# I Homogeneous Wave Equation and its Solutions

#### fog37

Hello,

There are many different wave equations that describe different wave-like phenomena. Being a differential equation, the WE is a pointwise relation and applies to the wavefield at spatial points.
• The equation is homogeneous when the source term is zero. That means that the solution functions satisfy the WE at spatial points where the source does not exist, correct? The wavefield is propagating away from the source location and we are mostly interested in the wavefield at those other location where the source does not exist and therefore be interested in solving the homogeneous WE. In which cases do we need to worry about and solve the inhomogeneous WE?
• If a wavefield solution f(x,y,z,t) to the WE is separable, i.e. f(x,y,z,t) = p(x,y,z) g(t) , does it always automatically mean that the wavefield is stationary and not traveling? Or can a traveling wavefields be described by a separable functions?
Thanks
fog37

Related Differential Equations News on Phys.org

#### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

#### Delta2

Homework Helper
Gold Member
1) Solving the homogeneous WE even with boundary conditions yields a wide range of generic functions. You need a source term, i.e solving the inhomogeneous WE with boundary conditions to narrow down the generic functions or uniquely determine a function.

2) if f, p and g are all real valued then answer is yes. if they are complex-valued they can be separable and yet describe a travelling wave, e.g $f(x,y,z,t)=e^{i ({xk_x+yk_y+zk_z}-\omega t)}$ describes a traveling plane wave with wave vector $k=k_x\hat{x}+k_y\hat{y}+k_z\hat{z}$ and angular frequency $\omega$ and it is separable.

The real part of f $\Re({f})=cos(xk_x+yk_y+zk_z-\omega t)$ also satisfies the WE (because WE is linear)but its not separable.

So what we usually do is to work with complex valued separable functions, and then take their real part.

Last edited:

#### fog37

Thank you. That was helpful.

For instance, given a 2D elastic membrane, stationary waves can be excited on the membrane. The boundary conditions impose that the wavefield must be zero at the membrane's edge. The wave equation becomes a separable equation assuming a wavefield solution that is separable since it the wavefield has to be stationary (cannot escape out of the edges). Using separation of variables, We find the eigenmodes which are also stationary, orthogonal and monochromatic solutions. We then build the specific solution to the problem by a linear superposition of weighted eigenmodes. The initial conditions ICs determine which eigenmodes will be needed and their amplitude. There is no source term in the wave equation.

In diffraction problems, there isn't really an edge. The solution wavefield is required to decay to zero at infinity but the field is known on a specific portion of the boundary (the diffracting aperture) and the homogeneous Helmholtz equation is solved inside the volume enclosed by the boundary. No source term here either.

-- As far as real-valued or complex-valued, traveling waves and stationary waves: if we work with real-valued functions, a separable solution is always and only a stationary (standing) wave. But we we work with complex-valued functions, if the real part of the complex-valued function is separable, then the solution will be stationary and not traveling. If not, it will be a traveling wave. At the end of the day, the solution must be real-valued, correct? Silly question: why does it have to be like that since it is just a mathematical concept?

#### Delta2

Homework Helper
Gold Member
Ok , I guess you are kind of right , depends what exactly boundary or initial conditions you have but you can uniquely determine solutions for the homogeneous WE.
About the concepts of real/complex valued and separability of functions: It is because the algebra is a bit different in C than R, while C inherits all the good properties from R. A function in C can be separable but its real part may not be and that is because if for example $f=f_1f_2$ is a separable function in C, its real part may not be necessarily separable because $\Re{(f)}=\Re{(f_1)}\Re{(f_2)}$ does not necessarily hold.

### Want to reply to this thread?

"Homogeneous Wave Equation and its Solutions"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving