# Eigenvalues and Eigenfunctions in Solving 2D Wave Equation in a Circle

• member 428835
In summary, the conversation discusses solving a 2D wave equation in a circle with given initial and boundary conditions. The solution involves two double sums and two constants, and using the initial conditions to solve for the constants. However, the initial condition ##u(t=0)=0## does not allow for the solution of either constant. The conversation also touches on using energy requirements and solvability conditions, and the importance of considering the eigenvalues and eigenfunctions carefully.
member 428835

## Homework Statement

Solve 2D wave eq. ##u_tt=c^2 \nabla^2u## in a circle of radius ##r=a## subject to $$u(t=0)=0\\ u_t(t=0)=\beta(r,\theta)\\u_r(r=a)=0\\$$and then symmetry for ##u_\theta(\theta=\pi)=u_\theta(\theta=-\pi)## and ##u(\theta=\pi)u(\theta=-\pi)##.

Lot's I'm sure.

## The Attempt at a Solution

So I find a solution for ##u## after applying the above boundary conditions where I have two double sums and two constants to solve for. To solve for these, I use the initial conditions. However, ##u(t=0)=0## does not allow me to solve for either of the two constants since ##u(t=0)## vanishes. Thus I have one initial condition yet two constants, one for each double sum. Any ideas?

I was thinking since the boundary was not moving at ##r=a## perhaps there is some energy requirement solvability condition that should be satisfied but I don't know. Any ideas?

joshmccraney said:
However, u(t=0)=0u(t=0)=0u(t=0)=0 does not allow me to solve for either of the two constants since u(t=0)u(t=0)u(t=0) vanishes.
This is similar to saying you cannot solve for A in the equation A=0.

Orodruin said:
This is similar to saying you cannot solve for A in the equation A=0.
Can you elaborate? I feel it's asking me to solve ##y=ax## given the line passes through ##(0,0)##.

Why don't you write out what you have and what the condition you get is?

joshmccraney said:
Can you elaborate? I feel it's asking me to solve ##y=ax## given the line passes through ##(0,0)##.
Actually, this is not what it is doing. It is asking you to solve ##x\vec e_1 + y\vec e_2 = 0## where ##\vec e_i## is a set of linearly independent basis vectors.

Here is what i have!

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Orodruin said:
Actually, this is not what it is doing. It is asking you to solve ##x\vec e_1 + y\vec e_2 = 0## where ##\vec e_i## is a set of linearly independent basis vectors.
I know, but this is how I see it though, so I need your help

joshmccraney said:
I know, but this is how I see it though, so I need your help
So what would be the solution to that equation?

Orodruin said:
So what would be the solution to that equation?
Equation (17) I think. But I only have one initial condition for both constants since the other vanishes ##u##!

• Most of your solution is fine, but you forgot the inner derivative when doing the time derivative in the end.
• Note that ##J_m(\sqrt{\lambda_{nm}} r) \cos(m\theta)## and ##J_m(\sqrt{\lambda_{nm}} r) \sin(m\theta)## are all linearly independent functions.
• You have already used the initial condition ##u(r,\theta,0) = 0## correctly to get rid of the cosine terms in time.
• You are missing the ##m = 0## term that leads to a constant function.

To be honest, I would write the angular solution on the form ##e^{im\theta}## with ##m\in \mathbb Z## instead - even if these functions are complex and lead to complex coefficients. It is just much easier to keep track.

I would also save writing some square roots by exchanging ##\lambda_{nm}## for the appropriate expression in terms of the zeros of the Bessel function derivatives.

member 428835
Orodruin said:
• Most of your solution is fine, but you forgot the inner derivative when doing the time derivative in the end.
Oops, totally spaced the chain rule!
Orodruin said:
• Note that ##J_m(\sqrt{\lambda_{nm}} r) \cos(m\theta)## and ##J_m(\sqrt{\lambda_{nm}} r) \sin(m\theta)## are all linearly independent functions.
So in the ##u_t## initial condition are you suggesting I multiply by ##\cos(p \theta):p\in\mathbb{N}## and use orthogonality of cosine and sine to solve for ##A## and then multiply by ##\sin(p \theta):p\in\mathbb{N}## to find ##B##?
Orodruin said:
• You have already used the initial condition ##u(r,\theta,0) = 0## correctly to get rid of the cosine terms in time.
• You are missing the ##m = 0## term that leads to a constant function.
Have I used this initial condition correctly though? Thinking about it now, if I consider the constant function for ##m=0## couldn't the time dependent cosine terms not vanish since those sums could add to the opposite of some arbitrary constant?

joshmccraney said:
So in the ##u_t## initial condition are you suggesting I multiply by ##\cos(p \theta):p\in\mathbb{N}## and use orthogonality of cosine and sine to solve for ##A## and then multiply by ##\sin(p \theta):p\in\mathbb{N}## to find ##B##?

Well, ##p\in \mathbb N## for ##\cos(p\theta)## and ##p\in \mathbb N^+## for ##\sin(p\theta)##. You will also need the orthogonality relations for the Bessel functions.

joshmccraney said:
Have I used this initial condition correctly though? Thinking about it now, if I consider the constant function for ##m=0## couldn't the time dependent cosine terms not vanish since those sums could add to the opposite of some arbitrary constant?

Consider the eigenvalues carefully. If your eigenvalue is zero, the time solution will not be sines and cosines. Also, there are eigenfunctions of the form ##J_0(\sqrt{\lambda}r)## (for suitable choices of ##\lambda > 0##).

member 428835
Orodruin said:
Well, ##p\in \mathbb N## for ##\cos(p\theta)## and ##p\in \mathbb N^+## for ##\sin(p\theta)##. You will also need the orthogonality relations for the Bessel functions.
Consider the eigenvalues carefully. If your eigenvalue is zero, the time solution will not be sines and cosines. Also, there are eigenfunctions of the form ##J_0(\sqrt{\lambda}r)## (for suitable choices of ##\lambda > 0##).

## 1. What is a PDE Wave Equation in a circle?

A PDE (Partial Differential Equation) Wave Equation in a circle is a mathematical equation that describes the behavior of a wave propagating in a circular domain. It is a type of PDE that involves both space and time variables, and is commonly used in fields such as physics, engineering, and mathematics.

## 2. How does a PDE Wave Equation in a circle differ from other PDE Wave Equations?

A PDE Wave Equation in a circle differs from other PDE Wave Equations in that it is specifically defined within a circular domain. This means that the equation is solved for points within a circle rather than a rectangular or irregular domain. The boundary conditions and solutions for a PDE Wave Equation in a circle also differ from those of other PDE Wave Equations.

## 3. What are some applications of PDE Wave Equations in circles?

PDE Wave Equations in circles have many applications in various fields such as acoustics, electromagnetics, and fluid dynamics. They are used to model and analyze the behavior of waves in circular systems, which can help in designing and optimizing structures and devices such as circular antennas, musical instruments, and circular pipes.

## 4. What are the main challenges in solving PDE Wave Equations in circles?

The main challenges in solving PDE Wave Equations in circles include finding appropriate boundary conditions, dealing with singularities at the center of the circle, and handling the complex geometry of circular domains. These challenges require advanced mathematical techniques and computational methods to accurately solve the equations.

## 5. Are there any real-world examples of PDE Wave Equations in circles?

Yes, there are many real-world examples of PDE Wave Equations in circles. For instance, the Schrödinger equation, which is used to describe the behavior of quantum particles, can be written in a circular form. Other examples include the Helmholtz equation for acoustic waves in circular cavities and the wave equation for electromagnetic waves in circular waveguides.

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