0
$\begingroup$

The wave equation

$$u_{xx}(x,t)=\frac {1}{c^2}u_{tt}(x,t) $$

requires two initial conditions because the equation is second order:

IC1: $$u(x,0)= f(x)$$

IC2: $$u_{t}(x,0)= g(x)$$

But when it is factored:

$$u_{tt} - c^2 u_{xx} = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u(x,t) = 0$$

it only requires one initial condition when each of the factors is set equal to zero:

$$\bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) u(x,t)= 0 $$ $$\bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg)u(x,t)=0 $$

IC1: $$u(x,0)= f(x)$$

because the 'factors' are first order.

So I understand mathematically why in the first case two ICs are needed whereas in the second case only one IC is needed. But if the factored and unfactored wave equations are equivalent--containing all of the same information-- , I don't understand intuitively or physically the difference in the number of required ICs.

My question is: Intuitively and physically, why does the wave equation need two ICs when the 'factored' wave equation needs only one?

See also:

Intuition into why the wave equation needs the second initial condition (e.g. velocity)

Intuitively, why are only two initial conditions needed for the wave equation? Why not 3 or 4?

https://math.stackexchange.com/q/2706776/147776

https://physics.stackexchange.com/a/403761/45664

EDIT 6/2/18 SEE @jcandy ANSWER BELOW FOR CORRECTIONS TO THIS QUESTION AND FOR THE ANSWER

The equations with the factors should have been written

$$\bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) u(x,t)= v $$ $$\bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg)v(x,t)=0 $$

with an IC given for each equation.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ A general solution to the wave equation is the sum of a solution to each of the two first-order equations you have. That means you need an initial condition for each of the two first-order equations, and $1 + 1 = 2$. $\endgroup$
    – knzhou
    May 29, 2018 at 20:28
  • $\begingroup$ So then basically my IC1 is used twice? :) $\endgroup$
    – user45664
    May 29, 2018 at 21:43
  • 1
    $\begingroup$ Not quite! Fundamentally, you have two ICs, but you're splitting them in different ways. In the second-order case, you split them into "initial position" and "initial velocity". In the first-order case, you split them into "initial amount of right-moving wave" and "initial amount of left-moving wave". So they are not the same, but they ultimately yield the same information. $\endgroup$
    – knzhou
    May 29, 2018 at 21:45
  • 1
    $\begingroup$ See the answer below. Your factorization is wrong. You have two fields, a doublet, when you factor. Thus, you have 1 IC for each = 2 IC for the field. $\endgroup$
    – user196418
    Jun 1, 2018 at 11:57
  • 1
    $\begingroup$ Because that is essentially what you've done, created a doublet of fields, a 2-dim field vector [u, v], where each satisfy a first order equation. In PE theory these may be related to the forward and backward propagating modes and can be transformed into other representations that decouple these modes in refractive media. $\endgroup$
    – user196418
    Jun 2, 2018 at 21:20

1 Answer 1

Reset to default
2
$\begingroup$

Your factorization is not correct. Think about the simpler equation $\partial_{xx} u = 0$. This can be "factored" according to $\partial_x \left( \partial_x u\right) = 0$. One would then say that the factored system can be written as \begin{align} \partial_x v &~= 0 \\ \partial_x u &~= v \end{align} where $v$ is an auxiliary function. In each line above, the derivative accounts for one undetermined coefficient. Thus, there are a total of two initial conditions required. A more detailed explanation is here: https://math.stackexchange.com/a/84268/275678

Improve this answer
$\endgroup$
5
  • $\begingroup$ I don't understand "Your factorization is not correct". Mine looks identical to what is in the reference you supplied. Although I understand "In each line above, the derivative accounts for one undetermined coefficient. Thus, there are a total of two initial conditions required" However look at my above comments to @knzhou . $\endgroup$
    – user45664
    Jun 1, 2018 at 17:53
  • 1
    $\begingroup$ You wrote $D_- u =0$ and $D_+ u = 0$, which are two uncoupled equations for the same field $u$. You should write $D_- v = 0$ and $D_+ u = v$, which are two coupled equations for a pair of fields, $u$ and $v$. $\endgroup$
    – jcandy
    Jun 1, 2018 at 18:35
  • $\begingroup$ So then in your answer can $\partial_x \left( \partial_x u\right) = 0$ be written as $\partial_x \left( v\right) = 0$ ? $\endgroup$
    – user45664
    Jun 2, 2018 at 17:08
  • $\begingroup$ In my answer, $\partial_x \left( \partial_x u\right)=0$ is written as $\partial_x v = 0$ where $v = \partial_x u$. You need both first-order equations to represent the original second-order equation. $\endgroup$
    – jcandy
    Jun 2, 2018 at 19:13
  • $\begingroup$ I have edited my question to refer to your answer. $\endgroup$
    – user45664
    Jun 2, 2018 at 21:12

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.