A doubt about the heat equation solution

Question

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.

$$ \frac{\partial}{\partial t}(e^{\lambda t}S(x)) = \gamma \cdot \frac{\partial}{\partial x^{2}}(e^{\lambda t}S(x)) $$

$$ \lambda e^{\lambda t}S(x) = \gamma \cdot e^{\lambda t} \frac{\partial^{2} S(x)}{\partial x^{2}}$$

$$ \therefore \; \lambda S(x) = \gamma \cdot \frac{\partial^{2} S(x)}{\partial x^{2}} \Longrightarrow S_{k}(x) = sin(k \pi x) \Longrightarrow \lambda = - \gamma k^{2} \pi^{2}$$

$$ \therefore \; T(t, x) = \sum_{k = 1}^{\infty} B_{k} e^{-\gamma k^{2} \pi^{2} t}S_{k}$$

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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.

Answer 1

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.

Answer 2

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.