# How to solve the integro - differential equation with complex value involved?

I am trying to solve the PDEs which contains Integro-differential and complex values. For example, the following PDE can be easily solved.

s = NDSolveValue[{D[TeV[t, z], t] == Cos[t*z], TeV[0, z] == 1,
TeV[t, 0] == 1}, TeV, {t, 0, 10}, {z, 0, 10}]
TeV1[t_, z_] := s[t, z]
Plot[{TeV1[0, z], TeV1[5, z], TeV1[10, z]}, {z, 0, 10},
PlotRange -> All]


But my PDE is like this

fun[t_?NumericQ, z_?NumericQ] := NIntegrate[TeV[t, z1], {z1, 0, z}];
fun1[t_, z_] := Exp[fun[t, z]]
s = NDSolveValue[{D[fun1[t, z], t] == Cos[t*z], TeV[0, z] == 1,
TeV[t, 0] == 1}, TeV, {t, 0, 10}, {z, 0, 10}]
TeV1[t_, z_] := s[t, z]
Plot[{TeV1[0, z], TeV1[5, z], TeV1[10, z]}, {z, 0, 10},
PlotRange -> All]


Or this

f[t_, z_] := (1 + I TeV[t, z]^2)^(1/3) // Abs
s = NDSolveValue[{D[f[t, z], t] ==Cos[t*z], TeV[0, z] == 1,
TeV[t, 0] == 1}, TeV, {t, 0, 10}, {z, 0, 10}]
TeV1[t_, z_] := s[t, z]
Plot[{TeV1[0, z], TeV1[5, z], TeV1[10, z]}, {z, 0, 10},
PlotRange -> All]


How to solve the last two equations? Thank you.

All three PDEs in the question have the form,

{D[g[t, z], t] == Cos[t*z], g[0, z] == gz[z]},


where g is a function of TeV, which is the quantity sought. (Note that the second boundary condition in each question, which could be represented as g[t, 0] == gt[t], is unnecessary, because the PDE does not contain derivatives with respect to z. In fact, including this extra boundary condition can cause NDSolve to return an incorrect answer!) The PDE system above can be solved symbolically without difficulty.

s = DSolve[{D[g[t, z], t] == Cos[t*z], g[0, z] == gz[z]},
g[t, z], {t, z}] // Flatten
(* {g[t, z] -> (z gz[z] + Sin[t z])/z} *)


with boundary condition,

bc = s /. t -> 0
(* {g[0, z] -> gz[z]} *)


as expected.

For the first PDE in the question, gz[z] is 1 and g[t, z] is TeV[t,z], so the solution is

{TeV[t, z] -> (z + Sin[t z])/z}
Plot3D[TeV[t, z] /. %, {z, 0, 10}, {t, 0, 10}, PlotRange -> All,
AxesLabel -> {z, t, TeV}, LabelStyle -> {12, Bold, Black}]


For the second PDE, gz[z] is z and TeV[t, z] is D[Log[g[t, z], z]] /. s, so the solution is

{TeV[t, z] -> Log[z]/Log[(z^2 + Sin[t z])/z]


For the third PDE, gz[z] is 2^(1/4} and

Abs[1 + I TeV[t, z]^2] == ((2^(1/4) z + Sin[t z])/z)^2


Thus, TeV[t, z] is underdetermined, because Arg[1 + I TeV[t, z]^2] is unknown. Without additional information there is no unique solution.