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Mathmatics

Static Math: ax^2 + bx + c = 0.
Fill-in Math: \sqrt{ \MathQuillMathField{x}^2 + \MathQuillMathField{y}^2 }

When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are \[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]

Problem:

{The one dimensional (linear) wave equation is given by $$ \pder[2]{u}{t^2} = c^2 \, \pder[2]{u}{x^2} \label{wave} $$ where $c$ is the (constant) wave speed. Show that this equation may be rewritten as the coupled system of two first order equations \begin{align*} u_t - c \, u_x &= w \\ w_t + c \, w_x &= 0. \end{align*} Hence show that the general solution of (\ref{wave}) is \[ u(x, \ t) = F(x-ct) + G(x+ct) \] for arbitrary functions $F$ and $G$.},

Solution:

Let $w=u_t-cu_x$. Then \[ w_t+cw_x = (u _t-cu_x)_t + c(u_t-cu_x)_x = u_{tt}-c^2\,u_{xx} = 0. \] Solving $w_t+cw_x=0 $ by the method of characteristics, gives $w(x, \ t) = P(x-ct)$ for some arbitrary function $P$. Thus \[ u_t - c \, u_x = P(x-ct). \] The characteristic variables for this equation are $t=\tau$ and $x = \xi -c \tau$. In characteristic variables, the equation is \[ U_\tau = P(\xi-c\tau-c\tau) = P(\xi-2c\tau) \] and so \[ U = F(\xi) + G(\xi-2c\tau) \] where $G’=-\frac{P}{2c}$ (an arbitrary function!). Finally we obtain \[ u(x, \ t) = F(x+ct) +G(x+ct-2ct) = F(x+ct) + G(x-ct) \] as required.}, keywords = {tutorial, characteristics}, owner = {msh51}, timestamp = {2009.03.02}

MathJax v3: TeX & MathML to HTML



Dynamic Equations in MathJax

Expand the following: $$ \require{action} \def\longest{x(x+1) + 1(x+1)} \def\click{\rlap{\enclose{roundedbox}{\small\text{next step}}}\hphantom{\longest}} \def\={\phantom{{} = {}}} (x+1)^2 \toggle {\begin{aligned}[t]& = \click\end{aligned}} {\begin{aligned}[t]& = (x+1)(x+1)\\[3px]&\=\click\end{aligned}} {\begin{aligned}[t]& = (x+1)(x+1)\\[3px]& = x(x+1) + 1(x+1)\\&\=\click\end{aligned}} {\begin{aligned}[t]& = (x+1)(x+1)\\[3px]& = x(x+1) + 1(x+1)\\[3px]& = (x^2+x) + (x+1)\\[3px]&\=\click\end{aligned}} {\begin{aligned}[t]& = (x+1)(x+1)\\[3px]& = x(x+1) + 1(x+1)\\[3px]& = (x^2+x) + (x+1)\\[3px]& = x^2 + (x + x) + 1\\[3px]&\=\click\end{aligned}} {\begin{aligned}[t]& = (x+1)(x+1)\\[3px]& = x(x+1) + 1(x+1)\\[3px]& = (x^2+x) + (x+1)\\[3px]& = x^2 + (x + x) + 1\\[3px]& = x^2 + 2x + 1\end{aligned}} \endtoggle $$

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