Finding the Greens Function of d^2/dx^2

Today I go over an example of finding the greens function for the operator d^2/dx^2 with boundary conditions f(0)=f(pi)=0
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[@ObserverAndObserved]

no BS. no beating around the bush. he cuts to the chase and just tutors. fucking lovely mate.

[@yagokf540]

Andrew, this quick video has done much more for my understanding of Green’s function than my professor was able to do in 3 weeks. Thank you for the great content.

[@ceroorec2153]

I cannot explain what an oasis this video was for me in a desert of frustration. Thank you for providing just a simple, clear example with only the necessary information. I finally feel like I have some scaffolding to understand these in general.

[@sinaelb2596]

Oh mein Gott! Thank you so much! So far no one has properly showed how the Green function actually workes within a problem, but thanks to your two videos I've finally understood.

[@mathunt1130]

This is an excellent introduction to Green's functions. The example chosen was simply enough to makes the ideas shine through. This is basically the method for calculating Greens functions for all cases. When you have infinite domains you can use transforms to get the Greens function but the added pain is obtaining the inverse transform from the transformed Green's function.

[@willnotsmith5058]

Beautifully explained, it appeared so easy here, while in class, I struggled to understand this. Thanks.

[@beez7528]

I looked up greens function on accident....... I’m not disappointed (Nice video)

[@AlchemistOfNirnroot]

Did you mean to search Green's Theorem? considering you said it was an "accident".

[@stephaniehowell5601]

I've spent the last hour or so searching through video after video about Green's functions and this was by far the best one! Thank you so much!

[@annakinderman8135]

Currently studying for a first grad E&M final. Thank you for this video!

[@Ensivion]

I'm currently trying to get my head around actually using these Greens functions and I'd like to see you do an example of a Neumann/Dirichlet boundary valued problem to solve for a potential.

[@zinzhao8231]

how is it going with physics?

[@mirzaammar3815]

​@@zinzhao8231not good

[@theunicornbay4286]

​@zinzhao8231 He failed lol

[@Ensivion]

@@theunicornbay4286ouch

[@KyleKabasares_PhD]

Awesome example! I know it was a rather simple case, but that made me feel way more confident about tackling Green’s function problems in the future!

[@CooperNicolaysen]

You inspired me! I'm now doing tutoring up to Diff EQ and I've already got a reoccuring client. I'm really excited.

[@TIENTI0000]

great, this is the best explanation that I have ever seen

[@joecoker7632]

Thank you so much for posting this. Your explanation was clear and your manner was very kind, which I greatly appreciated as someone who has to ask for a lot of help. If you aren't a teacher or a professor, you would make an excellent one.

[@MuhammadWaleed-f6n]

Weldone mate. You make it sound so easy. Infact you made it easy

[@sahdemattos]

Brilliant explanation ! It helped me in my mid term. Kind regards

[@possiblepilotdeviation5791]

Andrew!! I'm doing my Math Methods HW right now on GF, and you are much clearer than my text. Appreciate it!

[@samapanbhadury5148]

very very nice presentation. I also liked how you started explaining about the Green's function using matrix inversion method to solve for vector equations.

[@debajyotisg]

1:31 For those interested, the Green's function may be understood like so: If you have a differential equation of form D | f > = | p > ; where D is a differential operator (in this video, Andrew chooses d^2/dx^2) then you can write | f > = D^-1 | p > + Sum( ci | hi >); where D^-1, if it exists if the inverse of the operator D, and |hi> are the such that D |hi> = 0, which are the solution for the homogenous equation D | f > = 0. Then in a standard x basis, < x | f > = < x | D^-1 | x' > is your Green's function. (Propagator for the QFT fans)

[@vicen1090]

Clear and concise, perfect for my exam! ty ty

[@peteyoung3124]

So glad you made this video. I had to miss a makeup lecture from my professor on the day he taught this and his notes were so hard to follow

[@AlokmayDatta]

This is the thing I was looking for! Great 👍 👌

[@ianprado1488]

Wow, Green's function is an incredibly clever to solve DiffEqs

[@ritikpal1491]

Nice tutorial. It gave a good idea of how i can find green's function.

[@bhoopendragupta4782]

Awesome explanation, now it looks easy😎

[@nanocount09]

This was amazing! Can you please make another video where you find the Green's function for another operator? Or perhaps one with different boundary conditions. Thank you so much for this video, the explanation was super clear.

[@VineyKumar]

Thanks a lot! This really helped me understand Green's functions.

[@學習中的哈密瓜]

really appreciate and enjoy your video 🎉

[@Musiclover5258]

Excellent description, thank you!

[@Vcoja]

youre the best ting that has happened in the history of the universe after popeyes

[@AndrewDotsonvideos]

I'm glad that you're not undervaluing popeyes

[@Sherlock14-d6x]

Love your energy!

[@amograo3347]

Thanks, it helped us a lot!

[@seoktaelee1329]

good job ! It's a clear-cut summarizing

[@antejurcevic5855]

Hey Andrew, Thx for this great tutorial. I have only one question. Can you tell me how to integrate this Green's function now, are the limits of integral from 0 to 1, and do I need to split integral into two parts?

[@bairrao1303]

Amazing video but why does G satisfy the same boundary conditions as f?

[@srijanraghunath4642]

f=G when p=dirac delta. Therefore, it makes sense that the boundary conditions are the same because when p=dirac delta both G and f are the same. We're not making any assumptions either because the whole point is to go back and generalize for all p(x) after we find G. We just make p=dirac delta for just a moment so we can solve for G

[@maxguichard4337]

In this case in particular I would assume it's because in the boundary conditions f(x) = 0, so if f(x) = integral of G(x, y) * P(x)dx, at the boundary conditions that integral also needs to be 0, meaning the inner part G(x, y) * P(x) also has to be 0. Since P(x) is general and is not necessarily 0, then G(x, y) must be 0 at the boundary conditions as well. I realize this comment is old, however I hope this helps anyone else wondering the same thing.

[@supreethbhaskar3405]

@@maxguichard4337 you can't use this argument when f(x) = pi

[@demr04]

Thanks man. You save me alot :)

[@kateleaver9009]

I am so thankful for this video! Thank you so much :)

[@davkynaj]

The first whiteboard video from you that i actually understood quite well. Don't worry, it has nothing to do with your explaining abilities, I'm sure, and everything to do with me still being in highschool :D

[@pacolibre5411]

A good “exercise for the viewer” is to integrate the Green’s function over 0 to pi for your choice of function (say, x^2) to verify that your answer is a second antiderivative of your function that satisfies the boundary conditions.

[@NakintuLazia-mu1sj]

Is it possible to calculate the Green's function for the Poisson's equation for Newtonian gravity

[@kendytan91]

I dont quite understand why is the boundary conditions of f applicable/ equivalent to G. Or did I interpreted it wrongly?

[@srijanraghunath4642]

f=G when p=dirac delta. Therefore, it makes sense that the boundary conditions are the same because when p=dirac delta both G and f are the same. We're not making any assumptions either because the whole point is to go back and generalize for all p(x) after we find G. We just make p=dirac delta because thats when f=G. Then we can solve for G and then f.

[@edwardhartz1029]

@@srijanraghunath4642 Thank you for that intuitive explanation 🙏

[@jackhill4445]

From the definition of the Green's function, f(x) = \int G(x, y) p(y) dy, if we impose f(0) = 0 that means \int G(0, y) p(y) dy = 0. The only way to guarantee that this is always true for any definition of p(y) is to set G(0, y) = 0. Hence, f(0) = G(0, y) = 0 and so f and G have the same boundary condition. The same method also applies to f(pi) = 0.

[@tuo9433]

@@jackhill4445 What if the boundary conditions is non-zeros? Let say it is some constant c?

[@ur_khaan]

hey andrew great video, i have a question, let's say i know the homogenous solution do i still need the observation point (y) in your case, as for my case my function is not discontinuous so where do i use the point y if it is the case. thank you

[@s.g.johnson3403]

Did you live at home for undergrad? If so, can you do a video on what it was like adjusting to living away from home for grad school?

[@misnufkin1]

‏‪2:39‬‏ can you please explain why is it equal to the delta function?

[@user-pb8yw8cw3s]

There's a missing piece which is 1 for x=y Nice presentation dude !

[@pradyutmukherjee1110]

My question is what is the value of G(x,x') at x=x' ? If it is continuous then why not defined at x=x'?

[@stanleychan3212]

In general, the Green function satisfies the differential equation only in the distributional sense, it doesn't have to be defined everywhere. Moreover, many differential equations don't even have Green function.

[@christophas]

Thank you for this awesome introduction! However, I have one question. As you wrote, for a Green function ÔG(x,y) = 𝛿(x-y) must hold true. However, I do not really know how to derive 𝛿(x-y) from a piecewise defined function…? As someone who doesn’t really care about rigor, I would rewrite the piecewise definition using Heaviside step functions H(x -y), as their derivative is 𝛿(x-y). However I would like to know, if this is a proper method. Furthermore, using such an procedure would render G(y,y) being defined, contrary to the G given in your video.

[@stevedelchamps5113]

Great video; great lecture style. The only thing I would say is, try to learn to stand a little more to your right as you write on the board. It's tricky at first, but it will help your viewers see the stuff emerge as you speak.

[@amerkhoury8034]

Why could we substitute the boundary values of f(x) in the greens function tho..??

[@abhinavsaket1194]

Good video.

[@HotsumaSadamoto]

Should there be a factor of 2 floating around somewhere to make it so when you operate on your solution you get the delta function? I'm reading about Greens fxn's now and am a bit confused on that part.

[@micayahritchie7158]

How exactly would one find the solution for a specific p(x) using this function?

[@stefanopastore8886]

Micayah Ritchie Yoy just integrate for y going from 0 to pi G(x, y)*p(y)dy and you get your solution

[@yuma5297]

isnt it easier to Fourier Transform ÔG = delta(x-y) ?

[@barzipori4238]

A great video.

[@kierkegaard54]

Why isn't G(x, y) defined at x=y? is it because of the delta? like if u were to define G at x=y then the second derivative with respect to x, it blows to infinity

[@LakanBanwa]

But why is the integral of the dirac delta from the right and left of y equal to 1?(The real reason I'm presuming that tells us why we should set the difference of the derivative of the green's function at the same locations equal to 1)

[@timurpryadilin8830]

for this specific example can't we just integrate p(x) twice to get the result?

[@staticnullhazard6966]

Well yeah if your right hand side is just some easy function (polynomial) but if it is some distribution of Diracs -which is very common in physics- it isn't that simple. Therefore it is quite useful to know Green functions.

[@carlosalbertomorgancruz2247]

Bien explicado. Gracias

[@maazadnan117]

Thanks bro.

[@cesarmoreno987y]

Thanks!

[@markopolo6104]

Excellent

[@davinhocpinho]

Thanks.

[@nocontent3124]

Greens functions replace solving a hard problem with creating an even harder problem which you then have to solve in order to solve an easier problem which is made harder by solving a Green's function etc, QED and a Kronecker delta. Hence just just learn MATLAB and problems become not problems ie problem solved.

[@mokrodrigues4515]

Thanks a lot!

[@quahntasy]

200th video will be live of him studying

[@deeptochatterjee532]

Why does the Dirac Delta have x--1 in it?

[@AndrewDotsonvideos]

Deepto Chatterjee I guess my y’s look like -1’s

[@deeptochatterjee532]

@@AndrewDotsonvideos oh ok that makes more sense

[@timonph7073]

What does the d in d^2/dx^2 mean?

[@violymhi]

Differentiate I guess? The d^2/dx^2 as a whole is just a way of saying that we're taking the second derivative with respect to x

[@loganfisher3138]

Not offering tutoring services anymore? Your link is dead.

[@bintekhadam6776]

Excuse me! Sir.... Kindly solve greens function for laplace equation in partial differential equation for non-homogeneous equation

[@josecarlospesa]

Is really unfair that you are clever and handsome

[@urosgrandovec3409]

awesome

[@AlchemistOfNirnroot]

200th video should be campus tour!!!

[@duncanw9901]

Zeroeth view!

[@zoltankurti]

Impodsible, since the first view is the first view :D

[@ゾカリクゾ]

he never said he was the first view. He's the zeroth view. Deal with it.

[@AlchemistOfNirnroot]

Views exist? Thermodynamics**

[@markopolo6104]

Cool

[@mathlover2299]

It's Green function by the way.

[@danielrejat3370]

That moment when you start watching, understand a bit, a minute goes by you understand almost nothing. What do i watching this even though i am a highschool student

[@Gurke76_56]

Your underlined ONE looks like a TWO, thats really irritating

[@lvl3tensorboi929]

Why didn't you just use the look-up table for that integral?! lol

[@Muuumenn]

what

[@ianprado1488]

Hi Andrew, People who understand physics are cursed. If you research molten salt cooled reactors (FHRs, MSRs, etc), you will understand that this is *the technology* that can reverse climate change and end poverty. Ian

[@Legend_Hunter_Original]

curse is a matter of perspective, you could also view yourself to be blessed because you have the 'knowledge' to solve the mentioned problems.

[@oskarjung6738]

It's weird, this function is not even differentiable in it's domain. Then how can it be a solution to a differential equation. These things really tick me off in physics. I guess physics is not my cup of tea.

[@AndrewDotsonvideos]

Greens functions are not just used in physics, they can be rigorously defined. You have issue with them not being differentiable at all points, that's an artifact from the delta function source term that defines the greens function. They're no less rigorous than a delta function and there's the entire theory of generalized functions for those.

[@telotawa]

it feels like bullshit idk

[@jayjayf9699]

It would be good if you improved your handwriting as it looks like a 5 year old wrote that y

[@hasanhashemi4504]

Excellent