# 2.3: Oscillatory Solutions to Differential Equations

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##### Learning Objectives
• Explore the basis of the oscillatory solutions to the wave equation
• Understand the consequences of boundary conditions on the possible solutions
• Rationalize how satisfying boundary conditions forces quantization (i.e., only solutions with specific wavelengths exist)

The boundary conditions for the string held to zero at both ends argue that $$u(x,t)$$ collapses to zero at the extremes of the string (Figure 2.3.1 ).

Unfortunately, when $$K>0$$, the general solution (Equation 2.2.7) results in a sum of exponential decays and growths that cannot achieve the boundary conditions (except for the trivial solution); hence $$K<0$$. This means we must introduce complex numbers due to the $$\sqrt{K}$$ terms in Equation 2.2.5. So we can rewrite $$K$$:

$K = - p^2 \label{2.3.1}$

and Equation 2.2.4b can be

$\dfrac{d^2X(x)}{dx^2} +p^2 X(x) = 0 \label{2.3.2}$

The general solution to differential equations of the form of Equation \ref{2.3.2} is

$X(x) = A e^{ipx} + B e^{-ipx} \label{2.3.3}$

##### Example 2.3.1

Verify that Equation $$\ref{2.3.3}$$ is the general form for differential equations of the form of Equation $$\ref{2.3.2}$$, which when substituted with Equation $$\ref{2.3.1}$$ give

$X(x) = A e^{ipx} + B e^{-ipx} \nonumber$

###### Solution

Expand the complex exponentials into trigonometric functions via Euler formula ($$e^{i \theta} = \cos \theta + i\sin \theta$$)

$X(x) = A \left[\cos (px) + i \sin (px) \right] + B \left[ \cos (px) - i \sin (px) \right] \nonumber$

collecting like terms

$X(x) = (A + B ) \cos (px) + i (A - B) \sin (px) \label{2.3.6}$

Introduce new complex constants $$c_1=A+B$$ and $$c_2=i(A-B)$$ so that the general solution in Equation $$\ref{2.3.6}$$ can be expressed as oscillatory functions

$X(x) = c_1 \cos (px) + c_2 \sin (px) \label{2.3.7}$

Now let's apply the boundary conditions from Equation 2.2.7 to determine the constants $$c_1$$ and $$c_2$$. Substituting the first boundary condition ($$X(x=0)=0$$) into the general solutions of Equation $$\ref{2.3.7}$$ results in

\begin{align} X(x=0) = c_1 \cos (0) + c_2 \sin (0) &=0 \nonumber \\[4pt] c_1 + 0 &= 0 \nonumber \\[4pt] c_1 &=0 \label{2.3.8c} \end{align}

and substituting the second boundary condition ($$X(x=L)=0$$) into the general solutions of Equation $$\ref{2.3.7}$$ results in

$X(x=L) = c_1 \cos (pL) + c_2 \sin (pL) = 0 \label{2.3.9}$

we already know that $$c_1=0$$ from the first boundary condition so Equation $$\ref{2.3.9}$$ simplifies to

$c_2 \sin (pL) = 0 \label{2.3.10}$

Given the properties of sines, Equation $$\ref{2.3.9}$$ simplifies to

$pL= n\pi \label{2.3.11}$

with $$n=0$$ is the trivial solution that we ignore so $$n = 1, 2, 3...$$.

$p = \dfrac{n\pi}{L} \label{2.3.12}$

Substituting Equations $$\ref{2.3.12}$$ and $$\ref{2.3.8c}$$ into Equation $$\ref{2.3.7}$$ results in

$X(x) = c_2 \sin \left(\dfrac{n\pi x}{L} \right) \nonumber$

which can simplify to

$X(x) = c_2 \sin \left( \omega x \right) \nonumber$

with

$\omega=\dfrac{n\pi}{L} \nonumber$

A similar argument applies to the other half of the ansatz ($$T(t)$$).

##### Exercise 2.3.1

Given two traveling waves: $\psi_1 = \sin{(c_1 x+c_2 t)} \; \textrm{ and } \; \psi_2 = \sin{(c_1 x-c_2 t)} \nonumber$

1. Find the wavelength and the wave velocity of $$\psi_1$$ and $$\psi_2$$
2. Find the following and identify nodes: $\psi_+ = \psi_1 + \psi_2 \; \textrm{ and } \; \psi_- = \psi_1 - \psi_2 \nonumber$

Solution a:

$$\psi_1$$ is a sin function. At every integer $$n \pi$$ where $$n=0,\pm 1, \pm 2, ...$$, a sin function will be zero. Thus, $$\psi_1 = 0$$ when $$c_1 x + c_2 t = \pi n$$. Solving for the x, while ignoring trivial solutions:

$x = \frac{n \pi - c_2 t}{c_1} \nonumber$

The velocity of this wave is:

$\frac{dx}{dt} = -\frac{c_2}{c_1} \nonumber$

Similarly for $$\psi_2$$. At every integer $$n \pi$$ where $$n=0,\pm 1, \pm 2, ...$$, a sin function will be zero. Thus, $$\psi_2 = 0$$ when $$c_1 x - c_2 t = \pi n$$. Solving for x, for $$\psi_2$$:

$x = \frac{n \pi + c_2 t}{c_1} \nonumber$

The velocity of this wave is:

$\frac{dx}{dt} = \frac{c_2}{c_1} \nonumber$

The wavelength for each wave is twice the distance between two successive nodes. In other words,

$\lambda = 2(x_{n} - x_{n-1}) = \frac{2 \pi}{c_1} \nonumber$

Solution b:

Find $$\psi_+ = \psi_1 + \psi_2 \; \textrm{ and } \; \psi_- = \psi_1 - \psi_2$$.

\begin{align*} \psi_+ &= \sin (c_1 x + c_2 t) + \sin (c_1 x - c_2 t) \\[4pt] &= \sin (c_1 x ) \cos (c_2 t) + \cancel{\cos(c_1 x) \sin(c_1 t)} + \sin (c_1 x ) \cos (c_2 t) - \cancel{\cos(c_1 x) \sin(c_1 t)} \\[4pt] &= 2\sin (c_1 x ) \cos (c_2 t) \end{align*} \nonumber

This should have a node at every $$x= n \pi / c_1$$ and

\begin{align*} \psi_- &= \sin (c_1 x + c_2 t) - \sin (c_1 x - c_2 t) \\[4pt] &= \cancel{\sin (c_1 x ) \cos (c_2 t)} + \cos(c_1 x) \sin(c_1 t) - \cancel{\sin (c_1 x ) \cos (c_2 t)} + \cos(c_1 x) \sin(c_1 t) \\[4pt] &= 2\cos (c_1 x ) \sin (c_2 t) \end{align*} \nonumber

2.3: Oscillatory Solutions to Differential Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.