I would like to plot the solution of the Fredholm Equation $$f\left(x\right)+\frac{1}{\pi} \int_{-1}^{1} \frac{1}{1+\left(x-t\right)^2}f\left(t\right) dt=1, \ \ (|x|\leq 1)$$ I tried to use Mathematica to find a numerical solution:

PHI = DSolveValue[\[Phi][x] == 1 - 1/Pi*Integrate[\[Phi][t]/(1 + (x - t)^2), 
      {t, -1, 1}], \[Phi], x]
Plot[PHI, {x, -1, 1}]

But I obtained some errors and I can't plot the numerical solution. How can I fix this problem?

  • $\begingroup$ In your Latex line there is one equation and in the Mathematica code there is second one. What actually do you try to solve? $\endgroup$ Apr 4, 2021 at 23:03
  • $\begingroup$ Your code works for me for the DSolveValue part, but you won't get a plot without assigning a value for lambda. If you mean what you say in your code, you need to change your Latex to match. $\endgroup$
    – Bill Watts
    Apr 5, 2021 at 0:40
  • $\begingroup$ Sorry everyone, I was wrong to copy the function in Mathematica. I corrected the post. $\endgroup$
    – Mark
    Apr 5, 2021 at 6:34
  • $\begingroup$ Are you looking for analytical solution only? This equation can be solved by numerical method like colocation or iteration. $\endgroup$ Apr 5, 2021 at 9:50
  • 1
    $\begingroup$ Just to compare. The command of Maple 2021 intsolve(f(x) + int(f(t)/(1 + (x - t)^2), t = -1 .. 1)/Pi = 1, f(x), method = collocation, order = 3) performs $$f \! \left(x \right) = 0.09775266381 x^{2}+ 0.6581514078 .$$ Analytical output is too long to be cited here. $\endgroup$
    – user64494
    Apr 5, 2021 at 12:20

4 Answers 4


Perhaps NestList gives an iterated solution (Picard iteration)

sol = NestList[1 - 1/Pi*Integrate[#/(1 + (x - t)^2), {t, -1, 1},Assumptions -> -1 <= x <= 1] &, 0 , 10];    

Show[ Plot[sol  , {x, -1, 1}, PlotRange -> {0, 1}], 
Plot[sol[[-1]], {x, -1, 1}, PlotStyle -> {Thickness[.005], Black}]]

enter image description here

This gives the symbolic Neumann-Series .

  • $\begingroup$ As we know, for some Fredholm equations the iteration method is not converge at n->Infinity. It has minimal error at n about 6. What do you think about this equation? $\endgroup$ Apr 5, 2021 at 12:48
  • $\begingroup$ @ AlexTrounev How did you check the error? If I compare the difference of two succesive iterations the numerical error seems to degree monoton. $\endgroup$ Apr 5, 2021 at 12:58
  • 1
    $\begingroup$ The error is the difference between the true solution and an iterative one, not between two iterations. $\endgroup$ Apr 5, 2021 at 14:17
  • $\begingroup$ @AlexTrounev But the true solution isn't known so far? $\endgroup$ Apr 5, 2021 at 14:34
  • $\begingroup$ Yes it is. The true solution is known in theory only. $\endgroup$ Apr 5, 2021 at 15:51

You have a wonderfully behaving smooth almost flat kernel. Since you want a numerical solution, then why not simply solve it with a quadrature?

To matrix equation with quadrature

Here's a sample code in WolframAlpha notebook:

K=Table[1/(1 + (2/n*(i - j))^2), {i,n}, {j, n}];
ListPlot[N[Table[{2/n*(i -1/2)-1,f[[i]]}, {i,n}]]]

enter image description here

Here's how it works. Convert the integral equation into a matrix equation replacing the integral with the dumbest rectangle quadrature: $$\int K(x,t) f(t)dt=\sum_{i=1}^NK(x,t_i)f(t_i)\frac 2 N$$ where $t_i=2/N*(i-1/2)-1$.

Now, after plugging the same knots into $x$ variable you get a linear system: $$f_{t_i}+\frac 2 N \sum_jK_{t_it_j}f_{t_j}= \sum_j\left(\delta_{ij}+\frac 2 N K_{t_it_j}\right)f_{t_j}=1$$ $$\left(I+\frac 2 N K\right)f=\mathbf 1$$ which can be solved trivially with high precision as shown above.

Gauss-Legendre quadrature

You can go fancy by employing Gaussian quadrature. Here's an example:

n = 27
<< NumericalDifferentialEquationAnalysis`;
knot = 0.5
gwL = GaussianQuadratureWeights[n/3, -1, -knot];
gwC = GaussianQuadratureWeights[n/3, -knot, knot];
gwR = GaussianQuadratureWeights[n/3, knot,1];
gw = Join[gwL,gwC,gwR];
n = Length[gw]

K=Table[1/(1 + ((gw[[i,1]] -gw[[j,1]]))^2)*gw[[j,2]], {i,n}, {j, n}];
ListPlot[N[Table[{gw[[i,1]],f[[i]]}, {i,n}]]]


Out:     0.657412

enter image description here

Spectral decomposition

There is a different solution that gives an insight into the integral kernel. Apply the spectral decomposition to the kernel: $$-\frac 1 \pi \frac 1 {1+(x-t)^2}=\sum_{i=1}^\infty\lambda_i \psi_i(x)\psi_i(t)$$ Once you get the eigen values $\lambda_i$ and vectors $\psi_i(x)$, the solution to your problem is: $$f(x)\approx\sum_{i=1}^K \frac 1 {1-\lambda_i}\left(\int_{-1}^1\psi_i(t)dt\right)\psi(x)$$ , where $K$ is the number of eigenvectors used for approximation.

enter image description here

Usually, we order eigenvalues descending by absolute value. Because the kernel is smooth and almost flat only the first few eigenvalues are required because they decay so quickly: enter image description here

The first few eigen vectors are interesting to observe. The first one gets very close to the solution:

enter image description here

... and clearly only odd ones are symmetric and should have nonzero values of $\int_{-1}^1\psi_i(t)dt$ integral as shown next. Also, these integrals quickly decay too, helping to cut $K$, the number of required eigen vectors for high precision approximation:

enter image description here enter image description here enter image description here

We can see now that if the right hand side wasn't simply a constant, 1 in your case, then a different - potentially larger - number $K$ of eigenvectors may have been required for a good approximation.

Here's the full code:

x=N[Table[2*(i -1/2)/n-1, {i,n}]];
d=N[Table[1/(1 +(2/n* (i -j))^2)/Pi,{i,n},{j,n}]];
K = -d;
{vals,v} = Eigensystem[A,11];

v1 = v[[1]]*Sum[v[[1,i]],{i,n}]/(1-vals[[1]]);
v3 = v[[3]]*Sum[v[[3,i]],{i,n}]/(1-vals[[3]]);
v5 = v[[5]]*Sum[v[[5,i]],{i,n}]/(1-vals[[5]]);
v7 = v[[7]]*Sum[v[[7,i]],{i,n}]/(1-vals[[7]]);
v9 = v[[9]]*Sum[v[[9,i]],{i,n}]/(1-vals[[9]]);
v11 = v[[11]]*Sum[v[[11,i]],{i,n}]/(1-vals[[11]]);
vapp = v1+v3+v5+v7+v9+v11;

ListPlot[Table[{x[[i]],v[[1,i]]}, {i,n}]]
ListPlot[Table[{x[[i]],v[[2,i]]}, {i,n}]]
ListPlot[Table[{x[[i]],v[[3,i]]}, {i,n}]]
ListPlot[Table[{x[[i]],v[[4,i]]}, {i,n}]]

One more colocation method based on Bernoulli (or Euler) polynomials with using NMinimize and FindRoot

 nN = 12; xcol = Table[-1 + 2 j/nN, {j, 0, nN}]; 
v[x_] := Table[BernoulliB[n, x], {n, 0, nN}]; A = Array[a, {nN + 1}]; 
u[x_] = A . v[x];
eqs = Table[
   u[xcol[[i]]] - 1 + 
    1/Pi A . 
      NIntegrate[v[t]/(1 + (xcol[[i]] - t)^2), {t, -1, 1}, 
       AccuracyGoal -> 8], {i, Length[xcol]}]; 

First method

sol = NMinimize[Norm[eqs], A]

Second method

sol1 = FindRoot[Table[eqs[[i]] == 0, {i, Length[eqs]}], 
  Table[{a[i], 1/10}, {i, Length[A]}]]


Plot[{u[x] /. sol1, u[x] /. sol[[2]]}, {x, -1, 1}]

Figure 1

  • $\begingroup$ For x=0 both methods yield 0.65741153600362792,but using method user @Ulrich Neumann give us 2/3.Which is correct one? $\endgroup$ Apr 5, 2021 at 10:20
  • $\begingroup$ @MariuszIwaniuk Actually 2/3=0.66666..., and colocation method gives solution with some error of $h^2$. In my code h=1/12, so we have error of 1/144. $\endgroup$ Apr 5, 2021 at 10:57
  • $\begingroup$ I tried with nN = 36 and give me: 0.6574115... ? $\endgroup$ Apr 5, 2021 at 11:30
  • $\begingroup$ @MariuszIwaniuk Ah, I see, that your method also gives 0.657412 for n=10. May be true value is not 2/3? $\endgroup$ Apr 5, 2021 at 12:07
  • 1
    $\begingroup$ Looks like at x=0 value is not 2/3.Try: N[Expand[DSolveValue[\[Phi][x] == 1 - 1/Pi* Integrate[\[Phi][ t]*(Series[1/(1 + (x - t)^2), {x, 0, 4}] // Normal) // Simplify // Expand, {t, -1, 1}], \[Phi][x], x]], 20] $\endgroup$ Apr 5, 2021 at 15:50

Long ago I borrowed the code from site, but I don't know exactly where.

ClearAll["`*"]; Remove["`*"];

n = 5;
a1 = -1;
b1 = 1;
B[x_, i_] := Binomial[n, i]*((x - a1)^i*(b1 - x)^(n - i))/(b1 - a1)^n;
k[t_, x_] := -(1/(1 + (-t + x)^2));
a[x_] := 1;
f[x_] := 1;
\[Lambda] = 1/Pi;

nIntegrate[most__] := NIntegrate[most, Method -> {Automatic, SymbolicProcessing -> 0}];

K2[j_?NumericQ] := nIntegrate[B[x, j] f[x], {x, a1, b1}] // Quiet;
mid[x_?NumericQ, i_?NumericQ] := nIntegrate[\[Lambda] k[t, x] B[t, i], {t, a1, b1}] // Quiet;

K3[j_?NumericQ, i_?NumericQ] := 
nIntegrate[(a[x] B[x, i] - mid[x, i]) B[x, j], {x, a1, b1}] // Quiet;

sol = Table[Total@Table[M[i] K3[j, i], {i, 0, n}] == K2[j], {j, 0, n}];

func = Total@Table[M[i] B[x, i], {i, 0, n}] /. NSolve[sol, Table[M[i], {i, 0, n}]][[1]]

Plot[func, {x, -1, 1}, PlotRange -> {Automatic, {0, 1}}]

enter image description here

  • $\begingroup$ Some hours ago there was a tricky numerical solution at the end of your answer , why did you delete it? $\endgroup$ Apr 5, 2021 at 13:05
  • $\begingroup$ @Ulrich Neumann I'm not sure it was correct. $\endgroup$ Apr 5, 2021 at 13:59
  • $\begingroup$ What a pity I was hoping it would be something like the "Method of lines for DAE systems"! What was wrong? $\endgroup$ Apr 5, 2021 at 14:03

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