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[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?
z
in your example, which is the independent variable inf(z)
for it to be called integral equation? How did you handle the derivative of the absolute term there? $\endgroup$x-z
appears in the integrand. I split the integral in two parts (between 0 and z, and z and d). $\endgroup$