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Let's define function $u(x,t)$ where $x \in [0, 1], t \ge 0$.
We know that $u$ satisfies the equation $$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2},$$ with boundary condition $u(x, 0) = f(x)$.

Let $f \in C^2[0, 1]$. Moreover $f$ has period $1$.

I know that $$f(x) = u(x, 0) = \sum \limits_{n=0}^{\infty} \big( C_n \sin(2 \pi n x) + D_n \cos(2 \pi n x) \big) \tag{1}.$$

Of course $(1)$ says that $C_n$ and $D_n$ are Fourier coefficients of function $f$.

Thus $$C_n = \int \limits_{0}^{1} f(y) \sin(2 \pi n y) \, dy, \\ D_n = \int \limits_{0}^{1} f(y) \cos(2 \pi n y) \, dy.$$

That leads to $$u(x, t) = \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \big( C_n \sin(2 \pi n x) + D_n \cos(2 \pi n x) \big) \\ = \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \bigg( \int \limits_{0}^{1} f(y) \sin(2 \pi n y) \, dy \, \sin(2 \pi n x) + \int \limits_{0}^{1} f(y) \cos(2 \pi n y) \, dy \, \cos(2 \pi n x) \bigg) \\ = \int \limits_{0}^{1} f(y) \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \bigg(\sin(2 \pi n x) \sin(2 \pi n y) + \cos(2 \pi n x)\cos(2 \pi n y) \bigg) \, dy \\ = \int \limits_{0}^{1} f(y) \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \cos\big(2 \pi n (x-y)\big) \, dy \\ = \int \limits_{0}^{1} f(y) \bigg(1 + \sum \limits_{n=1}^{\infty} e^{-4 \pi^2 n^2 t} \cos\big(2 \pi n (x-y)\big) \bigg) \, dy.$$

However I know that the the kernel should be equal $$1 + 2\sum \limits_{n=1}^{\infty} e^{-4 \pi^2 n^2 t} \cos\big(2 \pi n (x-y)\big).$$

Where am I wrong?

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The Fourier coefficients of a function $f$ having period $p$ are

$$C_n =\frac2p\int_P f(t)\cos(\tfrac{2\pi n t}{p})\,dt \qquad \text{and} \qquad D_n =\frac2p\int_P f(t)\sin(\tfrac{2\pi n t}{p})\,dt,$$

where $P$ denotes some interval of measure $p$. In your case, the integrals are missing the coefficient of $\frac21=2$.


Reference: See the formula for coefficients here, and a proof on the site here.

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    $\begingroup$ Thanks. Can you provide a proof of give some references, please? $\endgroup$ – Hendrra Jun 9 at 12:35
  • $\begingroup$ @Hendrra I updated my answer to include reference. $\endgroup$ – Luke Collins Jun 9 at 12:50
  • $\begingroup$ Thank you very much! I truly appreciate your help! :) $\endgroup$ – Hendrra Jun 9 at 13:35
  • $\begingroup$ @Hendrra No problem :) $\endgroup$ – Luke Collins Jun 9 at 13:39

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