Hi Uwe,
for me these specialized notations and input interfaces are too difficult to learn. If your question is concerned with their use and reliability, thats not my field.
I break the question down to the transfomation of the T-function by adding the boundary function sin t:
eq1 = {D[T[t, x], t] == D[T[t, x], x, x], T[0, x] == 0,
Derivative[0, 1][T][t, 1] == 0, T[t, 0] == Sin[8 \[Pi] t]}
eq2 = eq1 /. {T :> ({t, x} |-> (U[t, x] + Sin[8 \[Pi] t]))} //
Simplify
Out:={8*Pi*Cos[8*Pi*t] + Derivative[1, 0][U][t, x] ==
Derivative[0, 2][U][t, x], U[0, x] == 0,
Derivative[0, 1][U][t, 1] == 0, U[t, 0] == 0}
Integrate both equations
T1[t_, x_] =
T[t, x] /. NDSolve[eq1, T[t, x], {t, 0, 16}, {x, 0, 1}][[1]]
T2[t_, x_] =
U[t, x] /. NDSolve[eq2, U[t, x], {t, 0, 16}, {x, 0, 1}][[1]]
The solutions are identical
Manipulate[
Plot[{T1[t, x], T2[t, x] + Sin[8 \[Pi] t]}, {x, 0, 1},
PlotRange -> {-1, 1},
PlotStyle -> {{Red, Thickness[0.02]}, {Black,
Thickness[0.01]}}], {t, 0, 16}]
The second equation has series solution that does not seem to have a closed form. Its the typical fourier sum in x of solutions of the heat equation in a finite interval with the index k solution
( D[#,t] -D[#,x,x]&) Exp[- k^2 t + I k x] == 0
{{U[t, x] ->
Sum[(128 E^(-(1/4) (1 - 2 k)^2 \[Pi]^2 t) ((1 - 2 k)^2 \[Pi] -
E^(1/4 (1 -
2 k)^2 \[Pi]^2 t) ((1 - 2 k)^2 \[Pi] Cos[8 \[Pi] t] +
32 Sin[8 \[Pi] t])) Sin[
1/2 (-1 + 2 k) \[Pi] x])/(1024 (-1 + 2 k) \[Pi] + (-1 +
2 k)^5 \[Pi]^3), {k, 1, \[Infinity]}]