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Heat equation, Fourier cosine transform 
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#19
Feb1312, 07:23 PM

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PF Gold
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Hint:
$$\int_0^k \cos kp\,dk = \frac{\sin kp}{p}$$ 


#20
Feb1312, 08:03 PM

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#21
Feb1512, 01:13 PM

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I edited my hint to make it more suggestive. Use it to evaluate the integral in post #17.



#22
Feb1512, 01:17 PM

PF Gold
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Ok vela, I've been helped on that particular integral (though I still have doubts) by JJacquelin.
I think the main idea was to consider the integrand as a function of k, then differentiate it with respect to k. Then integrate this result (from 0 to infinity) with respect to p (it's slightly less "ugly" than the original integral, though I'm still unsure how to perform that integral but I will ask for help). Then integrate this result with respect to k and take the constant of integration equal to 0. All in all this gives [itex]\int _0 ^{\infty }\frac{\sin (kp)e^{px}}{p}dp=\arctan \left ( \frac{k}{x} \right )[/itex]. Hence [itex]u(x,y)=\frac{1}{\pi} \arctan \left ( \frac{y+a}{x} \right )+\frac{1}{\pi} \arctan \left ( \frac{ay}{x} \right )[/itex]. I'd like to know whether your method is the same. If it's a different, I'd like to try it but I don't see how I can use your hint so far. Edit: I just see that you just replied to my last post. I'll check this out. 


#23
Feb1512, 01:19 PM

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It's slightly different but yields the same result.



#24
Feb1512, 01:32 PM

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[itex]I=\int _0 ^{\infty } \int _0^{k} \cos (kp)dke^{xp}dp=\int _0^k \int _0 ^{\infty } \cos (kp)e^{px}dpdk[/itex]. I'm stuck on that integral with respect to p (integral by parts failed and I don't see any useful substitution so I used Wolfram Alpha for it). [itex]I=\int _0 ^k \frac{e^{xp}[k \sin (kp)x \cos (kp)]}{k^2+x^2} \big  _0 ^{\infty } dk =\int _0 ^k \frac{x}{k^2+x^2}dk=\arctan \left ( \frac{k}{x} \right )[/itex].
Edit: Apparently there are at least 2 ways to tackle down that integral (the one I used wolfram alpha). A double integration by parts or a rewriting of the cos term in function of exponentials. I'll be looking forward this. Thank you very much for all vela. So the answer is post #22 is correct, right? 


#25
Feb1512, 01:52 PM

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#26
Feb2112, 11:14 PM

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Instead of starting a new thread about sine/cosine Fourier transform I'll ask here.
Basically I wonder why the problem statement made it clear to take a cosine Fourier transform instead of just a Fourier transform. Is it because it is obvious that the solution to the PDE, u(x,y) is an even function for any fixed x? So that a common Fourier transform would lead to a cosine Fourier transform anyway. P.S.: Just for fun, I think I've plotted the solution function (see post 9 and maybe 10 of http://www.physicsforums.com/showthread.php?t=577993). 


#27
Feb2212, 05:41 AM

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It may have been simply for pedagogical reasons.



#28
Feb2212, 02:19 PM

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#29
Feb2212, 03:16 PM

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Yes. That's where the Fourier cosine transform comes from, right? You plug the function into the regular Fourier transform integral and then simplify it using the fact the function is even.



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