How to 'shift' Fourier series to match the initial condition of this PDE?

In summary, the conversation discusses using a known Fourier series to find a solution without integration. The suggested method includes reflecting the series about the x-axis, shifting and scaling it to have a full period range from -L to L, and using an affine transformation. The speaker also asks for clarification on how to handle a phase shift in the original wave described by the series.
  • #1
Master1022
611
117
Homework Statement
How can I match the following Fourier series to match the initial condition for a PDE?
Relevant Equations
Fourier Series
Hi,

Question: If we have an initial condition, valid for [itex] -L \leq x \leq L [/itex]:
[tex] f(x) = \frac{40x}{L} [/tex] how can I utilise a know Fourier series to get to the solution without doing the integration (I know the integral isn't tricky, but still this method might help out in other situations)?

We are given the following Fourier series:
[tex] \frac{2A}{\pi} \left( sin(\phi) + \frac{1}{2} sin(2 \phi) + \frac{1}{3} sin(3\phi) + ... \right) [/tex] where [itex] \phi = \frac{2 \pi x}{T} [/itex] and [itex] A [/itex] is the amplitude. This represents a function that looks like this (this would be sharper with more terms, but I just put a few into Desmos):
Screen Shot 2020-08-24 at 9.08.11 PM.png


My attempt:
I think we want to do the following steps:
1. Reflect the series about the x-axis (multiply by -1)
2. Shift/scale the series have a zero at the origin and have a full period range from [itex] -L [/itex] to [itex] L [/itex]

It is the second step that confuses me and I am not completely sure how to do it, but this is how I approached it:
- we want the point where [itex] \pi [/itex] is to shift to the origin (so [itex] \phi + \pi [/itex])
then we get that [itex] sin(\phi + \pi) = -sin(\phi) [/itex].
- then we let T = 2L to get and A = 40 to get:
[tex] \frac{-2\times 40}{\pi} \left(-sin(\frac{\pi x}{L}) + \frac{1}{2} sin(\frac{2 \pi x}{L}) - \frac{1}{3} sin(\frac{3 \pi x}{L}) + ... +\frac{(-1)^n}{n}sin(\frac{n \pi x}{L}) \right) [/tex]

Does this method seem correct?

However, what if there is a shift and scale required? How does that change the order of the transformation? Should we always do the shift first, for out the trig double angle simplification, and then do the scaling in the x-direction?

Any help is greatly appreciated.
 
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  • #2
If you need an affine transformation, then you should do that in a single step: [tex]
[a,b] \to [0, 2\pi]: x \mapsto \frac{2\pi}{b - a}(x - a).[/tex] Thus here [tex]\phi = \frac{\pi}L (x + L), \qquad x = -L + \frac{L}{\pi} \phi[/tex] and [tex]
\frac{40x}{L} = -40 + \frac{40}{\pi} \phi.[/tex]
 
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  • #3
pasmith said:
If you need an affine transformation, then you should do that in a single step: [tex]
[a,b] \to [0, 2\pi]: x \mapsto \frac{2\pi}{b - a}(x - a).[/tex] Thus here [tex]\phi = \frac{\pi}L (x + L), \qquad x = -L + \frac{L}{\pi} \phi[/tex] and [tex]
\frac{40x}{L} = -40 + \frac{40}{\pi} \phi.[/tex]

Thank you for your reply! Okay, this method makes sense. However, why did you substitute in the expression for [itex] x [/itex] into the initial condition? I suppose that might be feasible if the [itex] \phi [/itex] was defined within the problem?

I think for this case, I would just use the expression [itex]\phi = \frac{\pi}L (x + L) [/itex] and substitute into the Fourier series to have a match to my initial condition? I think you know that though, I just wanted to confirm. By substituting in your expression for [itex] \phi [/itex] to the Fourier series, it yields the same trig identity so that is very re-assuring.

Thanks
 
  • #4
pasmith said:
If you need an affine transformation, then you should do that in a single step: [tex]
[a,b] \to [0, 2\pi]: x \mapsto \frac{2\pi}{b - a}(x - a).[/tex] Thus here [tex]\phi = \frac{\pi}L (x + L), \qquad x = -L + \frac{L}{\pi} \phi[/tex] and [tex]
\frac{40x}{L} = -40 + \frac{40}{\pi} \phi.[/tex]
Sorry to bring this back up, but what if the original wave described by the Fourier series was shifted by ## \pi ## (or any other phase)? Would I then:
1) use a phase shift ## \phi = \phi + \pi ## to shift the Fourier series wave to where I want
2) expand with double angle formula and simplify
3) use the affine transformation to map from ## \phi ## to ## x ##

Would that process be correct?

Thank you
 
  • #5
Master1022 said:
Thank you for your reply! Okay, this method makes sense. However, why did you substitute in the expression for [itex] x [/itex] into the initial condition? I suppose that might be feasible if the [itex] \phi [/itex] was defined within the problem?

It's the natural thing to do: To evaluate [tex]\int_a^b f(x) \sin\left(\frac{2n\pi(x - a)}{b - a}\right)\,dx = \frac{b - a}{2\pi}\int_0^{2\pi} f\left(a + \frac{(b-a)}{2\pi}\phi\right) \sin(n\phi)\,d\phi[/tex] one has to. Also, in this special case of a linear boundary condition, it is obvious from [tex]
\frac{40x}{L} = -40 + \frac{40}{\pi}\phi[/tex] that the initial condition is -40 plus [itex](40/\pi)[/itex] times the unshifted series for [itex]\phi[/itex]. This is easier to calculate than the shifted series.
 

1. How do you shift a Fourier series to match the initial condition of a PDE?

The Fourier series can be shifted by adding a constant term to the argument of the trigonometric functions in the series. This constant term is known as the phase shift and it can be calculated by equating the initial condition of the PDE to the Fourier series and solving for the phase shift.

2. What is the purpose of shifting a Fourier series?

The purpose of shifting a Fourier series is to match the initial condition of a PDE. This ensures that the solution to the PDE satisfies the given initial condition, making it a more accurate representation of the physical system being studied.

3. Can a Fourier series be shifted by any amount?

No, a Fourier series can only be shifted by integer multiples of the period of the function. This is because the Fourier series is periodic and any shift by a non-integer multiple of the period would result in a different function.

4. How does shifting a Fourier series affect the coefficients?

Shifting a Fourier series does not affect the coefficients of the series. The coefficients remain the same, only the argument of the trigonometric functions is changed by the addition of the phase shift.

5. Are there any limitations to shifting a Fourier series?

Yes, there are some limitations to shifting a Fourier series. The initial condition must be a smooth function and the Fourier series must converge to the initial condition. Additionally, the PDE must be linear and the boundary conditions must also be satisfied for the shifted Fourier series to be a valid solution.

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