This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
vpde_lecture23 [2020/03/26 13:07] trinh |
vpde_lecture23 [2020/03/26 17:05] (current) trinh |
||
|---|---|---|---|
| Line 1: | Line 1: | ||
| - | ====== MA20223 Lecture | + | ====== MA20223 Lecture |
| + | |||
| + | < | ||
| + | <iframe width=" | ||
| + | </ | ||
| ==== The 1D heat equation with zero Dirichlet conditions ==== | ==== The 1D heat equation with zero Dirichlet conditions ==== | ||
| Line 20: | Line 24: | ||
| The hardest part is to understand how to calculate the bn 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 bn 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. | ||
| < | < | ||
| Line 73: | Line 84: | ||
| ==== 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)=T0+(T1−T0)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' | ||