It is very intuitive that any function of the form $y=f(x+vt)$ would describe a wave in two spatial dimensions and time.

From that it is easy to use the chain rule, letting $w=x+vt$ and doing:

$$ \frac{{\partial} y}{{\partial} x}=\frac{dy}{dw}\frac{{\partial}w}{{\partial}x}=\frac{dy}{dw} $$

$$ \frac{{\partial} y}{{\partial} t}=\frac{dy}{dw}\frac{{\partial}w}{{\partial}t}=v \frac{dy}{dw} $$

And then:

$$ \frac{{\partial} y}{{\partial} x}-\frac{1}{v} \frac{{\partial} y}{{\partial} t}=0 $$

But instead the chain rule can be used twice to get the form of the wave equation you usually see in physics:

$$ \frac{{\partial}^2 y}{{\partial}^2 x}-\frac{1}{v^2} \frac{{\partial}^2 y}{{\partial}^2 t}=0 $$

I understand the derivation of the second order form from the tension on a vibrating string (as in http://oyc.yale.edu/physics/phys-201/lecture-14), but what I don't understand is what different meaning the second-order equation carries, if any, from the first order equation.

Are there cases where the first-order equation would be satisfied but the second order equation would not, or vice versa?

  • 3
    $\begingroup$ For one thing, the first order equation only has waves travelling in one direction. $\endgroup$ – Mariano Suárez-Álvarez Jul 18 '15 at 6:13
  • 2
    $\begingroup$ More technically, the differential equation has to satisfy certain geometrical invariance properties (which are consequences of the homogeneity of space, isotropy and so on) which impose rather strong restrictions on its form. One can prove that the wave equation is the simplest equation among all which satisfy those invariance properties. $\endgroup$ – Mariano Suárez-Álvarez Jul 18 '15 at 6:17
  • $\begingroup$ Your question is quite relevant to quantum theory and its history; if you're interested in this connection, you can read up on it in the Wikipedia articles on the Dirac equation and the Klein–Gordon equation, both of which also contain some historical information (in the sections "Dirac's coup" and "History", respectively). $\endgroup$ – joriki Jul 18 '15 at 17:07

The first-order equation $$\frac{{\partial} y}{{\partial} x}-\frac{1}{v} \frac{{\partial} y}{{\partial} t}=0$$ is also an important PDE, known as the transport equation. All its solutions consist of the initial values moving to the left with velocity $v$ (with the $+$ sign they would move to the right). This is not what we expect a vibrating string to do: if pulled in one direction (say, up), it vibrates, moving both up and down. Also, a wave created from a splash does not exactly retain the form of that splash, and it spreads in all space directions available to it (which in $\mathbb{R}^1$ is left and right).

Here is a more direct connection between the order and the object modeled.

  • If a PDE involves only the first derivative with respect to $t$, then its solution is determined by the initial value $u(x,0)$. Indeed, the PDE and initial value dictate what $u_t(x,0)$ should be, and the solution evolves from there.
  • If a PDE involves the second derivative with respect to $t$, then its solution is not determined by the initial value $u(x,0)$; one also needs to specify the initial velocity $u_t(x,0)$. Indeed, the PDE, initial value, and initial velocity together determine what $u_{tt}(x,0)$ should be, and the solution evolves from there.

Vibration is a process in which both the position and velocity matter: if a string is initially at rest position but has nonzero velocity, it will not stay at rest. Therefore, this process cannot be described by an equation that is of first order with respect to $t$.

  • $\begingroup$ In some applied fields the PDE is also known as the advection equation. $\endgroup$ – user856 Jul 21 '15 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.