Understanding how to apply the method of images to the wave equation

In summary, the problem is to find the general solution for the wave equation in a straight open magnetic field tube with a fixed bottom boundary condition. The solution involves deriving the d’Alembert solution for this case (semi-infinite). The domain is (0,∞) and the boundary condition is f(0,t)=0. Two initial conditions are given: f(x,0)=ϕ(x) and ft(x,0)=ψ(x). The method of images, as described in Introduction to Partial Differential Equations by Strauss, is used to find the final solution. The book states that the final solution is given by a pair of equations, depending on the values of x and t. The issue is
  • #1
JD_PM
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Homework Statement
Find the general solution for the wave equation ##f_{tt} = v^2 f_{zz}## in the straight open magnetic field tube. Assume that the bottom boundary condition is fixed: there is no perturbation of the magnetic field at or below the photosphere. Solve means deriving the d’Alembert solution for this case (semi-infinite).
Relevant Equations
##f_{tt} = v^2 f_{zz}##

$$f(0, t) = 0 \ \ \ \ \ x=0$$

$$f(x, 0) = \phi(x)$$

$$f_t(x, 0) = \psi(x)$$
Exercise statement

Find the general solution for the wave equation ftt=v2fzzftt=v2fzz in the straight open magnetic field tube. Assume that the bottom boundary condition is fixed: there is no perturbation of the magnetic field at or below the photosphere. Solve means deriving the d’Alembert solution for this case
(semi-infinite).


Some context on the problem:

Screenshot (947).png


We have an open magnetic flux tube that originates in the solar interior and continues on towards the outer solar system (to be precise at some point it will have to return because there is no net loss of magnetic field lines from the Sun, all lines have to come back, but for our purposes we can assume it goes off to infinity).

What I have done:

I have pictured in my mind the magnetic field line (when observed over time) like an elastic band that can carry waves.

We know that the domain is (0,+∞)(0,+∞).

Thus I have set up the problem as follows:


ftt=v2fzz 0<x<∞ftt=v2fzz 0<x<∞​
The (Dirichlet) boundary condition (just one because just one end is fixed):


f(0,t)=0 x=0f(0,t)=0 x=0​
Let's also state two initial conditions:


f(x,0)=ϕ(x)f(x,0)=ϕ(x)

ft(x,0)=ψ(x)ft(x,0)=ψ(x)​
OK so I am studying the solution method (method of images) given by the nice book Introduction to Partial Differential Equations by Strauss. I will post it and then ask questions.

Let's go step by step:

1) Let's get D'Alembert's solution for the wave equation on the whole line:

Let me skip the derivation of the general solution. It's in many books, such as the one I am reading and also online; for instance, check out:



The solution is:


u(x,t)=12[ϕ(x+ct)+ϕ(x−ct)]+12c∫x+ctx−ctψ(y)dyu(x,t)=12[ϕ(x+ct)+ϕ(x−ct)]+12c∫x−ctx+ctψ(y)dy​
2) Odd extension (I won't explain why odd and not even; I'll just state that when you're dealing with Dirichlet BCs you have to use odd extensions, and when you're dealing with Neumann BCs you have to use even ones; more details see section 5.2 of PDE book by Strauss) and final solution

The odd extensions are defined as follows:


ϕodd(x)=​
⎧⎪⎨⎪⎩ϕ(x), for x>0−ϕ(−x), for x<00, for x=0​

ϕodd(x)={ϕ(x), for x>0−ϕ(−x), for x<00, for x=0

ψodd(x)=​
⎧⎪⎨⎪⎩ψ(x), for x>0−ψ(−x), for x<00, for x=0​

ψodd(x)={ψ(x), for x>0−ψ(−x), for x<00, for x=0​
The book says that the final solution is given by the following pair:


f(x,t)=12[ϕodd(x+ct)+ϕodd(x−ct)]+12c∫x+ctx−ctψodd(y)dyf(x,t)=12[ϕodd(x+ct)+ϕodd(x−ct)]+12c∫x−ctx+ctψodd(y)dy

f(x,t)=12[ϕodd(ct+x)−ϕodd(ct−x)]+12c∫ct−xct+xψodd(y)dyf(x,t)=12[ϕodd(ct+x)−ϕodd(ct−x)]+12c∫ct+xct−xψodd(y)dy​
For x>v|t|x>v|t| and for 0<x<vt0<x<vt respectively.

My issue is that I do not understand why. I get lost in what Strauss calls the 'unwinding of D'Alembert solution' (pages 59, 60 of the book).

Please let me know if you need me to include more details.

EDIT: Suddenly some equations got messy (also the code). Let me know if I can do something about it.
 
Last edited:
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  • #2
JD_PM said:
Homework Statement:: Find the general solution for the wave equation ##f_{tt} = v^2 f_{zz}## in the straight open magnetic field tube. Assume that the bottom boundary condition is fixed: there is no perturbation of the magnetic field at or below the photosphere. Solve means deriving the d’Alembert solution for this case (semi-infinite).
Homework Equations:: ##f_{tt} = v^2 f_{zz}##

$$f(0, t) = 0 \ \ \ \ \ x=0$$

$$f(x, 0) = \phi(x)$$

$$f_t(x, 0) = \psi(x)$$

Exercise statement

Find the general solution for the wave equation ftt=v2fzzftt=v2fzz in the straight open magnetic field tube. Assume that the bottom boundary condition is fixed: there is no perturbation of the magnetic field at or below the photosphere. Solve means deriving the d’Alembert solution for this case
(semi-infinite).


Some context on the problem:

View attachment 253978

We have an open magnetic flux tube that originates in the solar interior and continues on towards the outer solar system (to be precise at some point it will have to return because there is no net loss of magnetic field lines from the Sun, all lines have to come back, but for our purposes we can assume it goes off to infinity).

OK let me rewrite the what I called 'What I have done' part:

What I have done:

I have pictured in my mind the magnetic field line (when observed over time) like an elastic band that can carry waves.

We know that the domain is ##(0, \infty)##

Thus I have set up the problem as follows:

$$f_{tt} = v^{2}f_{zz} \ \ \ 0 < x < \infty$$

$$f_{tt} = v^{2}f_{zz} \ \ \ 0 < x < \infty$$

BC (just one because just one end is fixed):

$$f(0, t) = 0 \ \ \ x = 0$$

Initial conditions:

$$f(x, 0) = \phi(x)$$

$$f_t(x, 0) = \psi(x)$$

OK so I am studying the solution method (method of images) given by the nice book Introduction to Partial Differential Equations by Strauss.

Let me attach the images:

Screenshot (948).png

Screenshot (949).png

Screenshot (950).png

I get lost in what Strauss calls the 'unwinding of D'Alembert solution', as I do not really see why he's checking out the cases ##x > c|t|## and ##0 < x < ct##

NOTATION NOTE: he uses ##c## instead of ##v## and ##u(x, t)## instead of ##f(z, t)##
 

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  • Screenshot (948).png
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Related to Understanding how to apply the method of images to the wave equation

What is the method of images?

The method of images is a mathematical technique used to solve boundary value problems involving wave equations. It involves creating a "phantom" image that reflects the boundary conditions of the problem, allowing for simpler and more efficient calculations.

How is the method of images applied to the wave equation?

To apply the method of images, the wave equation is first written in terms of the distance from the image, rather than the original source. Then, the image is introduced and the equation is solved for both the original source and the image, effectively doubling the number of solutions.

What are the advantages of using the method of images?

The method of images can simplify complex wave equations and reduce the computation time required to solve them. It is also useful for finding solutions in scenarios where traditional methods may not work, such as in the presence of boundaries or obstacles.

Are there any limitations to using the method of images?

Yes, the method of images can only be applied to problems with certain types of boundary conditions, such as Dirichlet or Neumann boundaries. It also assumes that the medium in which the wave is propagating is homogeneous and isotropic.

Can the method of images be applied to other types of equations?

Yes, while the method of images is most commonly associated with the wave equation, it can also be applied to other types of differential equations, such as the heat equation and Laplace's equation. The key is to identify a suitable "phantom" image that reflects the boundary conditions of the problem.

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