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vpde_lecture23 [2020/03/26 13:31] trinh |
vpde_lecture23 [2020/03/26 16:42] trinh [The 1D heat equation with a steady state temperature] |
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| The hardest part is to understand how to calculate the $b_n$ coefficients via an odd extension of the initial condition. In many ways, this is somewhat backwards (usually we ask "How do I compute a Fourier series for an odd extension" | The hardest part is to understand how to calculate the $b_n$ coefficients via an odd extension of the initial condition. In many ways, this is somewhat backwards (usually we ask "How do I compute a Fourier series for an odd extension" | ||
| + | |||
| + | Anyways, this we do in the video, and there we show that | ||
| + | $$ | ||
| + | b_n = -\frac{2}{n\pi}[(-1)^n - 1] | ||
| + | $$ | ||
| + | |||
| + | We'll then share a numerical simulation of this heat flow problem in the lecture. The code is below. | ||
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| ==== The 1D heat equation with a steady state temperature ==== | ==== The 1D heat equation with a steady state temperature ==== | ||
| + | We will now examine the methodology for solving the non-homogeneous heat equation with Dirichlet conditions. | ||
| + | |||
| + | $$ | ||
| + | \begin{gathered} | ||
| + | u_t = \kappa u_{xx}, \quad x\in[0, L], t \geq 0 \\ | ||
| + | u(0, t) = T_0, \quad u(L, t) = T_1 \\ | ||
| + | u(x, 0) = f(x). | ||
| + | \end{gathered} | ||
| + | $$ | ||
| + | |||
| + | The trick is to seek a steady state solution. Seek a solution that does not depend on time. Then $u(x, t) = U(x)$ and we must satisfy: | ||
| + | $$ | ||
| + | \begin{gathered} | ||
| + | 0 = \kappa u_{xx}, \quad x\in[0, L], t \geq 0 \\ | ||
| + | U(0) = T_0, \quad U(L) = T_1. | ||
| + | \end{gathered} | ||
| + | $$ | ||
| + | The solution is then $U(x) = T_0 + (T_1 - T_0)x/L$. | ||
| + | |||
| + | Next, we set the solution as | ||
| + | $$ | ||
| + | u(x,t) = U(x) + \hat{u}(x, | ||
| + | $$ | ||
| + | |||
| + | Why do this? Substitute into the system now to see that | ||
| + | $$ | ||
| + | \begin{gathered} | ||
| + | \hat{u}_t = \kappa \hat{u}_{xx}, | ||
| + | \hat{u}(0, t) = 0, \quad \hat{u}(L, t) = 0. \\ | ||
| + | \hat{u}(x, 0) = f(x) - U(x). | ||
| + | \end{gathered} | ||
| + | $$ | ||
| + | |||
| + | In other words, the effect of the trick of writing the solution using the steady-state $U(x)$ has effectively zero' | ||
| + | The algorithm is summarised in the video, and we will complete the demonstration in Friday' | ||