I'm trying to solve the following integral equation:

Integrate[-k/2 Exp[-k Abs[x-z]] f[x], {x,0,d}] == a f[z]

Differentiating twice with respect to z should yield the following differential equation if i'm not mistaken (using Leibniz integral rule):

a f[z] == Integrate[-k/2 Exp[k (x-z)] f[x], {x,0,z}] + 
    Integrate[-k/2 Exp[-k (x-z)] f[x], {x,z,d}]

a f'[z] == (-k/2 f[z] + Integrate[k^2/2 Exp[k (x-z)] f[x], {x,0,z}]) + 
    (k/2 f[z] + Integrate[-k^2/2 Exp[-k (x-z)] f[x], {x,z,d}]

a f''[z] == (k^2/2 f[z] + Integrate[-k^3/2 Exp[k (x-z)] f[x], {x,0,z}] +     
    (k^2/2 f[z] + Integrate[-k^3/2 Exp[-k (x-z)] f[x], {x,z,d}]

a*f''[z] == k^2 (a+1) f[z]


f[z] = C[1] Exp[Sqrt[1+1/a] k z] + C[2] Exp[-Sqrt[1+1/a] k z]

However, when I specify this function as the input, i.e.

f[z_]:= Exp[Sqrt[1+1/a] k z] + Exp[-Sqrt[1+1/a] k z]

Integrate[-k/2 Exp[-k Abs[x-z]] f[x], {x,0,d}, Assumptions->{z \[Element] Reals, 
    z > 0, z < d, k \[Element] Reals, k > 0, d \[Element] Reals} ]

I dont get a*f[z] back, nor anything similar. Is there anything I am missing?

  • $\begingroup$ This is not really an integral equation. The upper limit of the integral has to be z in your example, which is the independent variable in f(z) for it to be called integral equation? How did you handle the derivative of the absolute term there? $\endgroup$
    – Nasser
    Commented May 27, 2015 at 19:56
  • $\begingroup$ the z limit is there implicitly, as the absolute value of x-z appears in the integrand. I split the integral in two parts (between 0 and z, and z and d). $\endgroup$
    – Noel
    Commented May 27, 2015 at 20:11
  • $\begingroup$ I fail to see how differentiating the integral twice should yield k^2 (a+1) f[z]. Could you show how you derived that? $\endgroup$ Commented May 27, 2015 at 21:41
  • $\begingroup$ @SjoerdC.deVries I edited the question to show the derivation steps $\endgroup$
    – Noel
    Commented May 27, 2015 at 22:13

1 Answer 1


I found the problem. C[1] and C[2] are not arbitrary, but they are determined by the boundary conditions, f[0] and f[d]. Taking this into account the solution is C[1]=C[2]=0, and therefore f[z]==0.

  • $\begingroup$ i was just about to ask how the solution can not depend on d $\endgroup$
    – george2079
    Commented May 28, 2015 at 12:04

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