Confusion for the solution of the heat equation in a one dimensional bar

Question

The heat equation for a one dimensional bar is given by $$\frac{\partial}{\partial t} u = k \frac{\partial^2}{\partial x^2}u$$ where $k$ is the diffusion coefficient, and the boundary conditions are given by $$u(0,t) = u(L, t) = 0\ \forall t \geq 0$$ where $L$ is the length of the bar. Moreover at time $t = 0$, the temperature is given by a known function $f(x)$, i.e., $$u(x, 0) = f(x)\ \forall x \in [0, L].$$

Assuming that $u(x, t) = X(x) \cdot T(t)$, we can reduce the heat equation to the following system $$\begin{cases}X''(x) + \lambda^2 X(x) &= &0,\\ T'(t) + \lambda^2 k T(t) &= &0. \end{cases}$$ Both equations can be solved, and we have $$X(x) = A_{+} e^{i \lambda x} + A_{-}e^{-i\lambda x}$$ for the first one, and $$T(t) = B e^{-\lambda^2 k t}$$ for the second one. Thus we get $$u(x, t) = (C_1 e^{i \lambda x} + C_2 e^{-i\lambda x})e^{-\lambda^2 k t}.$$ Assuming $C_1 = -C_2$ and since $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$ and $\sin(\lambda L) = 0 \iff \lambda = \frac{m \pi}{L}$, $m \in \mathbb{Z}$, we finally get $$u(x, t) = C_1 2 i \sin\left( \frac{m \pi x}{L} \right) e^{-\frac{m^2 \pi^2 kt}{L^2}}.$$ Letting $b_m := C_1 2 i$, the general solution is given by $$u(x, t) = \sum_{m = -\infty}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) e^{-\frac{m^2 \pi^2 kt}{L^2}}.$$ Because of the boundary condition $u(x, 0) = f(x)$, we also have $$u(x, 0) = \sum_{m = -\infty}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) = f(x)$$ which is a Fourier serie in sine of period $2L$ and where $b_m$ is a constant.

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Sorry for the long introduction, this is taken from a class I followed, and I have a few questions about it.

1. Since $b_m$ is constant, do we not have $$u(x, 0) = \sum_{m = -\infty}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) = \sum_{m = 1}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) - \sum_{m = 1}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) = 0?$$

2. I thought that the bar was heated at a certain $x$ until time $t=0$, and then one would just observe how the heat was diffusing. To simplify the question, assume $L = \pi$, $k = 1$ and $C_1 2 i = 1$, so that the solution is $$u(x, t) = \sin\left( mx \right) e^{-m^2 t}.$$ Plugging $m = 2$ above leads to $$u(x, 0) = \sin\left(2x\right),$$ meaning that some parts of the bar have a negative temperature. Am I understanding this correctly, or do I make a mistake? Because this is super weird to me that after being heated, some parts of the bar can have subzero temperature.

3. What happens if we assume that $u(0,t) = u(L,t) \neq 0$?

4. What happens if we assume that $u(0, t) \neq u(L, t)$?

5. Can someone provide intuition on what $X(x)$ and $T(t)$ represent in the variable separation?

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Answer 1

Proposal 1. : Since $b_m$ is constant, do we not have $$u(x, 0) = \sum_{m = -\infty}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) = \sum_{m = 1}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) - \sum_{m = 1}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) = 0?$$

NO. Don't confuse ($b_m$ in $m>0$) and ($b_m$ in $m<0$). $$u(x, 0) = \sum_{m = -\infty}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) = \sum_{m = 1}^{\infty} b_m \sin\left( \frac{m \pi x}{L} \right) + \sum_{m = 1}^{\infty} b_{-m} \sin\left( \frac{m \pi x}{L} \right)$$ With $B_m=b_m+b_{-m}$ , better use this form of Fourier series : $$u(x, 0) = \sum_{m = 1}^{\infty} B_m \sin\left( \frac{m \pi x}{L} \right)$$

Proposal 2. : I thought that the bar was heated at a certain $x$ until time $t=0$, and then one would just observe how the heat was diffusing. To simplify the question, assume $L = \pi$, $k = 1$ and $C_1 2 i = 1$, so that the solution is $$u(x, t) = \sin\left( mx \right) e^{-m^2 t}.$$

NO. One cannot separate one term of the series. The Fourrier series a whole : $$u(x, t) =\sum_{m = 1}^{\infty} B_m \sin\left( mx \right) e^{-m^2 t}.$$ Some $B_m$ can be positive and others negative, depending on the function $f(x)$ on a given range $0<x<L$. Of course, at various $t$ some terms are negative but the sum will be such as $u$ remains positive on this limited range insofar $f(x)$ is positive on this range.

Outside the range where the Fourrier series is considered and where the coefficients are computed, the Fourier series is not valide and the computed $u(x,t)$ has no signifiance (so, doesn't mind if it becomes negative).