A doubt about the heat equation solution I was watching to Professor Gilbert Strang solving the heat equation and I did not understand one point. Let me guide you to my thinking:
Starting with the 1D heat equation
$$ \frac{\partial T}{\partial t} = \gamma \cdot \frac{\partial^{2}T}{\partial x^{2}} $$
In order to solve this differential equation, we can say that T(t,x) is of the form
$$ T(t, x) = e^{\lambda t}S(x) $$
Where $\lambda$ is the eigenvalue and $S(x)$ is the eigenfunction. Plugging this expression for T in the heat equation, we will be able to find what is the eigenfunction and then the eigenvalue.
$$ \frac{\partial}{\partial t}(e^{\lambda t}S(x)) = \gamma \cdot \frac{\partial}{\partial x^{2}}(e^{\lambda t}S(x)) $$
$$ \lambda e^{\lambda t}S(x) = \gamma \cdot e^{\lambda t} \frac{\partial^{2} S(x)}{\partial x^{2}}$$
$$ \therefore \; \lambda S(x) = \gamma \cdot \frac{\partial^{2} S(x)}{\partial x^{2}} \Longrightarrow S_{k}(x) = sin(k \pi x) \Longrightarrow \lambda = - \gamma k^{2} \pi^{2}$$
$$ \therefore \; T(t, x) = \sum_{k = 1}^{\infty} B_{k} e^{-\gamma k^{2} \pi^{2} t}S_{k}$$

What I did not understand was why is there a $k \pi$ inside the $sin$ of $S_{k}(x)$?
In my head, it would mean that imagining a rod has 10 meters of length, at $x = 0, x = 1, x = 2, x = 3, ... , x = 10$, the temperature would be 0, and that's insane to think about it because if I get a 10 meters rod right now and put a flame below these x points I will feel these points hot. How is temperature 0 at these points?
I have read once it was because of the boundary conditions but the temperature is being 0 not only at the beginning and the end of the rod but also in many points between them. 
 A: The $k\pi$ comes from the assumption that the temperature is held at zero at both ends (at $x=0$ and $x=1$) plus the assumption that the solution is separable (a product of a function of $t$ and a function of $x$). This boundary condition type and the domain determine the eigenfunctions; with a different BC and/or domain you would have different eigenfunctions and indeed different dynamics.
The overall solution is only formed by summing over $k$. So the fact that some of the $S_k$ have interior zeros for those points at all times does not mean that the whole solution has zeros at those points, unless the initial condition was prepared as a multiple of $\sin(k\pi x)$ in the first place. 
To get a feel for it, try computing the coefficients in the case where the initial condition is $T(x,0)=x-x^2$ and draw some plots.
A: The $\pi$ is just there for convenience.  If you leave it out you just rescale $x$.  The $k$ factor indexes the eigenfunctions.  There are $k$ half waves along the bar.  The heat equation washes out temperature differences.  The modes with large $k$ wash out more quickly because the curvature of the temperature is higher by a factor $k^2$.  
This set of eigenfunctions assumes the temperature is zero at the ends of the bar.  If it were not, you would have to add in functions of the form $\cos (k \pi x)$ as well.  
You can certainly have the temperature nonzero at points in the middle of the bar.  If you take $k$ to be an integer, as is usual, the length of the bar is $1$ (with the $\pi$ factor included) so the lowest mode is a half sine wave.  The point you are thinking of as $x=1$ on a $10$ meter bar is really $x=0.1$ because it is $\frac 1{10}$ of the way down the bar.  Many of the sine functions are nonzero at that point.
