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Consider the Fredholm Equation of the second kind,

$$\phi(x) = 3 + \lambda \int_{0}^{\pi} \text{cos}(x-s) \, \phi(s) \,ds$$

Where the analytical solution is found as,

$$\phi(x) = 3 + \frac{6\lambda}{1 - \lambda \frac{\pi}{2}}\,\text{sin}(x)$$

How could one use Mathematica to find a numerical solution to the same integral equation by using the method of successive approximations (i.e. the Neumann series approach)?

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2 Answers 2

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Use DSolve:

PHI = 
  DSolveValue[ϕ[x] == 3 + λ Integrate[ Cos[x - s] ϕ[s], {s, 0, Pi}], ϕ, x]
(*Function[{x}, (3 (-2 + π λ - 4 λ Sin[x]))/(-2 + π λ)]*)

The solution can be further used in the form PHI[x].

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  • $\begingroup$ Thank you, but how can I use the function Mathematica returns to, say, investigate the convergence of the new $\phi (x)$ function? $\endgroup$
    – LightningStrike
    Mar 1, 2019 at 19:40
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    $\begingroup$ @ user57401 I modified my answer! $\endgroup$
    – Ulrich Neumann
    Mar 1, 2019 at 19:57
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Following Weisstein, Eric W. "Integral Equation Neumann Series." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html, the Neumann series approximation is:

n = 10;  (* for example *)
ϕ[x_, 0] = 3;
Do[ϕ[x_, j_] = 3 + λ Integrate[Cos[x - p] ϕ[p, j - 1], {p, 0, π}], {j, n}]

The last term in the series ϕ[x,n] is the approximation to ϕ[x].

Here is what Mathematica returns for ϕ[x,10].

phi_of_ten

To investigate convergence, I guess we could look at the difference ϕ[x,n] - ϕ[x] as n gets large, since you know ϕ[x].

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  • $\begingroup$ Thank you! When I try to run this, my output is returning the value of 3? How did you get Mathematica to return the series above for [Phi][10]? $\endgroup$
    – LightningStrike
    Mar 1, 2019 at 20:27
  • $\begingroup$ Please clear out your variables, perhaps with Evaluation: Quit Kernel: Local. To print the final (nth) value: \[Phi][x, n]. $\endgroup$
    – mjw
    Mar 1, 2019 at 20:52
  • $\begingroup$ Made some edits to my answer. Had a couple of typos. Within a function definition it is x_, otherwise x. Also, \[Phi][x,j] needs two arguments, one for x and one for the jth approximation. Hope its clear. $\endgroup$
    – mjw
    Mar 1, 2019 at 20:55
  • $\begingroup$ @m_goldberg, how do you post symbols rather than the clutzy [Phi] type of notation here? $\endgroup$
    – mjw
    Mar 3, 2019 at 1:57
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    $\begingroup$ I use halirutan's plug-in. You can learn more about it here $\endgroup$
    – m_goldberg
    Mar 3, 2019 at 2:53

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