3.1.5 Boundary Conditions and Uniqueness Theorems

Given some boundary conditions, do we have enough to find exactly 1 solution? For Laplace's Equation, if we have the boundaries of a region specified, we have exactly one solution inside the region.

[CrushOfSiel]

These are so great to watch after reading the book. Then I go to class and I can just absorb it even more. Thanks!

[hugovalenzuela]

you saved my life, dude! I wasn't sure how this theorem also applies to Poisson's equation... Thanks, good video

[AsamaQureshi]

"The uniqueness theorem is a license to the imagination" sounds like some sort of inspirational quote.

[Mathswithali]

Thank you Dr. for your lecture . My question is can you show me how to show the uniqueness theorem holds for sobolev space?

[ninjajesus]

Thanks for uploading these videos. This class is killing me.

[simonacurkoska]

haha... good times.
Thanks for these videos. It's a good way to review before an exam.

[nehahahahahahahaha]

very well .... seriously it is very confusing .. but now I think my concept is clear

[AmanYadav-ovwl]

Same in David J Griffith Book of electrodynamics(word to word).

[ninjajesus]

I don't know if I can be helped. It seems like the only information I'm getting in this class is "here's 10 formulas, here's how we got these 10 formulas, now here's a problem, figure out which formula to use, figure out how to use the formula, and figure out how to customize the formula to fit the problem."

[atreyaacharyya]

@jg394 Your videos have really transformed the once difficult to understand Griffiths into my favorite physics book. Could u plz help me out with the first chapter, im having trouble with the part on orthogonal coordinates and the Delta Dirac equation.

[vishwasshankar]

At 8:08, I still can't wrap head around that. How can the relationship between the boundary conditions of the 3 potentials given by that? Like is it a definition?

[josephksajan]

Sir i didnt get last part where if there is some charges inside cavity does v1 and v2 in griffith shows potential at present inside the cavity=v1 and v2=in boundary or are both of them just mathematical abstraction

[govindrathi]

its necessary that charge density inside is zero for laplace's equation to hold(obviously).

[jg]

I thought I uploaded this? You'll have to check my video queue back 8 months ago.

[jg]

Let me know what questions you have, and I'll do my best.

[krishanudebnath]

But if there is a charge outside the region then V inside the region must depend on the distance from the charge inside the region how can there be a particular value of V inside the region please explain this sir. Please I am stuck in this for a long time.

[jg]

Yes, this is not a good way to approach physics.

Let me put together a video to kind of give you my approach to mastering physics. Poke me if I forget to do it in the next few days.

[PhysicsMath]

Sir can I have. V3= v1+v2
It also satisfy Laplace equation .. then it implies v1=-v2 . It is not unique. Please ?

[AmanYadav-ovwl]

Why V3 is zero?
you are saying V1 and V2 are same so V3 should be zero and it implies V1 equal to V2?

[pkushagra]

Ur videos are nothing but the audio(video?)book version of the writings of Griffiths.
You really don't add anything of value to it.