Fourier transform to solve diff equation

In summary, the student is trying to find the solution of a differential equation but is stuck because the equation is incorrect. They use the Fourier transform to find the solution, but the equation is incorrect. They are then able to solve the equation by using separation of variables.
  • #1
dankaroll
13
0

Homework Statement



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.

Homework Equations



Using differentiation properties of Fourier transform,

[tex]\frac{\mathrm{d^n}y }{\mathrm{d} x^n} = {(ik)^n }Y[f][/tex]

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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  • #2
dankaroll said:
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.
 
  • #3
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
Klockan3 said:
This equation is wrong, look it up.

[tex]
F\left[ \frac{\partial f}{\partial x} \right] = ikF[f]
[/tex]

http://www.thefouriertransform.com/transform/properties.php#derivative

where 2*f*pi = k

vela said:
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...
 
  • #5
How are you defining the Fourier transform?
 
  • #6
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
Hint: What's F'(y) equal to?
 
  • #8
vela said:
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].
 
  • #9
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
vela said:
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.
 
  • #11
Yes, that's right.
 
  • #12
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
I got the opposite sign in the exponent, but there's still the problem when k goes to infinity.
 
  • #14
vela said:
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...
 
  • #15
Found a sign error. Now my answer matches yours.
 

What is a Fourier transform?

A Fourier transform is a mathematical tool used to decompose a function into its individual frequency components. It allows us to represent a function as a sum of sine and cosine waves of different frequencies.

How is a Fourier transform used to solve differential equations?

By taking the Fourier transform of a differential equation, we can convert it into an algebraic equation in the frequency domain. This makes it easier to solve, as we can use standard algebraic techniques to find the solution.

What are the benefits of using a Fourier transform to solve differential equations?

The main benefit of using a Fourier transform is that it simplifies the process of solving differential equations. It allows us to solve complex equations that may not have an analytical solution in the time domain, and it also provides a way to visualize the frequency components of a function.

Are there any limitations to using a Fourier transform to solve differential equations?

While a Fourier transform is a powerful tool, it is not suitable for all types of differential equations. It is most effective for linear, time-invariant equations with constant coefficients. It may also introduce errors if the function being transformed has discontinuities or is not periodic.

Can a Fourier transform be used to solve partial differential equations?

Yes, a Fourier transform can also be applied to solve partial differential equations. It is commonly used in fields such as fluid dynamics and heat transfer to convert the equations into simpler forms that can be solved using standard techniques.

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