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Fourier transform to solve diff equation

by dankaroll
Tags: diff, equation, fourier, solve, transform
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dankaroll
#1
Apr26-11, 03:08 PM
P: 13
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 ik-x) = 0 [/tex]

so I'm slightly stuck here, since its zero on the right hand side...
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Klockan3
#2
Apr26-11, 05:48 PM
P: 614
Quote Quote by dankaroll View Post
Using differentiation properties of Fourier transform,

[tex]\frac{\mathrm{d^n}y }{\mathrm{d} x^n} = {(ik)^n }Y[f][/tex]
This equation is wrong, look it up.
vela
#3
Apr27-11, 04:24 AM
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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.

dankaroll
#4
Apr27-11, 11:33 AM
P: 13
Fourier transform to solve diff equation

Quote Quote by Klockan3 View Post
This equation is wrong, look it up.
[tex]
F\left[ \frac{\partial f}{\partial x} \right] = ikF[f]
[/tex]

http://www.thefouriertransform.com/t...php#derivative

where 2*f*pi = k

Quote Quote by vela View Post
Note that the Fourier transform of xy isn't xY(k). You can't treat x like a constant.
Looking over some properties, I can't seem to find what this would transform to. I think we can agree that x needs to go on the RHS and y needs to go away somehow...
vela
#5
Apr27-11, 02:34 PM
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How are you defining the Fourier transform?
fluidistic
#6
Feb26-12, 02:31 PM
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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?
vela
#7
Feb26-12, 02:40 PM
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Hint: What's F'(y) equal to?
fluidistic
#8
Feb26-12, 02:53 PM
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Quote Quote by vela View Post
Hint: What's F'(y) equal to?
Hmm I see I made a mistake, there shouldn't be any k's.
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].
vela
#9
Feb26-12, 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).
fluidistic
#10
Feb26-12, 10:27 PM
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Quote Quote by vela View Post
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).
Ok!
[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.
vela
#11
Feb27-12, 03:28 AM
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Yes, that's right.
fluidistic
#12
Feb27-12, 11:14 AM
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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?
vela
#13
Feb27-12, 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.
fluidistic
#14
Feb27-12, 05:39 PM
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Quote Quote by vela View Post
I got the opposite sign in the exponent, but there's still the problem when k goes to infinity.
You mean you have [itex]F(k)=Ae^{\frac{k^2}{2}\left ( \frac{k^2}{2}-\lambda \right ) }[/itex]?
Hmm and what does this problem mean? That the Fourier transform isn't appropriate or...
vela
#15
Feb27-12, 10:04 PM
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Found a sign error. Now my answer matches yours.


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