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Fourier transform to solve diff equation 
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#1
Apr2611, 03:08 PM

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1. The problem statement, all variables and given/known data
Use Fourier transform to find the solution of the following differential equation: [tex]\frac{\mathrm{d^3}y }{\mathrm{d} x^3}+ \lambda \frac{\mathrm{dy} }{\mathrm{d} x}  xy = 0, \lim_{x \to \infty } y(x)=0[/tex] Find the asymptotic of the solution for lambda>> 1. Normalize the solution so y(0) =1. 2. Relevant equations Using differentiation properties of Fourier transform, [tex]\frac{\mathrm{d^n}y }{\mathrm{d} x^n} = {(ik)^n }Y[f][/tex] 3. The attempt at a solution using the property [tex](ik)^3Y[f]+\lambda (ik)Y[f]xY[f] =0 [/tex] [tex]Y[f]((ik)^3+\lambda ikx) = 0 [/tex] so I'm slightly stuck here, since its zero on the right hand side... 


#2
Apr2611, 05:48 PM

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#3
Apr2711, 04:24 AM

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PF Gold
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Note that the Fourier transform of xy isn't xY(k). You can't treat x like a constant.
Also, how is the Fourier transform defined in your class? There are several common conventions in use. It would help to know which one you're using to avoid confusion. 


#4
Apr2711, 11:33 AM

P: 13

Fourier transform to solve diff equation
F\left[ \frac{\partial f}{\partial x} \right] = ikF[f] [/tex] http://www.thefouriertransform.com/t...php#derivative where 2*f*pi = k 


#5
Apr2711, 02:34 PM

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How are you defining the Fourier transform?



#6
Feb2612, 02:31 PM

PF Gold
P: 3,188

Let me try this problem.
Let's say I use the definition of the Fourier transform [itex]f(x)=\frac{1}{\sqrt {2\pi }}\int _{\infty}^{\infty } g(y)e^{ixy} dy[/itex], [itex]g(y)=\frac{1}{\sqrt{2\pi}}\int _{\infty}^{\infty } f(x)e^{ixy}dx[/itex]. I want to take the Fourier transform of the DE with respect to x. So I get that it's worth [itex]ik^3 \mathbb{F}(y)+\lambda \mathbb{F}(y)\mathbb{F}(xy)=0[/itex]. [itex]\mathbb{F}(xy)=\frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}xye^{ixy}dx[/itex]. I'm stuck here. Not sure how to solve this integral. Any idea? 


#7
Feb2612, 02:40 PM

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PF Gold
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Hint: What's F'(y) equal to?



#8
Feb2612, 02:53 PM

PF Gold
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I'm a bit confused because y depends on x. [itex]\mathbb{F}(y')=\frac{1}{\sqrt{2\pi }} \int _{\infty}^{\infty} y'e^{ixy}dx[/itex]. I was thinking about integration by parts but the fact that y depends on x troubles me. I know I'm almost sure I should express this in terms of [itex]\mathbb{F}(y)=\frac{1}{\sqrt{2\pi }} \int _{\infty}^{\infty} ye^{ixy}dx[/itex]. 


#9
Feb2612, 10:11 PM

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I didn't even notice the k's. Since the original problem uses y(x), let's use k to be the variable conjugate to x:
\begin{align*} f(x) &= \frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty} F(k)e^{ikx}\, dk \\ F(k) &= \frac{1}{\sqrt{2\pi}} \int_{\infty}^{\infty} f(x)e^{ikx}\,dx \end{align*} So consider F'(k). 


#10
Feb2612, 10:27 PM

PF Gold
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[itex]F'(k)=\frac{i}{\sqrt {2\pi}} \int _{\infty}^{\infty}xf(x)e^{ikx}dx=i \mathbb{F}(xf(x))[/itex]. Does this mean that if [itex]f(x)=y[/itex], I get that [itex]F'(k)=i \mathbb{F}(xy)[/itex]? I must say the notation of Mathews and Walker's book somehow confuses me, because the variable conjugate of x is y there. 


#11
Feb2712, 03:28 AM

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Yes, that's right.



#12
Feb2712, 11:14 AM

PF Gold
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Ok thank you.
I get: [itex]\mathbb{F}[xf(x)]=\frac{d}{idk}\mathbb{F}[f(x)][/itex]. I take [itex]f(x)=y(x)=y[/itex] (for notation). Thus [itex]\mathbb{F}(xy)=\frac{d}{idk}\mathbb{F}(y)[/itex]. Using this, I tansform the original DE into [itex]ik^3\mathbb{F}(y)+\lambda ik \mathbb{F}(y)+\frac{1}{i}\frac{d\mathbb{F}(y)}{dk}=0[/itex]. I multiply by i and factorize some terms to get [itex]F'(k)+(k^3\lambda k)F(k)=0[/itex]. I used separation of variables to get [tex]F(k)=Ae^{\frac{k^2}{2}\left ( \lambda  \frac{k^2}{2} \right ) }[/tex]. I doubt this is right because the question says to get the solution for large lambda. If that is right, now the next step is to invert that expression to get y(x), right? 


#13
Feb2712, 05:07 PM

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I got the opposite sign in the exponent, but there's still the problem when k goes to infinity.



#14
Feb2712, 05:39 PM

PF Gold
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Hmm and what does this problem mean? That the Fourier transform isn't appropriate or... 


#15
Feb2712, 10:04 PM

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Found a sign error. Now my answer matches yours.



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