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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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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 –  Nasser Nov 10 '13 at 8:20
    
Thanks @Nasser for the trace info. I will contact wolfram to confirm that shortly. –  Guo Nov 10 '13 at 9:19
    
@b.gatessucks Thanks for your reply. But that u is what I am trying to solve. –  Guo Nov 10 '13 at 14:46
    
I see, I had misunderstood your question. –  b.gatessucks Nov 10 '13 at 14:59
    
Thanks for all your generous help. I've just contacted wolfram technical support, and 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]. –  Guo Nov 12 '13 at 0:48
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1 Answer 1

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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