I Wave equation and the d'Alembert solution

[Robin04]

I have a few questions about the wave equation and the D'Alambert solution:

0) First of all, I'm a bit confused with the terminology. Wikipedia says that THE wave equation is a PDE of the form: $\frac{\partial^2 u}{ \partial t^2 } = c^2 \nabla^2 u$, however there are other PDEs that have "wave-like" solutions like sine-Gordon, Klein-Gordon, etc., and I've seen books where these were called wave equations too. Is there an accepted definition of what a wave equation is?

1) Most of the text I've found about the derivation of the D'Alambert solution (for the equation mentioned above) starts with the idea of changing coordinates. In 1+1 dimensions $(x,t) \rightarrow (u,s)$, where $u = x+ct$, and $s=x-ct$. I understand the physical motivation of this (it is like "jumping" on the wave and in that system the motion is the simplest), but are there any mathematical methods to determine what coordinate transformation do I have to do in order to get the simplest form of a PDE, without considering the physical meaning of the equation?

2) The final form of the D'Alambert formula is stated as (by WIkipedia): $u(x,t) = \frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds$. I'm confused about how to apply the boundary conditions I choose. My idea is that the boundary value problem cannot be solved for any boundary conditions so it cannot be part of the formula for the general solution. Therefore this has to be some operation that I do on the solution after evaluating the D'Alambert formula. For example for the fixed boundary condition (lets say $u(0,t)= u(L,t)=0$) I image I would have to multiply it with a function which has the value 0 at x=0, and x=L and some non-zero value (not sure what exactly) for the interval which is in between and far enough from the boundaries. How to do this properly with the most common boundary conditions?

3) One way to derive the wave equation(s?) is to take the continuum limit of a discrete system. If this discrete system is linear then its dynamics can be represented by a matrix and the symmetry transformations of the system can be determined by finding the matrices that commute with the dynamic matrix. Also, these matrices form a group. Is there an analogous phrasing of this in the case of (linear or maybe non-linear as well) PDEs?

 

[jasonRF]

I'm not an expert, but since other folks aren't answering here is what I know.

0) When folks refer to "THE wave equation" that is the equation they are referring to. But it is also common to refer to other equations that have wave solutions as wave equations. My training is in plasma physics, where there is a whole zoo of waves even when the media is homogeneous and we have linearized the equations. While I never hear people refer to a system of equations as "wave equations", when the equations can be combined into a single equation with wave solutions it was often called "A wave equation". Examples include the Klein-Gordon and the Nonlinear Schrodinger, but of course sometimes they are equations that haven't been given names.

1) The general version of this approach is called the Method of Characteristics. It can work for linear and nonlinear equations, systems of equations, etc. Part of the method also classifies the PDE, so that you know what kinds of solutions you might expect.

I don't think I can be much help for the rest of your questions. Hopefully someone else jumps in here.

Jason

 

[Robin04]

I'm not an expert, but since other folks aren't answering here is what I know.

0) When folks refer to "THE wave equation" that is the equation they are referring to. But it is also common to refer to other equations that have wave solutions as wave equations. My training is in plasma physics, where there is a whole zoo of waves even when the media is homogeneous and we have linearized the equations. While I never hear people refer to a system of equations as "wave equations", when the equations can be combined into a single equation with wave solutions it was often called "A wave equation". Examples include the Klein-Gordon and the Nonlinear Schrodinger, but of course sometimes they are equations that haven't been given names.

1) The general version of this approach is called the Method of Characteristics. It can work for linear and nonlinear equations, systems of equations, etc. Part of the method also classifies the PDE, so that you know what kinds of solutions you might expect.

I don't think I can be much help for the rest of your questions. Hopefully someone else jumps in here.

Jason

Thank you very much! It helped a lot! :)

 

[coquelicot]

0. Yes, it is what is called the "classical wave equation".

1. There are plenty of mathematical methods to solve PDE and in particular wave equations. That covers several branches of mathematics, and methods range from the simplest usual tricks to the highest research level. You have probably heard about Lie theory of differential equations, Noether theorem, D-modules etc. Lie theory for example tells us how to use the symmetries of the differential equations to solve them.

2. In the Wikipedia article, the boundary condition is a whole function u(x, 0) = f(x), u_t(x, 0) = g(x), that is, f(x) and g(x) are the shape of the wave and its partial t-derivative at time 0. This whole shape IS the boundary condition, and is given by the nature of the problem.

3. Your phrasing is not sufficiently precise for me, and probably refers to a domain I'm unacquainted with.

 

[Robin04]

2. In the Wikipedia article, the boundary condition is a whole function u(x, 0) = f(x), u_t(x, 0) = g(x), that is, f(x) and g(x) are the shape of the wave at time 0 and t resp. This whole shape IS the boundary condition, and is given by the nature of the problem.

But isn't this the initial condition you are referring to? $u(x,0)$ and $u_t(x, 0)$ are functions that describe the system at t=0. These functions have to be consistent with the boundary conditions of course because if my boundary condition is such that $u(0,t) = 0$ then I can't have $f(x),g(x)$ such that they are not equal to 0 at x = 0. But if I want my system to behave consistently with my boundary conditions for all t then the state of a system at a certain t is not a sufficient condition.

3. Your phrasing is not sufficiently precise for me, and probably refers to a domain I'm unacquainted with.

Let's have a dynamical system where a series of balls uneffeced by gravity can move in 1D on a rod and they are connected to each other with a spring. The system of differential equations of this system can be written in the form $\ddot{u} = kDu$, where u is a vector that has the displacement of each ball relative to their equilibrium as its components, k is a constant, and D is a matrix. It turns out that the symmetry group of this system can be described by the following formula: $[D,S_i] = 0$, where S_i are the matrices that correspond to the symmetry transformations.

If we take the continuum limit of the system as the number of balls goes to infinity while keeping some properties constant (linear density, length of the system) then we get the classical wave equation written above. The symmetry group shouldn't change with this limit but we lose its description with matrices.

So my question is: how do we get the same symmetry group from the classical wave equation than the one we got from the discrete system?

https://www.reed.edu/physics/courses/Physics411/html/page2/files/Lecture.16.pdf
Section 16.1.1 shows how the continuum limit is done with the wave equation at the end.

http://delta.cs.cinvestav.mx/~mcintosh/comun/algebra/node70.html
http://delta.cs.cinvestav.mx/~mcintosh/comun/algebra/node71.html
These two texts show how the symmetry group of the discrete system is constructed. I guess the dynamical matrix they use here is the same as for the system I just mentioned above.

 

[coquelicot]

As I said, I'm not very acquainted with this domain (you have not precised what are the symmetry matrices, well this may be obvious for persons knowing this domain). But this strongly recalls me the theory of Sophus Lie for differential equations. Here, a group of symmetries analogous to Galois group in algebra is defined and used to solve or obtain insights about partial differential equations.

 

[Robin04]

But this strongly recalls me the theory of Sophus Lie for differential equations. Here, a group of symmetries analogous to Galois group in algebra is defined and used to solve or obtain insights about partial differential equations.

You're right. I just found quite a few papers about it. It's much more advanced than I thought so I'll just save it for later.

I only have problem 2) left to understand.

 

[coquelicot]

I thought I've answered to your question 2). So, it appears I haven't understood it, and if you want I try to answer, please, reformulate it once more.

 

[Robin04]

I thought I've answered to your question 2). So, it appears I haven't understood it, and if you want I try to answer, please, reformulate it once more.

I reacted to your answer in #5. You didn't notice it, or you want me to elaborate that one?

 

[coquelicot]

Indeed, I haven't noticed that.
Your boundary condition, that is u(0,t) = 0 for all t, is of different type from the boundary condition in Wikipedia. This boundary condition alone does not determine the problem and you need other additional conditions.

 

[Robin04]

Indeed, I haven't noticed that.
Your boundary condition, that is u(0,t) = 0 for all t, is of different type from the boundary condition in Wikipedia. This boundary condition alone does not determine the problem and you need other additional conditions.

What other conditions would I need? (I assumed the initial position and velocity functions to be given)

 

[coquelicot]

What I am trying to explain is that the condition u(0,t) = 0 = u(L,t) for all t is not sufficient to determine the wave for all t (and even at t = 0). In fact, it completely changes the nature of the solution: you now deal with what is called "condition at the limits", and D'Alembert solution you wrote above does not apply anymore. What works in general for this type of conditions is to analyze the equation according to its modes, and to superpose the solutions (that is the way Fourier found the general solution of the equation of heat). In our case, it is easy to check that any function of the form
$$u_n(x,t) = B_n \sin \left( n \pi { x \over L}\right) \cos \left(n \pi {vt \over L }\right)$$ is a particular solution of your problem, and the general solution is a sum of these solution, the B_n being arbitrary (but the series must converge). This gives the solution of your problem, and also show that there are infinitely many solution that satisfy your boundary conditions. So, to entirely determine the problem, you need other conditions: the shape of the wave at time t = 0 for example. Let it be denoted by f(x). Then there hold $$f(x) = \sum_{n=1}^{\infty} B_n \sin \left(n \pi {x \over L}\right),$$
so the B_n are the Fourier coefficients in the decomposition of f into a Fourier series.

 

[Orodruin]

and D'Alembert solution you wrote above does not apply anymore.

This is not necessarily true. You can solve it using the D'Alembert solution. It will just require some additional tweaking and extending the solution to the full real line by choosing appropriate periodic $f(x)$ and $g(x)$.

$$u_n(x,t) = B_n \sin \left( n \pi { x \over L}\right) \cos \left(n \pi {vt \over L }\right)$$ is a particular solution of your problem, and the general solution is a sum of these solution, the B_n being arbitrary (but the series must converge).

This is not the most general solution. It is the most general solution for the particular case of $u_t(x,0) = 0$, but it does not hold if there is a non-zero initial velocity.

 

[coquelicot]

Yes I was just about to retract what I wrote, that D'alembert solution does not apply anymore. In fact, we have trigonometric relations of the form
$$2\cos(kx)\cos(wt) = \cos(wt +kx) + \cos(wt-kx),$$ so D'alembert solution works even in this case, with the appropriate periodic function derived from what I wrote above. But it is still true that the intuition behind this formula, is, in my opinion, not very useful here.
Regarding your last remark, this is what I have assumed implicitely.
I add that the only change necessary to take into account the possible initial velocity of the wave is:
$$u_n(x,t) = B_n \sin \left( n \pi { x \over L}\right) \cos \left(n \pi {vt \over L } + \varphi_n \right)$$

 

[Orodruin]

To take the D'Alembert solution in the case of Dirichlet boundary conditions:

The general solution to the wave equation is of the form
$$
u(x,t) = f_1(x-ct) + f_2(x+ct).
$$
The boudary condition at $x = 0$ now requires
$$
u(0,t) = f_1(-ct) + f_2(ct) = 0 \quad \Longrightarrow \quad f_1(-ct) = - f_2(ct).
$$
Since $t$ is arbitrary, this means that
$$
u(x,t) = f_1(x-ct) - f_1(-x-ct).
$$
The other boundary condition leads to
$$
u(L,t) = f_1(L-ct) - f_1(-L-ct) = 0 \quad \Longrightarrow \quad f_1(L-ct) = f_1(-L-ct).
$$
In other words, since $ct$ is again arbitrary, $f_1$ must be $2L$ periodic. With those requirements, we can look at the D'Alembert solution
$$
u(x,t) = \frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds.
$$
By design, this solution will satisfy $u(x,0) = f(x)$ and $u_t(x,0) = g(x)$ in $0 < x < L$. The question is how we can extend $f(x)$ and $g(x)$ to functions on the real line while satisfying the boundary conditions. To start off, they must obviously satisfy the boundary conditions themselves, i.e., $f(0) = f(L) = 0$ etc. Furthermore, for $u(0,t) = 0$, we must have
$$
u(0,t) = \frac{f(-ct) + f(ct)}{2} + \frac{1}{2c} \int_{-ct}^{+ct} g(s) ds = 0.
$$
Both terms here must vanish separately and therefore $f(-ct) = -f(ct)$, i.e., $f(x)$ is an odd function, and $g(s)$ must also be an odd function for the integral to vanish. For the boundary condition at $x = L$, we find
$$
u(L,t) = \frac{f(L-ct) + f(L+ct)}{2} + \frac{1}{2c} \int_{L-ct}^{L+ct} g(s) ds = 0.
$$
Now, in particular, we can make the change of variables $ct = L + \ell$, leading to $f(2L + \ell) = f(\ell)$ as well as
$$
\int_{-\ell}^{2L + \ell} g(s) ds = 0.
$$
The first requirement tells us that $f$ is $2L$ periodic and the second that $g$ must integrate to zero over any $2L$ interval, which it will do if it is also $2L$ periodic (added to the constraint of being odd around $x = 0$.

 

[Orodruin]

But it is still true that the intuition behind this formula, is, in my opinion, not very useful here.

I would disagree with this. I think it is very useful to think about the wave form in this manner. It gives a feeling for how standing waves in a finite interval are composed of the same wave being reflected against the boundaries repeatedly.

 

[coquelicot]

That's true too. But I still believe that harmonic analysis is much more suitable here. And also generalizes to higher dimensions.

 

[Orodruin]

That's true too. But I still believe that harmonic analysis is much more suitable here. And also generalizes to higher dimensions.

What makes you think d’Alembert’s formula does not have a generalisation to higher dimensions?

It is true that eigenfunction expansion would be the typical method applied, but that does not necessarily make other methods obsolete or less illuminating.

 

[Robin04]

To take the D'Alembert solution in the case of Dirichlet boundary conditions:

The general solution to the wave equation is of the form
$$
u(x,t) = f_1(x-ct) + f_2(x+ct).
$$
The boudary condition at $x = 0$ now requires
$$
u(0,t) = f_1(-ct) + f_2(ct) = 0 \quad \Longrightarrow \quad f_1(-ct) = - f_2(ct).
$$
Since $t$ is arbitrary, this means that
$$
u(x,t) = f_1(x-ct) - f_1(-x-ct).
$$
The other boundary condition leads to
$$
u(L,t) = f_1(L-ct) - f_1(-L-ct) = 0 \quad \Longrightarrow \quad f_1(L-ct) = f_1(-L-ct).
$$
In other words, since $ct$ is again arbitrary, $f_1$ must be $2L$ periodic. With those requirements, we can look at the D'Alembert solution
$$
u(x,t) = \frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds.
$$
By design, this solution will satisfy $u(x,0) = f(x)$ and $u_t(x,0) = g(x)$ in $0 < x < L$. The question is how we can extend $f(x)$ and $g(x)$ to functions on the real line while satisfying the boundary conditions. To start off, they must obviously satisfy the boundary conditions themselves, i.e., $f(0) = f(L) = 0$ etc. Furthermore, for $u(0,t) = 0$, we must have
$$
u(0,t) = \frac{f(-ct) + f(ct)}{2} + \frac{1}{2c} \int_{-ct}^{+ct} g(s) ds = 0.
$$
Both terms here must vanish separately and therefore $f(-ct) = -f(ct)$, i.e., $f(x)$ is an odd function, and $g(s)$ must also be an odd function for the integral to vanish. For the boundary condition at $x = L$, we find
$$
u(L,t) = \frac{f(L-ct) + f(L+ct)}{2} + \frac{1}{2c} \int_{L-ct}^{L+ct} g(s) ds = 0.
$$
Now, in particular, we can make the change of variables $ct = L + \ell$, leading to $f(2L + \ell) = f(\ell)$ as well as
$$
\int_{-\ell}^{2L + \ell} g(s) ds = 0.
$$
The first requirement tells us that $f$ is $2L$ periodic and the second that $g$ must integrate to zero over any $2L$ interval, which it will do if it is also $2L$ periodic (added to the constraint of being odd around $x = 0$.

Thank you very much for this derivation!

So if I have no boundary conditions, $f$ and $g$ functions are just arbitrary continuous functions that stand for the initial conditions. But if I have a boundary condition then this shows as constraints on $f$ and $g$, and also expresses that the initial conditions cannot be arbitrary with a specified boundary conditions.

Every time I want to solve a boundary value problem I have to follow this derivation and determine what constraints I have on $f$ and $g$, right?

I'll also take time to understand the method with Fourier analysis.