# Wave Equation

For a quick intuitive derivation of the wave equation we consider an elastic string, that does not resist bending. This means if we bend the string into some shape it remains in this shape and does not bend back. Suppose we displace the string by pulling, how does this displacement of the string evolves in time as we free the string? If we assume that the force acting on a string element is proportinal to the concavity $F=u_{xx}$ of the string element with mass $m$ is accelerated with $a=u_{tt}$, we get the wave equation by Newtons law $F=ma$ as. \begin{align} m \cdot u_{tt}= k \cdot u_{xx} \end{align} and we arrive at the familiar form of the wave equation \begin{align} \frac{\partial^2 u}{\partial x^2} &= c^2 \frac{\partial^2 u}{\partial t^2}\\ u_{xx}&=c^2 u_{tt} \end{align} For a more rigoros physical derivation watch the video lectures listed at the end of the article. We are now interested in solutions of the wave equation.

## d'Alembert Formula

The wave equation has a general solution formula called d'Alembert formula, which is usually not the case for PDEs. One we know the initial conditions to a wave equation, we can write down the solution immediately. Let us again state the problem we want to solve \begin{align} \left( \frac{\partial^2 }{\partial x^2}-c^2 \frac{\partial^2 }{\partial t^2} \right) u &=0\\ \\ u(x,0)&=f(x) \dots \text{initial position}\\ u_t(x,0)&=g(x) \dots \text{initial velocity} \end{align} Notice, the linear operator \begin{align} \left( \frac{\partial^2 }{\partial x^2}-c^2 \frac{\partial^2 }{\partial t^2} \right)=\left( \frac{\partial }{\partial x}-c \frac{\partial }{\partial t} \right) \cdot \left( \frac{\partial }{\partial x}+c \frac{\partial }{\partial t} \right) \\ \end{align} nicely factors in the operators of the transport equation. Let us use this fact to construct a general solution to the wave equation. We know that the solution of the forward transport equation ($v>0$) is some function $p(x+ct)$ and for the backward transport equation ($v<0$) the solution is $q(x-ct)$. Now let us rewrite the wave equation by assuming that the solution to the an operator acting on $u$ gives the solution $h(x,t)$ \begin{align} \left( \frac{\partial }{\partial x}-c \frac{\partial }{\partial t} \right) \cdot \underbrace{\left( \frac{\partial }{\partial x}+c \frac{\partial }{\partial t} \right) u(x,t)}_{h(x,t)}=0\\ \end{align} and similarly \begin{align} \left( \frac{\partial }{\partial x}+c \frac{\partial }{\partial t} \right) \cdot \underbrace{\left( \frac{\partial }{\partial x}-c \frac{\partial }{\partial t} \right) u(x,t)}_{h(x,t)}=0\\ \end{align} We can write the wave equation as the system \begin{align} u_{t}-c \cdot u_{x}=w\tag{1} \end{align} and \begin{align} w_{t}+c \cdot w_{x}=0 \tag{2} \end{align} because plugging (1) into (2) gives \begin{align} (u_{t}-c u_{x})_{t}+c (u_{t}-c u_{x})_{x}=u_{tt}-c^2u_{xx}=0 \end{align} Thus the system of partial differential equations (1)-(2) is equivalent to the wave equation. The same thing is true for the system \begin{align} u_{t}+c \cdot u_{x}=v \end{align} \begin{align} v_{t}-c \cdot v_{x}=0 \end{align} with different signs. Since we now the solution to the (forward and backward) transport equation, we can calculate the solution to the wave equation as follows.

The solution to the transport equations are $w=f(x-ct)$ and $v=g(x+ct)$ respectively. So we have the two equations \begin{align} u_{t}-c \cdot u_{x}=f(x+ct) \tag{3} \end{align} \begin{align} u_{t}+c \cdot u_{x}=g(x-ct) \tag{4} \end{align} Adding equation (3) and (4) gives \begin{align} u_{t}=\frac{1}{2} [f(x+ct)+g(x-ct)] \end{align} Subtracting equation (3) from (4) gives \begin{align} u_{x}=\frac{1}{2c} [f(x+ct)-g(x-ct)] \end{align} Integrating these equations gives \begin{align} u(x,t)=\frac{1}{2c} [F(x+ct)-G(x-ct)]+ k_1(x) \end{align} and \begin{align} u(x,t)=\frac{1}{2c} [F(x+ct)-G(x-ct)]+k_2(t) \end{align} The two equations for $u(x,t)$ must be equal. The only way the integrating constants $k_1(x)$ and $k_2(t)$ can be equal is when they are constant. We can choose these constants zero, that will make no difference, because $F$ and $G$ are arbitrary functions and can include constants. This way we find, that the general solution of the wave equation is \begin{align} u(x,t)=\frac{1}{2c} [F(x+ct)-G(x-ct)]=p(x+ct)+q(x-ct) \end{align} where we renamed $p(x-ct)=-\frac{1}{2c} G(x+ct)$ and $q(x+ct)=\frac{1}{2c} F(x+ct)$.

Let us now use the initial conditions and the general solution to derive d'Alemberts formula. \begin{align} u(x,t)&=p(x-ct)+q(x+ct)\\ u(x,0)&=f(x)=p(x)+q(x) \end{align} and \begin{align} u_t(x,t)&=-c \cdot p'(x-ct)+c \cdot q'(x+ct)\\ u_t(x,0)&=g(x)=-cp'(x)+cq'(x) \end{align} Integrating $g(x)$ we get \begin{align} G(x)=-cp(x)+cq(x) \end{align} Now, if we add and subtract the equations \begin{align} G(x)/c=-p(x)+q(x) f(x)=p(x)+q(x) \end{align} from each other, this results in \begin{align} q(x)=\frac{1}{2}f(x)+\frac{1}{2c}G(x)\\ p(x)=\frac{1}{2}f(x)-\frac{1}{2c}G(x) \end{align} and from the general solution we can calculate \begin{align} u(x,t)=p(x+ct)+q(x-ct)=\frac{1}{2}f(x+ct)+\frac{1}{2c}G(x+ct)+\frac{1}{2}f(x-ct)-\frac{1}{2c}G(x-ct) \end{align} This is already what we aimed for, but let us bring it in a more convenient form \begin{align} u(x,t)&=\frac{1}{2} \left( f(x+ct)+f(x-ct) \right) +\frac{1}{2c} \left( G(x+ct)+-G(x-ct) \right)\\ &=\frac{1}{2} \left( f(x+ct)+f(x-ct) \right) +\frac{1}{2c} \left( \int_{0}^{x+ct} g(s) ds - \int_0^{x-ct} g(s) ds \right)\\ &=\frac{1}{2} \left( f(x+ct)+f(x-ct) \right) +\frac{1}{2c} \left( \int_{0}^{x+ct} g(s) ds + \int_{x-ct}^0 g(s) ds \right)\\ &=\frac{1}{2} \left( f(x+ct)+f(x-ct) \right) +\frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds \\ \end{align}

This is the celebrated d'Alembert formula \begin{align} u(x,t)&=\frac{1}{2} \left( f(x+ct)+f(x-ct) \right) +\frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds \\ \end{align}

solution to the wave equation \begin{align} \left( \frac{\partial^2 }{\partial x^2}-c^2 \frac{\partial^2 }{\partial t^2} \right) u &=0\\ \\ u(x,0)&=f(x) \dots \text{initial position}\\ u_t(x,0)&=g(x) \dots \text{initial velocity} \end{align}

Example

Solve the wave equation for initial conditions \begin{align} u(x,0)=\phi(x)=0\\ u_t(x,0)=\psi =cos(x) \end{align} \begin{align} u(x,t)&=\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(s) ds \\ u(x,t)&=\frac{1}{2c} \left( sin(x+ct)-sin(x-ct) \right) \\ \end{align} If we use the right trigonometric theorem, we see that \begin{align} u(x,t)&=\frac{1}{c} cos(x)sin(ct) \\ \end{align} the solution is a standing wave of the form $cos(x)$, that oscillates with the factor $sin(ct)$.

Video Lectures:

• PDE 7 - Wave equation: intuition 
• PDE 8 - Wave equation: derivation 
• PDE 9 - Wave equation: general solution 
• PDE 10 - Wave equation: d'Alembert's formula 
• NPTEL Mechanical Engineering - Vibrations of Strings 
• NPTEL Mechanical Engineering - Vibrations of Bars