Wave equation

At this stage you might be thinking that you have seen the new method (the method of variables) to solve a wave equation. But that is not true. Prior to this cause on PDE you were unknowing using it. Let us see how.

Plane harmonic wave (Monochromatic wave): Suppose that we want to find a solution of a wave equation

$$u_{tt}=c^2u_{xx}$$

in the form

$$u(x,t)=f(x)\exp(\pm \imath \omega t),$$ (3.48)

where $c$ is velocity, $\omega$ is angular frequency and $f(x)$ is amplitude function and also such solution are called time harmonic wave. Now form Eq. (), one gets

$$u_x=f'(x)\exp(\pm \imath \omega t),~u_{xx}=f''(x)\exp(\pm \imath \omega t),$$

and

$$u_t=\pm\imath \omega t~u_{tt}=-\omega^2 u.$$

Putting all these in Eq. (), one have

$$-\omega^2 u=c^2f''(x)\exp(\pm \imath \omega t),$$

or

$$f''(x)+\frac{\omega^2}{c^2}f(x)=0.$$

This equation is know as Helmholtz equation. This an ODE, whose solution is

$$f(x)=c_1\exp(\frac{\imath \omega}{c}x)+c_2\exp(\frac{-\imath \omega}{c}x)$$

and the required solution of the wave equation is

$$u(x,t)=\big(c_1\exp(\frac{\imath \omega}{c}x)+c_2\exp(\frac{-\imath \omega}{c}x)\big)\exp(\pm \imath \omega t)$$

Express $k=\frac{\omega}{c}=$ wave number. Thus we have the solution of form

$$u(x,t)=A\exp(\imath(\pm kx\pm\omega t)).$$

These solutions you already know as a plane wave solutions which have number of application in electromagnetic optics.

$$\begin{cases} \exp(\imath( kx+\omega t)) & \\ \exp(\imath( kx-... ...a t) \\ \exp(\imath(-kx-\omega t))&=\exp-\imath(kx+\omega t)\\ \end{cases}$$

THe wave equation Eq. () can also be solved by transforming it to its canonical form.