Heat Equation¶
- 1-D heat equation
\[
\begin{gather}
\frac{\partial U}{\partial t} = k \frac{\partial^{2}U}{\partial x^{2}}
\end{gather}
\]
- 3-D heat equation
\[
\begin{gather}
\frac{\partial U}{\partial t} = k( \frac{\partial^{2}U}{\partial x^{2}}+\frac{\partial^{2}U}{\partial y^{2}}
+\frac{\partial^{2}U}{\partial z^{2}})
\end{gather}
\]
Principle¶
- 沒微分兩端固定 → \(\sin\)
- 兩端固定一次微分等於0 → \(\cos\)
Finite Medium¶
take
\[
\begin{gather}
U(x, 0) = f(x)
\\\\
U_{t}(x, 0) = g(x)
\\\\
X'' + \lambda X = 0
\\\\
T' + \lambda k T = 0
\end{gather}
\]
沒微分兩端固定 = 0¶
- 沒微分兩端固定 → \(\sin\)
- \(U(0, t) = U(L, t) = 0\)
\[
\begin{align}
&\lambda_{n} = \frac{n^{2}\pi^{2}}{L^{2}} & n=1, 2, 3 , \dots
\end{align}
\]
\[
\begin{gather}
X_{n} = \sin(\sqrt{\lambda_{n}}x)
\\\\
T_{n} = c_{n}e^{-\lambda t}
\end{gather}
\]
\[
\begin{gather}
c_{n} = \frac{2}{L}\int_{0}^{L}{f(x)\sin(\sqrt\lambda x)dx}
\\\\
U(x, t) = \sum{c_{n}\sin(\sqrt{\lambda }x)\ e^{^{-k\lambda t}}}
\end{gather}
\]
兩端固定一次微分等於0¶
- 兩端固定一次微分等於0 → \(\cos\)
- \(U_{x}(0, t) = U_{x}(L, t) = 0\)
\[
\begin{align}
&\lambda_{n} = \frac{n^{2}\pi^{2}}{L^{2}} & n=0, 1, 2, 3 , \dots
\end{align}
\]
\[
\begin{gather}
X_{n} = \cos(\sqrt{\lambda_{n}}x)
\\\\
T_{n} = c_{n}e^{-\lambda t}
\\\\
c_{n} = \frac{2}{L}\int_{0}^{L}{f(x)\cos(\sqrt{\lambda}x)dx}
\\\\
U(x, t) = \frac{1}{2}c_{0} + \sum_{n=1}^{\infty}{c_{n}\cos(\sqrt{\lambda}x)\ e^{-k\lambda t}}
\end{gather}
\]
有 A¶
-
\(U(0, t) = 0; \quad U_{x}(L, t) = -AU(L, t)\)
-
用作圖解 \(\tan(\alpha L) = -\frac{\alpha}{A}\), 令解為 \(A_{1}, A_{2}, \dots\)
-
解 case 3 (case 1, case 2 are trivial)
-
\[f(x) = \sum_{n=1}^{\infty}{c_{n}\sin(\sqrt{\lambda}x)}\]
-
\[c_{n} = \frac{(f(x), \sin(\sqrt{\lambda}x))}{||\sin(\sqrt{\lambda}x)||}\]
-
\[ \begin{gather} U(x, t) = \sum_{n=1}^{\infty}{c_{n}\sin(\sqrt{\lambda}x)e^{-k\lambda t}} \end{gather} \]
兩端固定不為 0¶
-
\(U(0, t) = T_{1}, \quad U(L, t) = T_{2}\)
-
Set \(U(x, t) = u(x, t) + \phi(x)\)
- Then \(u_{t} = k(u_{xx}+\phi''(x))\)
-
Try let \(\phi''(x) = 0\)
-
solve \(u(x, t)\)
有外力¶
- \(U_{t} = k U_{xx}+F(x, t)\)
-
且兩端固定為 0
-
\(U(x, t) =\) 外力修正 + 沒外力解
-
\[B_{n}(t) = \frac{2}{L}\int_{0}^{L}{F(x, t)\sin(\sqrt{\lambda}x)dx}\]
-
\[ \begin{gather} b_{n} = \int_{0}^{t}{e^{-k\lambda (t-\tau)}B_{n}(\tau)\ d\tau} \end{gather} \]
-
\[ \begin{gather} U(x, t) = \sum_{n=1}^{\infty}{b_{n}\sin(\sqrt{\lambda_{n}}x)}+ \sum_{n=1}^{\infty}{c_{n}\sin{\sqrt{\lambda_{n}}}\ e^{-k\lambda_{n}t}} \end{gather} \]
Infinite Media¶
-
\[ \begin{gather} \lambda_{n} = \omega^{2} \end{gather} \]
-
\[ \begin{gather} U(x, t) = \int_{o}^{\infty}{[a_{\omega}\cos(\omega x)+b_\omega \sin(\omega x)]e^{-k\omega^{2}t}\ d\omega} \\\\ a_{\omega} = \frac{1}{\pi} \int_{-\infty}^{\infty}{f(x)\cos(\omega x) dx} \\\\ b_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty}{f(x)\sin(\omega x)dx} \end{gather} \]
Half infinite Media¶
- \(x \in (0, \infty)\)