Partial Differential Equations (PDEs)

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

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

(Ex01) 1st order PDEs

ut+a(x)ux=f(x,t) u_t + a(x)u_x = f(x,t) ut+a(x)ux=f(x,t)

(Ex02) 2nd order PDEs

utt−C2uxx=fa wave equationu(0,t)=0u(L,t)=0 \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} uttC2uxxu(0,t)u(L,t)=fa wave equation=0=0

Other Second-Order PDEs

ut−kuxx=fone-dimensional heat equationuxx+uyy=0two-dimensional Laplacian equationuxx+uyy=f \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} utkuxxuxx+uyyuxx+uyy=fone-dimensional heat equation=0two-dimensional Laplacian equation=f

(Ex03) Darcy Flow

{J=−k∇uu: the pressure∇J=0 \begin{cases} J = -k \nabla u \hspace{1cm} \text{u: the pressure}\\ \nabla J = 0 \end{cases} {J=kuu: the pressureJ=0

If k≡1k\equiv 1k1: −∇(∇u)=0⟶uxx+uyy=0-\nabla(\nabla u) = 0 \longrightarrow u_{xx} + u_{yy} = 0(u)=0uxx+uyy=0

(Definition) of Laplacian Operator

Δu=uxx+uyy=∇∇u \Delta u = u_{xx} + u_{yy} = \nabla\nabla u Δu=uxx+uyy=u

(Ex04) Wave Equation

utt−C2uxx=0 u_{tt} - C^2 u_{xx} = 0 uttC2uxx=0

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

⟶{utt=−C2sin⁡(Ct)cos⁡(x)uxx=−sin⁡(Ct)cos⁡(x) \longrightarrow \begin{cases} u_{tt} &= -C^2 \sin(Ct) \cos(x) \\ u_{xx} &= -\sin(Ct)\cos(x) \\ \end{cases} {uttuxx=C2sin(Ct)cos(x)=sin(Ct)cos(x)

(Ex05) Heat Equation

ut−kuxx=0 u_t - ku_{xx} = 0 utkuxx=0

⟶u(x,t)=e−ktsin⁡(x) \longrightarrow u(x,t) = e^{-kt} \sin(x) u(x,t)=ektsin(x)

⟶{ut=−ke−ktsin⁡(x)uxx=−e−ktsin⁡(x) \longrightarrow \begin{cases} u_t &= -k e^{-kt}\sin(x) \\ u_{xx} &= -e^{-kt}\sin(x) \end{cases} {utuxx=kektsin(x)=ektsin(x)

(Ex06) Laplacian Equation

uxx+uyy=0 u_{xx} + u_{yy} = 0 uxx+uyy=0

⟶u(x,y)=x2−y2 \longrightarrow u(x,y) = x^2 - y^2 u(x,y)=x2y2

⟶{uxx=2uyy=2 \longrightarrow \begin{cases} u_{xx} = 2\\ u_{yy} = 2\\ \end{cases} {uxx=2uyy=2

Thm 01 (Superposition)

If u1u_1u1, u2u_2u2 are solutions of a linear PDEs, then C1u1+C2u2C_1u_1+C_2u_2C1u1+C2u2 for C1,C2∈RC_1,C_2 \in RC1,C2R is also a solution of the PDE.

12.3. Solution of Variables

Wave Equation

utt−C2uxx=0 u_{tt} - C^2u_{xx} = 0 \\ uttC2uxx=0

Boundary Condtion:

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

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

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

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

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

utt=C2uxx u_{tt} = C^2 u_{xx} utt=C2uxx

⟶F(x)G′′(t)C2F(x)G(t)=F′′(x)G(t)F(x)G(t) \longrightarrow \frac{F(x)G''(t)}{C^2 F(x)G(t)} = \frac{F''(x)G(t)}{F(x)G(t)} C2F(x)G(t)F(x)G(t)=F(x)G(t)F(x)G(t)

⟶G′′(t)G(t)=C2F′′(x)F(x)=k (a constant) \longrightarrow \frac{G''(t)}{G(t)} = C^2 \frac{F''(x)}{F(x)} = k \text{ (a constant)} G(t)G(t)=C2F(x)F(x)=k (a constant)

  1. F′′F=k\frac{F''}{F}= kFF=k iff F′′−kF=0F'' - kF = 0FkF=0
  2. G′′C2G=K\frac{G''}{C^2G}= KC2GG=K iff G′′−kC2G=0G'' - kC^2G = 0GkC2G=0

Boundary Condition:

u(0,t)=F(0)G(t)=0 for any t u(0,t) = F(0)G(t) = 0 \text{ for any } t u(0,t)=F(0)G(t)=0 for any t

⟶F(0)=0 \longrightarrow F(0) = 0 F(0)=0

u(L,t)=F(L)G(t)=0 for any t u(L,t) = F(L)G(t) = 0 \text{ for any } t u(L,t)=F(L)G(t)=0 for any t

⟶F(L)=0 \longrightarrow F(L) = 0 F(L)=0

Then we get the question w.r.t. only FFF

{F′′−KF=0F(0)=0F(L)=0 \begin{cases} F'' - KF = 0 \\ F(0) = 0 \\ F(L) = 0 \\ \end{cases} FKF=0F(0)=0F(L)=0

It’s a Sturm-Liouville System

F(x)=erx⟹r2−k=0⟹r=±kF(x) = e^{rx} \Longrightarrow r^2-k=0 \Longrightarrow r = \plusmn\sqrt{k}F(x)=erxr2k=0r=±k

(a) k>0⟹F(x)=C1ekx+C2e−kxk>0 \Longrightarrow F(x) = C_1 e^{\sqrt{k}x} + C_2 e^{-\sqrt{k}x}k>0F(x)=C1ekx+C2ekx

{F(0)=0F(L)=0⟹{C1=0C2=0⟹F≡0 \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 {F(0)=0F(L)=0{C1=0C2=0F0

(b) k=0⟹F(x)=C1+C2x⟹{C1=0C2=0⟹F≡0k = 0 \Longrightarrow F(x) = C_1 + C_2 x \Longrightarrow \begin{cases} C_1 = 0\\ C_2 = 0\\ \end{cases} \Longrightarrow F\equiv 0k=0F(x)=C1+C2x{C1=0C2=0F0

© k&lt;0k &lt; 0k<0 and let k=−p2k=-p^2k=p2 (p&gt;0)(p&gt;0)(p>0)

r=±piF(x)=C1cos⁡(px)+C2sin⁡(px)F(0)=C1=0⟹F(x)=C2sin⁡(px)⟹F(L)=C2sin⁡(pL)=0 r = \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 r=±piF(x)=C1cos(px)+C2sin(px)F(0)=C1=0F(x)=C2sin(px)F(L)=C2sin(pL)=0

So we have pL=nπpL = n\pipL=nπ iff p=nπLp=\frac{n\pi}{L}p=Lnπ

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

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