Differentiate Fourier Sine Series

[Mindscrape]

The problem is that if f(x) is continuous function, except for a jump discontinuity at $$x = x_0$$, where $$f(x_0^-) = \alpha$$ and $$f(x_0^+) = \beta$$, and df/dx is piecewise smooth, determine the Fourier cosine series of df/dx in terms of the Fourier sine series coefficients of f(x).

Let me preface this by saying that I split f(x) into f1 from 0 to x0 and f2 from x0 to L. So that would mean, which might not necessarily be true, that f'(x) would f1' from 0 to x0 and f2' from x0 to L. Is this alright so far? I could see the derivative gaining extra discontinuities that I am not accounting for properly, in which case I don't know how to account for them. But if it is okay, then read on.

Start off by saying that f'(x) is a FCS

$$f'(x) = \hat{A}_0 + \sum_1^\infty \hat{A}_n cos \frac{n \pi x}{L}$$

*I put hats on just to clarify these are not regular FCS coefficients. So that would mean that

$$\hat{A}_0 = \frac{1}{L}[ \int_0^{x_0} f_1' dx + \int_{x_0}^L f_2' dx] = \frac{1}{L} [(f_1(x_0) - f(0)) + (f_2(L) - f_2(x_0)]$$

so that would end up being

$$\hat{A}_0 = \frac{1}{L} [(\alpha - f(0)) + (f_2(L) - \beta)]$$

Then do a similar treatment for other coefficient

$$\hat{A}_n = \frac{2}{L} \int_0^L f'(x) cos\frac{n \pi x}{L} dx$$

Here is another part I am uncertain about, but I'm pretty sure it is okay. Integration by parts can only be used on continuous functions with continuous derivatives, but I should be alright using IBP on f1' from 0 to x0 and f2' from x0 to L. So, use integration by parts (u = cosnπx/L dv = f'(x)) on each of the functions to get

$$\hat{A}_n = \frac{2}{L} [ (\alpha cos\frac{n\pi\alpha}{L} - f_1(0)) + (f_2(L) - \beta cos\frac{n\pi\beta}{L}) + \int_0^{x_0} \frac{n\pi}{L} sin\frac{n\pi x}{L} f_1(x) + \int_{x_0}^L \frac{n\pi}{L} sin\frac{n\pi x}{L} f_2(x)]$$

Quick detour to the FSS coefficient of f(x)

$$f(x) = \sum_0^\infty B_n sin\frac{n\pi x}{L}$$

which means that

$$B_n = \frac{L}{2}[\int_0^{x_0} {L} sin\frac{n\pi x}{L} f_1(x) + \int_{x_0}^L sin\frac{n\pi x}{L} f_2(x)]$$

Substitute B_n into A_n and it should be

$$\hat{A}_n = \frac{2}{L} [ (\alpha cos\frac{n\pi\alpha}{L} - f_1(0)) + (f_2(L) - \beta cos\frac{n\pi\beta}{L})] + \frac{n\pi}{L} B_n$$

Would this be correct?

 

[mighty2000]

Thank you for your question. I appreciate your approach to solving this problem and it seems like you have a good understanding of Fourier series. However, I would like to offer some clarification and potential corrections to your solution.

Firstly, splitting f(x) into f1 and f2 is not necessary. We can still use the same approach without this splitting. Also, there is no need to introduce new coefficients (with hats) for the Fourier cosine series. We can still use the regular coefficients, but we will have to take into account the discontinuity at x=x0.

To start off, we can write f'(x) as a Fourier cosine series:

f'(x) = \hat{B}_0 + \sum_1^\infty \hat{B}_n cos \frac{n \pi x}{L}

where \hat{B}_n = \frac{2}{L} \int_0^L f'(x) cos\frac{n \pi x}{L} dx.

Now, we can use the fact that f(x) is continuous except for a jump discontinuity at x=x0. This means that f'(x) is also continuous except at x=x0. So, we can write:

\hat{B}_n = \frac{2}{L} \int_0^L f'(x) cos\frac{n \pi x}{L} dx = \frac{2}{L} \int_0^{x_0} f'(x) cos\frac{n \pi x}{L} dx + \frac{2}{L} \int_{x_0}^L f'(x) cos\frac{n \pi x}{L} dx

The first integral on the right-hand side can be evaluated using integration by parts, as you mentioned. However, the second integral cannot be evaluated using integration by parts because f'(x) has a discontinuity at x=x0. Instead, we can use a special case of integration by parts called the Leibniz rule, which is used for functions with jump discontinuities. This rule states that for a function f(x) with a jump discontinuity at x=x0, the integral of f'(x) from a to b is equal to f(b)-f(a) + the sum of the jumps at x=x0.

Applying this to our problem, we get:

\hat{B}_n =