# 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? – Nasser May 27 '15 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 '15 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? – Sjoerd C. de Vries May 27 '15 at 21:41
• @SjoerdC.deVries I edited the question to show the derivation steps – Noel May 27 '15 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 – george2079 May 28 '15 at 12:04