Wave equation in free space

[solanojedi]

Hi everyone,

I'm reading about the solution of the wave equation in free space on Stratton - Electromagnetic Theory and Snider - PDE and I got a little confused. The wave equation in 3D (plus time) is the following $$\frac{\partial^{2} \Psi (x,y,z,t)} {\partial t^{2}}=\nabla ^{2}\Psi (x,y,z,t)$$ but the doubt can be carried out also from the "1D" (plus time) wave equation $$\frac{\partial^{2} \Psi (x,t)} {\partial t^{2}}=\nabla ^{2}\Psi (x,t).$$ My intent is to solve the equation in free space, hence $-\infty<x<+\infty$, but also on positive and negative time. If we apply the separation of variables to the last equation, from $\Psi(x,t)=X(x)T(t)$ we get two separated equation in space and time, i.e. $$\frac{X''}{X}=\lambda,~~~ -\infty<x<+\infty \\ \frac{T''}{T}=\lambda,~~~ -\infty<t<+\infty \\$$ The two equations are identical and are two singular Sturm-Liouville problem, so, as I can see on Snider pagg.474 for the same type of equations, the solution should be the infinite superposition of complex exponentials $\int_{-\infty}^{+\infty} A(\omega) e^{i \omega x} e^{i \omega t} d\omega = \int_{-\infty}^{+\infty} A(\omega) e^{i\omega(x+t)} d\omega$, if $\omega=\sqrt{\lambda}$. However, with this approach, I don't find the two traveling packet waves. Instead, the correct solution is the usual two-traveling-wave equation $$\Psi (x,t)=\int_{-\infty}^{+\infty} [A(\omega) e^{i \omega x}+B(\omega) e^{-i \omega x}] e^{-i\omega t} d\omega.$$ Also Snider's book, when discussing the wave equation in 3D + time, for the spatial variables keeps the singular Sturm Liouvilee approach using complex exponentials, while for the time equation, that has the same form, insert sa $\sin(\omega t)$ as a solution, obtaining $$\Psi (x,y,z,t)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} A(\omega_1, \omega_1, \omega_3) e^{i\omega_1 x}e^{i\omega_2 x}e^{i\omega_3 x} \sin(t \sqrt{\omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2}}) d\omega_{1} d\omega_{2} d\omega_{3}$$
Of course I get that this is the correct solution, but what is wrong with my approach? Thank you in advance.

 

[Orodruin]

You are missing half of the separated solutions.

 

[solanojedi]

You are missing half of the separated solutions.

Thank you for your answer.
But how come (in the 3D case) it is ok to keep just $\int_{-\infty}^{+\infty}e^{i\omega x}$ instead of $\int_{-\infty}^{+\infty} (e^{i\omega x}+e^{-i\omega x})$ for the spatial variables and not for the time variable? And why in the 1D case I get the two exponentials for the spatial variable and just one for the time variable?

 

[Orodruin]

Thank you for your answer.
But how come (in the 3D case) it is ok to keep just $\int_{-\infty}^{+\infty}e^{i\omega x}$ instead of $\int_{-\infty}^{+\infty} (e^{i\omega x}+e^{-i\omega x})$ for the spatial variables and not for the time variable?

If you want the most general solution it isn’t.

And why in the 1D case I get the two exponentials for the spatial variable and just one for the time variable?

You do get two for the time as well. Note that you are integrating over both positive and negative $\omega$.

 

[solanojedi]

This now confuses me. From what I read in Snider, the solution to $\frac{X''}{X}=\lambda$ over an infinite domain is given by the superposition of single exponentials (that is the Fourier transform), as in this picture: https://ibb.co/ekFq7G

 

[Orodruin]

You have two choices. You can either use $(A(\omega) e^{i\omega x} + B(\omega) e^{-i\omega x}) e^{i\omega t}$ or $(C(\omega) e^{i\omega t} + D(\omega) e^{-i\omega t}) e^{i\omega x}$. They are equivalent.

Note that the terms that are "missing" can be rewritten in terms of the ones you have already! It is all in the integral.
$$
\int_{-\infty}^\infty [C(\omega) e^{i\omega x} + D(\omega) e^{-i\omega x}] e^{i\omega t} d\omega
= \{\omega \to -\omega\}
=
\int_{-\infty}^\infty [C(-\omega) e^{-i\omega x} + D(-\omega) e^{i\omega x}] e^{-i\omega t} d\omega.
$$
This is why you only need to write two of the terms.

 

[solanojedi]

I think I expressed myself badly. My doubt is about the difference between the solution that I would write if I'd follow the rule written in Snider (i.e. using the Fourier transform as a solution of every separated equation) - so obtaining $$\Psi(x,t)=\int_{-\infty}^{+\infty} A(\omega) e^{i \omega x} e^{i \omega t} d\omega,$$ and the usual solution of the wave equation (or the telegraph equation) that is given in other books, i.e. $$\Psi(x,t)=\int_{-\infty}^{+\infty} [B(\omega) e^{i \omega x} + C(\omega) e^{-i \omega x}] e^{i \omega t} d\omega.$$ It seems to me that for the time variable the Fourier transform is applied while for the spatial variable it's something else, even if the starting differential equations in time and in space are the same.

PS: the approach that gives the second equation is usually always the same in many books. First it is supposed that $\Psi (x,t)$ is Fourier-transformable in the $t$ domain and so it can be written as a superposition of complex exponential in time, $\Psi (x,t)= X (x) e^{i\omega t}$. After that, this espression is inserted in the PDE and we get the separated spatial equation $X'' / X=\omega$: then they say that the solution of this one is $[B(\omega) e^{i \omega x} + C(\omega) e^{-i \omega x}]$ and the superposition of space and time gives the complete solution $\Psi(x,t)=\int_{-\infty}^{+\infty} [B(\omega) e^{i \omega x} + C(\omega) e^{-i \omega x}] e^{i \omega t} d\omega.$

 

[Orodruin]

It seems to me that for the time variable the Fourier transform is applied while for the spatial variable it's something else, even if the starting differential equations in time and in space are the same.

It is more instructive to apply the FT in the spatial domain. That is usually where you have the initial conditions and so an actual singular SL problem because the domain of interest is $-\infty < x < \infty$. In time you are typically interested in a domain that is bounded in at least one direction. Note that
$$
\int_{-\infty}^\infty [B(\omega) e^{i\omega x} + C(\omega) e^{-i\omega x}]e^{i\omega t} d\omega
= \int_{-\infty}^\infty [B(\omega) e^{i\omega t} + C(-\omega) e^{-i\omega t}] e^{i\omega x} d\omega,
$$
which suggests a FT in $x$. The typical way of going about to do things is to expand the function in the eigenfunctions of a SL problem, in this case the SL problem is in the $x$ direction.

From what you say about Snider, that is not the most general solution. It is only a wave moving to the left.

 

[solanojedi]

I'm really sorry Orodruin, but I'm stucked with this idea that a singular SL problem solution is written as a Fourier transform, hence using a superposition of single positive exponentials (with the integral going from $-\infty$ to $+\infty$). That's what I read on many books. And since both time and space equations have infinite domains, it seems to me that both solutions should be written as superposition of a only positive complex exponential.

However, I also know that a general second order DE has as a general solution the sum of two opposite complex exponential. In the solutions that I wrote above, it seems that there is a mixture of these solutions, even if the two starting DEs (time and space) I think are of the same type, domains and boundary conditions included.
I'm really struggling with this different writing of the solutions for space and time problems, I still can't see a justification.

 

[Orodruin]

And since both time and space equations have infinite domains, it seems to me that both solutions should be written as superposition of a only positive complex exponential.

This is simply not true in most actual problems you will encounter. You will typically have initial conditions on the time part, effectively making the time domain bounded. You also need yo get rid of the notion that a singular SL problem implies a Fourier transform. First of all, there are many different singular SL problems, all leading to different eigenfunctions. Second, just because you have a SL operator does not mean you have a SL problem. A SL problem also requires homogeneous boundary conditions or bounding behaviour. The way of solving a pde using separation of variables is to identify what directions you have a SL problem in and then expanding in the corresponding eigenfunctions. This will generally lead to a set of pdes with one variable less for the expansion coefficients.

In your case, you will not get further unless you specify more than the PDE. You need to specify what your time domain of interest is and what your initial conditions are.

If you have problems with this, I would suggest first working out what happens with the wave equation on a finite domain with given homogeneous boundary conditions and given initial conditions.

This is an extensive subject that one could easily write a book on...

 

[solanojedi]

Second, just because you have a SL operator does not mean you have a SL problem. A SL problem also requires homogeneous boundary conditions or bounding behaviour. The way of solving a pde using separation of variables is to identify what directions you have a SL problem in and then expanding in the corresponding eigenfunctions. This will generally lead to a set of pdes with one variable less for the expansion coefficients.

I think that boundary conditions are the key to my problem. From your reply and from what I read, the separation of variable technique starts by separating variables and then analyze AS FIRST EQUATION the one that has homogeneous boundary conditions. The wave equation for $\Psi (x,t)$ where $x$ is doubly unbounded and $t>0$, the separated function on $t$ has usually inhomogenous condition on the first derivative $\partial {\Psi (x,t)} / \partial t$ while homogenous in the initial state $\Psi(x,0)=0$. In this situation, the separated equation on $x$ is a singular SL problem and generates a Fourier transform with the single complex exponential $e^{i\omega x}$. For the time equation, for $t>0$ paired with the initial conditions, it is not a singular SL problem, hence the solution is $sin(\omega t)$ and not another Fourier transform. The global solution will be the integral of the product of the two functions $\int_{-\infty}^{+\infty} A(\omega) \sin(\omega t) e^{i\omega t}$. If I expand the $\sin$, I eventually get the two opposite exponentials and hence the two traveling wave!

Up to this everything seemed now coherent to me, but please correct me if I got wrong ideas.
Then I saw this https://ocw.mit.edu/courses/mathema...quations-fall-2006/lecture-notes/fourtran.pdf , where the heat equation is studied with the separation of variable. The space equation seemed the same as the wave equation to me (second order with "no" boundary conditions since the domain $x$ is doubly unbounded), so I was expecting also in this case a Fourier transform in $x$. The solution in my mind, with that paper notation, would have been $\int_{-\infty}^{+\infty} A(\omega) e^{i \omega x} e^{-\omega^{2} \kappa t}$. Instead they get $(A \cos(\omega t) + B \sin (\omega t)) e^{-\omega^{2} \kappa t}$, and the final integral is calculated from 0 to $+\infty$, hence for positive $\omega$ only.

From what I understand, I forgot another boundary condition. NOT ONLY the solution $X(x)$ of the space equation must be bounded at $\pm \infty$, but also the COMPLETE solution $X(x) T(t)$ must be bounded for $x$ at $\pm \infty$ and $t$ at $+\infty$. In fact, this condition will exclude negative eigenvalues from the final solution since the negative exponential inside the solution of $T(t)$ would explode for $t=+\infty$.
However, now that I noticed this condition, how I get the $A \cos(\omega t) + B \sin (\omega t)$ from the complex exponential $A e^{i \omega x}$ that I found as a solution of the equation in $x$? Or is it something that I have to acknowledge in the middle of the procedure, hence saying that the solution of the singular SL problem in $x$ is NOT a Fourier transform? Can you help me on this point?
Thank you again for the patience.

 

[Orodruin]

the separated function on ttt has usually inhomogenous condition on the first derivative ∂Ψ(x,t)/∂t∂Ψ(x,t)/∂t\partial {\Psi (x,t)} / \partial t while homogenous in the initial state Ψ(x,0)=0Ψ(x,0)=0\Psi(x,0)=0.

This is not generally the case. In many cases you will also have an inhomogeneity in the initial displacement as well. Both are possible.

hence the solution is sin(ωt)sin(ωt)sin(\omega t) and not another Fourier transform.

In the case where you only have an inhomogeneity in the time derivative, yes. In general you will also have a cosine term that depends on the initial displacement.

would have been ∫+∞−∞A(ω)eiωxe−ω2κt∫−∞+∞A(ω)eiωxe−ω2κt\int_{-\infty}^{+\infty} A(\omega) e^{i \omega x} e^{-\omega^{2} \kappa t} . Instead they get (Acos(ωt)+Bsin(ωt))e−ω2κt(Acos⁡(ωt)+Bsin⁡(ωt))e−ω2κt(A \cos(\omega t) + B \sin (\omega t)) e^{-\omega^{2} \kappa t}, and the final integral is calculated from 0 to +∞+∞+\infty, hence for positive ωω\omega only.

Both are correct. The A and B in the second depend on the real and imaginary parts of $A(\omega)$ from the first. In order for the first to be real, there must be a relation between $A(\omega)$ and $A(-\omega)$. Essentially you get the second if you take the negative $\omega$ part of the first and make the variable substitution $\omega \to -\omega$ in it. It is a standard rewriting of the Fourier transform as the sum of a Fourier sine and a Fourier Cosine transform, each representing the anti-symmetric and symmetric part of the function, respectively.

In fact, this condition will exclude negative eigenvalues from the final solution since the negative exponential inside the solution of T(t)T(t)T(t) would explode for t=+∞t=+∞t=+\infty.

The negative eigenvalues are excluded already by the functions being bounded at $x \to \pm \infty$.

 

[solanojedi]

Both are correct. The A and B in the second depend on the real and imaginary parts of $A(\omega)$ from the first. In order for the first to be real, there must be a relation between $A(\omega)$ and $A(-\omega)$. Essentially you get the second if you take the negative $\omega$ part of the first and make the variable substitution $\omega \to -\omega$ in it. It is a standard rewriting of the Fourier transform as the sum of a Fourier sine and a Fourier Cosine transform, each representing the anti-symmetric and symmetric part of the function, respectively.

You mean that we will simply have, starting from the Fourier transform form, $A(\omega) e^{i\omega x} e^{-\omega^2 \kappa t} = [A(\omega) \cos(\omega x) + iA(\omega) \sin(\omega x)]e^{-\omega^2 \kappa t} = [A(\omega) \cos(\omega x) + B(\omega) \sin(\omega x)]e^{-\omega^2 \kappa t}$ and those are the coefficients of the solution on that paper?

 

[Orodruin]

Well, you are forgetting the negative frequency contributions to the coefficients of the A and B in the last step, but the idea is the same.

 

[solanojedi]

I'm sorry but I don't understand, but I admit I'm not very used "thinking" and working with the sine and cosine transforms, my bad. Even in the discussions before I wasn't exactly following you when you introduced the $\omega$ sign change for the coefficients...

 

[Orodruin]

Consider an expression on the form
$$
I = \int_{-\infty}^\infty A(\omega) e^{i\omega x} d\omega.
$$
By splitting it in two you would obtain
$$
I = \int_0^\infty A(\omega) e^{i\omega x} d\omega + \int_{-\infty}^0 A(\omega) e^{i\omega x} d\omega
= \int_0^\infty [A(\omega) e^{i\omega x} + A(-\omega)e^{-i\omega x}] d\omega,
$$
where we exchanged $\omega \to -\omega$ in the second term. Writing the exponentials in terms of sine and cosine, you would find
$$
I = \int_0^\infty [(A(\omega) + A(-\omega))\cos(\omega x) + i(A(\omega) - A(-\omega)) \sin(\omega x)] d\omega,
$$
which is exactly the same as your second form with coefficients $A(\omega)+A(-\omega)$ and $i(A(\omega) - A(-\omega))$, respectively.

 

[solanojedi]

Wonderful, now I got what you meant! Your solution is in the general case where I start from a complete Fourier transform and I don't have constraint on the spectrum.

In the case discussed inside the paper, we agreed that the solution, before applying the boundary conditions and keeping the same notation, can also be written as $\int_{-\infty}^{+\infty} A(\sqrt{\lambda}) e^{i\sqrt{\lambda} x} e^{-\kappa \lambda t} d\lambda$. Since the abs value of the function inside the integral must be bounded for $x \rightarrow \pm \infty$ and $t \rightarrow \infty$ and the complex exponential doesn't contribute, the real exponential cannot have a positive coefficient otherwise it will diverge, hence $\lambda$ must be positive. In this way we limit the integral lower extremis $\int_{0}^{+\infty} A(\sqrt{\lambda}) e^{i\sqrt{\lambda} x} e^{-\kappa \lambda t} d\lambda$. In this particular situation, if I expand the complex exponential coefficient that I found in the previous reply ( $A(\sqrt{\lambda})$ and $B(\sqrt{\lambda})=iA(\sqrt{\lambda})$) is correct?
(Of course this would be a special situation)

 

[solanojedi]

Hi Orodruin,
I got what you were saying and I realize that I made a mistake in my last reply. I supposed that the solution of the heat equation was already written with a single exponential, while I think it should be written as $$\Psi (x,t)= \int_{0}^{+\infty} [c_{1}(\omega) e^{i\omega x}+ c_{2}(\omega) e^{-i\omega x} ] e^{-\kappa \omega^{2}t} d\omega.$$ From the theory of second order ODE, this should be an equivalent writing of the solution shown by the paper $\Psi (x,t)= \int_{0}^{+\infty} [A(\omega) \cos(\omega x)+ B(\omega) sin( \omega x) ] e^{-\kappa \omega^{2}t} d\omega$. The only difference should be that $c_1(\omega)$ and $c_2(\omega)$ are complex, while $A(\omega)$ and $B(\omega)$ are real (since the solution $\Psi(x,t)$ is real). Now, I can decompose the solution in the sum of two integrals
$$\Psi (x,t)= \int_{0}^{+\infty} c_{1}(\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega + \int_{0}^{+\infty} c_{2}(\omega) e^{-i\omega x} e^{-\kappa \omega^{2}t} d\omega$$ and changing the sign on the second integral we get $$\int_{0}^{+\infty} c_{1}(\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega + \int_{-\infty}^{0} c_{2}(-\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega = \int_{-\infty}^{+\infty} C(\omega) e^{i\omega x} e^{-\kappa \omega^{2}t} d\omega,$$ where $C(\omega)=\begin{cases} c_{1}(\omega), ~~\omega \geq 0 \\ c_{2} (-\omega), ~~\omega<0 \end{cases}.$ Hence we get the full Fourier transform. Can you please give me your opinion?

Supposing that this approach is correct, how can I now show that $C(\omega)$ is complex conjugate? From Fourier theory, supposing the signal $\Psi(x,t)$ is real, its transform should be complex conjugate. Maybe I'll create a separate post about this.

Thank you in advance!