# 2.3: Oscillatory Solutions to Differential Equations

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 $$\PageIndex{1}$$).

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 ($$T(t)$$).