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

The problem is related to this question: How to find eigenfunctions of a linear operator (follow-up question) I posted earlier.

Suppose I want to solve the following integral equation: $$\int_0^1 K(x,t)y(t)dt=\sqrt{2x^2-2x+1}$$ where $$K(x,t)=\max((1-x)t,(1-t)x),0<x<1,0<t<1.$$

Eigenfunctions of $ K(x,t)$ was found by @oenamen in the answer to the above-mentioned question. I thought one should be able to use the eigenfunctions to find a solution to the above equation but I fail to see if this is the case. I would appreciate any suggestions or comments on this.

share|cite|improve this question
Why not take the second derivative of both sides and see what happens ... – user26872 Apr 20 '12 at 4:36
To @oenamen: There is a theorem by Picard which states that equation of the above type has a solution if and only if the right hand side can be expanded into a mean-square convergent series with respect to eigenfunctions of the kernel $K(x, t)$. I was trying to use that instead of trying to solve the equation by differentiation. – Mikael Anderson Apr 20 '12 at 12:12
Would not your argument imply that the equation in my question does not have a solution since Picard's theorem is formulated as ``if and only if''? – Mikael Anderson Apr 20 '12 at 20:13
To @oenamen: I do not know how I can start a chat but if you have a minute to chat please let me know. I just finished my calculations in Mathematica and I am puzzled by your previous comment that $y=-f''$ because I think $y=f''$ was correct. – Mikael Anderson Apr 20 '12 at 22:25
let us continue this discussion in chat – Mikael Anderson Apr 20 '12 at 22:27
up vote 2 down vote accepted

Explicit solution

Take the second derivative of each side of the integral equation, $$y = f''.$$ Plugging this back into the integral equation we find that $f$ must satisfy the Robin boundary conditions obeyed by the eigenfunctions, $$\begin{eqnarray*} f'(0) + f(0) &=& 0 \\ f'(1) - f(1) &=& 0. \end{eqnarray*}$$ (We should expect to be able to expand a function obeying these boundary conditions in terms of the eigenfunctions.) Since $f(x) = \sqrt{2x^2-2x+1}$ satisfies these boundary conditions the solution exists. Thus, $$y = \frac{1}{(2x^2-2x+1)^{3/2}}.$$

Eigenfunction expansion

We write the integral equation schematically as $$f = K y.$$ Let $y_n$ denote the $n$th eigenfunction of $K$, $$y_n = \lambda_n K y_n.$$ These have been found for the operator pertinent to this question here. The eigenfunctions are orthogonal and we assume they have been normalized $$y_m \cdot y_n = \delta_{mn}.$$ (The inner product is $f\cdot g = \int_0^1 d t\, f(t)g(t)$.)

Picard's theorem mentioned in the comments states that $f$ can be expanded in terms of the eigenfunctions, $$f = \sum f_n y_n$$ where $f_n = y_n \cdot f$. Then $$\begin{eqnarray*} y &=& K^{-1} \sum f_n y_n \\ &=& \sum \lambda_n f_n y_n \\ &=& \sum c_n y_n \end{eqnarray*}$$ Thus, the coefficients of the expansion for $y$ are $c_n = \lambda_n f_n$.

Some details

Define $$y_0 = A_0(\sqrt{\lambda_0} \cosh\sqrt{\lambda_0} x - \sinh\sqrt{\lambda_0} x)$$ and $$y_n = A_n(\sqrt{\mu_n} \cos\sqrt{\mu_n} x - \sin\sqrt{\mu_n} x)$$ for $n\ge 1$. Note that $\lambda_n = -\mu_n$ for $n\ge 1$. Normalizing we find $$A_0 \approx 0.769,\hspace{3ex} A_1 \approx 0.672, \hspace{3ex} A_2 \approx 0.241, \hspace{3ex} A_3 \approx 0.154, \hspace{3ex} \ldots$$ The coefficients $c_n = \lambda_n \int_0^1 d t\, y_n(t) f(t)$ are $$c_0 \approx 1.94, \hspace{3ex} c_1 \approx 0, \hspace{3ex} c_2 \approx -0.757, \hspace{3ex} c_3 \approx 0, \hspace{3ex}\ldots $$ The function $y = \sum_{n=0}^3 c_n y_n$ already provides a very good approximate solution to the integral equation.

share|cite|improve this answer
To @oenamen: Many thanks for taking the time to provide the detailed answer. I appreciate all your help. – Mikael Anderson Apr 21 '12 at 11:38
@MikaelAnderson: Glad to help, Mikael. – user26872 Apr 21 '12 at 15:58
To @oenamen: I have followed all the details of the calculations and I get the same results as in your answer above. For $n=3$, I get the following approximation of the solution: $$y(t)= 1.49516 (1.5434 \cosh (1.5434 t)-\sinh (1.5434 t))-0.182402 (5.95017 \cos (5.95017 t)-\sin (5.95017 t))$$ However, when I plot this together with the exact answer $\frac{1}{\left(2 t^2-2 t+1\right)^{3/2}}$ I see great difference between the exact and approximation close to the boundaries at 0 and 1. I wonder if you get the same answer as above for $n=3$. – Mikael Anderson Apr 22 '12 at 18:12
Further, I thought one could improve the approximation by increasing $n$ to say 50 or 100 but something which puzzles me is that the approximation gets worse by increasing $n$. This does not sound right so I would appreciate to hear your comments on this. – Mikael Anderson Apr 22 '12 at 18:14
@MikaelAnderson: Hi Mikael. Your $y$ looks fine. At $x=0$ and $1$ I find the fit and exact solution disagree by about $20\%$. However, you should base the goodness of fit on how well $y$ solves the integral equation. On that basis the fit is very good indeed. I recommend doing the integral to see this. – user26872 Apr 22 '12 at 19:11

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.