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

 

2U/x2 – (1/c2)2U/t2 = 0                               (eq.1)

 

such that

 

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

 

Ut(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)

 

2U/x2 – (1/c2)2U/t2 = 0                              

 

subject to the auxiliary conditions

 

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

 

Ut(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 = Uxx + Uyy  + Uzz                                                    (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 R3. Also let

 

∆U - (1/c2)Utt  =  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)

 

Ut(x,y,z,0) = F(x,y,z)                                  (eq.12)

 

is given by

 

U = Ft + G +

   (n=1 to ){c2n[(1/(2n+1)!)(∆nF)t2n+1 + (1/(2n)!)(∆nG)t2n]} 

 

                                                       (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)

 

2U/x2 – (1/c2)2U/t2 = 0                              

 

subject to the same auxiliary conditions of Problem 1a

 

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

 

Ut(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/c2)Utt  =  0                                    

 

will reduce to simply

 

2U/x2 – (1/c2)2U/t2 = 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” = d2F/dx2    

 

and

 

∆G = G” = d2G/dx2

 

respectively. We should also note that for a function

F of x alone, we will have,

 

∆(∆F) = ∆2F= d4F/dx4

 

∆(∆(∆F)) = ∆(∆2F) = ∆3F = d6F/dx6 , 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

 

2U/x2 - (1/c2)Utt  =  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)

 

Ut(x,0) = F(x)                                          (eq.18)

 

is given by

 

U = Ft + G +

   (n=1 to ){c2n[(1/(2n+1)!)(d2nF/dx2n)t2n+1 + (1/(2n)!)

 

    (d2nG/dx2n)t2n]},                                    (eq.19)

 

(where d2nF/dx2n  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)

 

Ut(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 d2nF/dx2n for positive

integers n

 

integer n                 d2nF/dx2n

 

  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 d2nG/dx2n for positive

integers n

 

integer n                 d2nG/dx2n

 

  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)+{c2[(1/3!)(-Sin(x))t3 +

   

    (1/2!)(-1/c)Cos(x)t2] +

 

   c4[(1/5!)(Sin(x))t5 +(1/4!)(1/c)Cos(x)t4] +

 

   c6[(1/7!)(-Sin(x))t7 +(1/6!)(-1/c)Cos(x)t6] +

 

   c8[(1/9!)(Sin(x))t9 +(1/8!)(1/c)Cos(x)t8]+ ... }

 

From which we have

 

U = Sin(x)[t – (1/3!)t3c2 + (1/5!)t5c4  – (1/7!)t7c6 + ...] +

 

    ((1/c)Cos(x))[1 – (1/2!)t2c2 + (1/4!)t4c4  – (1/6!)t6c6 + ...]

 

which gives us

 

U = (1/c)Sin(x)[tc – (1/3!)t3c3 + (1/5!)t5c5  – (1/7!)t7c7 + ...] +

 

    ((1/c)Cos(x))[1 – (1/2!)t2c2 + (1/4!)t4c4  – (1/6!)t6c6 + ...]

 

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)Cos(ct)

 

from which we have

 

U(x,t) = (1/c)[ Sin(x)Sin(ct) + (Cos(x)Cos(ct)]        (eq.24)

 

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)Cos(B)]          (eq.26)

 

Referring to a table of trigonometric identities

the reader will see that

 

Sin(A)Sin(B) + (Cos(A)Cos(B) = Cos(A – B)

 

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.