,u_i\rangle = 1$, $$\begin{eqnarray*} c_1 &=& \frac{2}{\sqrt{(2-\sin 2) \alpha^2-4 \alpha\sin ^2 1 +\sin 2+2}} \\ c_2 &=& \frac{2}{\sqrt{(2+\sin 2) \alpha^2-4 \alpha\sin ^2 1 -\sin 2+2}}. \end{eqnarray*}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.