Fourier transform to solve PDE (2nd order)

  • #1
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Homework Statement:
Find the transformed solution to the 2nd order PDE uxx + uxt + utt = 0
Relevant Equations:
Fourier transform equation
I just want to make sure I am on the right track here (hence have not given the other information in the question). In taking the Fourier transform of the PDE above, I get:
F{uxx} = iω^2*F{u},
F{uxt} = d/dt F{ux} = iω d/dt F{u}
F{utt} = d^2/dt^2 F{u}
Together the transformed PDE gives a second order ODE which is: iω^2*F{u} + iω d/dt F{u} + d^2/dt^2 F{u} = 0.
Are these transformations correct??
Thanks!
 

Answers and Replies

  • #2
dRic2
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Don't seem correct to me. First of all: do you want to Fourier-transform for the space coordinate ##x##, or the time coordinate ##t##? Or both?

By the way, it is not difficult to work out the correct transformation starting from the definition of Fourirer transform. Starting with the spatial component:

By definition
$$u(x, t) = \int_{R} \frac {dk} {2 \pi} e^{ikx}F(k, t)$$
So
$$\partial_x u(x, t) = \partial_x \int_{R} \frac {dk} {2 \pi} e^{ikx}F(k, t)$$
Hoping that your function is not a pathological case you can bring the derivative inside the integral since it act only on ##x## and you find:
$$\partial_x u(x, t) = \int_{R} \frac {dk} {2 \pi} \partial_x(e^{ikx}F(k, t)) = ik \int_{R} \frac {dk} {2 \pi} e^{ikx}F(k, t)$$
Thus you find the 'rule'
$$\partial_x u(x, t) \rightarrow ik F(k, t)$$

Starting from this example can you work out the 'rules' for ##\partial_{xx}##, ##\partial_{xt}## and ##\partial_{tt}##?

Ps: I think it is common practice to define the Fourier transform for the time domain with a negative sign, that is with ##e^{-i \omega t}##. It doesn't really matter as long as you stick with your convention (https://physics.stackexchange.com/questions/308234/fourier-transform-standard-practice-for-physics)
 
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  • #3
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Don't seem correct to me. First of all: do you want to Fourier-transform for the space coordinate ##x##, or the time coordinate ##t##? Or both?

By the way, it is not difficult to work out the correct transformation starting from the definition of Fourirer transform. Starting with the spatial component:

By definition
$$u(x, t) = \int_{R} \frac {dk} {2 \pi} e^{ikx}F(k, t)$$
So
$$\partial_x u(x, t) = \partial_x \int_{R} \frac {dk} {2 \pi} e^{ikx}F(k, t)$$
Hoping that your function is not a pathological case you can bring the derivative inside the integral since it act only on ##x## and you find:
$$\partial_x u(x, t) = \int_{R} \frac {dk} {2 \pi} \partial_x(e^{ikx}F(k, t)) = ik \int_{R} \frac {dk} {2 \pi} e^{ikx}F(k, t)$$
Thus you find the 'rule'
$$\partial_x u(x, t) \rightarrow ik F(k, t)$$

Starting from this example can you work out the 'rules' for ##\partial_{xx}##, ##\partial_{xt}## and ##\partial_{tt}##?

Ps: I think it is common practice to define the Fourier transform for the time domain with a negative sign, that is with ##e^{-i \omega t}##. It doesn't really matter as long as you stick with your convention (https://physics.stackexchange.com/questions/308234/fourier-transform-standard-practice-for-physics)
I am wanting to transform the space coordinate.

Where you said by definition u(x, t) = ...., what is that definition you are referring to? The Fourier transform? Because the way I have learned Fourier transforms is the following:
The Fourier transform of f(t) is the integral over R of f(t)e^-iωt dt.

Also, taking the Fourier transform of the n-th derivative = (iω)^2•F{f(t)}. In terms of partial derivatives, this holds when you are taking the Fourier transform of the derivative with respect to the variable being integrated w.r.t (if that makes sense? Not sure how to word it correctly).

As for the Fourier transform of the other variable, the derivative operator w.r.t the other variable comes out n times.

Using this I deduced by initial result. What have I gotten wrong or mixed up?
Thanks.
 
  • #4
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My result seems to be right to me, when you solve via definition of Fourier transform you get F{uxx} + F{uxt} + F{utt} = 0
=> Ʉtt + iωɄt - ω^2Ʉ = 0 (2nd order ODE) (Hence PDE has reduced to ODE) where F{u} = Ʉ.
Ʉtt and Ʉt are the second and first order derivatives with respect to t (respectively).
 
  • #5
dRic2
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Also, taking the Fourier transform of the n-th derivative = (iω)^2•F{f(t)}
Yes, but you initially wrote
iω^2*F{u},
which is different! and got me confused.

Also, when you define the Fourier transform for the space coordinate, you usually use the "wave vector" ##k## instead of the frequency ##\omega##. That was also confusing to me.

So now that you clarified some points, yes you're good.

ps:
Where you said by definition u(x, t) = ...., what is that definition you are referring to? The Fourier transform? Because the way I have learned Fourier transforms is the following:
The Fourier transform of f(t) is the integral over R of f(t)e^-iωt dt.
I used the inverse-fourier transform:

If
$$F(k, t) = \int_{R} dx e^{-ikx}u(x, t)$$
then, by inverse-fourier transform, $u(x, t)$ is given by the equation I wrote.
 
  • #6
Charles Link
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I also think the ## \omega ##'s need to be ## k ##'s. The original differential equation is dimensionally incorrect/inconsistent, and that may be the reason why you might think it needs to be an ## \omega ##. IMO ## \omega ## is incorrect.
 
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  • #7
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I also think the ## \omega ##'s need to be ## k ##'s. The original differential equation is dimensionally incorrect/inconsistent, and that may be the reason why you might think it needs to be an ## \omega ##. IMO ## \omega ## is incorrect.
In my textbook I have the Fourier transform defined as follows (for f(t)):
Screen Shot 2021-05-30 at 12.39.23 PM.png

The question asks me to find an expression for U(ω, t), where U(ω, t) is the transformed equation (probably an ODE).
Why would using omega be wrong? Isn't it just a dummy variable anyway? we have the transformed space as frequency (or with ω, angular frequency).
 
  • #8
Charles Link
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Usually, the coordinate x gets transformed into k space. The approach they are using seems to be rather clumsy.
 
  • #9
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Usually, the coordinate x gets transformed into k space. The approach they are using seems to be rather clumsy.
But in this case they are asking for it to be transformed into the omega space, they ask for an expression for U(ω, t) where F{u} = U.
In this case, do you think my expression for the resulting ODE (after taking Fourier transforms) is correct?
 
  • #10
Charles Link
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You did the best you could do with it. I question whether the textbook is first-rate.
 
  • #12
Charles Link
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The one-dimensional function in time gets transformed to the frequency domain. When the function is both space and time, the spatial part normally gets transformed into k-space. There is a simple relation ## \omega=c k ##, with ## c ## being the propagation velocity. Perhaps I shouldn't question the textbook, but over the years I have encountered some good books, and some that have room for improvement. When the question began with a differential equation whose dimensions were inconsistent, I do think they could do better there. Some advanced theoreticians use a format where the speed of light is unity, but I think they need to teach engineering with consistent units.
 

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