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Chemistry LibreTexts

2.3: Oscillatory Solutions to Differential Equations

  • Page ID
    92349
  • Skills to Develop

    • Explain the following laws within the Ideal Gas Law:

    The boundary conditions for the string held to zero at both end argue that \(u(x,t)\) collapses to zero at the extremes of the string (Figure \(\PageIndex{1}\)).

    Standing_waves_on_a_string.gif

    Figure \(\PageIndex{1}\): Standing waves in a string (both spatially and temporally). Image used with permission from Wikipedia.

    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

    \[X(x) = A e^{ix} + B e^{-ix} \label{2.3.3}\]

    Exercise \(\PageIndex{1}\)

    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

    \[X(x) = A e^{ipx} + B e^{-ipx} \label{2.2.4}\]

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

    \[X(x) = A \left[\cos (px) + i \sin (px) \right] + B \left[ \cos (px) - i \sin (px) \right] \label{2.3.5}\]

    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

    \[ X(x=0)= c_1 \cos (0) + c_2 \sin (0) =0 \,\,\, at \; x=0 \label{2.3.8a}\]

    \[ c_1 + 0 = 0 \label{2.3.8b}\]

    \[c_1=0 \label{2.3.8c}\]

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

    \[ X(x=L) = c_1 \cos (pL) + c_2 \sin (pL) = 0 \,\,\, at \; x=L \label{2.3.9}\]

    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

    \[X(x) = c_2 \sin \left(\dfrac{n\pi x}{L} \right) \label{2.3.13}\]

    which can simplify to

    \[X(x) = c_2 \sin \left( \omega x \right) \label{2.3.14}\]

    with

    \[\omega=\dfrac{n\pi}{L}\]

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

    Exercise \(\PageIndex{1}\)

    Given two traveling waves:

    \[ \psi_1 = \sin{(c_1 x+c_2 t)} \; \textrm{ and } \; \psi_2 = \sin{(c_1 x-c_2 t)} \]

    1. Find the wavelength and the wave velocity of \( \psi_1 \) and \( \psi_2 \)
    2. Find the following and identify nodes:

    \[ \psi_+ = \psi_1 + \psi_2 \; \textrm{ and } \; \psi_- = \psi_1 - \psi_2 \]

    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 \( \psi_+ = \psi_1 + \psi_2 \; \textrm{ and } \; \psi_- = \psi_1 - \psi_2 \).

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

    This should have a node at every \( x= n \pi / c_1 \)

    And

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