We will now demonstrate that the General Solution process for the 3D wave equation is also applicable to the well known one

spatial dimensional case. A general requirement for all generalizations is that the general procedure also applies to the (original) special case. The reader will soon see that the

general solution of the 3D wave equation previously presented (see link above) is also applicable to the one spatial dimensional case.

It is well known that the D’Alembert Solution U = U(x,t)

of the equation

∂^{2}U/∂x^{2 }– (1/c^{2})∂^{2}U/∂t^{2} = 0 (eq.1)

such that

U(x,0) = b(x) (eq.2a)

U_{t}(x,0) = a(x) (eq.2b)

is given by

U(x,t) = (1/2)[b(x-ct) + b(x+ct)] +

+ (1/2c)*_{} (eq.3)

Now we will first find the solution of (eq.1),

that satisfies a specific set of auxiliary conditions,

by using D’Alembert’s formula and then we will find

the same solution by specializing the General Solution

of the 3D Wave Equation to the 1-spatial dimensional

case.

__Problem 1a__

Solve (eq.1)

∂^{2}U/∂x^{2 }– (1/c^{2})∂^{2}U/∂t^{2} = 0

subject to the auxiliary conditions

U(x,0) = (1/c)Cos(x) (eq.4a)

U_{t}(x,0) = Sin(x) (eq.4b)

So then comparing (eqs.2a & 2b) to (eqs. 4a & 4b)

we have

b(x) = (1/c)Cos(x) (eq.5a)

a(x) = Sin(x) (eq.5b)

Therefore

b(x-ct) = (1/c)Cos(x-ct) (eq.6a)

b(x+ct) = (1/c)Cos(x+ct) (eq.6b)

We also have

_{} = _{} = - [cos(x+ct) –
cos(x-ct)]

so

_{} = - [cos(x+ct) – cos(x-ct)] (eq.7)

After substituting (eq.6a), (eq.6b) and (eq.7) into

(eq.3) and simplifying we obtain

U(x,t) = (1/c)[cos(x-ct)] (eq.8)

The reader is most likely familiar with the

Laplacian Partial Differential Operator. It may

be defined as follows:

∆U =
U_{xx} + U_{yy} + U_{zz }(eq.8a)

We will now recall, from the link above, the_{}

following (using new equation # designations):_{}

Theorem I

Let

F = F(x,y,z)
(eq.9a)

G = G(x,y,z) (eq.9b)

be infinitely differentiable functions of x, y and z

in some domain D of R^{3}.
Also let

∆U -
(1/c^{2})U_{tt} = 0
(eq.10)

then a solution U = U(x,y,z,t) of (eq.10) that

satisfies the auxiliary conditions

U(x,y,z,0) = G(x,y,z) (eq.11)

U_{t}(x,y,z,0)
= F(x,y,z)
(eq.12)

is given by

U = Ft + G +

∑(n=1 to ∞){c^{2n}[(1/(2n+1)!)(∆^{n}F)t^{2n+1}
+ (1/(2n)!)(∆^{n}G)t^{2n}]}

(eq.13)

provided
that the indicated infinite series on the

right side
of (eq.13) converges in the domain D. Note

that the
notation ∑(n=1 to ∞) indicates that the quantity

in the
curly braces (on the right side of (eq.13) is to

be summed
from n = 1 to n = ∞ .

_{}

An informal proof is given in the link above.

The Theorem (I) above will enable us to solve

__Problem 1b__

Solve (eq.1)

∂^{2}U/∂x^{2 }– (1/c^{2})∂^{2}U/∂t^{2} = 0

subject to the same auxiliary conditions of __Problem 1a__

U(x,0) = (1/c)Cos(x) (eq.14a)

U_{t}(x,0) = Sin(x) (eq.14b)

by using the Theorem I stated above.

These auxiliary conditions have been re-labeled for easy

reference.

__Solution__

First we will fix both y and z, in the domain of the

function U = U(x,y,z,t), at zero so that in actuality U

becomes a function of x and t only. In that event the

equation

∆U -
(1/c^{2})U_{tt} = 0

will reduce
to simply

∂^{2}U/∂x^{2 }– (1/c^{2})∂^{2}U/∂t^{2} = 0

Likewise the functions F and G, from (eqs. 9a & 9b) will

become functions of x alone. Therefore the Laplacian

Partial Differential operator applied to F or G (which are

functions of x alone) will reduce to

∆F =
F” = d^{2}F/dx^{2 }

^{ }

and^{ }

^{ }

∆G =
G” = d^{2}G/dx^{2}

respectively. We should also note that for a function

F of x alone, we will have,

∆(∆F)
= ∆^{2}F= d^{4}F/dx^{4}

^{ }

∆(∆(∆F))
= ∆(∆^{2}F) = ∆^{3}F = d^{6}F/dx^{6
}, etc.^{}

Now we present a special case of Theorem I as follows:

Theorem I.a

Let

F = F(x)
(eq.15a)

G = G(x)
(eq.15b)

be infinitely differentiable functions of x

in some domain D of R.
Also let

∂^{2}U/∂x^{2} - (1/c^{2})U_{tt} =
0
(eq.16)

then a solution U = U(x,t) of (eq.16) that

satisfies the auxiliary conditions

U(x,0) = G(x) (eq.17)

U_{t}(x,0)
= F(x)
(eq.18)

is given by

U = Ft + G +

∑(n=1 to ∞){c^{2n}[(1/(2n+1)!)(d^{2n}F/dx^{2n})t^{2n+1}
+ (1/(2n)!)

(d^{2n}G/dx^{2n})t^{2n}]},
(eq.19)

(where d^{2n}F/dx^{2n }indicates the derivative of
order

2n of F with respect to x),

provided
that the indicated infinite series on the

right side
of (eq.19) converges in the domain D. Note

that the
notation ∑(n=1 to ∞) indicates that the quantity

in the
curly braces (on the right side of (eq.19) is to

be summed
from n = 1 to n = ∞ .

Now let

G(x) = (1/c)cos(x) (eq.20)

and let

F(x) = Sin(x) (eq.21)

then, referring to (eqs. 17 & 18), our auxiliary conditions

become

U(x,0) = (1/c)cos(x) (eq.22)

U_{t}(x,0)
= Sin(x)
(eq.23)

It will suffice to only compute a finite number of

the derivatives of G(x) and F(x).

At this time why don’t we take a finite break for

humor?

If there is one and if there is another

then surely one of them must be the other

but if you add a third one, then a trio

you will obtain.

But if there is one and there is no other

and if you then delete the one without

replacing it with another

then nothing will remain.

Now to compute a few of the derivatives of F and G.

For F(x) = Sin(x), we need to compute d^{2n}F/dx^{2n }for
positive

integers n

__integer n__ ^{ } __d ^{2n}F/dx^{2n}__

1 - Sin(x) = F”

2 Sin(x) = F””

3_{} - Sin(x)
= F^{(6)}^{}

4 Sin(x) = F^{(8)}

etc.........................

For G(x) = (1/c)Cos(x), we need to compute d^{2n}G/dx^{2n }for
positive

integers n

__integer n__ ^{ } __d ^{2n}G/dx^{2n}__

1 - (1/c)Cos(x) = G”

2 (1/c)Cos(x) = G””

3_{} - (1/c)Cos(x)
= G^{(6)}

4 (1/c)Cos(x) = G^{(8)}

etc., ...........................

After explicitly calculating the terms of the infinite series, on the right side of (eq.19), that corresponds to n =1, 2, 3

and 4, and also substituting for F and G, (eq.19) may be

written as

U = Sin(x)*t+(1/c)Cos(x)+{c^{2}[(1/3!)(-Sin(x))t^{3}
+

(1/2!)(-1/c)Cos(x)t^{2}] +

c^{4}[(1/5!)(Sin(x))t^{5}
+(1/4!)(1/c)Cos(x)t^{4}] +

c^{6}[(1/7!)(-Sin(x))t^{7} +(1/6!)(-1/c)Cos(x)t^{6}] +

c^{8}[(1/9!)(Sin(x))t^{9}
+(1/8!)(1/c)^{8}]+ ... }

From which
we have

U = Sin(x)[t – (1/3!)t^{3}c^{2}
+ (1/5!)t^{5}c^{4 }–
(1/7!)t^{7}c^{6 }+ ...] +

((1/c)Cos(x))[1 – (1/2!)t^{2}c^{2} + (1/4!)t^{4}c^{4 }– (1/6!)t^{6}c^{6 }+
...]

which gives
us

U =
(1/c)Sin(x)[tc – (1/3!)t^{3}c^{3}
+ (1/5!)t^{5}c^{5 }–
(1/7!)t^{7}c^{7 }+ ...] +

((1/c)Cos(x))[1 – (1/2!)t^{2}c^{2} + (1/4!)t^{4}c^{4 }– (1/6!)t^{6}c^{6 }+
...]

The reader
will recognize, in the equation above, the power series in the odd powers of tc
(or ct) immediately as Sin(ct) and the power series in the even powers of tc
(or ct) as Cos(ct). Accordingly we may write the equation above as

U =
(1/c)Sin(x)Sin(ct) + (1/c)Cos(x)

from which
we have

U(x,t) =
(1/c)[ Sin(x)Sin(ct) + (Cos(x)

Now let

A = x
(eq.25a)

and

B = ct
(eq.25b)

and then
substitute these results into (eq.24). We

have

U(x,t) =
(1/c)[ Sin(A)Sin(B) + (Cos(A)

Referring
to a table of trigonometric identities

the reader
will see that

Sin(A)Sin(B)
+ (Cos(A)

therefore
(eq.26) becomes

U(x,t) =
(1/c)[Cos(A – B)]

or

U(x,t) =
(1/c)[Cos(x – ct)]
(eq.27)

This
satisfies the objective of the current problem

since (eq.8)
and (eq.27) are identical.

The
numbers never slumber

as they increase in size

and the bumble bee is not a bumbler

as it elegantly flies.

But when they approach infinity

will the integers then rest?

And as surely as the nectars

attract the honey bee

the angles and the vectors

studied in geometry

will appear on some high school test.