# Integral equation

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]


Or

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

• 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? May 27, 2015 at 19:56
• 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).
– Noel
May 27, 2015 at 20:11
• I fail to see how differentiating the integral twice should yield k^2 (a+1) f[z]. Could you show how you derived that? May 27, 2015 at 21:41
• @SjoerdC.deVries I edited the question to show the derivation steps
– Noel
May 27, 2015 at 22:13

I found the problem. C and C are not arbitrary, but they are determined by the boundary conditions, f and f[d]. Taking this into account the solution is C=C=0, and therefore f[z]==0.
• i was just about to ask how the solution can not depend on d May 28, 2015 at 12:04