I am attempting to solve a homogeneous heat equation
\begin{equation} u_t = \alpha^2 u_{xx}, \end{equation} with an initial temperature $u_0$, and time-varying boundary conditions $u(0,t) = u(L,t) = u_s(t)$. The challenge is that the function $u_s (t)$ isn't a continuous function that can be expressed neatly, but rather a series of temperatures given for every minute. I followed the method laid out here
Solution methods for heat equation with time-dependent boundary conditions,
and got the result
\begin{equation} u_c(t) = u_s (t) - \frac{4}{\pi} \sum_{n=0}^\infty \frac{(-1)^n}{2 n + 1} e^{- (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \int_0^t e^{(2n + 1)^2 \pi^2 \alpha^2 / L^2 \tau} u_s ' (\tau) d\tau + u_s (0) - u_0 \right), \end{equation}
hoping that I could still approximate the integral with a sum like so
\begin{equation} u_c(t) \approx u_s (t) - \frac{4}{\pi} \sum_{n=0}^\infty \frac{(-1)^n}{2 n + 1} e^{- 60 (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \sum_{m=0}^{t - 1} e^{60(2n + 1)^2 \pi^2 \alpha^2 / L^2 m} \Delta u_s^m + u_s (0) - u_0 \right), \quad \Delta u_s^m \equiv u_s^{m+1} - u_s^m. \end{equation} ($m$ represents minutes, and in any real calculation $t$ would be floored down to the nearest integer) I am able to compare the above result to measured values for the temperature, and it does not hold up at all.
My question is then, do anyone know of a better way to handle boundary conditions that are put in by hand every minute? Ultimately, I would like to be able to extract the size $L$, which is why I am not doing this numerically.
If you would like to see my derivation, I post it below:
We have
\begin{equation} u_t = \alpha^2 u_{xx}, \end{equation} with boundary and initial conditions
\begin{align*} u(x,0) & = f(x), \quad u(0,t) = u(L,t) = u_s (t)\\ f(0) & = f(L) = u_s(0), \quad f(0<x<L) = u_0, \end{align*} and assume
\begin{equation} s_{xx}(x,t) = 0, \end{equation} so that
\begin{equation} s(x,t) = u_s (t). \end{equation} Then we get the inhomogenous equation
\begin{equation} v_t = \alpha^2 v_{xx} - u_s'. \end{equation} The homogeneous equation is solved by
\begin{equation} v(x,t) = \sum_{n=1}^\infty T_n (t) X_n (x), \quad X_n (x) = \sin \left( \frac{n \pi x}{L} \right). \end{equation} For the inhomogeneous equation, we make the ansatz
\begin{equation} u'_s (t) = - \sum_{n=1}^\infty Q_n (t) X_n (x), \end{equation} so that the heat equation becomes
\begin{equation} \sum_{n=1}^\infty \left( T_n' (t) + \lambda_n T_n (t) - Q_n (t) \right) X_n (x) = 0, \quad \lambda_n = \left( \frac{n \pi \alpha}{L} \right)^2, \end{equation} which means that
\begin{equation} T_n (t) = e^{-\lambda_n t} \int_0^t e^{\lambda_n \tau} Q_n (\tau) d \tau + C_n e^{-\lambda_n t}. \end{equation} The $C_n$'s are found by checking the initial conditions
\begin{equation} v(x,0) = u(x,0) - s(x,0) = u_0 - u_s (0) = \sum_{n=1}^\infty T_n (0) X_n (x) = \sum_{n=1}^\infty C_n X_n (x), \end{equation} so
\begin{equation} C_n = \frac{2 \left( 1 - (-1)^n \right) \left( u_0 - u_s (0) \right)}{n \pi}. \end{equation} $Q_n (t)$ is found from the orthogonality of the $X_n (x)$'s
\begin{equation} Q_n (t) = 2 \frac{(-1)^n - 1}{n \pi} u_s' (t), \end{equation} or, since only terms with $n$ odd are non-zero,
\begin{align*} C_n & = \frac{4}{(2n+1) \pi} \left( u_0 - u_s (0) \right),\\ Q_n (t) & = - \frac{4}{(2 n + 1) \pi} u_s ' (t). \end{align*} Then
\begin{equation} T_n (t) = - \frac{4}{(2 n + 1) \pi} e^{- (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \int_0^t e^{(2n + 1)^2 \pi^2 \alpha^2 / L^2 \tau} u_s ' (\tau) d\tau + u_s(0) - u_0 \right) \end{equation} and
\begin{equation} v(x,t) = - \frac{4}{\pi} \sum_{n=0}^\infty \frac{1}{(2 n + 1) \pi} e^{- (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \int_0^t e^{(2n + 1)^2 \pi^2 \alpha^2 / L^2 \tau} u_s ' (\tau) d\tau + u_s(0) - u_0 \right) \sin \left( \frac{(2n+1) \pi x}{L} \right). \end{equation} The temperature is then
\begin{equation} u(x,t) = u_s (t) - \frac{4}{\pi} \sum_{n=0}^\infty \frac{1}{(2 n + 1) \pi} e^{- (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \int_0^t e^{(2n + 1)^2 \pi^2 \alpha^2 / L^2 \tau} u_s ' (\tau) d\tau + u_s(0) - u_0 \right) \sin \left( \frac{(2n+1) \pi x}{L} \right). \end{equation} In the core the temperature is
\begin{equation} u_c(t) = u_s (t) - \frac{4}{\pi} \sum_{n=0}^\infty \frac{(-1)^n}{2 n + 1} e^{- (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \int_0^t e^{(2n + 1)^2 \pi^2 \alpha^2 / L^2 \tau} u_s ' (\tau) d\tau + u_s (0) - u_0 \right). \end{equation} Numerically, this is
\begin{equation} u_c(t) \approx u_s (t) - \frac{4}{\pi} \sum_{n=0}^\infty \frac{(-1)^n}{2 n + 1} e^{- 60 (2n + 1)^2 \pi^2 \alpha^2 / L^2 t} \left( \sum_{m=0}^{t - 1} e^{60(2n + 1)^2 \pi^2 \alpha^2 / L^2 m} \Delta u_s^m + u_s (0) - u_0 \right), \quad \Delta u_s^m \equiv u_s^{m+1} - u_s^m. \end{equation}
