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



∆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.







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]} 




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 +




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



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 + 




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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