Fourier transform to solve PDE (2nd order)

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SUMMARY

The discussion centers on the application of the Fourier transform to solve a second-order partial differential equation (PDE). The transformation yields the equation: iω²*F{u} + iω d/dt F{u} + d²/dt² F{u} = 0, which simplifies to a second-order ordinary differential equation (ODE). Participants clarify the correct application of Fourier transforms, emphasizing the use of wave vector k for spatial components and angular frequency ω for temporal components. The conversation highlights the importance of consistent dimensional analysis in mathematical definitions and transformations.

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  • Understanding of Fourier transforms, specifically the definitions for spatial and temporal components.
  • Familiarity with partial differential equations (PDEs) and their transformation into ordinary differential equations (ODEs).
  • Knowledge of complex numbers and their application in differential equations.
  • Basic principles of dimensional analysis in mathematical physics.
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  • Study the definition and properties of the Fourier transform, focusing on both spatial and temporal domains.
  • Learn about the relationship between wave vector k and angular frequency ω in the context of wave equations.
  • Explore dimensional analysis techniques to ensure consistency in mathematical equations.
  • Investigate common practices in transforming PDEs to ODEs using Fourier methods.
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Mathematicians, physicists, and engineers involved in solving partial differential equations, particularly those utilizing Fourier analysis in their work.

Haku
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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!
 
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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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dRic2 said:
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.
 
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).
 
Haku said:
Also, taking the Fourier transform of the n-th derivative = (iω)^2•F{f(t)}
Yes, but you initially wrote
Haku said:
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:
Haku said:
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.
 
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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Charles Link said:
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).
 
Usually, the coordinate x gets transformed into k space. The approach they are using seems to be rather clumsy.
 
Charles Link said:
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?
 
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You did the best you could do with it. I question whether the textbook is first-rate.
 
  • #11
Charles Link said:
You did the best you could do with it. I question whether the textbook is first-rate.
This is a link to the textbook https://www.google.co.nz/books/edit...for_Engineers_and_S/orOTiguKIR4C?hl=en&gbpv=0

We covered this in class also though. Fourier transforms, transform things in one domain to the frequency domain. So why would omega (angular frequency) be a subpar representation of how a Fourier transform is defined?
 
  • #12
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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