数学物理方程学习笔记5

2.热传导方程

2.2Cauchy问题的解

Cauchy问题：

n=1
{ut−a2uxx=0,x∈R,t>0u(x,0)=φ(x),x∈R\begin{cases} u_t-a^2u_{xx}=0,x \in R,t>0\\ u(x,0)=\varphi(x),x \in R \end{cases} {ut−a2uxx=0,x∈R,t>0u(x,0)=φ(x),x∈R
方法：

Flourier变换法
step1：对方程和初始条件关于x进行Flourier变换{u^t−a2(iλ)2u^=0u^(λ,0)=φ^(λ)PDE→ODEstep2:求解ODE方程u^=φ^(λ)e−a2λ2tstep3:对u^进行反演若g^(λ)=e−a2λ2t,此时u^=φ^(λ)g^(λ)根据卷积的性质：(φ∗g)∧=2πφ^g^u=(φ^(λ)g^(λ))∨=12πφ∗gstep4:求g,使g^(λ)=e−a2λ2t根据之前的结论：(e−Ax2)∧=12Ae−λ24A令A=14a2t,g=2Ae−Ax2=1a2te−x24a2tu=12πφ∗g=12π∫−∞+∞φ(ξ)1a2te−(x−ξ)24a2tdξ=12aπt∫−∞+∞φ(ξ)e−(x−ξ)24a2tdξstep1：对方程和初始条件关于x进行Flourier变换\\ \begin{cases} \hat u_t-a^2(i\lambda)^2\hat u=0\\ \hat u(\lambda,0)=\hat\varphi(\lambda) \end{cases}\\ PDE\rightarrow ODE\\ \\ step2:求解ODE方程\\ \hat u=\hat \varphi(\lambda)e^{-a^2\lambda^2t}\\ \\ step3:对\hat u进行反演\\ 若\hat g(\lambda)=e^{-a^2\lambda^2t},此时\hat u=\hat \varphi(\lambda)\hat g(\lambda)\\ 根据卷积的性质：(\varphi *g)^\wedge=\sqrt{2\pi}\hat \varphi\hat g\\ u=(\hat \varphi(\lambda)\hat g(\lambda))^\vee=\frac{1}{\sqrt{2\pi}}\varphi *g\\ \\ step4:求g,使\hat g(\lambda)=e^{-a^2\lambda^2t}\\ 根据之前的结论：(e^{-Ax^2})^\wedge=\frac{1}{\sqrt{2A}}e^{-\frac{\lambda^2}{4A}}\\ 令A=\frac{1}{4a^2t},g=\sqrt{2A}e^{-Ax^2}=\frac{1}{a\sqrt{2t}}e^{-\frac{x^2}{4a^2t}}\\ \\ u=\frac{1}{\sqrt{2\pi}}\varphi *g=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\varphi(\xi)\frac{1}{a\sqrt{2t}}e^{-\frac{(x-\xi)^2}{4a^2t}}d\xi=\frac{1}{2a\sqrt{\pi t}}\int_{-\infty}^{+\infty}\varphi(\xi)e^{-\frac{(x-\xi)^2}{4a^2t}}d\xi step1：对方程和初始条件关于x进行Flourier变换{u^t−a2(iλ)2u^=0u^(λ,0)=φ^(λ)PDE→ODEstep2:求解ODE方程u^=φ^(λ)e−a2λ2tstep3:对u^进行反演若g^(λ)=e−a2λ2t,此时u^=φ^(λ)g^(λ)根据卷积的性质：(φ∗g)∧=2πφ^g^u=(φ^(λ)g^(λ))∨=2π1φ∗gstep4:求g,使g^(λ)=e−a2λ2t根据之前的结论：(e−Ax2)∧=2A1e−4Aλ2令A=4a2t1,g=2Ae−Ax2=a2t1e−4a2tx2u=2π1φ∗g=2π1∫−∞+∞φ(ξ)a2t1e−4a2t(x−ξ)2dξ=2aπt1∫−∞+∞φ(ξ)e−4a2t(x−ξ)2dξ
非齐次
{ut−a2uxx=f(x,t),x∈R,t>0u(x,0)=φ(x),x∈R\begin{cases} u_t-a^2u_{xx}=f(x,t),x \in R,t>0\\ u(x,0)=\varphi(x),x \in R \end{cases} {ut−a2uxx=f(x,t),x∈R,t>0u(x,0)=φ(x),x∈R
使用常数变易法φ(ξ)→f(ξ,τ),t→t−τ,并对τ在(0,t)上积分\varphi(\xi)\rightarrow f(\xi,\tau),t\rightarrow t-\tau,并对\tau 在(0,t)上积分φ(ξ)→f(ξ,τ),t→t−τ,并对τ在(0,t)上积分
u=12aπt∫−∞+∞φ(ξ)e−(x−ξ)24a2tdξ+∫0t12aπ(t−τ)∫−∞+∞f(ξ,τ)e−(x−ξ)24a2(t−τ)dξdt=12aπt∫−∞+∞φ(ξ)e−(x−ξ)24a2tdξ+∫0t∫−∞+∞12aπ(t−τ)f(ξ,τ)e−(x−ξ)24a2(t−τ)dξdtK(x,t)={12aπte−x24a2t,t>00,t≤0−−基本解u=∫−∞+∞φ(ξ)K(x−ξ,t)dξ+∫0tdτ∫−∞+∞f(ξ,τ)K(x−ξ,t−τ)dξ−−Possion公式u=\frac{1}{2a\sqrt{\pi t}}\int_{-\infty}^{+\infty}\varphi(\xi)e^{-\frac{(x-\xi)^2}{4a^2t}}d\xi+\int_0^t\frac{1}{2a\sqrt{\pi (t-\tau)}}\int_{-\infty}^{+\infty}f(\xi,\tau)e^{-\frac{(x-\xi)^2}{4a^2(t-\tau)}}d\xi dt \\=\frac{1}{2a\sqrt{\pi t}}\int_{-\infty}^{+\infty}\varphi(\xi)e^{-\frac{(x-\xi)^2}{4a^2t}}d\xi+\int_0^t\int_{-\infty}^{+\infty}\frac{1}{2a\sqrt{\pi (t-\tau)}}f(\xi,\tau)e^{-\frac{(x-\xi)^2}{4a^2(t-\tau)}}d\xi dt\\ K(x,t)=\begin{cases} \frac{1}{2a\sqrt{\pi t}}e^{-\frac{x^2}{4a^2t}},t>0\\ 0,t\leq 0\\ \end{cases}--基本解 \\ u=\int_{-\infty}^{+\infty}\varphi(\xi)K(x-\xi,t)d\xi+\int_0^td\tau\int_{-\infty}^{+\infty}f(\xi,\tau)K(x-\xi,t-\tau)d\xi--Possion公式 u=2aπt1∫−∞+∞φ(ξ)e−4a2t(x−ξ)2dξ+∫0t2aπ(t−τ)1∫−∞+∞f(ξ,τ)e−4a2(t−τ)(x−ξ)2dξdt=2aπt1∫−∞+∞φ(ξ)e−4a2t(x−ξ)2dξ+∫0t∫−∞+∞2aπ(t−τ)1f(ξ,τ)e−4a2(t−τ)(x−ξ)2dξdtK(x,t)={2aπt1e−4a2tx2,t>00,t≤0−−基本解u=∫−∞+∞φ(ξ)K(x−ξ,t)dξ+∫0tdτ∫−∞+∞f(ξ,τ)K(x−ξ,t−τ)dξ−−Possion公式
巩固：用Flourier变换法求解弦振动方程

问题：
{utt−a2uxx=0,x∈R,t>0u(x,0)=φ(x),x∈Rut(x,0)=ψ(x),x∈R\begin{cases} u_{tt}-a^2u_{xx}=0,x\in R,t>0\\ u(x,0)=\varphi(x),x\in R \\ u_t(x,0)=\psi(x),x\in R \end{cases} ⎩⎪⎨⎪⎧utt−a2uxx=0,x∈R,t>0u(x,0)=φ(x),x∈Rut(x,0)=ψ(x),x∈R
解答：
step1：对方程和初始条件关于x进行Flourier变换{u^tt−a2(iλ)2u^=0u^(λ,0)=φ^(λ)u^∣t(λ,0)=ψ^(λ)PDE→ODEstep2:求解ODE方程u^=c1eiaλt+c2e−iaλtc1+c2=φ^(λ)iaλ(c1−c2)=ψ^(λ)u^=12(φ^(λ)−iψ^(λ)aλ)eiaλt+12(φ^(λ)+iψ^(λ)aλ)e−iaλtstep3:对u^进行反演(φ^(λ)eiaλt)∨=12πlim⁡N→+∞∫−N+Nφ^(λ)eiaλteiλxdλ=12πlim⁡N→+∞∫−N+Nφ^(λ)eiλ(x+at)dλ=φ(x+at)同理：(φ^(λ)e−iaλt)∨=φ(x−at)[ψ^(λ)λ(eiaλt−e−iaλt)]∨=12πlim⁡N→+∞∫−N+Nψ^(λ)λ(eiaλt−e−iaλt)eiλxdλ=12πlim⁡N→+∞∫−N+Nψ^(λ)λ(eiλ(x+at)−e−iλ(x−at))dλ=12πlim⁡N→+∞∫−N+Nψ^(λ)λeiλξ∣x−atx+atdλ=12πlim⁡N→+∞∫−N+Nψ^(λ)λ∫x−atx+ateiλξiλdξdλ=i∫x−atx+at12πlim⁡N→+∞∫−N+Nψ^(λ)eiλξdλdξ=i∫x−atx+atψ(ξ)dξu=12(φ(x+at)+φ(x−at))+12a∫x−atx+atψ(ξ)dξstep1：对方程和初始条件关于x进行Flourier变换\\ \begin{cases} \hat u_{tt}-a^2(i\lambda)^2\hat u= 0\\ \hat u(\lambda,0)=\hat\varphi(\lambda)\\ \hat u|_t(\lambda,0)=\hat\psi(\lambda)\\ \end{cases}\\ PDE\rightarrow ODE\\ \\ step2:求解ODE方程\\ \hat u=c_1e^{ia\lambda t}+c_2e^{-ia\lambda t}\\ c_1+c_2=\hat\varphi(\lambda)\\ ia\lambda (c_1-c_2)=\hat\psi(\lambda)\\ \hat u=\frac{1}{2}(\hat\varphi(\lambda)-i\frac{\hat\psi(\lambda)}{a\lambda})e^{ia\lambda t}+\frac{1}{2}(\hat\varphi(\lambda)+i\frac{\hat\psi(\lambda)}{a\lambda})e^{-ia\lambda t}\\ \\ step3:对\hat u进行反演\\ (\hat\varphi(\lambda)e^{ia\lambda t})^\vee=\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\hat\varphi(\lambda)e^{ia\lambda t}e^{i\lambda x}d\lambda\\ =\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\hat\varphi(\lambda)e^{i\lambda (x+at)}d\lambda=\varphi(x+at)\\ 同理：(\hat\varphi(\lambda)e^{-ia\lambda t})^\vee=\varphi(x-at)\\ [\frac{\hat\psi(\lambda)}{\lambda}(e^{ia\lambda t}-e^{-ia\lambda t})]^\vee=\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\frac{\hat\psi(\lambda)}{\lambda}(e^{ia\lambda t}-e^{-ia\lambda t})e^{i\lambda x}d\lambda\\ =\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\frac{\hat\psi(\lambda)}{\lambda}(e^{i\lambda (x+at)}-e^{-i\lambda (x-at)})d\lambda\\ =\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\frac{\hat\psi(\lambda)}{\lambda}e^{i\lambda\xi}|_{x-at}^{x+at}d\lambda\\ =\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\frac{\hat\psi(\lambda)}{\lambda}\int_{x-at}^{x+at}e^{i\lambda\xi}i\lambda d\xi d\lambda\\ =i\int_{x-at}^{x+at}\frac{1}{\sqrt{2\pi}}\lim\limits_{N\rightarrow+\infty}\int_{-N}^{+N}\hat\psi(\lambda)e^{i\lambda\xi}d\lambda d\xi \\ =i\int_{x-at}^{x+at}\psi(\xi)d\xi\\ \\ u=\frac{1}{2}(\varphi(x+at)+\varphi(x-at))+\frac{1}{2a}\int_{x-at}^{x+at}\psi(\xi)d\xi step1：对方程和初始条件关于x进行Flourier变换⎩⎪⎨⎪⎧u^tt−a2(iλ)2u^=0u^(λ,0)=φ^(λ)u^∣t(λ,0)=ψ^(λ)PDE→ODEstep2:求解ODE方程u^=c1eiaλt+c2e−iaλtc1+c2=φ^(λ)iaλ(c1−c2)=ψ^(λ)u^=21(φ^(λ)−iaλψ^(λ))eiaλt+21(φ^(λ)+iaλψ^(λ))e−iaλtstep3:对u^进行反演(φ^(λ)eiaλt)∨=2π1N→+∞lim∫−N+Nφ^(λ)eiaλteiλxdλ=2π1N→+∞lim∫−N+Nφ^(λ)eiλ(x+at)dλ=φ(x+at)同理：(φ^(λ)e−iaλt)∨=φ(x−at)[λψ^(λ)(eiaλt−e−iaλt)]∨=2π1N→+∞lim∫−N+Nλψ^(λ)(eiaλt−e−iaλt)eiλxdλ=2π1N→+∞lim∫−N+Nλψ^(λ)(eiλ(x+at)−e−iλ(x−at))dλ=2π1N→+∞lim∫−N+Nλψ^(λ)eiλξ∣x−atx+atdλ=2π1N→+∞lim∫−N+Nλψ^(λ)∫x−atx+ateiλξiλdξdλ=i∫x−atx+at2π1N→+∞lim∫−N+Nψ^(λ)eiλξdλdξ=i∫x−atx+atψ(ξ)dξu=21(φ(x+at)+φ(x−at))+2a1∫x−atx+atψ(ξ)dξ
解的性质：

基本解K(x,t)K(x,t)K(x,t)
K(x,t)={12aπte−x24a2t,t>00,t≤0−−基本解K(x,t)=\begin{cases} \frac{1}{2a\sqrt{\pi t}}e^{-\frac{x^2}{4a^2t}},t>0\\ 0,t\leq 0\\ \end{cases}--基本解 \\ K(x,t)={2aπt1e−4a2tx2,t>00,t≤0−−基本解
(1)K(x,t)K(x,t)K(x,t)是(x,t)(x,t)(x,t)的解析函数，即无穷次可微

(2)非负性：K(x,t)>0,∀t>0K(x,t)>0 ,\forall t>0K(x,t)>0,∀t>0

(3)lim⁡∣x∣→+∞K(x,t)=0,lim⁡∣x∣→+∞∂mK(x,t)∂xm=0,∀t>0\lim\limits_{|x|\rightarrow +\infty}K(x,t)=0,\lim\limits_{|x|\rightarrow +\infty}\frac{\partial ^{m}K(x,t)}{\partial x^m}=0,\forall t>0∣x∣→+∞limK(x,t)=0,∣x∣→+∞lim∂xm∂mK(x,t)=0,∀t>0

(4)lim⁡t→0K(x,t)=0,∀δ>0,∣x∣<δ\lim\limits_{t\rightarrow 0}K(x,t)=0,\forall \delta>0,|x|<\deltat→0limK(x,t)=0,∀δ>0,∣x∣<δ

(5)规范性：∫−∞+∞K(x,t)dx=1\int_{-\infty}^{+\infty}K(x,t)dx=1∫−∞+∞K(x,t)dx=1

Possion公式

u=∫−∞+∞φ(ξ)K(x−ξ,t)dξ+∫0tdτ∫−∞+∞f(ξ,τ)K(x−ξ,t−τ)dξu=\int_{-\infty}^{+\infty}\varphi(\xi)K(x-\xi,t)d\xi+\int_0^td\tau\int_{-\infty}^{+\infty}f(\xi,\tau)K(x-\xi,t-\tau)d\xiu=∫−∞+∞φ(ξ)K(x−ξ,t)dξ+∫0tdτ∫−∞+∞f(ξ,τ)K(x−ξ,t−τ)dξ

(1)有界性
φ∈C1(R),f=0⟶u是Cauchy问题的有界解\varphi \in C^1(R),f=0\longrightarrow u是Cauchy问题的有界解 φ∈C1(R),f=0⟶u是Cauchy问题的有界解
证明：
设∣φ(x)∣≤M∣u∣=∣∫−∞+∞φ(ξ)K(x−ξ,t)dξ∣≤∫−∞+∞∣φ(ξ)∣∣K(x−ξ,t)∣dξ≤M∫−∞+∞K(x−ξ,t)dξ=M⋅1设|\varphi(x)|\leq M\\ |u|=|\int_{-\infty}^{+\infty}\varphi(\xi)K(x-\xi,t)d\xi|\leq\int_{-\infty}^{+\infty}|\varphi(\xi)||K(x-\xi,t)|d\xi\\ \leq M\int_{-\infty}^{+\infty}K(x-\xi,t)d\xi=M\cdot1\\ 设∣φ(x)∣≤M∣u∣=∣∫−∞+∞φ(ξ)K(x−ξ,t)dξ∣≤∫−∞+∞∣φ(ξ)∣∣K(x−ξ,t)∣dξ≤M∫−∞+∞K(x−ξ,t)dξ=M⋅1
(2)奇偶性、周期性

φ是x\varphi是xφ是x的奇/偶/周期函数，则u是xu是xu是x的奇/偶/周期函数。

(3)无限传播速度

∀x∈(x0−δ,x0+δ),φ(x)>0,则u(x,t)>0,t>0\forall x\in(x_0-\delta,x_0+\delta),\varphi(x)>0,则u(x,t)>0,t>0∀x∈(x0−δ,x0+δ),φ(x)>0,则u(x,t)>0,t>0

解释：在顷刻之间，热量传递到杆上的任意一点，在x0x_0x0附近的点受到的影响较大，离x0x_0x0较远的点受到的影响较小。

对于弦振动方程来说，有“影响已达到的点”和“影响未达到的点”之分，这是两者的本质区别。

(4)无穷次可微

φ连续有解，则u(x,t)∈C∞(R+2)\varphi 连续有解，则u(x,t)\in C^{\infty}(R_{+}^2)φ连续有解，则u(x,t)∈C∞(R+2)