Fourier transform to solve PDE (2nd order)

In summary, 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?
  • #1
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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  • #2
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
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.
 
  • #4
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
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.
 
  • #6
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
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).
 
  • #8
Usually, the coordinate x gets transformed into k space. The approach they are using seems to be rather clumsy.
 
  • #9
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?
 
  • #10
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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FAQ: Fourier transform to solve PDE (2nd order)

What is a Fourier transform?

A Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It converts a function from its original domain (usually time or space) to a representation in the frequency domain.

How does a Fourier transform help solve PDEs?

A Fourier transform can be used to convert a partial differential equation (PDE) from the time or space domain into the frequency domain. This allows for a simpler and more efficient solution process, as well as providing insight into the behavior of the solution.

What types of PDEs can be solved using a Fourier transform?

A Fourier transform can be used to solve linear, homogeneous PDEs with constant coefficients. This includes second-order PDEs such as the heat equation, wave equation, and Laplace's equation.

Are there any limitations to using a Fourier transform to solve PDEs?

Yes, there are limitations to using a Fourier transform. It can only be applied to PDEs with constant coefficients, and the boundary conditions must be compatible with the transformed equation. Additionally, the solution must be periodic or have a finite domain.

How is the inverse Fourier transform used in the solution process?

The inverse Fourier transform is used to convert the solution from the frequency domain back to the original time or space domain. This allows for the final solution to be expressed in terms of the original variables.

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