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Good Day, I am struggling to connect 2 parts of the lecture. Consider the Eigenvalue Problem with periodic conditions $$y''+ky = 0, y(0) = y(1), y'(0) = y'(1) $$

$$\text{Solution is in the form:} \\ y =A\sin(\sqrt{k}x) + B\cos(\sqrt{k}x) \\ y' =A\sqrt{k}\cos(\sqrt{k}x) - B\sin(\sqrt{k}x) \\\text{from the first condition, we have } \\ A\sin\sqrt(k) - B(1-\cos\sqrt(k)) = 0 \\ \sin(\frac{\sqrt{k}}{2})(A\cos(\frac{\sqrt{k}}{2}) - B\sin(\frac{\sqrt{k}}{2})) = 0 \\ \text{from the second condition} \\ \sin(\frac{\sqrt{k}}{2})(A\cos(\frac{\sqrt{k}}{2}) + B\sin(\frac{\sqrt{k}}{2})) = 0 \\ \text {I then get} \\ \sin(\frac{\sqrt{k}}{2}) = 0, \frac{\sqrt{k}}{2} = n\pi => k = 4n^2pi^2 => \sin(\sqrt{k}x) = \sin(2n\pi x)$$

I solved this problem and got the following formula for the eigenvalue and eigenfunctions $$ k = 4n^2\pi^2 \\\sin(2n\pi x),cos(2n\pi x)$$ But the internet says that it is $n\pi x$ instead of $2n \pi x$, is there a difference and my asnwer is just wrong? for example, I want to use the inner product for the general Fourier series for these eigenfunctions and $f(x) = x$. Will there be a difference in the coefficients?

SuperMage1
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  • Please, provide the calculations, so that we can see, what and where goes wrong – Daigaku no Baku Jan 17 '24 at 13:11
  • I edited it, but I am little confident in the calculations and think its a matter of convention? I am not sure about the eigenfunction I should work with in the first place – SuperMage1 Jan 17 '24 at 13:32
  • Could you be more specific with respect to your source? "The internet" can be anyone. – Gonçalo Jan 17 '24 at 13:33
  • https://math.stackexchange.com/questions/2854069/eigenvalue-problem-with-periodic-boundary-conditions-are-these-solutions-consis. https://math.libretexts.org/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/11%3A_Boundary_Value_Problems_and_Fourier_Expansions/11.01%3A_Eigenvalue_Problems_for_y''__y__0

    These are some sources that I have found.

     – SuperMage1 Jan 17 '24 at 13:38
  • If $n$ is odd, $\cos(n\pi x)$ and $\sin(n\pi x)$ satisfy one of the boundary conditions, but not both. – Gonçalo Jan 17 '24 at 13:41
  • Your first source is wrong, and the second considers different boundary conditions $(y(-L)=y(L), y'(-L)=y'(L))$. – Gonçalo Jan 17 '24 at 13:50
  • Your result is correct. – Gonçalo Jan 17 '24 at 13:51

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