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Bug introduced in 9.0 or earlier and fixed in 10.3.0

I am trying to get the Green's function of a toy diffusion equation

$$\frac{\partial^2 u(x,t)}{\partial x^2} = \frac{1}{\alpha^2}\frac{\partial u(x,t)}{\partial t}$$

with Mathematica 9. Then solve it by inverse Fourier transform:

u[x_, t_] := InverseFourierTransform[U[k, t], k, x]
D[u[x, t], {x, 2}]  
D[D[u[x, t], x], x] 

(* 0 *)
(* InverseFourierTransform[-k^2 U[k, t], k, x] *)

But can someone tell me why Out[4] and Out[5] are different? Thanks for your kindness.

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  • $\begingroup$ I'd say it is a bug. You can see from the trace also where the different behavior is: Trace[D[u[x, t], {x, 2}]] vs. Trace[D[D[u[x, t], x], x]] shows this: !Mathematica graphics Mathematically speaking, both should give same result. Maple gives same result as you can see !Mathematica graphics $\endgroup$
    – Nasser
    Nov 10, 2013 at 8:20
  • $\begingroup$ Thanks @Nasser for the trace info. I will contact wolfram to confirm that shortly. $\endgroup$
    – Guo
    Nov 10, 2013 at 9:19
  • $\begingroup$ @b.gatessucks Thanks for your reply. But that u is what I am trying to solve. $\endgroup$
    – Guo
    Nov 10, 2013 at 14:46
  • $\begingroup$ I see, I had misunderstood your question. $\endgroup$
    – b.gates.you.know.what
    Nov 10, 2013 at 14:59
  • $\begingroup$ @Guo would you care to post that as an official answer to your question, so that we can consider this issue closed and mark it as a bug for our site? $\endgroup$
    – Oleksandr R.
    Nov 25, 2013 at 2:45

1 Answer 1

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Thanks for all your generous help. The wolfram technical support has just confirmed the issue originates from the first derivative of inverse Fourier transform. Actually D[u[x, t], x] should output InverseFourierTransform[I k U[k, t], k, x] rather than InverseFourierTransform[-I k U[k, t], k, x].

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