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积分变换法

2. Laplace变换法

张瑞
中国科学技术大学数学科学学院

Laplace变换法

Laplace变换定义为

\[F(p)=L[f(t)]=\int_0^{+\infty} f(t)e^{-pt}dt \]

Laplace逆变换\[L^{-1}[F(p)]=\frac1{2\pi i}\int_{\beta-i\infty}^{\beta+i\infty}F(p)e^{pt}dp , t>0 \]

在使用Laplace变换解题时,必须选取方程中在$(0,+\infty)$上变化的自变量(通常是时间变量)作积分变量

微分关系

\[L[f^{(n)}(t)]=p^nF(p)-p^{n-1}f(0+)-p^{n-2}f'(0+) \\ -\cdots-f^{(n-1)}(0+) \]

卷积

\[(f*g)(t)=\int_0^t f(t-x)g(s)ds \]

\[L[f*g]=L[f]\cdot L[g] , \\ L^{-1}[L[f]L[g]]=f*g \]

例 1. 解混合问题

\[\casefunc{ & u_{tt}=a^2u_{xx}+f(t), x>0, t>0 \\ & u|_{t=0}=0, u_t|_{t=0}=0 \\ & u|_{x=0}=0 } \]

例 2. 解定解问题

\[\casefunc{ & u_t=a^2u_{xx} , x>0, t>0 \\ & u(0,x)=0 \\ & u(t,0)=f(t) } \]

example 1

例 3. 解定解问题

\[\casefunc{ & u_{tt}=a^2u_{xx} , 0<x<l, t>0 \\ & u(t,0)=0, u_x(t,l)=A\sin(\omega t) , \\ & u(0,x)=0 , u_t(0,x)=0 } \]

5.

例 4. 解定解问题

\[\casefunc{ & u_tt=u_xx+k\sin(\pi x) , x\in(0,1), t>0 \\ & u(t,0)=u(t,1)=0 , t>0 \\ & u(0,x)=u_t(0,x)=0 , x\in[0,1] } \]

4.$t$作Laplace变换,有

\[p^2\tilde u(p,x)-p(0,x)-u_t(0,x)-\frac{d^2}{dx^2}\tilde u(p,x)=\frac{k}p\sin(\pi x) \]$u(0,x)=u_t(0,x)=0$,有

\[p^2\tilde u(p,x)-\frac{d^2}{dx^2}\tilde u(p,x)=\frac{k}p\sin(\pi x) \] 解得

\[\tilde u(p,x)=C_1(p)e^{px}+C_2(p)e^{-px}+\frac{k}{p(p^2+\pi^2)}\sin(\pi x) \]$u(t,0)=u(t,1)=0$,可得$C_1(p)=C_2(p)=0$,则

\[\tilde u(p,x)=\frac{k}{p(p^2+\pi^2)}\sin(\pi x) =\frac{k}{\pi^2}(\frac1p-\frac{p}{p^2+\pi^2})\sin(\pi x) \]

所以

\[u(t,x)=\frac{k}{\pi^2}(1-\cos(\pi t))\sin(\pi x) \]

目录

本节读完

例 5. 用Laplace变换法,求解半无界弦的振动问题:

\[\casefunc{ & u_{tt}=u_{xx}, x>0, t>0 \\ & u(0,x)=0 , u_t(0,x)=1 \\ & u(t,0)=0, u_x(t,+\infty)=0 } \]

5.