Solving 2D convection-conduction equation via using Fourier integral transform: the disappearance of the convection term?(with code)

Question

I am currently solving a 2D convection-conduction equation. The convection is only working on the x direction. The governing equation and its associated conditions are given as

where T is the temperature, x and z are the spatial distances, t is the time, and v is the velocity.

Firstly, I applied the Laplace transform method to eliminate t. These equations then become

where T with a bar represents the function in the Laplace domain and s denotes the Laplace parameter.

Later, I used the function called FourierTransform in MMA12 to transform the aforementioned equations to the Laplace-Fourier domain. The results are obtained as

in which T with the double bars represents the function in the Laplace-Fourier domain, a is the Fourier parameter, i is the imaginary unit, and is the dirac delta function.

After applying the DSolve function, I got this result

It can be observed that the convection term is disappeared (i.e., v = 0). Am I missing something here? Or is the convection term not important in this PDE? Thank you for the help.

The associated code I used to obtain the function in Fourier domain is

FourierTransform[T''[x], x, a]

gives

-a^2 FourierTransform[T[x], x, a]

The convection term with applying Fourier transform

FourierTransform[-v*T'[x], x, a]

gives

I a v FourierTransform[T[x], x, a]

and

FourierTransform[1/s, x, a]

gives

(Sqrt[2 \[Pi]] DiracDelta[a])/s

Applying the DSolve

DSolve[{s*T[z] == -a^2*T[z] + I a v*T[z] + T''[z], 
   T[0] == (Sqrt[2 \[Pi]] DiracDelta[a])/s, T[1] == 0}, {T[z]}, 
  z] // Simplify

one gives

T[z] -> (E^(-Sqrt[s] z) (E^(2 Sqrt[s]) - E^(2 Sqrt[s] z)) Sqrt[
  2 \[Pi]] DiracDelta[a])/((-1 + E^(2 Sqrt[s])) s)

Answer 1

Let me extend my comment to an answer. I think the solution is correct. It's the boundary conditions at infinity that filter out influence of v. I'm not that familiar with the theory of integral transforms, but only functions that satisfy certain criterion can be FourierTransformed, which isn't satisfied by solution of PDE involving convective terms in many cases. Somewhat related:

https://math.stackexchange.com/q/2084704/58219

To support my point from another side, let's solve the problem with a slightly different method. First write down the equation:

With[{T = T[t, x, z]}, eq = D[T, t] == D[T, x, x] - v D[T, x] + D[T, z, z];
 ic = T == 0 /. t -> 0;
 bc = {T == 1 /. z -> 0, T == 0 /. z -> 1}]

Then impose Laplace transform just as you've done:

{teq, tbc} = 
 LaplaceTransform[{eq, bc}, t, s] /. Rule @@ ic /. 
  HoldPattern@LaplaceTransform[a_, __] :> a

(* {s*T[t, x, z] == Derivative[0, 0, 2][T][t, x, z] - v*Derivative[0, 1, 0][T][t, x, z] + 
       Derivative[0, 2, 0][T][t, x, z], {T[t, x, 0] == 1/s, T[t, x, 1] == 0}} *)

I've made replacement HoldPattern@LaplaceTransform[a_, __] :> a because LaplaceTransform will cause trouble in subsequent programming. Just keep in mind T[t, x, z] in teq and tbc actually means Laplace transform of $T$.

Next, we use finite Fourier sine transform to eliminate derivative of z

tteq = finiteFourierSinTransform[teq, {z, 0, 1}, n] /. Rule @@@ tbc /. 
  HoldPattern@finiteFourierSinTransform[a_, __] :> a

(*
s*T[t, x, z] == (-n)*Pi*(-(1/s) + n*Pi*T[t, x, z]) - v*Derivative[0, 1, 0][T][t, x, z] + 
     Derivative[0, 2, 0][T][t, x, z]
 *)

Similarly, I've made replacement HoldPattern@finiteFourierSinTransform[a_, __] :> a to avoid trouble in subsequent programming, keep in mind T[t, x, z] is actually finite Fourier sine transform of Laplace transform of $T$ in tteq.

Finally, directly solve tteq with DSolve:

DSolve[tteq, T[t, x, z], x][[1]]
(*
{T[t, x, z] -> -((n π)/((-n^2 π^2 - s) s)) + 
   E^(1/2 (v - Sqrt[4 n^2 π^2 + 4 s + v^2]) x) C[1] + 
   E^(1/2 (v + Sqrt[4 n^2 π^2 + 4 s + v^2]) x) C[2]}
 *)

It's clear that C[…] can only be 0 if the boundary conditions at infinity are satisfied i.e. v doesn't play a role in the solution.