SOLUTION OF 3D HELMHOLTZ WAVE EQUATION USING GREEN’S
FUNCTION
No geometric spreading losses, no multiples, a layered media and no transmission losses
are unrealistic for many purposes. So for arbitrary acoustic media, we must solve the wave
equation using Green’s function.
The 3D Helmholtz equation is given by,
[∇2 + 𝜔2 𝑠(𝑟)2]P(r’)=F(r’)
𝜔
Where k(r)=𝑐(𝑟)
eqn(3.4.1)
for an inhomogeneous medium with velocity c(r), F(r) is harmonically
oscillating source at r and P(r’) is the pressure field.
The Green’s function associated with the 3-D Helmholtz equation for an arbitrary medium is
given by
[∇2 + 𝑘 2 ]G(𝑟, 𝑟 ′ )=δ(r’-r)
eqn(3.4.2)
Where ,
δ(r’-r) = δ(x-x’) δ(y-y’) δ(z-z’)
eqn(3.4.3)
For this second order partial differential equation there are two independent solution. One is an
outgoing Green’s function G(𝑟 ′ , 𝑟) and another one is its complex conjugate, the incoming
Green’s function. The outgoing Green’s function for a homogeneous medium with velocity c0 is
given by
G(𝑟, 𝑟 ′ ) = 
1 ik0 ( r  r ' )
e
/( r  r ' )
4
eqn(3.4.4)
The denominator term is geometrical spreading factor the represents the geometrical spreading
factor while the exponential phase factor is proportional to the with distance between the
'
observer r and the source r . Green’s function is the acoustic response measured at r for a
'
harmonically oscillating point source located at r .The source and receiver locations can be
interchanged so that G(𝑟 ′ , 𝑟) = G(𝑟, 𝑟 ′ ) which is the reciprocity principle. This says that a trace
recorded at position A excited by a source at B will be the same as the trace located at B for a
source excitation at A.
Multiplying the Helmholtz equation by the Green’s function gives
G(𝑟, 𝑟 ′ ) [∇2 + 𝑘 2 ]P(r’)=F(r’) G(𝑟, 𝑟 ′ )
eqn(3.4.5)
The differentiation ∇ is with respect to the primed coordinates. Use the identity G∇2 𝑃 = 𝑃∇2 𝐺 +
∇. (G∇P − P∇G) into the above equation gives,
P(r’) [∇2 + k 2 ] G(r, r ′ )+ ∇. (GVP − P∇G) = F(r’) G(r, r ′ )
eqn(3.4.6)
Integrating above equation over entire volume yields,
∰[P(r’)[∇2 + k 2 ]G(r, r ′ ) + ∇. (GVP − P∇G)] = ∰ F(r’) G(r, r ′ ) dv ′
∰ P(r’)δ(r’ − r)dv ′ + ∰ ∇. (GVP − P∇G) = ∰ F(r’) G(r, r ′ ) dv ′
Or by using Divergence theorem ∰ ∇. Fdv = ∯(F. n). ds
P(r’) +∯(GVP − P∇G). dA = ∰ F(r’) G(r, r ′ ) dv ′
Rearranging gives the integral equation solution to the Helmholtz equation
P(r’) − ∯(GVP − P∇G). dA + ∰ F(r’) G(r, r ′ ) dv ′
If only outgoing waves are considered and the surface integral is at infinity then the surface
integration is zero in the above equation. Therefore the solution becomes,
P(r’) = ∰ F(r’) G(r, r ′ ) dv ′