Partial Differential Equations (PDEs)

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最新推荐文章于 2023-12-31 01:46:32 发布

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分类专栏： Mathematics 文章标签： Mathematics

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本文介绍了偏微分方程的基本概念与定义，并通过几个典型例子如波动方程、热传导方程及拉普拉斯方程展示了不同阶次的偏微分方程。此外，文章还探讨了分离变量法解决特定偏微分方程的方法。

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Partial Differential Equations (PDEs)

12.1 PDEs

(Defintion) of PDE

A PDE is an equation involving one or more partial derivatives of an unknown
The order of a PDE: the highest order of the partial derivatives of the unknown

(Ex01) 1st order PDEs

$u_t + a(x)u_x = f(x,t)$

(Ex02) 2nd order PDEs

$\begin{aligned} u_{tt} - C^2 u_{xx} &= f \hspace{1cm} \text{a wave equation} \\ u(0,t) &= 0\\ u(L,t) &= 0 \end{aligned}$

Other Second-Order PDEs

$\begin{aligned} u_t - ku_{xx} &= f \hspace{1cm} \text{one-dimensional heat equation} \\ u_{xx} + u_{yy} &= 0 \hspace{1cm} \text{two-dimensional Laplacian equation} \\ u_{xx} + u_{yy} &= f \end{aligned}$

(Ex03) Darcy Flow

$\begin{cases} J = -k \nabla u \hspace{1cm} \text{u: the pressure}\\ \nabla J = 0 \end{cases}$

If $k≡1k\equiv 1$ : $−∇(∇u)=0⟶uxx+uyy=0-\nabla(\nabla u) = 0 \longrightarrow u_{xx} + u_{yy} = 0$

(Definition) of Laplacian Operator

$\Delta u = u_{xx} + u_{yy} = \nabla\nabla u$

(Ex04) Wave Equation

$u_{tt} - C^2 u_{xx} = 0$

$\longrightarrow u(x,t) = \sin(Ct) \cos(x)$

$\longrightarrow \begin{cases} u_{tt} &= -C^2 \sin(Ct) \cos(x) \\ u_{xx} &= -\sin(Ct)\cos(x) \\ \end{cases}$

(Ex05) Heat Equation

$u_t - ku_{xx} = 0$

$\longrightarrow u(x,t) = e^{-kt} \sin(x)$

$\longrightarrow \begin{cases} u_t &= -k e^{-kt}\sin(x) \\ u_{xx} &= -e^{-kt}\sin(x) \end{cases}$

(Ex06) Laplacian Equation

$u_{xx} + u_{yy} = 0$

$\longrightarrow u(x,y) = x^2 - y^2$

$\longrightarrow \begin{cases} u_{xx} = 2\\ u_{yy} = 2\\ \end{cases}$

Thm 01 (Superposition)

If $u_1$ , $u_2$ are solutions of a linear PDEs, then $C_1u_1+C_2u_2$ for $C1,C2∈RC_1,C_2 \in R$ is also a solution of the PDE.

12.3. Solution of Variables

Wave Equation

$u_{tt} - C^2u_{xx} = 0 \\$

Boundary Condtion:

$\begin{cases} u(0,t) = 0\\ u(L,t) = 0 \\ u(x,0) = f(x) \\ u_t(x,0) = g(x) \\ \end{cases}$

Remark: This is a IBVP (Initial Boundary Value Problem)

$u (x, t)$ : the displacement at $x$ and $t$ .

(Idea): Assume that $u (x, t) = F (x) G (t)$ , then

$\begin{cases} u_{tt} = F(x)G''(t)\\ u_{xx} = F''(x)G(t) \end{cases}$

$u_{tt} = C^2 u_{xx}$

$\longrightarrow \frac{F(x)G''(t)}{C^2 F(x)G(t)} = \frac{F''(x)G(t)}{F(x)G(t)}$

$\longrightarrow \frac{G''(t)}{G(t)} = C^2 \frac{F''(x)}{F(x)} = k \text{ (a constant)}$

$F′′F=k\frac{F''}{F}= k$ iff $F^{''} - k F = 0$
$G′′C2G=K\frac{G''}{C^2G}= K$ iff $kC^2G = 0$

Boundary Condition:

$\text{ for any } t$

$\longrightarrow F(0) = 0$

$\text{ for any } t$

$\longrightarrow F(L) = 0$

Then we get the question w.r.t. only $F$

$\begin{cases} F'' - KF = 0 \\ F(0) = 0 \\ F(L) = 0 \\ \end{cases}$

It’s a Sturm-Liouville System

$e^{rx} \Longrightarrow r^2-k=0 \Longrightarrow r = \plusmn\sqrt{k}$

(a) $\Longrightarrow F(x) = C_1 e^{\sqrt{k}x} + C_2 e^{-\sqrt{k}x}$

$\begin{cases} F(0) = 0\\ F(L) = 0\\ \end{cases} \Longrightarrow \begin{cases} C_1 = 0\\ C_2 = 0\\ \end{cases} \Longrightarrow F \equiv 0$

(b) $\Longrightarrow F(x) = C_1 + C_2 x \Longrightarrow \begin{cases} C_1 = 0\\ C_2 = 0\\ \end{cases} \Longrightarrow F\equiv 0$

$\plusmn pi\\ F(x) = C_1 \cos(px) + C_2 \sin(px)\\ F(0) = C_1 = 0 \Longrightarrow F(x) = C_2\sin(px)\\ \Longrightarrow F(L) = C_2 \sin(pL) = 0$

So we have $n\pi$ iff $p=nπLp=\frac{n\pi}{L}$

Let $pn=nπLp_n = n\frac{\pi}{L}$ and $Fn(x)=sin⁡(nπLx)F_n(x) = \sin(n\frac{\pi}{L}x)$