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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

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.

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    Why not take the second derivative of both sides and see what happens ...2012-04-20
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    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.2012-04-20
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    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''?2012-04-20
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    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.2012-04-20
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3179/discussion-between-mikael-anderson-and-oenamen)2012-04-20

1 Answers 1

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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.

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    To @oenamen: Many thanks for taking the time to provide the detailed answer. I appreciate all your help.2012-04-21
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    @MikaelAnderson: Glad to help, Mikael.2012-04-21
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    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 $n=3$.2012-04-22
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    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.2012-04-22
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    @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.2012-04-22
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    @MikaelAnderson: The fit gets better as $n$ increases. I have checked this for $n=50$. Have a close look at your code and get back to me if you still have trouble. To get you started I find $A_4 \approx 0.114$ and $c_4 \approx -0.0429$.2012-04-22
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    To @oenamen: Hi. Yes, I agree. I calculated the integral and even for $n=3$ it is almost impossible to distinguish it from $f(x).$2012-04-22
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    To @oenamen:: Thanks. I get the same values as you, namely $A_4\approx 0.114371$ and $c_4\approx -0.042892.$ I am still puzzled why the approximation of $f''$ by $y$ seems to get worse by increasing $n$ but it could be due to the working precision.2012-04-22
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    @MikaelAnderson: Probably not. I recommend having a close look at your code.2012-04-22
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    To @oenamen: Everything works fine now but it seems that I need more terms to get a good approximation of $f''$ compared to $f$. This makes me wonder if there are some theoretical results on rate of convergence of these approximations. A related question is regarding $c_i$. I have calculated the first 100 of them and for odd $i$ they are zero and for even ones they converge to zero. Does this imply that this is a mean-square convergent series? Otherwise how can one show this when we do not have a closed form expressions for $c_i$?2012-04-25
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    @Mikael Anderson: Note that any difference between $f$ and the series is magnified by taking derivatives, and smoothed out by integrating. See any text on Fourier series where concrete and simpler examples than the one dealt with above are given.2012-04-26
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    @Mikael Anderson: Regarding convergence, it is known that if $f$ is in $C^2[a,b]$ and satisfies the boundary conditions of the eigenfunctions, than the series converges to $f$ uniformly on $[a,b]$. (Note that this is stronger than mean-square convergence.) See, for example, Theorem 12.5.3 of Lebovitz's online [ODEs textbook](http://people.cs.uchicago.edu/~lebovitz/eodesnotes.html). I suspect that if $f$ is in $C^4[a,b]$ that the series for $f''$ will converge uniformly to $f''$.2012-04-26
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    @Mikael Anderson: If you are interested in the rate of convergence you will have to do some digging. In terms of differential equations you are studying a regular Sturm-Liouville system. [This search](https://www.google.com/search?q=regular%20sturm%20liouville%20system%20convergence) should get you started.2012-04-26
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    To @oenamen: Hi oenamen. Many thanks for your comments and the references. Very useful indeed.2012-04-26