General Solution of 3D Wave Equation by Ronald H. Brady

 

Let U = U(x,y,z,t) be a function of the spatial coordinates, of a rectangular coordinate system, and also of the time t. Let Ux      

and Uxx  denote the first and second order partial derivatives of

U with respect to x respectively. Also let the partial derivatives of U with respect to y,z and t be denoted in a similar fashion.

 

Let us now consider the equation

 

Uxx + Uyy  + Uzz – (1/c2)Utt  =  0                           (eq.1)

                 

This equation will be recognized immediately as the wave

equation of mathematical physics. It describes wave motion

which has a constant speed of propagation c.

 

It is conventional to denote the constant speed of propa-

gation of light or electromagnetic radiation in vacuum by

c. If the wave motion involves a phenomenon other than light then the constant speed of the wave motion is denoted by v.

 

The reader will recall that the Laplacian partial dif-

ferential operator ∆ (applied to a function U) is defined

by

 

∆U = Uxx + Uyy  + Uzz                                                        (eq.2)

 

We may therefore re-write (eq.1) as

 

∆U - (1/c2)Utt  =  0                                    

 

For a given function F the Laplacian may be applied

multiple times. For example

 

2F = ∆(∆F)

 

3F = ∆(∆2F) = ∆(∆(∆F))

 

and more generally the symbol ∆nF denotes the appli-

cation of the Laplacian operator, to the operand F,

n successive times. 

 

Often PDE with constant coefficients are solved by

the method of Separation of Variables. But we now will present an alternative solution process that may be more straight-forward for many applications.

 

 

Theorem

 

Let

 

F = F(x,y,z)                                            (eq.3a)

 

G = G(x,y,z)                                            (eq.3b)

 

be infinitely differentiable functions of x,y and z

in some domain D of R3. Also let

 

∆U - (1/c2)Utt  =  0                                     (eq.4)                                     

 

then a solution U = U(x,y,z,t) of (eq.4) that

satisfies the auxiliary conditions

 

U(x,y,z,0) = G(x,y,z)                                   (eq.5)

 

Ut(x,y,z,0) = F(x,y,z)                                  (eq.6)

 

is given by

 

U = Ft + G +

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

 

                                                       (eq.7)

 

provided that the indicated infinite series on the

right side of (eq.7) 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.7) is to

be summed from n = 1 to n = ∞ .

 

The proof of the Theorem is as follows:

 

By the hypothesis that U is represented by a convergent power

series it follows that term by term differentiation is valid.

 

Applying the Laplacian ∆ to both sides of (eq.7)

will give us

 

∆U = (∆F)t + ∆G +

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

 

We obtain, after writing the terms of the summation that

corresponds to n = 1,2,3,4,..., the following

 

∆U = (∆F)t + ∆G + c2[(1/(3)!)(∆2F)t3+(1/2!)(∆2G)t2] +  

 

     c4[(1/5!)(∆3F)t5  + (1/4!)(∆3G)t4] + 

 

     c6[(1/7!)(∆4F)t7 + (1/6!)(∆4G)t6] + 

 

     c8[(1/9!)(∆5F)t9 + (1/8!)(∆5G)t8] +...       (eq.8) 

 

Now for each side of (eq.7) we will take the second

order partial derivative with respect to t and then

multiply the results by 1/c2 . We will obtain

 

(1/c2)Utt = (1/c2)∑(n=1 to ∞){c2n[((2n+1)2n)/(2n+1)!)(∆nF)t2n-1 +

 

           (2n(2n-1)/(2n)!)(∆nG)t2n-2]} 

 

We have, after writing the terms of the summation that

corresponds to n = 1,2,3,4,5,..., the following

 

 

(1/c2)Utt = (1/c2)[c2[((3*2)/3!)(∆F)t + (2(1)/2!)(∆G)t0] +

 

           c4[(5*4/5!)(∆2F)t3 + (4*3/4!)(∆2G)t2] +

 

           c6[(7*6/7!)(∆3F)t5 + (6*5/6!)(∆3G)t4] +

 

           c8[(9*8/9!)(∆4F)t7 + (8*7/8!)(∆4G)t6] +

 

           c10[(11*10/11!)(∆5F)t9 + (10*9/10!)(∆5G)t8] + ...] 

 

which gives us after simplification

 

(1/c2)Utt = [(∆F)t + (∆G) +

 

           c2[(1/3!)(∆2F)t3 + (1/2!)(∆2G)t2] +

 

           c4[(1/5!)(∆3F)t5 + (1/4!)(∆3G)t4] +

 

           c6[(1/7!)(∆4F)t7 + (1/6!)(∆4G)t6] +

 

           c8[(1/9!)(∆5F)t9 + (1/8!)(∆5G)t8] + ...]    (eq.9) 

 

 

Now if we substitute the expression ∆U from (eq.8)

and the expression for (1/c2)Utt  from (eq.9) into

the left side of (eq.4) and simplify we will

obtain 0 = 0, which indicates that U as defined by

(eq.7) satisfies the wave equation (eq.4) in a

domain D provided that the indicated infinite series

on the right side of (eq.7) converges in the

Domain D. Of course, the indicated infinite series will

always converge if F and G are polynomials: because of the

repeated application of the Laplacian ∆.

 

We will now show that the auxiliary conditions are

also satisfied. Set t = 0 on both sides of (eq.7) and

obtain

 

U(x,y,z,0) = G

 

U(x,y,z,0)= G(x,y,z)

 

which agrees with the auxiliary conditions of (eq.5).

 

Now take the partial derivative with respect to t of both sides of (eq.7). The resulting equation is

 

Ut =  F +  (n=1 to ){c2n[(1/(2n)!)(∆nF)t2n + 

    

     (1/(2n-1)!)(∆nG)t2n-1]}

 

now let t = 0 on both sides of the equation above. We will arrive at

 

Ut(x,y,z,0) = F

 

Ut(x,y,z,0) = F(x,y,z)

 

which agrees with (eq.6) of the auxiliary conditions.

 

This completes the proof.

 

Examples, illustrations and actual applications will

be available here

 

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