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 ofU 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 waveequation of mathematical physics. It describes wave motionwhich 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

LU = Uxx + Uyy + Uzz (eq.2)

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

LU - (1/c2)Utt = 0

For a given function F the Laplacian may be appliedmultiple times. For example

L2F = L(LF)

L3F = L(L2F) = L(L(LF))

and more generally the symbol LnF denotes the appli-cation of the Laplacian operator, to the operand F, n successive times.

Often PDE with constant coefficients are solved bythe 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 zin some domain D of R3. Also let

LU - (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)!)(LnF)t2n+1 + (1/(2n)!)(LnG)t2n]}

(eq.7)

provided that the indicated infinite series on the right side of (eq.7) converges in the domain D. Notethat the notation ∑(n=1 to ∞) indicates that the quantity in the curly braces (on the right side of (eq.7) is tobe summed from n = 1 to n = ∞ .

The proof of the Theorem is as follows:

By the hypothesis that U is represented by a convergent powerseries it follows that term by term differeniation is valid.

Applying the Laplacian L to both sides of (eq.7)will give us

LU = (LF)t + LG + ∑(n=1 to ∞){c2n[(1/(2n+1)!)(Ln+1F)t2n+1 + (1/(2n)!)(Ln+1G)t2n]}

so that

(Ed. Note: to be proof read)

LU = (LF)t + LG + c2[(1/(3)!)(L2F)t3+(1/2!)(L2G)t2] +

c4[(1/5!)(L3F)t5 + (1/4!)(L3G)t4] +

c6[(1/7!)(L4F)t7 + (1/6!)(L4G)t6] +

c8[(1/9!)(L5F)t9 + (1/8!)(L5G)t8] +... (eq.8)

Now for each side of (eq.7) we will take the secondorder partial derivative with respect to t and thenmultiply the results by 1/c2 . We will obtain

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

(2n(2n-1)/(2n)!)(LnG)t2n-2]}

which gives us

(1/c2)Utt = (1/c2)[c2[((3*2)/3!)(LF)t + (2(1)/2!)(LG)t0] +

c4[(5*4/5!)(L2F)t3 + (4*3/4!)(L2G)t2] +

c6[(7*6/7!)(L3F)t5 + (6*5/6!)(L3G)t4] +

c8[(9*8/9!)(L4F)t7 + (8*7/8!)(L4G)t6] +

c10[(11*10/11!)(L5F)t9 + (10*9/10!)(L5G)t8] + ...]

which gives us after simplification

(1/c2)Utt = [(LF)t + (LG) +

c2[(1/3!)(L2F)t3 + (1/2!)(L2G)t2] +

c4[(1/5!)(L3F)t5 + (1/4!)(L3G)t4] +

c6[(1/7!)(L4F)t7 + (1/6!)(L4G)t6] +

c8[(1/9!)(L5F)t9 + (1/8!)(L5G)t8] + ...] (eq.9)

Now if we substitute the expression LU from (eq.8)and the expression for (1/c2)Utt from (eq.9) intothe 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 adomain D provided that the indicated infinite serieson the rihgt side of (eq.7) converges in theDomain D. As stated before, the infinite series willalways converge if F and G are polynomials.

We will now show that the auxiliary conditions arealso satisfied. Set t = 0 pn both sides of (eq.7) andobtain

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)!)(LnF)t2n + (1/(2n-1)!)(LnG)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