1D Wave equation: Misc

  1. Half-line: method of continuation
  2. Finite interval: method of continuation

Half-line: method of continuation

Consider wave equation in domain $\{x>0, t>0\}$, initial conditions, and a boundary condition \begin{align} &u_{tt}-c^2 u_{xx}=f(x,t) \qquad &&x>0, \label{equ-7a.1}\\ &u|_{t=0}=g(x) && x>0,\label{equ-7a.2}\\ &u_t|_{t=0}=h(x) && x>0,\label{equ-7a.3} \\ &u|_{x=0}=0. \label{equ-7a.4} \end{align} Alternatively, instead of (\ref{equ-7a.4}) we consider \begin{align} & u_x |_{x=0}=0.\qquad&&\qquad\quad \tag*{$(4)'$}\label{equ-7a.4-'} \end{align}

Remark. It is crucial that we consider either Dirichlet or Neumann homogeneous boundary conditions.

To deal with this problem consider first IVP on the whole line: \begin{align} &U_{tt}-c^2 U_{xx}=F(x,t) \qquad &&x>0,\label{equ-7a.5}\\ &U|_{t=0}=G(x) && x>0,\label{equ-7a.6}\\ &U_t|_{t=0}=H(x) && x>0,\label{equ-7a.7} \\ \end{align} and consider $V(x,t)= \varsigma U(-x,t)$ with $\varsigma =\pm 1$.

Proposition 1. If $U$ satisfies (\ref{equ-7a.5})-(\ref{equ-7a.7}) then $V$ satisfies similar problem albeit with right-hand expression $\varsigma F(-x,t)$, and initial functions $\varsigma G(-x)$ and $\varsigma H(-x)$.

Proof. Plugging $V$ into equation we use the fact that $V_t(x,t)=\varsigma U_t (-x,t)$, $V_x(x,t)=-\varsigma U_x (-x,t)$, $V_{tt}(x,t)=\varsigma U_{tt} (-x,t)$, $V_{tx}(x,t)=-\varsigma U_{tx} (-x,t)$, $V_{xx}(x,t)=\varsigma U_{xx} (-x,t)$ etc and exploit the fact that wave equation contains only even-order derivatives with respect to $x$.

Note that if $F,G,H$ are even functions with respect to $x$, and $\varsigma=1$ then $V(x,t)$ satisfies the same IVP as $U(x,t)$. Similarly, if $F,G,H$ are odd functions with respect to $x$, and $\varsigma=-1$ then $V(x,t)$ satisfies the same IVP as $U(x,t)$.

However we know that solution of (\ref{equ-7a.5})-(\ref{equ-7a.7}) is unique and therefore $U(x,t)=V(x,t)=\varsigma U(-x,t)$.

Therefore

Corollary 1.

  1. If $\varsigma=-1$ and $F,G,H$ are odd functions with respect to $x$ then $U(x,t)$ is also an odd function with respect to $x$.

  2. If $\varsigma=1$ and $F,G,H$ are even functions with respect to $x$ then $U(x,t)$ is also an even function with respect to $x$.

However we know that a. odd function with respect to $x$ vanishes as $x=0$; b. derivative of the even function with respect to $x$ vanishes as $x=0$

and we arrive to

Corollary 2.

  1. In the framework of Proposition 1a $U$ satisfies (\ref{equ-7a.4});
  2. In the framework of Proposition 1b $U$ satisfies \ref{equ-7a.4-'}.

Therefore

Corollary 3. a. To solve (\ref{equ-7a.1})-(\ref{equ-7a.3}), (\ref{equ-7a.4}) we need to take an odd continuation of $f,g,h$ to $x<0$ and solve the corresponding IVP; b. To solve (\ref{equ-7a.1})-(\ref{equ-7a.3}), \ref{equ-7a.4-'} we need to take an even continuation of $f,g,h$ to $x<0$ and solve the corresponding IVP.

So far we have not used much that we have exactly wave equation (similar argments with minor modification work for heat equation as well etc). Now we apply D'Alembert formula (5.4): \begin{multline*} u(x,t)= \frac{1}{2}\bigl( G(x+ct)+G(x-ct) \bigr) + \frac{1}{2c}\int_{x-ct} ^{x+ct} H(x')\,dx' +\\ \frac{1}{2c} \int_0^t \int _{x-c(t-t')} ^{x+c(t-t')} F(x',t')\, dx' dt' \end{multline*} and we need to take $0< x< ct$, resulting for $f=0\implies F=0$ \begin{equation} u(x,t)= \frac{1}{2}\bigl( g(ct+x)- g(ct-x) \bigr) + \frac{1}{2c}\int_{ct-x} ^{ct+x} h(x')\,dx' , \label{equ-7a.8} \end{equation} and \begin{multline} u(x,t)= \frac{1}{2}\bigl( g(ct+x)+ g(ct-x) \bigr) +\\ \frac{1}{2c}\int_0 ^{ct-x} h(x')\,dx' + \frac{1}{2c}\int_0 ^{ct+x} h(x')\,dx'\qquad \tag*{(8)'}\label{equ-7a.8-'} \end{multline} for boundary condition (\ref{equ-7a.4}) and \ref{equ-7a.4-'} respectively.

Example. Consider wave equation with $c=1$ and let $f=0$,

  1. $g= \sin (x)$, $h=0$ and Dirichlet boundary condition. Obviously $G(x)=\sin (x)$ (since we take odd continuation and $\sin(x)$ is an odd function). Then \begin{equation} u(x,t)=\frac{1}{2}\bigl( \sin (x+t)+\sin (x-t)\bigr)= \sin (x)\cos (t). \label{equ-7a.9} \end{equation}
  2. $g= \sin (x)$, $h=0$ and Neumann boundary condition. Obviously $G(x)=-\sin (x)$ as $x<0$ (since we take even continuation and $\sin(x)$ is an odd function). Then $u$ is given by (\ref{equ-7a.9}) for $x\ >t\ >0$ and \begin{equation} u(x,t)=\frac{1}{2}\bigl( \sin (x+t)-\sin (x-t)\bigr)= \sin (t)\cos(x) \label{equ-7a.10} \end{equation} as $0< x< t$.
  3. $g= \cos (x)$, $h=0$ and Neumann boundary condition. Obviously $G(x)=\cos (x)$. Then \begin{equation} u(x,t)=\frac{1}{2}\bigl( \cos (x+t)+\cos (x-t)\bigr)= \cos (x)\cos (t). \label{equ-7a.11} \end{equation}
  4. $g= \cos (x)$, $h=0$ and Dirichlet boundary condition. Obviously $G(x)=-\cos (x)$ as $x<0$. Then $u$ is given by (\ref{equ-7a.9}) for $x>t>0$ and \begin{equation} u(x,t)=\frac{1}{2}\bigl( \cos (x+t)-\cos (x-t)\bigr)= \sin (t)\sin (x) \label{equ-7a.12} \end{equation} as $0< x< t$.

Finite interval: method of continuation

Consider the same problem albeit on interval $0< x< l$ with either Dirichlet or Neumann condition on each end. Then we need to take odd continuation through "Dirichlet end" and even continuation through "Neumann end". On figures below we have respectively Dirichlet conditions on each end (indicated by red), Neumann conditions on each end (indicated by green), and Dirichlet condition on one end (indicated by red) and Neumann conditions on another end (indicated by green). Resulting continuations are $2l$, $2l$ and $4l$ periodic respectively.

Fig-8.1