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This question is closely related to:

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

Given the wave equation

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

with initial conditions:

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

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

My question: Intuitively/physically, Why are only two initial conditions needed?

For example, Commonly initial conditions for displacement and velocity are given. Why not also acceleration--why is it not needed? It seems to me intuitively that if the initial acceleration was varied, it should have an effect on the wave that is propagated.

I know that mathematically since the equation is second order it needs two initial conditions but I don't completely understand it intuitively or physically.

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    $\begingroup$ Related: physics.stackexchange.com/q/18588/2451 and links therein. $\endgroup$ – Qmechanic Apr 13 '18 at 17:55
  • $\begingroup$ You don't need an initial condition to solve this algebraic equation: $2x + 10 = 22$. A differential equation is just an algebraic equation, but with derivatives essentially. Just as you solved for $x$ in the above algebraic equation, you could solve for $u_{tt}$ and be done. $u_{tt}$ would be the initial acceleration condition. It's already there. However with differential equations, we are always more interested in the original function, from which all these derivatives sprang. Otherwise, a Diff EQ class would be really simple. You'd just factor and whatnot as we did in algebra $\endgroup$ – DWade64 Apr 13 '18 at 19:00
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This is not the true mathematical answer which is much more straightforward but much less intuitive, but it is a strong suggestion towards the idea that $u(x,0)$ and $\partial_tu(x,0)$ embody all information sufficient and necessary to construct the solution of the problem.

Suppose you know $u(x,0)$ and $\partial_tu(x,0)$, the differential equations gives $\partial^2_tu(x,0)$ for free : $$\partial^2_tu(x,0) = c^2 \partial^2_xu(x,0)\:.$$ Actually, the differential equation also says $$\partial^3_tu(x,t) = c^2 \partial_t\partial^2_xu(x,t) =c^2\partial^2_x\partial_tu(x,t)\:.\tag{1}$$ In particular $$\partial^3_tu(x,0) = c^2 \partial^2_x\partial_tu(x,0)\:.$$ The right-hand side is known if we know $\partial_tu(x,0)$ (we have to derive twice this function with respect to $x$), and we know it. So knowing $u(x,0)$ and $\partial_tu(x,0)$, we also know $\partial^2_tu(x,0)$ and $\partial^3_tu(x,0)$. We can go further computing another derivative of (1): $$\partial^4_tu(x,t) = c^2 \partial^2_x\partial^2_tu(x,t)\:.\tag{2}$$ so that $$\partial^4_tu(x,0) = c^2 \partial^2_x\partial^2_tu(x,0)\:.$$ We know the right-hand side, so we also know $\partial^4_tu(x,0)$.

It should be evident that this series never ends: knowing $u(x,0)$ and $\partial_tu(x,0)$, the differential equation permits us to write down all time derivatives $\partial^n_tu(x,0)$ for $n=0,1,2,3,\ldots$ just computing many $x$-derivatives of $u(x,0)$ and $\partial_tu(x,0)$

At least formally speaking, assuming that the solution admits a Taylor expansion at $t=0$, the solution is $$u(x,t) = \sum_{n=0}^{+\infty} \frac{t^n}{n!}\partial^n_tu(x,0)$$ where the right-hand side is known as soon as we know nothing but the differential equation, $u(x,0)$ and $\partial_tu(x,0)$.

It also clear that, even if the series does not converge to the solution, knowing the first two initial conditions necessarily implies that you also know the third and the fourth one, and so on, using the equation. So you cannot fix them freely without facing some contradiction. At most two initial conditions are permitted. Whether or not they really determine a (unique) solution depends on many other mathematical regularity conditions.

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  • $\begingroup$ Also see physics.stackexchange.com/q/399647/45664 . The accepted answer there implies that the higher order derivative ICs can not be derived from the given lower order derivative ICs because only the values at a single point, t=0, are given (at least that's what I understand) so I am confused. $\endgroup$ – user45664 Apr 13 '18 at 20:08
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    $\begingroup$ @user45664 Yes, I took a mistake, 3rd and 4th, I have corrected the text, thanks! (I have some problems with dealing with numbers) $\endgroup$ – Valter Moretti Apr 14 '18 at 7:40
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    $\begingroup$ @user45664 The accepted answer you mention (physics.stackexchange.com/questions/391222/…) says that you cannot obtain the second initial condition from the first one: they are independent. This is OK. However you can get the $t$-derivatives of higher order from these two initial conditions using the form of the 2nd order PDE in the considered case. This is exactly what I proved in my answer. $\endgroup$ – Valter Moretti Apr 14 '18 at 11:13
  • $\begingroup$ @ Valter Moretti Should the extra c squared in equation one's middle term be associated with the term on the right? ( I greatly appreciate your answer and am still working my way through it. ) $\endgroup$ – user45664 Apr 16 '18 at 17:07
  • $\begingroup$ @user45664 Yes! Well, actually I also forgot a factor $c^2$ in the last term of the identity. I have now corrected everything. $\endgroup$ – Valter Moretti Apr 16 '18 at 18:15
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The intuitive explanation is this: the wave equation is one example of Newton's second law where the acceleration is determined by the position and velocity. The space derivatives are just describing the forces internal to $u$. There can also be external forces. For example, in $$u_{tt}=c^2u_{xx}+J(x,t)$$ $J$ is an external force. Just like with a simple harmonic oscillator, to find its behavior we need two pieces of information to fix the homogeneous parts of the equation because there are two derivatives.

You actually need more than just two conditions, though. You need what are known as boundary conditions. Specifying what conditions are needed to solve any particular equation is a part of the study of partial differential equations, in general.

Say, I'm trying to describe what happens in a medium $u$ that obeys the wave equation in a box from time $t_0$ to $t_f$. What do I have to specify to nail that down? There is not a unique answer to this question. Examples of what's needed are:

  1. $u$ on the edges of the box (can vary with time), $u$ at the start, $u$ at the end;
  2. $u$ on the edges of the box (can vary with time), $u$ at the start, $u_t$ at the start;
  3. $u$ on the edges of the box (can vary with time), $u$ at the end, $u_t$ at the end; and
  4. $u_x$ on the edges of the box (can vary with time), the average value of $u$ on the edge of the box (can vary with time), $u$ at the start, $u$ at the end.

That last one I'm not 100% sure will work because I've not seen it used outside of the static case (i.e. no time dependence).

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Once the first initial condition $u(x,0)=f(x)$ is given, you can Fourier transform it; from the wave equation you already know the speed $c$ with which each of the sinusoids in the Fourier transform propagates. The one thing you don't know is the direction in which each of them propagates. Specifying the time derivative as well gives a $(+)$ sign to all backward-propagating $x+ct$ Fourier components and a $(-)$ sign to all the forward-propagating $x-ct$ components.

And then there's nothing more to know.

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  • $\begingroup$ We're also making use of the fact that the wave equation is linear, so that superpositions of solutions are also solutions. $\endgroup$ – ostrichCamel Apr 13 '18 at 18:05
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Intuitively, without getting into mathematical detail, I think it all starts with how you got to the wave equation. Basically you can say the forces on each point are well known: Tension and if the rope has a mass also gravity. If you know what are the forces acting on a specific body you know what is the acceleration. By knowing the forces you can develop the wave equation. So the second derivative information is stored inside this equation. If it had a different net force acting on the rope probably it would have another term or it would look differently, maybe it would not be a linear equation at all. The equation tells you the behavior of a physical entity, but behavior is a general thing. To understand a specific rope you need more then acceleration. To jump from one time step to another you need to know how x is changing in time - that is the speed, but this of course is not enough because it can start from different locations/ configurations, so you need to know how it started. What about how the acceleration changes? As long there is no other forces acting on the rope the acceleration is not changing, so you have all your information. But what if it can? You have to know how it changes to add it when you developed the wave equation. as you will see this adds as external force, and now the wave equation has a source term. But you see? The information is all ready known...

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One need to assume certain kind of property as "intuitive/physical" to continue any kind of argument, because human language's "wave" is pretty loosely defined: for Schrodinger equation $i \partial_t \psi =H \psi$ only one initial condition needed.

You can ask the same question for the motion of particle. It's from Newton's law $F(x,x_t)=m x_{tt}$ that you know you need to know both the initial position and velocity to solve it, or you can have "intuition" from you daily life experience that only position is not enough, but throw ball with a constant velocity, the result is always the same.

(P.S.: I heard that for some field theory, higher order derivative might cause you trouble when one wants to do quantization, like non-unitarity, but I don't know whether you care about quantum mechanics.)

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protected by Qmechanic Apr 13 '18 at 17:48

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