Difference between FourierTransform and LaplaceTransform

Question

There is an equation for example：

eqn=D[c[x, t], t] == d D[c[x, t], x, x];

When I make a LaplaceTransform of it:

LaplaceTransform[eqn, t, s]

Then mathematica will impose the transform to both sides of the '=', and the result will be :

s*LaplaceTransform[c[x, t], t, s] - c[x, 0] == d*LaplaceTransform[Derivative[2, 0][c][x, t], t, s]

But when I make a FourierTransform it won't do that, and the output :

FourierTransform[Derivative[0, 1][c][x, t] == d*Derivative[2, 0][c][x, t], t, \[Omega]]

And I just wonder know why. What's the difference？

BTW a workaround can be found here: https://mathematica.stackexchange.com/q/56237/1871 — xzczd, Nov 10 '18 at 12:30

score 2 · Accepted Answer · answered Nov 10 '18 at 07:06

I think the only difference is that some developer told LaplaceTransform to do so while it came to nobody's mind to do the same for FourierTransform. So, no deep mathematics behind that.

If you like, you can add a the following definition to FourierTransform:

Unprotect[FourierTransform];
FourierTransform[eq_Equal, args__] := FourierTransform[#, args] & /@ eq;
Protect[FourierTransform];

Then you obtain

FourierTransform[Derivative[0, 1][c][x, t] == d*Derivative[2, 0][c][x, t], t, ω]

(-I)ωFourierTransform[c[x, t], t, ω] == FourierTransform[d*Derivative[2, 0][c][x, t], t, ω]

Difference between FourierTransform and LaplaceTransform

1 Answers1