PDE utt = c^2uxx 詳細求解過程

 求一維波動方程式 one-dimentional wave equation 之解，即 $$\qquad\qquad \displaystyle \frac{\partial ^{2}}{\partial t^{2}} u=c^{2}\frac{\partial ^{2}}{\partial x^{2}} u，\\滿足邊界條件  \begin{cases}u( 0,t) =0\\ u( L,t) =0\end{cases} ,\forall t 及初始條件  \begin{cases} u( x,0) =f( x) & \\ \left. \frac{\partial u}{\partial t}\right| _{t=0} =g( x) & \end{cases}$$

解答： $$令 u( x,t) =F( x) G( t)，則 \begin{cases} u_{tt} =FG''\\u_{xx} =F''G \end{cases}，代回原式可得 FG''=c^{2} F''G \Rightarrow \frac{G''}{c^{2} G} =\frac{F''}{F}\\由於 \frac{G''}{c^{2} G} 只含變數  t ，而 \frac{F''}{F} 只含變數 x ，兩者等值代表同為一常數 k ，即\frac{G''}{c^{2} G} =\frac{F''}{F} =k\\則邊界條件:\cases{u(0,t) = F(0)G(t)=0\\ u(L,t)=F(L)G(t)=0}，若G(t)=0，則u(x,t)=0為明顯解，不列入討論\\因此邊界條件轉變成F(0)=F(L)=0；\\F''-kF=0 \Rightarrow F=c_1e^{\sqrt kx}+ c_2e^{-\sqrt k x} ，代入初始值\Rightarrow \cases{F(0)=c_1+c_2=0 \cdots(1)\\ F(L)= c_1e^{\sqrt kL}+ c_2e^{-\sqrt k L} =0 \cdots(2)}\\ 由(1)得c_2=-c_1代入(2) \Rightarrow c_1(e^{\sqrt kL}-e^{-\sqrt k L})=0 \Rightarrow c_1=0或e^{\sqrt kL}＝e^{-\sqrt k L}\\ 若c_1=0 \Rightarrow c_2=0 \Rightarrow F=0為明顯解，不列入討論;因此只需考慮e^{\sqrt kL}＝e^{-\sqrt k L}\\ 而e^{\sqrt kL}\ne e^{-\sqrt k L},\forall k\gt 0;若k=0 \Rightarrow F=0為明顯解，不列入討論\\ 因此只有k\lt 0,此時F(x)= c_1(e^{i\sqrt{|k|}x}-e^{-i\sqrt{|k|}x}) =2ic_1 \sin (\sqrt{|k|}x) \\ \Rightarrow F(L)=2ic_1 \sin (\sqrt{|k|}L) =0 \Rightarrow \sqrt{|k|}L=n\pi,n\in \mathbb N \Rightarrow \sqrt{|k|}=n\pi /L \\ \Rightarrow F(x)=2ic_1 \sin (n\pi x/L) =A \sin\left({n\pi \over L}x\right),A為常數\\ k=-{n^2\pi^2 \over L^2} \Rightarrow G''+{n^2\pi^2 \over L^2}c^2G=0 \Rightarrow G(t)=c_3 \cos\left({n\pi \over L}ct\right) +c_4\sin \left({n\pi \over L}ct\right)\\因此u(x,t)=A \sin\left({n\pi \over L}x\right)\left( c_3 \cos\left({n\pi \over L}ct\right) +c_4\sin \left({n\pi \over L}ct\right)\right) \\=\sum_{n=1}^\infty \sin\left({n\pi \over L}x\right)\left( A_n \cos\left({n\pi \over L}ct\right) +B_n\sin \left({n\pi \over L}ct\right)\right)$$

$$初始條件1: u(x,0)=f(x) \Rightarrow f(x)=\sum_{n=1}^\infty A_n \sin({n\pi \over L}x) \\\Rightarrow \int_0^L f(x)\sin({k\pi \over L}x)\;dx = \sum_{n=1}^\infty \int_0^L A_n\sin({k\pi \over L}x)\sin({n\pi \over L}x)\;dx =\int_0^L A_k \sin^2 ({k\pi \over L}x)\;dx ={L\over 2}A_k\\ \Rightarrow A_k={2\over L}\int_0^L f(x)\sin({k\pi \over L}x)\;dx \Rightarrow A_n={2\over L}\int_0^L f(x)\sin({n\pi \over L}x)\;dx\\初始條件2: \left. {\partial u\over \partial t}\right|_{t=0} =g(x) \Rightarrow g(x) = \sum_{n=1}^\infty \left( -{cn\pi \over L}A_n \sin({cn\pi \over L}t) +{cn\pi \over L}B_n\cos ({cn\pi \over L}t) \right)\sin({n\pi \over L}x)|_{t=0} \\ =\sum_{n=1}^\infty {cn\pi \over L}B_n\sin({n\pi \over L}x) \Rightarrow \sin({k\pi \over L}x)g(t) =\sum_{n=1}^\infty {cn\pi \over L}B_n\sin({k\pi \over L}x)\sin({n\pi \over L}x) \\ \Rightarrow \int_0^L \sin({k\pi \over L}x)g(x)\;dx = \sum_{n=1}^\infty \int_0^L {cn\pi \over L}B_n\sin({k\pi \over L}x)\sin({n\pi \over L}x)\;dx \\={ck\pi \over L}B_k\int_0^L \sin^2({k\pi \over L}x)\; dx={ck\pi \over L}B_k \cdot {L\over 2} ={ck\pi \over 2}B_k \Rightarrow B_k= {2\over ck\pi }\int_0^L \sin({k\pi \over L}x)g(x)\;dx\\ \Rightarrow B_n= {2\over cn\pi }\int_0^L \sin({n\pi \over L}x)g(x)\;dx \\ \Rightarrow \bbox[red,2pt]{ u(x,t)=\sum_{n=1}^\infty \left(A_n \cos({cn\pi\over L }t) +B_n \sin({cn\pi\over L }t)\right) \sin({n\pi \over L}x)}，其中\\A_n={2\over L}\int_0^L f(x)\sin({n\pi \over L}x)\;dx,B_n= {2\over cn\pi }\int_0^L \sin({n\pi \over L}x)g(x)\;dx$$