D'Alembert's solution of wave equation

Consider the initial value problem of Cauchy type given by

$$u_{tt}-c^2u_{xx} =0,~-\infty<x<\infty,~t>0,$$ (3.50)

$$u(x,0) =u_0(x),$$ (3.51)

$$u_t(x,0) =v_0(x),$$ (3.52)

where initial displacement $u_0(x)$ and the initial velocity $v_0(x)$ are assumed to be twice continuously differentiable functions. As we have derived in the previous Sec. , the general solution of the wave equation is given by

$$u(x,t)=F(x-ct)+G(x+ct),$$ (3.53)

where $F$ and $G$ are arbitrary functions of their respective arguments. By using Eq. () and Eq. (), one gets

$$u(x,0)=F(x)+G(x)=u_0(x),$$ (3.54)

and by using Eq. () and Eq. (), one gets

$$u_t(x,0)=-cF'(x)+cG'(x)=v_0(x).$$ (3.55)

By integrating Eq. () w.r.t. $x$, one gets

$$-F(x)+G(x)=\frac{1}{c}\int_{0}^{x}v_0(x)dx.$$ (3.56)

Solving Eq. () and Eq. () for $F(x)$ and $G(x)$, one gets

$$G(x)=\frac{u_0(x)}{2}+\frac{1}{2c}\int_{0}^{x}v_0(x)dx,$$

and

$$F(x)=\frac{u_0(x)}{2}-\frac{1}{2c}\int_{0}^{x}v_0(x)dx,$$

Using this in Eq. (), one gets

$$u(x,t)=\frac{1}{2}\big[u_0(x-ct)+u_0(x+ct)\big]+\frac{1}{2c}\big[-\int_{0}^{x-ct}v_0(x)dx+\int_{0}^{x+ct}v_0(x)dx\big]$$

$$u(x,t)=\frac{1}{2}\big[u_0(x-ct)+u_0(x+ct)\big]+\frac{1}{2c}\big[\int_{x-ct}^{x+ct}v_0(x)dx\big]$$ (3.57)

This is known as the D'Alembert solution of $1-D$ wave equation.

Remark This result shows that the two initial conditions (initial velocity and initial displacement) determine the solution of wave equation.