How to Solve the Wave Equation in One Dimension

Step-by-Step Guide

1. 1

   Step 1: Begin with the wave equation.

   The wave equation can be derived from various fields in physics, but all wave equations in one dimension must satisfy the form below, where ψ=ψ(x,t){\displaystyle \psi =\psi (x,t)} is the displacement function and c{\displaystyle c} is a constant.
   
   As usual, ψ(±∞)=0.{\displaystyle \psi (\pm \infty )=0.} 1c2∂2ψ∂t2=∂2ψ∂x2{\displaystyle {\frac {1}{c^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}={\frac {\partial ^{2}\psi }{\partial x^{2}}}}
2. 2

   Step 2: Transform into the k{\displaystyle k} space.

   The point of integral transforms such as the Fourier transform is to simplify the problem in a different space.
   
   Once we are done, we then transform back into position space, as desired.
   
   In this article, we insert a normalization constant out front for symmetry with the inverse transform.
   
   Let's start with the left side.
   
   We can pull the time derivatives out of the integral because the integral does not depend on time. 1(2π)1/2∫−∞∞1c2∂2ψ∂t2e−ikxdx=1c2∂2∂t21(2π)1/2∫−∞∞ψe−ikxdx=1c2∂2ψ~(k,t)∂t2{\displaystyle {\begin{aligned}{\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }{\frac {1}{c^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}e^{-ikx}{\mathrm {d} }x&={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}{\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }\psi e^{-ikx}{\mathrm {d} }x\\&={\frac {1}{c^{2}}}{\frac {\partial ^{2}{\tilde {\psi }}(k,t)}{\partial t^{2}}}\end{aligned}}} Now we move on to the right.
   
   We can perform integration by parts twice to get the −k2{\displaystyle
   -k^{2}} constant out of the integral. 1(2π)1/2∫−∞∞∂2ψ∂x2e−ikxdx=1(2π)1/2∫−∞∞−k2ψe−ikxdx=−k2ψ~(k,t){\displaystyle {\begin{aligned}{\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }{\frac {\partial ^{2}\psi }{\partial x^{2}}}e^{-ikx}{\mathrm {d} }x&={\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }-k^{2}\psi e^{-ikx}{\mathrm {d} }x\\&=-k^{2}{\tilde {\psi }}(k,t)\end{aligned}}} We have transformed into the k{\displaystyle k} space.
   
   Now, we equate the two results we just obtained (we move the c2{\displaystyle c^{2}} to the right) to obtain the wave equation in this space. ∂2ψ~∂t2=−k2c2ψ~{\displaystyle {\frac {\partial ^{2}{\tilde {\psi }}}{\partial t^{2}}}=-k^{2}c^{2}{\tilde {\psi }}} , This is a linear, homogeneous, ordinary differential equation, which is considerably easier to solve.
   
   Set up the characteristic equation.
   
   For equations of this form, we try ψ~=erx,{\displaystyle {\tilde {\psi }}=e^{rx},} because derivatives of the exponential function are multiples of itself.
   
   Then, we see that (r2+k2c2)erx=0.{\displaystyle (r^{2}+k^{2}c^{2})e^{rx}=0.} Since erx{\displaystyle e^{rx}} can never be 0, we are looking for roots of the equation below. r2+k2c2=0{\displaystyle r^{2}+k^{2}c^{2}=0} Solve the characteristic equation for r.{\displaystyle r.} r=±ick{\displaystyle r=\pm ick} By linearity, the general solution is ψ~=F~e−ickt+G~eickt,{\displaystyle {\tilde {\psi }}={\tilde {F}}e^{-ickt}+{\tilde {G}}e^{ickt},} where F~{\displaystyle {\tilde {F}}} and G~{\displaystyle {\tilde {G}}} are arbitrary constants. , Now that we have ψ~,{\displaystyle {\tilde {\psi }},} all we need to do is to transform back to obtain the physically meaningful result. ψ(x,t)=1(2π)1/2∫−∞∞ψ~(k,t)eikxdk=1(2π)1/2∫−∞∞(F~e−ickt+G~eickt)eikxdk=1(2π)1/2∫−∞∞(F~eik(x−ct)+G~eik(x+ct))dk=F(x−ct)+G(x+ct){\displaystyle {\begin{aligned}\psi (x,t)&={\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }{\tilde {\psi }}(k,t)e^{ikx}{\mathrm {d} }k\\&={\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }({\tilde {F}}e^{-ickt}+{\tilde {G}}e^{ickt})e^{ikx}{\mathrm {d} }k\\&={\frac {1}{(2\pi )^{1/2}}}\int _{-\infty }^{\infty }({\tilde {F}}e^{ik(x-ct)}+{\tilde {G}}e^{ik(x+ct)}){\mathrm {d} }k\\&=F(x-ct)+G(x+ct)\end{aligned}}} , The general solution below, called d'Alembert's solution, describes one-dimensional waves, where F{\displaystyle F} and G{\displaystyle G} are any functions, as long as they contain their respective arguments x−ct{\displaystyle x-ct} and x+ct.{\displaystyle x+ct.} This solution describes a linear combination of waves propagating in the positive and negative x{\displaystyle x} directions, respectively.
   
   We can clearly see that c{\displaystyle c} denotes the velocity of the wave; for example, for light waves, c{\displaystyle c} denotes the speed of light. ψ(x,t)=F(x−ct)+G(x+ct){\displaystyle \psi (x,t)=F(x-ct)+G(x+ct)}
3. 3

   Step 3: Solve the above differential equation for ψ~{\displaystyle {\tilde {\psi }}}.

4. 4

   Step 4: Transform back into the x{\displaystyle x} space.

5. 5

   Step 5: Arrive at the solution.

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