Diferencia entre revisiones de «Approximation of the heat equation: Fourier method»
| Línea 24: | Línea 24: | ||
The main point is to observe that, if <math>\varphi(x)</math> is solution of the eigenvalue problem | The main point is to observe that, if <math>\varphi(x)</math> is solution of the eigenvalue problem | ||
| + | |||
<math> | <math> | ||
\left\{ \begin{array}{l} | \left\{ \begin{array}{l} | ||
| Línea 29: | Línea 30: | ||
\varphi(0)=0, \quad \varphi(L)=0, \end{array} \right. | \varphi(0)=0, \quad \varphi(L)=0, \end{array} \right. | ||
</math> | </math> | ||
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for some <math>\lambda</math>, then | for some <math>\lambda</math>, then | ||
| + | |||
<math> | <math> | ||
u(x,t)=\varphi(x) e^{-\lambda t} | u(x,t)=\varphi(x) e^{-\lambda t} | ||
</math> | </math> | ||
| + | |||
is a solution of the heat equation <math>u_t-u_{xx}=0</math> and the boundary conditions <math>u(0,t)=u(L,t)=0.</math> | is a solution of the heat equation <math>u_t-u_{xx}=0</math> and the boundary conditions <math>u(0,t)=u(L,t)=0.</math> | ||
| + | |||
| + | The eigenvalues of the above eigenvalue problem are <math>\lambda_k=\frac{k^2 \pi^2}{L^2}</math> with <math>k=1,2,3,...</math> and the associated eigenfunctions are | ||
| + | |||
| + | <math> | ||
| + | \varphi_k=\sin(\frac{k\pi x}{L}). | ||
| + | </math> | ||
| + | |||
| + | Therefore, all these functions are solutions of the heat equation and boundary conditions | ||
| + | |||
| + | <math> | ||
| + | u_k (x,t)= e^{-k^2\pi^2 /L^2 t}\sin(\frac{k\pi x}{L}) | ||
| + | </math> | ||
Revisión del 10:17 26 abr 2016
Let [math]{\bf u(x_1,x_2,x_3,t)}[/math] be the temperature at the point [math](x_1,x_2,x_3)\in [/math] bar, and time [math]t\gt0[/math].
Assume that the cross section is so small that we can consider the bar as an unidimensional object in the interval [math]x\in [0,L][/math], [math] u(x_1,x_2,x_3,t)=u(x_1,t)=u(x,t). [/math]
If we assume that the extreme are at zero temperature, the system of equations for u is given by [math] \left\{ \begin{array}{l} u_t-u_{xx}=0, \qquad x\in(0,L), \qquad t\gt0, \\ u(0,t)=0, \qquad t\gt0, \\ u(L,t)=0, \qquad t\gt0, \\ u(x,0)=u^0(x), \qquad x\in(0,L). \end{array} \right. [/math] where [math]u_0(x)[/math] is a function that describes the initial temperature of the bar.
We describe below the Fourier method to approximate the solutions of this system.
The main point is to observe that, if [math]\varphi(x)[/math] is solution of the eigenvalue problem
[math] \left\{ \begin{array}{l} \varphi''(x)+\lambda \varphi(x)=0, \\ \varphi(0)=0, \quad \varphi(L)=0, \end{array} \right. [/math]
for some [math]\lambda[/math], then
[math] u(x,t)=\varphi(x) e^{-\lambda t} [/math]
is a solution of the heat equation [math]u_t-u_{xx}=0[/math] and the boundary conditions [math]u(0,t)=u(L,t)=0.[/math]
The eigenvalues of the above eigenvalue problem are [math]\lambda_k=\frac{k^2 \pi^2}{L^2}[/math] with [math]k=1,2,3,...[/math] and the associated eigenfunctions are
[math] \varphi_k=\sin(\frac{k\pi x}{L}). [/math]
Therefore, all these functions are solutions of the heat equation and boundary conditions
[math] u_k (x,t)= e^{-k^2\pi^2 /L^2 t}\sin(\frac{k\pi x}{L}) [/math]
