3.1 Review of Partial Differential Equations |
51 |
3.1Review of Partial Differential Equations
3.1.1Complex Solutions of a Partial Differential Equation
The acoustic wave equation is a second-order partial differential equation. The acoustic wave equation for a one-dimensional plane wave in Cartesian coordinates is:
∂2 pðx, tÞ ¼ 1 ∂2 pðx, tÞ ∂x2 c2 ∂t2
where p is sound pressure, x is a dimension, t is time, and c is the propagation speed of sound.
It can be directly shown (see Example 3.1) that when ωk ¼ c, the following four basic complex solutions:
pðx, tÞ ¼ Ae jðωtþkxþϕÞ
pðx, tÞ ¼ Ae jðωtþkxþϕÞ
pðx, tÞ ¼ Ae jðωt kxþϕÞ
pðx, tÞ ¼ Ae jðωt kxþϕÞ
satisfy the one-dimensional acoustic wave equation of p(x, t):
∂2 pðx, tÞ ¼ 1 ∂2 pðx, tÞ
∂x2 c2 ∂t2
where A is an amplitude (real number) and ϕ is a phase shift. The detailed derivation will be shown in Sect. 3.2. The following example is proof of the solutions:
Example 3.1 A complex exponential function p(x, t) is given as:
pðx, tÞ ¼ Ae jðωtþkxþϕÞ
where A is the amplitude (real number) and ϕ is the phase (angle). Show that when ωk ¼ c, the above complex exponential function p(x, t) satisfies the one-dimensional acoustic wave equation:
∂2 pðx, tÞ ¼ 1 ∂2 pðx, tÞ
∂x2 c2 ∂t2
52 3 Solutions of Acoustic Wave Equation
Example 3.1 Solution Step 1: The left-hand side of the given acoustic wave equation is ∂∂x22 pðx, tÞ.
2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2 |
|
|
|
|
|
|||||||||
Calculate |
|
∂ |
p by substituting the given function p(x, t) into |
∂ |
pðx, tÞ to arrive at: |
||||||||||||||||||||||||
∂x2 |
∂x2 |
||||||||||||||||||||||||||||
|
|
|
|
|
∂2 |
|
∂ |
|
|
|
∂ |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
||||
|
|
|
|
∂x2 p ¼ |
|
∂x |
∂x Ae jðωtþkxþϕÞ |
|
|
|
|||||||||||||||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
h |
|
|
|
|
i |
|
|
|
||||||
|
|
|
|
|
|
|
|
∂ |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|||
¼ |
|
|
|
|
njkhAe jðωtþkxþϕÞio |
|
|
|
|||||||||||||||||||||
|
∂x |
|
|
|
|||||||||||||||||||||||||
|
|
|
|
|
|
¼ k2hAe jðωtþkxþϕÞi |
|
∂2 |
|
||||||||||||||||||||
Step 2: The2 right-hand side of the given acoustic wave equation2 |
is |
pðx, tÞ. |
|||||||||||||||||||||||||||
∂t2 |
|||||||||||||||||||||||||||||
Calculate |
∂ |
p by substituting the given function p(x, t) into |
∂ |
pðx, tÞ to get: |
|||||||||||||||||||||||||
∂t2 |
∂t2 |
||||||||||||||||||||||||||||
|
|
|
|
|
∂2 |
|
∂ |
|
|
|
∂ |
Ae jðωtþkxþϕÞ |
|
|
|
||||||||||||||
|
|
|
|
|
∂t2 p ¼ |
|
∂t |
∂t |
|
|
|
||||||||||||||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
h |
|
|
|
|
i |
|
|
|
||||||
|
|
|
|
|
|
|
|
∂ |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|||
¼ |
|
|
|
hjωAe jðωtþkxþϕÞi |
|
|
|
||||||||||||||||||||||
|
∂t |
|
|
|
|||||||||||||||||||||||||
|
|
|
|
|
|
¼ ω2hAe jðωtþkxþϕÞi |
|
|
|
||||||||||||||||||||
|
|
|
|
|
|
|
|
|
|
|
2 |
|
|
2 |
|
|
|
|
|
|
|
|
|
||||||
Step 3: Substituting the calculated |
∂ |
p and |
∂ |
p into the acoustic wave equation |
|||||||||||||||||||||||||
2 |
2 |
||||||||||||||||||||||||||||
will give: |
|
|
|
|
∂x |
|
|
∂t |
|
|
|
||||||||||||||||||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
||||||||
|
|
|
|
|
|
|
|
|
∂2 |
1 |
|
|
∂2 |
|
|
|
|||||||||||||
|
|
|
|
|
|
|
|
|
|
|
p ¼ c2 |
|
|
p |
|
|
|
||||||||||||
|
|
|
|
|
|
|
|
∂x2 |
∂t2 |
|
|
|
|||||||||||||||||
|
|
|
|
! k2hAe jðωtþkxþϕÞi |
1 |
ω2hAe jðωtþkxþϕÞi |
|
|
|
||||||||||||||||||||
|
|
|
|
¼ |
|
|
|
|
|||||||||||||||||||||
|
|
|
|
c2 |
|
|
|
||||||||||||||||||||||
|
|
|
|
|
|
|
! |
|
|
|
k2 |
1 |
|
|
|
|
|
|
|
|
|
|
|
|
|
||||
|
|
|
|
|
|
|
|
|
|
|
¼ c2 |
|
|
|
|
|
|
|
|
|
|
|
|
||||||
|
|
|
|
|
|
|
|
|
ω2 |
|
|
|
|
|
|
|
|
|
|
|
|
||||||||
|
|
|
|
|
|
|
! |
|
|
|
|
ω |
¼ c |
|
|
|
|
|
|
|
|
|
|
|
|
||||
|
|
|
|
|
|
|
|
|
|
|
k |
|
|
|
|
|
|
|
|
|
|
|
|
||||||
It shows that when ωk ¼ c, the complex exponential function p(x, t) satisfies the one-dimensional acoustic wave equation.
3.1 Review of Partial Differential Equations |
53 |
3.1.2Trigonometric Solutions of a Partial Differential Equation
Similar to the complex solutions of the partial differential equation covered in the previous section, it can be directly shown that when ωk ¼ c, the following four basic complex solutions:
pðx, tÞ ¼ Aþc cos
pðx, tÞ ¼ Aþs sin
pðx, tÞ ¼ A c cos
pðx, tÞ ¼ A s sin
ωt kx þ ϕþωt kx þ ϕþ
ðωt þ kx þ ϕ Þ ðωt þ kx þ ϕ Þ
satisfy the one-dimensional acoustic wave equation of p(x, t):
∂2 pðx, tÞ ¼ 1 ∂2 pðx, tÞ
∂x2 c2 ∂t2
where A is an amplitude (real number) and ϕ is a phase shift. The detailed derivation will be shown in Sect. 3.2.
Example 3.2 A backward traveling wave p (x, t) (Form 1:RIP) is given as:
p ðx, tÞ ¼ A s sin ðωt þ kxÞ
where A s is an amplitude. Show that if ωk ¼ c, the above trigonometric function p(x, t) satisfies the one-dimensional acoustic wave equation:
∂2 pðx, tÞ ¼ 1 ∂2 pðx, tÞ ∂x2 c2 ∂t2
Example 3.2 Solution Step 1: The left-hand side of the given acoustic wave equation is:
∂2
∂x2 pðx, tÞ
Calculate ∂2 p by substituting the given function p(x, t) into ∂2 pðx, tÞ to get:
∂x2 ∂x2