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Differential equations
The wave equation in higher dimensions: Huygens' principle

The wave equation in higher dimensions: Huygens' principle

In Sec. 2.6. we discussed the 1D wave equation and expressed its solution in terms of d'Alembert's solution; this may be applied to the classical example of the vibration of a one-dimensional object (like, for example, a string). Other applications of the 1D wave equation arise for physical models whose solutions may be described using one-dimensional waves; the latter are commonly referred to as plane waves. For example, sound waves that are sufficiently far away from the source may be modelled using the 1D1 \mathrm{D} wave equation. The 2D2 \mathrm{D} wave equation may be used to model the vibration of an ideal drum; twodimensional waves are referred to as cylindrical waves while one of the most famous examples modelled by the 3D wave equation is the propagation of sound in a fluid; three-dimensional waves are referred to as spherical waves.

The wave equation behaves differently in even and odd spatial dimensions. For odd dimensions, the solution satisfies a modified version of d'Alembert's solution discussed in the previous section. Further, a principle referred to as Huygens' principle holds. Consider the 3D wave equation:

utt=c2(uxx+uyy+uzz),u_{t t}=c^{2}\left(u_{x x}+u_{y y}+u_{z z}\right),

with cc constant and u=u(x,y,z,t)u=u(x, y, z, t) subject to the following initial conditions,

u(x,y,z,0)=ϕ(x,y,z),ut(x,y,z,0)=ψ(x,y,z).u(x, y, z, 0)=\phi(x, y, z), \quad u_{t}(x, y, z, 0)=\psi(x, y, z) .

The objective is to find an explicit formula for the solution of the initial-value problem like we did in the previous section arriving at the d'Alembert solution. Deriving the solution to the 3D3 \mathrm{D} wave equation is certainly a nontrivial task but we briefly discuss here the method of solution as well as an important property on how information propagates for the wave equation.

To solve the problem given by Eqs. (2.188) and (2.189), we consider the simpler problem given by Eqs. (2.188) and (2.189) but with ϕ=0\phi=0. The simpler problem can be solved using the Fourier transform and its solution is given by:

u(x,y,z,t)=tψˉu(x, y, z, t)=t \bar{\psi}

where ψˉ\bar{\psi} is the averaged initial disturbance of ψ\psi over the sphere of radius ct[cc t[c is the velocity in Eq. (2.188)], centred at the point (x,y,z)(x, y, z). This is given by the following integral,

ψˉ=0π02πψ(x+ctsinαcosθ,y+ctsinαsinθ,z+ctcosα)c2t2sinαdθdα\bar{\psi}=\int_{0}^{\pi} \int_{0}^{2 \pi} \psi(x+c t \sin \alpha \cos \theta, y+c t \sin \alpha \sin \theta, z+c t \cos \alpha) c^{2} t^{2} \sin \alpha d \theta d \alpha,

where spherical coordinates are used, i.e. x=rsinαcosθ,y=rsinαsinθx=r \sin \alpha \cos \theta, y=r \sin \alpha \sin \theta and z=rcosαz=r \cos \alpha with θ[0,2π]\theta \in[0,2 \pi] and α[0,π]\alpha \in[0, \pi]. This means that the solution at the centre of the sphere, i.e. u(x,y,z,t)u(x, y, z, t) to this part of the problem [i.e. with ϕ=0\phi=0 in Eqs. [2.189)] is given in terms of the average of the initial disturbance, ψ\psi on a sphere. It follows that ψ\psi radiates outward spherically with velocity cc.

To complete the solution to the original problem, we need to solve the other half of it given by the PDE (2.188) subject to (2.189) with ψ=0\psi=0 and some nonzero, initial ϕ\phi. With the help of Stokes' theorem, we obtain the solution to the second half of the problem in a similar way (using Fourier transform) as before; in particular:

u(x,y,z,t)=t(tϕˉ)u(x, y, z, t)=\frac{\partial}{\partial t}(t \bar{\phi})

where ϕˉ\bar{\phi} is the averaged ϕ\phi over the sphere, centred at (x,y,z)(x, y, z) with radius ctc t, as before. Due to the linearity of the PDE, we may write the solution to the original problem as:

u(x,y,z,t)=tψˉ+t(tϕˉ)u(x, y, z, t)=t \bar{\psi}+\frac{\partial}{\partial t}(t \bar{\phi})

Equation (2.193) is a generalization of d'Alembert's solution known as the Kirchoff formula (even though it was derived by Poisson).

The wave equation is classified as hyperbolic (see Sec. 2.8 for more details); for hyperbolic PDEs, all information propagates according to the speed, cc. In odd dimensions and for n3n \geq 3, it can be shown that all information moves at speed exactly equal to cc (never slower, never faster). This is the result known as Huygens' principle. Now, a wave represents a signal or a disturbance in a medium that gets propagated over time, carrying energy with it. Given an initial condition, the information propagates along sharp fronts and, once the front is passed a point, the information it was carrying, is forgotten. More specifically, consider Fig. 2.8(a): it shows a wave propagation for dimension n=3n=3 where the wave has a sharp leading edge and tail. Physically, one can think of sound waves in three-dimensions which reach an object (say, our ears) but die off instantaneously when the wave has passed the object. An explanation of this corresponding to the mathematical solution of the 3D equation is as follows: the solution shows that it depends only on the initial disturbances ϕ\phi and ψ\psi lying at a distance ctc t from the centre (x,y,z)(x, y, z).

Huygens' principle does not hold in two dimensions. Let us first obtain the mathematical solution of the 2D2 \mathrm{D} wave equation and then give an interpretation in terms of Huygens' principle. To solve the 2D2 \mathrm{D} equation, we let the initial disturbances ϕ\phi and ψ\psi depend only on xx and yy (i.e. two-dimensional disturbances that are constant in zz ). In other words, we have:

u(x,y,0)=ϕ(x,y),ut(x,y,0)=ψ(x,y).u(x, y, 0)=\phi(x, y), u_{t}(x, y, 0)=\psi(x, y) .

The solution to the 2D2 \mathrm{D} problem takes the form of Eq. (2.193) where the technique used is referred to as the method of descent and can be used to find the solution to the wave equation in even dimensions. We can evaluate Kirchoff's formula given by Eq. (2.193) at z=0z=0 and this gives rise to the 2D solution, as follows:

u(x,y,t)=t12πct(02π0ctψ(x1,y1)c2t2r2rdrdθ)ψˉ+t[t12πct02π0ct(ϕ(x1,y1)c2t2r2rdrdθ)ϕˉ]\begin{aligned} u(x, y, t) & =t \underbrace{\frac{1}{2 \pi c t}\left(\int_{0}^{2 \pi} \int_{0}^{c t} \frac{\psi\left(x_{1}, y_{1}\right)}{\sqrt{c^{2} t^{2}-r^{2}}} r d r d \theta\right)}_{\bar{\psi}} \\ & +\frac{\partial}{\partial t}[t \underbrace{\frac{1}{2 \pi c t} \int_{0}^{2 \pi} \int_{0}^{c t}\left(\frac{\phi\left(x_{1}, y_{1}\right)}{\sqrt{c^{2} t^{2}-r^{2}}} r d r d \theta\right)}_{\bar{\phi}}] \end{aligned}

where x1=x+rcosθx_{1}=x+r \cos \theta and y1=y+rsinθy_{1}=y+r \sin \theta. Note that in the 2D2 \mathrm{D} solution we are integrating ψ\psi and ϕ\phi over the interior of the circle [centred at (x,y)(x, y) ] i.e. from 0 to ctc t (recall ctc t represents the radius). Contrast this with the 3D3 \mathrm{D} case, where the integrals were obtained over the surface of the sphere. What this means is that the solution depends on all initial data within a circle of radius ctc t and not just on its circumference. This further implies that the solution propagates with a trailing wave that is not sharp since the disturbance does not die down instantaneously and the solution does not become zero abruptly as in the 3D3 \mathrm{D} case.

3{ }^{3} Note that, again, this is by no means a trivial task! As a physical example, waves generated by dropping a stone into a flat water surface, does not only generate a circular expanding wave, but also lots of concentric ripples behind the leading wave. This notion is also applied to catastrophical large waves in the ocean like a tsunami which leaves significant trailing action behind the sharp front. In terms of wave propagation this is sketched in Fig. [2.8(b): we still have a sharp front but the wave does not leave a sharp tail. Instead, a small fraction of energy travels slower than speed c. While the tail is not sharp however, it still decays and dampens to zero eventually [see Fig. 2.8(b)].

Figure 2.8: Wave propagation properties for (a) n=3n=3 and (b) n=2n=2.