Why Is the Source Term Crucial in Differential Equations?

  • Thread starter Thread starter fisico30
  • Start date Start date
  • Tags Tags
    Source Term
AI Thread Summary
The source term in ordinary and partial differential equations is crucial because it represents the origin of disturbances, such as waves or temperature changes. Omitting the source term simplifies the equations but can lead to oversimplification, particularly in complex scenarios with multiple sources. In cases like electrostatics, while a homogeneous equation may suffice for single point charges, the situation becomes more intricate with multiple charges, necessitating careful consideration of boundary conditions. Ultimately, understanding the source term is essential for accurately modeling and solving real-world phenomena. The complexity of the problem remains constant, regardless of the approach taken.
fisico30
Messages
362
Reaction score
0
Hello everyone,

my question is regarding the source term in ODE and PDE.
If the region where the phenomenon (wave field, temperature,...) is observed is circumscribed to a volume not containing its source, then the differential equation becomes homogeneous (no source term) and easier.
So why solve the inhomogeneous eqn ever, unless we are inside the source, since our volume of observation can always omit the source?
Clearly, a source must exist somewhere to create the dusturbance.
thanks
 
Physics news on Phys.org
Let's consider electrostatics with point charges.
Then you are right: solving without source term is all that is needed.
But you should realize that the domain to be considered will become more complicated if many charges are involved. And you will need to use boundary conditions around each of these charges. The simplification is a pure illusion. But there are indeed many methods to solve these problems, each with there specific advantages in specific situations.

In the end, the quantity of information to be taken into acount remains the same.
 

Thread 'Off resonance driving of a MW/RF cavity'

Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
View full post »

Similar threads

I Where to find this uniqueness theorem of electrostatics?
Replies
27
Views
2K
What is a "source term" & what is it physically in QFT?
Replies
4
Views
3K
I Homogeneous Wave Equation and its Solutions
Replies
3
Views
3K
MHB Heat equation with annoying source term
Replies
1
Views
2K
Why 2nd order differential equation has two solutions
Replies
4
Views
8K
Solving a PDE in spherical with source term
Replies
2
Views
1K
Linear differential equations: source term constant
Replies
3
Views
3K
From differential equations to transfer functions
Replies
5
Views
2K
How did they get to this differential equation?
Replies
16
Views
2K
Should i be worried? (DE Course)
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top