DSolve solution for 1st order PDE involving complex number does not match initial condition

Question

Bug introduced in 11.3 or earlier, persisting through 13.2.1.

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[Mathematica 12.0.0.0, MacOS X x86 (64bit)]

Trying to solve

ClearAll["Global`*"];
DSolveValue[
  D[F[a, t], t] == 2 I a F[a, t] + 2 (a + 2 I a^2) D[F[a, t], a] && 
   F[a, 0] == E^(I a Q), F[a, t], {a, t}] // FullSimplify

returns

(E^((a E^(2 t) Q)/(I - 2 a (1 + E^(2 t)))) Sqrt[(
 1 + 2 I a)/(-I + 2 a (1 + E^(2 t)))])/Sqrt[I - 2 a]

For $t=0$ this evaluates to $$\frac{\sqrt{\frac{1+2 i a}{4a-i}}e^{\frac{a Q}{i-4a}}}{\sqrt{i-2a}}$$

which does not match the initial condition $F[a,0]=e^{i a Q}$.

Any idea where the problem lies?

Answer 1

This interesting question can be answered as follows. First, obtain the solution without an initial condition.

DSolveValue[D[F[a, t], t] == 2 I a F[a, t] + 2 (a + 2 I a^2) D[F[a, t], a], 
     F[a, t], {a, t}]
(* C[1][1/4 (4 t - 2 I ArcTan[2 a] + 2 Log[a] - Log[1 + 4 a^2])]/Sqrt[I - 2 a] *)

Not surprisingly, DSolve has used the Method of Characteristics to obtain this answer. See the Introduction to Symbolic Solutions to PDEs. The solution is given by constant values along characteristics:

cn = Numerator[%][[1]]
(* 1/4 (4 t - 2 I ArcTan[2 a] + 2 Log[a] - Log[1 + 4 a^2]) *)

divided by Sqrt[I - 2 a]. So, to obtain the solution for any {a, t}, trace back along the corresponding characteristic to t = 0, find the value there, and divide it by Sqrt[I - 2 a]. To relate (a, t} to {a0, 0}, solve

Solve[cn == (cn /. t -> 0 /. a -> a0), a0] // FullSimplify // Flatten
(* {a0 -> (a E^t)/((-1 - 4 I a) Cosh[t] + Sinh[t]), 
    a0 -> (a E^t)/(Cosh[t] + (-1 - 4 I a) Sinh[t])} *)

The first expression leads to the incorrect result identified in the question. The second expression, however, leads to the correct result.

sol = Exp[I Q a0] Sqrt[I - 2 a0]/Sqrt[I - 2 a] /. %[[2]]
(* (E^((I a E^t Q)/(Cosh[t] + (-1 - 4 I a) Sinh[t])) 
   Sqrt[I - (2 a E^t)/(Cosh[t] + (-1 - 4 I a) Sinh[t])])/Sqrt[I - 2 a] *)

That this result is correct can be verified by

FullSimplify[Unevaluated[{D[F[a, t], t] == 
    2 I a F[a, t] + 2 (a + 2 I a^2) D[F[a, t], a], F[a, 0] == E^(I a Q)}] 
    /. {F[a, t] -> sol, F[a, 0] -> (sol /. t -> 0)}]
(* {True, True} *)

I suspect that DSolve obtained the incorrect result in the question, because it did not handle branch cuts in the complex plane correctly. By the way, in the course of investigating this problem, I obtained a simpler but equivalent solution,

Sqrt[I - ((2*I)*a*E^(2*t))/(I - 2*a + 2*a*E^(2*t))]/
    E^((a*E^(2*t)*Q)/(I - 2*a + 2*a*E^(2*t)))/Sqrt[I - 2 a]]

The derivation is a bit long.