Three dimensional wave equation

Three dimensional wave equation

Henceforth we will define a wave generally as any solution to the wave equation. We now try to find a three-dimensional equation governing wave motion in time and space i.e., the three-dimensional wave equation. The goal then is to derive a connection between partial derivatives of a wave disturbance. Before we begin, it is worth noting that nothing in the one-dimensional wave equation gave preference to the x-direction. That is, the wave equation we derive should be symmetric in spatial coordinates or in other words, we should be able to replace any spatial coordinate by any other spatial coordinate and leave the equation unchanged.

Deriving the 3D Wave Equation

Lets start with our three-dimensional representation of a plane wave

$\psi \left(\overset{\rightharpoonup }{r},t\right)=e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}$

and take partial derivatives

$\partial _x^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-k_x^2 e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}, \\ \partial _y^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-k_y^2 e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}, \\ \partial _z^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-k_z^2 e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}, \\ \partial _t^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-\omega^2 e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}$

adding the three spatial derivatives and simplifying we get

$\partial _x^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _y^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _z^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-e^{i \left(k_x x+k_y y+k_z z-\omega t \right)} (k_x^2+k_y^2+k_z^2)$

we recognize the exponential as the wave itself

$\partial _x^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _y^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _z^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-\psi \left(\overset{\rightharpoonup }{r},t\right)(k_x^2+k_y^2+k_z^2)$

we also note that the sum in the parenthesis is the square modulus of the wave vector i.e.,

$\left| \overset{\rightharpoonup }{k}\right| ^2\equiv k^2=(\sqrt{\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{k}})^2=k_x^2+k_y^2+k_z^2$

inserting this information

$\partial _x^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _y^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _z^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-k^2 \psi \left(\overset{\rightharpoonup }{r},t\right)$

remembering that ω = v k, the temporal derivative becomes

$\partial _t^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-(v k)^2 e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}$

or, rearranged

$\frac{1}{v^2}\partial _t^2\psi \left(\overset{\rightharpoonup }{r},t\right)=-k^2 e^{i \left(k_x x+k_y y+k_z z-\omega t \right)}$

inserting this into our differential equation, we arrive at

$\partial _x^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _y^2\psi \left(\overset{\rightharpoonup }{r},t\right)+\partial _z^2\psi \left(\overset{\rightharpoonup }{r},t\right)=\frac{1}{v^2}\partial _t^2\psi \left(\overset{\rightharpoonup }{r},t\right)$

the three-dimensional wave equation. Note that we have, albeit very heuristically derived a three-dimensional wave equation whose spatial coordinates are symmetric, as desired. The three-dimensional wave equation is usually written in more concise form by introducing the Laplacian operator given by the divergence of the gradient operator, where the gradient is defined as

$\overset{\rightharpoonup }{\nabla }=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y} \hat{j}+\frac{\partial }{\partial z}\hat{k}$

the Laplacian is

$\overset{\rightharpoonup }{\nabla }\cdot \left(\overset{\rightharpoonup }{\nabla }\right)\equiv \nabla ^2=\partial _x^2+\partial _y^2+\partial _z^2$

Using this notation, the wave equation is now compactly written as

$\nabla ^2\psi =\frac{1}{v^2}\partial _t^2\psi$

D'Alembert's Solution

At this point it is worth noting that back in one-dimension e.g.

$\partial _x^2\psi =\frac{1}{v^2}\partial _t^2\psi$

the wave equation may be factored as follows: First we move all terms to one side

$v^2\partial _x^2\psi -\partial _t^2\psi =0$

and then we can write

$\left(v \partial _x-\partial _t\right)\left(v \partial _x+\partial _t\right)\psi =0$

the proof of this is given by simply writing out the operator

$\left(v \partial _x-\partial _t\right)\left(v \partial _x+\partial _t\right)=v \partial _x\left(v \partial _x+\partial _t\right)-\partial _t\left(v \partial _x+\partial _t\right)$

which is

$\left(v \partial _x-\partial _t\right)\left(v \partial _x+\partial _t\right)=v^2\partial _x^2+v\partial_x \partial_t-v\partial _t\partial _x-\partial _t^2$

note that as long as the function has a continuous second derivative in both t and x we can switch the order of partial derivatives and cancel the two central terms leaving

$\left(v \partial _x-\partial _t\right)\left(v \partial _x+\partial _t\right)= v^2\partial _x^2-\partial _t^2$

which is the desired result. In this case, the factored one-dimensional wave equation is

$\left(v \partial _x-\partial _t\right)\left(v \partial _x+\partial _t\right)\psi =0$

now define

$u\equiv \frac{\partial \psi }{\partial t}+v \frac{\partial \psi }{\partial x}$

then the wave equation is

$v \frac{\partial u}{\partial x}-\frac{\partial u}{\partial t}=0$

(again, this can shown to be true by simply inserting u again). From this we see that u must have the form

$u(x,t)=h(x+v t)$

performing the substitution $\xi \equiv x+ v t$, changes partial derivatives in the following way

$\frac{\partial }{\partial \xi }=\frac{\partial }{\partial x}\frac{\partial x}{\partial \xi }=\frac{\partial }{\partial x}$

and

$\frac{\partial }{\partial \xi }=\frac{\partial }{\partial t}\frac{\partial t}{\partial \xi }=\left(\frac{\partial \xi }{\partial t}\right)^{-1}\frac{\partial }{\partial t}=\frac{1}{v}\frac{\partial }{\partial t}$

so that

$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi }=\frac{\partial h(\xi )}{\partial \xi }$

and similarly

$\frac{\partial u}{\partial t}=v \frac{\partial h(\xi )}{\partial \xi }$

plugging this in we see

$v \frac{\partial h(\xi )}{\partial \xi }-v \frac{\partial h(\xi )}{\partial \xi }=0$

which is true. Plugging in the result for u into our equation for ψ, we have the inhomogeneous partial differential equation

$\frac{\partial \psi }{\partial t}+v \frac{\partial \psi }{\partial x}=h(x+v t)$

now we seek a solution in the form

$\psi =\psi _{\hom }+\psi _{\text{par}}$

where $\psi _{\hom }$ is the solution to the homogeneous equation

$\frac{\partial \psi _{\hom }}{\partial t}+v \frac{\partial \psi _{\hom }}{\partial x}=0$

which, in this case, by the same reasoning above, is given by another function

$\psi _{\hom }=g(x-v t)$

(note the sign change on the v t). We can now "guess" a particular solution of the form

$\psi _{\text{par}}=f(x+v t)$.

Plugging this into the first order partial differential equation for ψ yields

$\frac{\partial \psi }{\partial t}+v \frac{\partial \psi }{\partial x}=\frac{\partial \left(\psi _{\hom }+\psi _{\text{par}}\right)}{\partial t}+v \frac{\partial \left(\psi _{\hom }+\psi _{\text{par}}\right)}{\partial x}=\frac{\partial \psi _{\hom }}{\partial t}+v \frac{\partial \psi _{\hom }}{\partial x}+\frac{\partial \psi _{\text{par}}}{\partial t}+v \frac{\partial \psi _{\text{par}}}{\partial x}=h (x+v t)$

with the homogeneous part identically zero, we have

$\frac{\partial \psi _{\text{par}}}{\partial t}+v \frac{\partial \psi _{\text{par}}}{\partial x}=h(x+v t)$

or, using our guess $\psi _{\text{par}}=f(x+v t)$

$\frac{\partial \psi _{\text{par}}}{\partial t}+v \frac{\partial \psi _{\text{par}}}{\partial x}=h(x+v t)$

and again, using a variable substitution $\xi \equiv x+v t$, which implies the change of derivatives from before, we get

$v \frac{\partial f(\xi )}{\partial \xi }+v \frac{\partial f(\xi )}{\partial \xi }=h(\xi )$

which reduces to

$\frac{\partial f(\xi )}{\partial \xi }=\frac{h(\xi )}{2 v}$

which means, that at least in principle f($\xi$) could be found by integration. Then, the full solution is

$\psi =\psi _{\hom }+\psi _{\text{par}}=f(x+v t)+g(x-v t)$

Without specifying boundary or initial conditions, we can surmise that the solution to the one-dimensional wave equation is simply a function 'g' traveling to the right with speed v added to another function 'f' traveling to the left with identical speed v. This is D'Alembert's solution to the wave equation.

Unfortunately factoring the wave equation in higher dimensions is not so simple and requires something called a Dirac operator, which essentially represents the 'square-root' or 'half-iterate' of the Laplacian.

The point here is that the most general solution to the one-dimensional wave equation is nothing more than some 'spatial-shape' traveling to the left added(superposition) to some other spatial-shape traveling to the right.