How can I reference a specific point of my function inside NDSolve?

Question

The problem:

I am trying to solve this diffrential equation:

K[x_, x1_] := 1;
NDSolve[{A''[x] == Integrate[K[x, x1] A[x1], {x1, 0, 1}], 
         A[0] == 0, A'[1] == 1}, A[x], x]

and I'm getting errors (Function::slotn and NDSolve::ndnum)
(it should return a numeric function that is equal to 3/16 x^2 + 5/8 x)

I am looking for a way to solve this differential equation: Is there a way to write it in a better form, such that NDSolve will understand it? Is there another function or package that can help?

Note 1: In my full problem, K[x, x1] is not 1 -- it depends (in a complex way) on x and x1.
Note 2: Naively deriving the two sides of the equation with respect to x won't work, because the integral limits are definite.

My first impression:

It seems that Mathematica doesn't like me referencing a point in A[x] -- the same errors occur when I'm doing this simplified version:

NDSolve[{A''[x] == A[0.5], A[0] == 0, A'[1] == 1}, A[x], x]

(it should return a numeric function that is equal to 2/11 x^2 + 7/11 x)

In this case one can avoid this problem by analytically solving A''[x] == c, and then finding c, but in my first problem it seems to not work -- it only transform the differential equation to an integral one, which (N)DSolve doesn't solve afterwards.

Answer 1

I can suggest a way to reduce your equation to an integral equation, which can be solved numerically by approximating its kernel with a matrix, thereby reducing the integration to matrix multiplication.

First, it is clear that the equation can be integrated twice over x, first from 1 to x, and then from 0 to x, so that:

We can now discretize this equation, putting it on a equidistant grid:

Here, the A[x] becomes a vector, and the integrated kernel iniIntK becomes a matrix, while integration is replaced by a matrix multiplication. The problem is then reduced to a system of linear equations.

The easiest case (that I will consider here) is when the kernel iniIntK can be derived analytically - in this case this method will be quite fast. Here is the function to produce the integrated kernel as a pure function:

Clear[computeDoubleIntK]
computeDoubleIntK[kernelF_] :=
  Block[{x, x1},
   Function[
    Evaluate[
      Integrate[
         Integrate[kernelF[y, x1], {y, 1, x}] /. x -> y, {y, 0, x}] /. 
         {x -> #1, x1 -> #2}]]];

In our case:

In[99]:= K[x_,x1_]:=1;

In[100]:= kernel = computeDoubleIntK[K]
Out[100]= -#1+#1^2/2&

Here is the function to produce the kernel matrix and the r.h,s vector:

 computeDiscreteKernelMatrixAndRHS[intkernel_, a0_, aprime1_ , 
    delta_, interval : {_, _}] :=
  Module[{grid, rhs, matrix},
    grid = Range[Sequence @@ interval, delta];
    rhs = a0 + aprime1*grid; (* constant plus a linear term *)
    matrix = 
      IdentityMatrix[Length[grid]] - delta*Outer[intkernel, grid, grid];
    {matrix, rhs}]

To give a very rough idea how this may look like (I use here delta = 1/2):

In[101]:= computeDiscreteKernelMatrixAndRHS[kernel,0,1,1/2,{0,1}]

Out[101]= {{{1,0,0},{3/16,19/16,3/16},{1/4,1/4,5/4}},{0,1/2,1}}

We now need to solve the linear equation, and interpolate the result, which is done by the following function:

Clear[computeSolution];
computeSolution[intkernel_, a0_, aprime1_ , delta_, interval : {_, _}] :=
With[{grid = Range[Sequence @@ interval, delta]},
  Interpolation@Transpose[{
    grid,
    LinearSolve @@ 
      computeDiscreteKernelMatrixAndRHS[intkernel, a0, aprime1, delta,interval]
  }]]

Here I will call it with a delta = 0.1:

In[90]:= solA = computeSolution[kernel,0,1,0.1,{0,1}]
Out[90]= InterpolatingFunction[{{0.,1.}},<>] 

We now plot the result vs. the exact analytical solution found by @Sasha, as well as the error:

I intentionally chose delta large enough so the errors are visible. If you chose delta say 0.01, the plots will be visually identical. Of course, the price of taking smaller delta is the need to produce and solve larger matrices.

For kernels that can be obtained analytically, the main bottleneck will be in the LinearSolve, but in practice it is pretty fast (for matrices not too large). When kernels can not be integrated analytically, the main bottleneck will be in computing the kernel in many points (matrix creation. The matrix inverse has a larger asymptotic complexity, but this will start play a role for really large matrices - which are not necessary in this approach, since it can be combined with an iterative one - see below). You will typically define:

intK[x_?NumericQ, x1_?NumericQ] := NIntegrate[K[y, x1], {y, 1, x}]
intIntK[x_?NumericQ, x1_?NumericQ] := NIntegrate[intK[z, x1], {z, 0, x}]

As a way to speed it up in such cases, you can precompute the kernel intK on a grid and then interpolate, and the same for intIntK. This will however introduce additional errors, which you'll have to estimate (account for).

The grid itself needs not be equidistant (I just used it for simplicity), but may (and probably should) be adaptive, and generally non-uniform.

As a final illustration, consider an equation with a non-trivial but symbolically integrable kernel:

In[146]:= sinkern =  computeDoubleIntK[50*Sin[Pi/2*(#1-#2)]&]

Out[146]= (100 (2 Sin[1/2 \[Pi] (-#1+#2)]+Sin[(\[Pi] #2)/2] 
     (-2+\[Pi] #1)))/\[Pi]^2&

In[157]:= solSin = computeSolution[sinkern,0,1,0.01,{0,1}]
Out[157]= InterpolatingFunction[{{0.,1.}},<>]

Here are some checks:

In[163]:= Chop[{solSin[0],solSin'[1]}]
Out[163]= {0,1.}

In[153]:= 
diff[x_?NumericQ]:=
  solSin''[x] - NIntegrate[50*Sin[Pi/2*(#1-#2)]&[x,x1]*solSin[x1],{x1,0,1}];

In[162]:= diff/@Range[0,1,0.1]

Out[162]=  {-0.0675775,-0.0654974,-0.0632056,-0.0593575,-0.0540479,-0.0474074,
   -0.0395995,-0.0308166,-0.0212749,-0.0112093,0.000369261}

To conclude, I just want to stress that one has to perform a careful error - estimation analysis for this method, which I did not do.

EDIT

You can also use this method to get the initial approximate solution, and then iteratively improve it using FixedPoint or other means - in this way you will have a relatively fast convergence and will be able to reach the required precision without the need to construct and solve huge matrices.

Answer 2

This is complementary to Leonid Shifrin's approach. We start with a linear function that interpolates the value and first derivative at the starting point. We use that in the integration with the given kernel function. We can then iterate, using each previous approximation in the integrated kernel that is used to make the next approximation.

I show an example below, using a more complicated kernel than just a constant function. I'll take it through two iterations and show tables of discrepancies.

kernel[x_, y_] := Sqrt[x]/(y^2 + 1/5)*Sin[x^2 + y]
intkern[x_?NumericQ, aa_] := 
 NIntegrate[kernel[x, y]*aa[y], {y, 0, 1}, MinRecursion -> 2, 
  AccuracyGoal -> 3]

Clear[a];
a0 = 0;
a1 = 1;
a[0][x_] := a0 + a1*x

soln1 = a[1][x] /. 
   First[NDSolve[{(a[1]^\[Prime]\[Prime])[x] == intkern[x, a[0], y], 
      a[1][0] == a0, a[1][1] == a1}, a[1][x], {x, 0, 1}]];
a[1][x_] = soln1;

In[283]:= Table[a[1]''[x] - intkern[x, a[1]], {x, 0., 1, .1}]

Out[283]= {4.336808689942018*10^-19, 0.01145100326794241, \
0.01721655945379122, 0.02313249302884235, 0.02990900241909161, \
0.03778448183557359, 0.04676409320217928, 0.05657128568058478, \
0.06665818935524814, 0.07624149919589895, 0.08412643746245929}

In[285]:= 
soln2 = a[2][x] /. 
   First[NDSolve[{(a[2]^\[Prime]\[Prime])[x] == intkern[x, a[1]], 
      a[2][0] == a0, a[2][1] == a1}, a[2][x], {x, 0, 1}]];
a[2][x_] = soln2;

In[287]:= Table[a[2]''[x] - intkern[x, a[2]], {x, 0., 1, .1}]

Out[287]= {-2.168404344971009*10^-19, -0.001009606971360516, \
-0.00152476679745811, -0.002045817184941901, -0.002645356229312557, \
-0.003343218015068372, -0.004121109614310836, -0.004977453722712966, \
-0.005846840469889258, -0.006731367269472544, -0.007404971586975062}

So we have errors of less than .01 at this stage. Not too bad. One drawback is that it was fairly slow to get the second approximation. There may be ways to tune NDSolve to improve on that.

This is complementary to Leonid's method for two reasons.

(1) If this did not converge well because the initial linear approximation was not sufficiently close to the true result, one might instead begin with an approximation found by a finite differencing scheme. That would be akin to what he did.

(2) He pretty much indicated this himself, as a method that might follow his and produce refinements.

Daniel Lichtblau