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 U_x

and U_xx 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

U_xx + U_yy + U_zz – (1/c²)U_tt = 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 = U_xx + U_yy + U_zz (eq.2)

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

∆U - (1/c²)U_tt = 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)U_tt = 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)

U_t(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)U_tt = (1/c²)∑(n=1 to ∞){c²ⁿ[((2n+1)2n)/(2n+1)!)(∆ⁿF)t^2n-1 +

(2n(2n-1)/(2n)!)(∆ⁿG)t^2n-2]}

We have, after writing the terms of the summation that

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

(1/c²)U_tt = (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)U_tt = [(∆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)U_tt 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

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

(1/(2n-1)!)(∆ⁿG)t^2n-1]}

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

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

U_t(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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