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Use Fourier Transform to solve the BVP: \begin{cases} u_t + a u_x - b u_{xx} = 0, & \mbox{for } x \in [-1,1] \\ u(x,0) = f(x) \\ u(x+2,t) = u(x,t) \end{cases}

I solved the problem (attached); however, I do not know how to apply the periodic boundary condition. I know how to do it with separation of variable, but not for any Integral Transform method. Please help. Thank you. enter image description here enter image description here

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The way I view the use of the FT using periodic BCs is that the problem is really about expressing the solution in terms of Fourier series with the "transform" being the Fourier coefficients.

The periodic BC's imply that

$$u(x,t) = \sum_{n=-\infty}^{\infty} c_n(t) e^{i n \pi x} $$

where

$$f(x) = \sum_{n=-\infty}^{\infty} c_n(0) e^{i n \pi x} $$

Thus,

$$c_n(t) = \frac12 \int_{-1}^1 dx \, u(x,t) \, e^{-i n \pi x} $$

Now,

$$u_t + a u_x - b u_{xx} = 0 $$

implies that

$$c_n'(t) + (i n \pi a +n^2 \pi^2 b) c_n(t) = 0 $$

or

$$c_n(t) = c_n(0) e^{-i n \pi a t} e^{-n^2 \pi^2 b t} $$

where

$$c_n(0) = \frac12 \int_{-1}^1 dx \,f(x) e^{-i n \pi x}$$

The solution to the equation is therefore

$$u(x,t) = \sum_{n=-\infty}^{\infty} c_n(0) e^{i n \pi (x-a t)} e^{-n^2 \pi^2 b t}$$

If $f(x) \in \mathbb{R}$, then $c_{-n}(0) = c_n(0)^*$ and

$$u(x,t) = c_0(0) + 2 \sum_{n=1}^{\infty} \left [\operatorname{Re}{[c_n(0)]} \cos{[n \pi (x-a t)]}+\operatorname{Im}{[c_n(0)]} \sin{[n \pi (x-a t)]}\right ] e^{-n^2 \pi^2 b t}$$

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