Wave equation in Cartesian coordinates
• The wave equation is given in one spatial dimension
∂2u
1 ∂2u
=
∂x 2
v 2 ∂t 2
• We again use separation of variables u(x, t) = X (x)T (t), and
then we can write the wave equation as,
X 00
1 T 00
= 2
= −k 2
X
v T
• Hence we wind up with two Helmholtz equations to solve,
X 00 + k 2 X = 0
T 00 + k 2 v 2 T = 0
Wave equation in Cartesian coordinates, continued
• We find solutions T (t) = sin ωt, T (t) = cos ωt, X (x) = sin kx,
and X (x) = cos kx
• Here the angular frequency ω = kv , and k = 2π/λ
• From y (x, t) = X (x)T (t), we have four different basic solutions
y (x, t) = sin kx sin ωt
y (x, t) = sin kx cos ωt
y (x, t) = cos kx sin ωt
y (x, t) = cos kx cos ωt
• We can have linear combinations of solutions of this kind
depending on the boundary conditions and initial conditions
(superposition principle!)
Example: Waves on a string with fixed ends
• Imagine a string (e.g. a guitar string) fixed at the ends, so that
y (x = 0, t) = y (x = l, t) = 0
• We then see that y (x, t) = cos kx sin ωt and
y (x, t) = cos kx cos ωt are not acceptable, since cos 0 = 1
• To satisfy the boundary condition at x = l, we require that
sin kl = 0. There are many values k that can do this!
• We take kn = nπ/l with n = 1, 2, 3, ..., since sin kn l = sin nπ = 0
• From ω = kv , we get ωn = kn v = nπv
l and we have two solutions
y (x, t) = sin
nπx
nπvt
sin
l
l
y (x, t) = sin
nπx
nπvt
cos
l
l
Example: Waves on a string with fixed ends
• Then because we can apply the principle of superpostion, the
most general solution we can write is,
y (x, t) =
∞
X
n=1
sin
nπx h
nπvt
nπvt i
an cos
+ bn sin
l
l
l
• To determine the an and the bn we need initial conditions!
• We will get a Fourier series for the initial displacement and
velocity of the string
• Because the wave equation is second-order in time, we need two
initial conditions (and hence we have an and bn
Example: Waves on a string with fixed ends, initial
conditions
• We can solve if we know the displacement at two times, or the
velocity at two times
• More often, we specify the displacement and velocity at some
time, usually t = 0 since the origin of time we can always specify
• Assume we observe y (x, t = 0) = y0 (x) and dy
dx |t=0 = v0 (x)
y (x, t) =
∞
X
n=1
sin
nπx h
nπvt
nπvt i
an cos
+ bn sin
l
l
l
• At t = 0 the displacement is then
y (x, t = 0) = y0 (x) =
∞
X
n=1
an sin
nπx
l
Example: Waves on a string with fixed ends, initial
conditions
• We take the time-derivative of our expansion, and then set t = 0,
∞
X nπv dy
nπx
|t=0 = v0 (x) =
bn
sin
dt
l
l
n=1
• We see that both y0 (x) and v0 (x) are given by a Fourier series
• Notice compared to before, when we expanded periodic
functions, the y0 (x) and v0 (x) are not periodic with period l
• However, we can show that the sin nπx
l make a complete,
orthogonal set over the interval 0 < x < l
Z
l
sin
0
mπx
nπx
l
sin
dx = δm,n
l
l
2
Example: Waves on a string with fixed ends, initial
conditions
• So from the initial conditions we have
∞
X
y0 (x) =
an sin
n=1
v0 (x) =
∞
X
bn
n=1
nπx
l
nπv l
sin
nπx
l
• Using the orthogonality of the sin nπx/l terms,
an =
2
l
l
Z
2
bn =
nπv
y0 (x) sin
0
Z
nπx
dx
l
l
v0 (x) sin
0
nπx
dx
l