> [!NOTE] > Consider the following initial value (or *Cauchy*) problem for the [[Wave Equation in 1 Spatial Dimension|wave equation in 1 spatial dimension]] $\begin{align} > \partial_{tt}u(x,t) &= c^2 \partial_{xx}u(x,t), & (x,t)\in \mathbb{R} \times (0,\infty), \\ > u(x,0) &= \Phi(x), & x\in \mathbb{R}, \\ > \partial_{t}u(x,0) &= V(x), & x\in \mathbb{R}. > \end{align}$where $u:\mathbb{R}\times(0,\infty)\to \mathbb{R}.$ > > If $\Phi\in C^2(\mathbb{R})$ and $V\in C^1(\mathbb{R})$ then the solution to the IVP is given by $u(x,t)=\frac{1}{2} \left(\Phi(x+ct)+\Phi(x-ct)\right) + \frac{1}{2c} \int_{x-ct}^{x+ct}V(r) dr.$ ###### Proof We know that [[General Solution to 1D Wave Equation|general solution]] to the wave equation is given by $u(x,t)=f(x+ct)+g(x-ct)$for some functions $f,g:\mathbb{R}\to \mathbb{R}.$ Assuming differentiability, the initial conditions yield $\begin{align} \Phi(x) = u(x,0) = f(x)+g(x) &\implies f'(x)+g'(x)=\Phi'(x), \\ V(x) = \partial_{t}u(x,0)=cf'(x)-cg'(x) & \implies f'(x)-g'(x) = \frac{1}{c}V(x). \end{align}$ Adding and subtracting the two equations yields $\begin{align} f'(x) &= \frac{1}{2} \left( \Phi'(x)+\frac{1}{c} V(x) \right), \\ g'(x) &= \frac{1}{2}\left( \Phi'(x)- \frac{1}{c} V(x) \right). \end{align}$ Integration yields $\begin{array}{rcl}f(x)&=&\frac{1}{2}\Phi(x)+\frac{1}{2c}\int_0^xV(r)dr+C_{1},\\g(x)&=&\frac{1}{2}\Phi(x)-\frac{1}{2c}\int_0^xV(r)dr+C_2\end{array}$ with integration constants $C_1$ and $C_2.$ Since $f(x)+g(x)=\Phi(x)$ we see that $C_1+C_2=0.$ So $f$ and $g$ may not be uniquely determined but the solution $u$ is:$\begin{aligned}u(x,t)&=f(x+ct)+g(x-ct)\\&=\frac{1}{2}\Phi(x+ct)+\frac1{2c}\int_0^{x+ct}V(r)dr+C_{1}+\frac12\Phi(x-ct)-\frac{1}{2c}\int_0^{x-ct}V(r)dr+C_2\\&=\frac12\Big(\Phi(x+ct)+\Phi(x-ct)\Big)+\frac{1}{2c}\int_{x-ct}^{x+ct}V(r)dr.\end{aligned}$ To see that this $u(x,t)$ does satisfy the IVP, ... Using [[Leibniz' Integral Rule (Differentiation under the Integral Sign)|Leibniz' rule]] yields $\partial_{x}\int_{x-ct}^{x+ct} V(r) \; dr = V(x+ct)-V(x-ct)$ .... # Properties Consider the following Cauchy problem $\begin{align} \partial_{tt}u(x,t) &= 36\partial_{xx}u(x,t), & (x,t)\in \mathbb{R}\times(0,\infty ), \\ u(x,0) &= e^{-4x^{2}}, & x\in \mathbb{R} \\ \partial_{t} u(x,0) &= 0 & x\in FR \end{align}$The solution is given by $\begin{align} u(x,t) &= \frac{e^{-4(x-6t)^{2}}+e^{-4(x+6t)^{2} }}{2} \\ &= \frac{e^{-4x^{2} +48xt -144t^{2}} + e^{-4x^{2} -48xt -144t^{2} }}{2} \\ &= e^{-4x^{2} -144t^{2}} \cosh(48xt) \end{align}$ See the solution over time: https://www.desmos.com/calculator/k5xozgbvgw. If the initial condition has a compact support, then the solution of the wave equation has a compact support at all times. Try plucked string example. # Sources 1. [[MA265-10 Methods of Modelling 3 Notes]].