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

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 𝛿 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 == (Sqrt[2 \[Pi]] DiracDelta[a])/s, T == 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)

• Can you show your code to obtain these? LaTeX is good, but if we can see the code you used it is really helpful in helping us help you :) which is the convection term originally also? If it is time dependent that may be your answer, as you eliminated t—but if it is a result of DSolve, then one could surmise that it disappeared due to it being dx/dt? I’m not as learned on this as I should be, admittedly. Aug 21 '19 at 3:43
• @CATrevillian hi, I added the code I used to obtain the result of the ODE in the Laplace domain. The convection term is the function multiplied by v. It seems that the use of the Laplace transform does not lead to such a result. The disappearance of v is confusing me. Aug 21 '19 at 3:56
• s is signal yeah? Is that not like the velocity? But also what about your T not having the other terms in it? Aug 21 '19 at 4:06
• @CATrevillian Actually, s in the model is a Laplace parameter which can be found after applying the Laplace transform. Also, the convection term is -v dT/dx in the equation. T=0 means the temperature in the investigated domain is zero initially. Aug 21 '19 at 4:26
• Thank you for the clarifications, in your final step, what you input into DSolve, the T(x=0) term does not include all of the items you list prior, could this be the cause? Aug 21 '19 at 4:30

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][]
(*
{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 +
E^(1/2 (v + Sqrt[4 n^2 π^2 + 4 s + v^2]) x) C}
*)


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.

• This is the result of incompatible boundary conditions and initial data. It is necessary to solve the problem using FEM and see what happens when the scale of the region increases along x. Aug 21 '19 at 13:02
• @Alex According to my (limited) experience, integral transform should be able to find the solution in the sense of limitation when b.c.s are inconsistent, here is an example: mathematica.stackexchange.com/q/127081/1871 Aug 21 '19 at 14:11
• After applying the Fourier sine transform to dT/dx, it will become the function in the Fourier cosine domain. Can we solve the ODE consisting of the function of T in the Fourier since and cosines domains simultaneously? Aug 27 '19 at 7:46
• @LingLong I'm not aware of such technique. Also, notice imposing Fourier sine transform is amount to implicitly setting b.c. at infinity i.e. the system will be over-determined if you use it in $t$ direction. Aug 27 '19 at 8:30