Wave equation

Introduction

It belongs to the collection of famous equations.

The goal is to find $u (t, x, y, z)$ such that:

u_{t t} = c^{2} Δ u

where $Δ$ is the Laplacian operator.

Keep an eye: by abuse of language, a wave equation is called any equation that admits wave-like solutions, which take the form $f (\vec{x} \pm \vec{v} t)$ . The equation $\frac{\partial^{2} f}{\partial t^{2}} = c^{2} \nabla^{2} f$ , despite being called "the" wave equation, is not the only equation that does this.

Exponential solutions

To find exponential solutions to this wave equation, we can try a solution of the form:

u (t, x, y, z) = e^{i (k x + l y + m z - ω t)}

where $k, l, m$ are the wave numbers in the $x, y, z$ directions, respectively, and $ω$ is the angular frequency of the wave. The $i$ is the imaginary unit. This form of the solution is chosen because it represents a wave-like solution, which you mentioned as a characteristic of equations that can be called wave equations.

Let's calculate the second time derivative and the spatial derivatives of $u$ and substitute them into the wave equation:

$u_{t t} = - ω^{2} e^{i (k x + l y + m z - ω t)}$
$Δ u = (- k^{2} - l^{2} - m^{2}) e^{i (k x + l y + m z - ω t)}$

Substituting these into the wave equation, we get:

- ω^{2} e^{i (k x + l y + m z - ω t)} = c^{2} (- k^{2} - l^{2} - m^{2}) e^{i (k x + l y + m z - ω t)}

For this equation to hold for all $t, x, y,$ and $z$ , the coefficients of the exponentials on both sides must be equal. This leads to the dispersion relation:

ω^{2} = c^{2} (k^{2} + l^{2} + m^{2})

Thus, the exponential functions of the form $e^{i (k x + l y + m z - ω t)}$ are solutions to the wave equation provided that their wave numbers and frequency satisfy the dispersion relation $ω^{2} = c^{2} (k^{2} + l^{2} + m^{2})$ . This represents a wide range of possible solutions, as any $k, l,$ and $m$ that satisfy this relation will give a valid solution.

As a Cauchy problem

Wave equation stated as a Cauchy problem. With initial and boundary conditions

See this link

Appears D'Alembert formula:
The Cauchy problem for the one-dimensional homogeneous wave equation is given by

\begin{aligned} \begin{array}{r} u_{t t} - c^{2} u_{x x} = 0 - \infty < x < \infty, t > 0, \\ u (x, 0) = f (x), u_{t} (x, 0) = g (x), - \infty < x < \infty . \end{array} \end{aligned}

and the solution is given by d'Alembert's formula:

\begin{aligned} u (x, 0) = F (x) + G (x) = f (x) . \\ u_{t} (x, 0) = c F^{'} (x) - c G^{'} (x) = g (x) \\ u (x, t) = \frac{f (x + c t) + f (x - c t)}{2} + \frac{1}{2 c} \int_{x - c t}^{x + c t} g (s) d s, \end{aligned}

Derivation

See anotacioneslatex.tex

\subsection

The process above that explain the \textit{brainwave} to obtain the expression for the eigenvector $ξ (j)$ (go to equation \eqref{eigenvectorrelation}) can be performed with the original equation of motion for one of the central masses of several coupled oscillators.

Consider

m {\ddot{Ψ}}_{j} (t) = - k (Ψ_{j} (t) - Ψ_{j + 1} (t)) - k (Ψ_{j} (t) - Ψ_{j - 1} (t)) =

= k Ψ_{j - 1} (t) - 2 k Ψ_{j} (t) + k Ψ_{j + 1} (t)

We can try to augment the number of \textit{beads} to infinity, while reducing the \textit{distance} to 0. We will obtain a \textit{continuous} system, whose elements will be tagged with $x$ instead with $j$ . During the process we will call $d$ to the distance of the beads, so

x = j d

and

Ψ (x, t) = Ψ_{j} (t)

Also, the mass $m$ has to decrease when the distance gets shorter, because otherwise we would arrive to infinite density. This way, $m = μ d$ . For the same reason, the Hook constant $k$ must get greater, since because continuity the differences $Ψ_{j} - Ψ_{j - 1}$ trends to 0 and we wouldn't get oscillation at all.

μ d {\ddot{Ψ}}_{j} (t) = \frac{κ}{d} (Ψ_{j + 1} (t) - Ψ_{j} (t)) - \frac{κ}{d} (Ψ_{j} (t) - Ψ_{j - 1} (t))

or also

μ d \ddot{Ψ} (x, t) = \frac{κ}{d} (Ψ (x + d, t) - Ψ (x, t)) - \frac{κ}{d} (Ψ (x, t) - Ψ (x - d, t))

Taking $d \to 0$ ``in two steps'':

μ d {\ddot{Ψ}}_{j} (t) = κ \deriv Ψ (x + d / 2, t) x - κ \deriv Ψ (x - d / 2, t) x

and

\ddot{Ψ} (x, t) = \frac{κ}{μ} \deriv^{2} x^{2} Ψ (x, t)

which is known as the wave equation. It represents the \textit{continuous case} of infinite oscillating masses.

Usually is taken $v = \sqrt{\frac{κ}{μ}}$ , and the equation is rewritten as

\deriv^{2} t^{2} Ψ (x, t) = v^{2} \deriv^{2} x^{2} Ψ (x, t)

If we observe the previous dispersion relation in the form

ω_{k} = 2 \sqrt{\frac{T}{m d}} \sin (\frac{k d}{2})

and since $m = μ d$ and $k d$ is very small:

ω_{k} \approx 2 \sqrt{\frac{T}{μ d^{2}}} \frac{k d}{2} = \sqrt{\frac{T}{μ}} k = v k

But this last expression could be derived directly like the original dispersion relation: we are inspired in the discrete case and consider that a "normal mode" solution would be

Ψ (x, t) = e^{i ω t} ξ (x)

where the complex function $ξ (x)$ take the role of the eigenvector. We infer $ξ (x) = e^{i k x}$ and force $Ψ$ to verify the equation. From here we would obtain

ω_{k}^{2} = v^{2} k^{2}

But observe two things:

\begin

\item For every $k$ we get two ``basis'' solutions

Ψ^{k} (x, t) = e^{i ω_{k} t} e^{i k t}

and

{\tilde{Ψ}}^{k} (x, t) = e^{- i ω_{k} t} e^{i k t}

the combinations of them ( $k \in \RR$ ) produces all the solutions

Ψ (x, t) = \int_{- \infty}^{\infty} C (k) e^{i ω_{k} t} e^{i k x} + \tilde{C} (k) e^{- i ω_{k} t} e^{i k x} d k

\item Begin with

Ψ (x, t) = \int_{- \infty}^{\infty} C (k) Ψ^{k} + \tilde{C} (k) {\tilde{Ψ}}^{k} d k

Observe that $Ψ^{k} = (Ψ^{- k})^{*}$ and ${\tilde{Ψ}}^{k} = ({\tilde{Ψ}}^{- k})^{*}$ , so if want for a real solution $Ψ$ , i.e., $Ψ = Ψ^{*}$ , since:

Ψ (x, t)^{*} = \int_{- \infty}^{\infty} C (k)^{*} (Ψ^{k})^{*} + \tilde{C} (k)^{*} ({\tilde{Ψ}}^{k})^{*} d k = \int_{- \infty}^{\infty} C (k)^{*} Ψ^{- k} + \tilde{C} (k)^{*} {\tilde{Ψ}}^{- k} d k =

= \int_{- \infty}^{\infty} C (- k)^{*} Ψ^{k} + \tilde{C} (- k)^{*} {\tilde{Ψ}}^{k} d k

we conclude, by ``Fourier unicity'' (I'm not sure about this):

C (- k) = C (k)^{*}

\tilde{C} (- k) = \tilde{C} (k)^{*}

So we can write the real solution as

Ψ (x, t) = \int_{- \infty}^{0} C (k) Ψ^{k} + \tilde{C} (k) {\tilde{Ψ}}^{k} d k + \int_{0}^{+ \infty} C (k) Ψ^{k} + \tilde{C} (k) {\tilde{Ψ}}^{k} d k =

= \int_{0}^{+ \infty} C (k)^{*} (Ψ^{k})^{*} + \tilde{C} (k)^{*} ({\tilde{Ψ}}^{k})^{*} d k + \int_{0}^{+ \infty} C (k) Ψ^{k} + \tilde{C} (k) {\tilde{Ψ}}^{k} d k =

\begin{equation}\label

=\int_{0}^{+\infty} 2Re[C(k) \Psi^k +\tilde{C}(k) \tilde{\Psi}^k] dk

\end

\subsection

Suppose we want a, for the moment, complex solution that verifies the initial condition $Ψ (x, 0) = g (x) \in \CC$ for every $x \in \RR$ , we force:

g (x) = \int_{- π}^{π} (C (k) + \tilde{C} (k)) e^{i k x} d k

and therefore $C (k) + \tilde{C} (k)$ will be the Fourier transform of $g$ , $\hat{g}$ .

But we still have more freedom degrees, we can fit an ``initial velocity profile'' $h (x)$ :

\deriv Ψ t (x, 0) = h (x)

h (x) = \int_{- \infty}^{\infty} i ω_{k} C (k) e^{i k x} - i ω_{k} \tilde{C} (k) e^{i k x} d k = \int_{- \infty}^{\infty} i (ω_{k} C (k) - ω_{k} \tilde{C} (k)) e^{i k x} d k

and therefore $i (ω_{k} C (k) - ω_{k} \tilde{C} (k))$ is the Fourier transform of $h$ , $\hat{h}$ .

We then solve the equations:

{\begin{aligned} C (k) + \tilde{C} (k) = & \hat{g} \\ i (ω_{k} C (k) - ω_{k} \tilde{C} (k)) = & \hat{h} \end{aligned}

and we have finished.

\subsection

On the onther hand, observe that the general solution is:

Ψ (x, t) = \int_{- \infty}^{\infty} C (k) Ψ^{k} + \tilde{C} (k) {\tilde{Ψ}}^{k} d k

If we take $t = 0$ ,

Ψ (x, 0) = \int_{- \infty}^{\infty} C (k) e^{i k x} d k + \int_{- \infty}^{\infty} \tilde{C} (k) e^{i k x} d k = F (x) + G (x)

where the functions $F$ and $G$ are inverse Fourier transform of $C$ and $\tilde{C}$ respectively. Then

F (x + v t) = \int_{- \infty}^{\infty} C (k) e^{i k (x + v t)} = \int_{- \infty}^{\infty} C (k) e^{i k x + ω_{k} t}

and

G (x - v t) = \int_{- \infty}^{\infty} \tilde{C} (k) e^{i k (x - v t)} = \int_{- \infty}^{\infty} \tilde{C} (k) e^{i k x - ω_{k} t}

and therefore

Ψ (x, t) = F (x + v t) + G (x - v t)

\section

First, we are going to consider the case where we have two extremes are atached to a wall:

\begin

\includegraphics[width=8cm]

\end

So we have $Ψ (0, t) = Ψ (L, t) = 0$ . In particular,

Ψ (0, t) = F (v t) + G (- v t) = 0

for every $t$ . So we conclude that $G (y) = - F (- y)$ and so

Ψ (x, t) = F (x + v t) - F (- x + v t)

On the other hand, imposing $Ψ (L, t) = 0$ :

Ψ (L, t) = \int_{- \infty}^{\infty} C (k) e^{i k v t + i k L} d k - \int_{- \infty}^{\infty} C (k) e^{i k v t - i k L} d k =

= \int_{- \infty}^{\infty} C (k) e^{i k v t} [e^{i k L} - e^{- i k L}] d k = 0

Let $H (y)$ be the inverse Fourier transform of $C (k) [e^{i k L} - e^{- i k L}]$ . Then

H (v t) = Ψ (L, t) = 0

for every $t$ . So $H \equiv 0$ , and then

C (k) [e^{i k L} - e^{- i k L}] = 0

If we demand that the protagonist functions fulfill good properties then

e^{i k L} - e^{- i k L} = 0

and therefore

2 k L = 2 n π

and

k = \frac{n π}{L}

So the most general solution becomes

Ψ (x, t) = \int_{- \infty}^{\infty} C (k) e^{i k v t} [e^{i k x} - e^{- i k x}] d k = \int_{- \infty}^{\infty} C (k) e^{i k v t} 2 i \sin (k x) d k =

= \sum_{n = - \infty}^{n = \infty} C_{n} e^{i ω_{n} t} \sin (\frac{n π x}{L})

If we are looking for real solutions, mixing this ideas with the ones of equation \ref{realsolution2}:

Ψ (x, t) = C_{0} + \sum_{n = 1}^{+ \infty} 2 R e [C_{n} e^{i ω_{n} t}] \sin (\frac{n π x}{L}) =

= C_{0} + \sum_{n = 1}^{+ \infty} A_{n} \cos (\frac{n π v}{L} t + ϕ_{n}) \sin (\frac{n π x}{L})

\section

In the simplest wave equation

\deriv^{2} Ψ t^{2} = v^{2} \deriv^{2} Ψ x^{2}

we have a basis solutions of the form

e^{i (k x - ω t)}

where necessarily

v = \frac{ω}{k}

which is known as the dispersion relation.

But there are other equations, more complicated, that accept solutions of this form. They are sometimes called wave equation, too. But now, when we substitute the solution in the equation we obtain a more complicated relation between $ω$ and $k$ . In other words, the basis waves don't move at the same velocity. The relation is known as dispersive relation because it explain how the different components of the Fourier decomposition of a general solution move. The result is that every comonent move at a different speed so the original shape spreads out when time evolves.

\section

We consider that distance of the slit to the screen is far bigger than the slit separation. We will have constructive interference (bright point) when the difference of the path from both slits is a multiple of th wavelength $λ$ .

But since the screen is far away, we get a right angle:

\begin

\includegraphics[width=12cm]

\end

From here we have:

s i n (θ) = \frac{n λ}{d}

where $n \in \ZZ$ and $d$ is the slit separation.

The destructive interference occurs at

s i n (θ) = \frac{n λ / 2}{d}

A more useful setup is a \textbf{diffraction grating}, because we have more constructive interference and the bright spots are brighter. But the formula is the same.

\section

When a wave goes through a slit it creates an interference pattern in a screen with several fringes. The fringes respond to a formula. They form an angle $θ$ with the slit such that:

s i n (θ) = \frac{m λ}{w}

with $m \in \ZZ$ , $λ$ being the wavelength and $w$ the slit length.

\begin

\includegraphics[width=12cm]

\end