in Higher Dimensions (2D and 3D)
Starting with the
wave equation in one dimension, we can extend it into greater
dimensions. Due to graphical limitations, it is difficult to draw
multi-dimensional waves. As such, they are often schematically drawn as
The motion of the wave is drawn in red. In a multi-dimensional
environment, the wave will also have depth (that extends into the
computer screen). Instead of drawing the entire wave, we just draw the
blue lines above to denote the crests of the wave as they move through
the 3D environment. As this point, we introduce the concept of affine
transformations and affine functions. They will prove to be very useful
in describing and generating the wave equation in higher dimensions.
Transformations and Functions
Given a function Ax+By = C, how can we graph it? In general, we can
plot out a set of values. However, this is very tedious, and is too
specific to be of any general use. We can modify the formula into the
form y = mx +b , where m equals the slope, and b is the y-intercept.
When we try to solve for the function in y, we run into a distinct
= C - Ax
= -(A/B)x + (C/B) = mx + b
For most choices of A,B, and C the function is well defined. However,
if B equals zero, then this equation fails since you cannot divide
through by zero. We would then have to augment the equation as follows:
Ax+(0)y = C
Ax = C
x = (C/A)
For any general equation, it is inconvienent to constantly need to
consider two cases. Another method is required, we can first
change the given function of Ax+By = C by
re-writing as Ax+By - C = 0.
At this point, we'll consider the general case where C = 0.
When C = 0, the function reduces to Ax + By = 0. We can plot this
function by first graphing the line (or vector) [A,B], and then drawing
the line perpendicular to it that goes through the origin. We can see that this
general method works based on the first equation for the line that we
= -(A/B)x + (C/B). The vector [A,B]
has slope B/A, and the slope of the curve is -(A/B). The product
of these slopes is -1, which is the condition for perpendicular lines.
This method has the additional advantage in that it corrects against
the case where B = 0 because the vector [A,0] is simply the x axis, and
the vector perpendicular to it is the y axis. If A and B equal zero,
then the equation represents the origin, and is the degenerate case
where the line becomes a single point. The following is a graphic where
the blue line represents the vector [2,1], and the red line is graph of
2x + y = 0 (You can also see that the red line has slope of -2, and is
indeed perpendicular to the blue line).
While we have
considered the simple case of C = 0, we also need consider cases where C
is non-zero. The first thing that we should notice is that the formula y
= -(A/B)x + (C/B) guarantees that as
long as we can draw vector [A,B], the line of any Ax+ By = C will be
perpendicular to it. We need to see how the constant value C will
augment the original line where C = 0. Here's a graph of the lines 2x + y +C = 0, where
C = ±15, ±10, ±5, ±4, ±3, ±2, ±1 and 0. From left to
right, the value of C increases, starting from -15 on the far left to 15
on the far right. Each of lines remains parallel (as predicted by the
slope equation). It also appears as through the lines are spaced apart
relative to the value of the constant value, C.
This allows to interpret the general relation where we can graph
Ax+By=0, and then apply a correction factor for the constant.The value
of the distance (through the perpendicular) between each pairing of
lines is given by:
Distance = C2-C1
/[sqrt(A2+B2)] or when C1 equals 0, C2 /[sqrt(A2+B2)]
In this formula, if C is positive the graph will be above the graph of
Ax+By=0 towards the right, and in the direction of [A,B]. If C is
negative, then the graph is left of C = 0.
The Wave Equation in Higher Dimensions
We can use the result of the wave equation to solve the wave
equation in higher dimensions.
As above the red
curve represent a part of a three dimension wave that we can view. The
blue represent the crests of waves are hidden behind the front one.
There is a marker for the wavelength λ, which represents one
period of the wave. We can see that the crests of the wave are very
similar to the parallel lines from the example that we used for the
constant determination. In fact, λ happens to be
perpendicular to the waves, in a fashion similar to [A.B] in the
previous case, and is regarded as a multiple of [A.B].
We can write down some relations:
= k [A,B]
where k is a
constant, and [A,B] is perpendicular to the direction of motion (from
the definition of λ)
take the length of λ as defined by the
direction of motion to get:
k [sqrt (A2+B2) ]
If we calculate the signed distance of λ based on its
similar to C (for a given C) we
║λ║ = C / [sqrt (A2+B2)]
Now we equate the two
expressions for the length of λ:
║λ║ = C / [sqrt (A2+B2)]
[sqrt (A2+B2) ]
If we solve for k we
k = C / A2+B2
As a result, the
equation for λ is given by:
= k [A,B] = C / A2+B2×
At time =
0, the wave height equation for a 2D time independent curve is given by h(x,y) = A cos (ax + by). This is
the same as in 1D case , but with an extra factor for the y-dimension.
As in the 1D case, we can now apply the radian frequency correction
factor to generate our final equation of:
h(x,y) = A cos (ax + by - ωt),
where a and b are constants, and ω is the radial frequency function (2πc/λ).
This equation can be raised into any number
of dimensions by adding more dimension terms with constants.