Відео

Hi. No, I am just using this equation as an example (because it is the most often used example). The FEM can be applied to a variety of different problems.

Thanks! :)

Thanks :D

Thanks for the kind words! :)

This was discussed also in another comment. I don't know how to link it here. Maybe you can find it under this video or under the other video on FEM. But your are right, I am not telling the full story in the video and I understand the confusion. Even when using quadratic shape functions, the derivatives of u are not continuous at the nodes. Therefore, the strong form could not be fulfilled at the nodes. To understand why this is not a problem in the weak formulation, one would need to study which function space u belongs to and study a bit of measure theory, which was not the purpose of the video. There are indeed methods that work with the strong form. These are called collocation methods. You assume an ansatz, insert it in the strong form and minimize the residuals of the strong form a set of points x. Such methods have in general not as nice properties as the FEM based on the weak form.

@@DrSimulate Ah I think I understand, it would be great to see a video on that at some point!

For anyone else interested, if you go to the FEM video ua-cam.com/video/1wSE6iQiScg/v-deo.html and crtl+F for this: "I have a question though. If you had chosen 2nd or higher order polynomials for the shape functions N(x), u''(x) would not necessarily be 0 everywhere." The comment should come up

@@robm624 Yes, I hope I can do a video that is more mathematically rigorous and covers e.g. the function spaces in the future :)

That's right. Do I say something contradicting in the video? If yes, can you tell me were? :)

@@DrSimulate sorry, you did not say anything contradicting. I added this info so that if someone wants to know why the displacement tensor is zero for the translation, they can find it in the comment section.

@@Prashanth-yn9zd Ahh, I see. Thanks, appreciate it! :D

@@RSLT Thanks :)

@@cleisonarmandomanriqueagui9176 I plan to cover all these questions in future videos :)

@@stitaswain7349 Thanks :)

Thank you! :)

@@5eurosenelsuelo Good question! If the additional test functions are linearly dependent on the other test functions, then nothing changes. If not, the system may have no solution and the problem needs to be approached by minimizing the sum of the squared residuals of the system. I don't know, how this will affect the solution 🙄

@@DrSimulate Interesting. My intuition would be that the additional equations will be linearly dependent because it wouldn't add any new information to the system of equations but I don't have enough experience in the topic to test the hypothesis myself. If the new equation makes the system unsolvable it'd be so strange... That new solution found by minimizing residuals would be a more accurate solution compared to the real analytical solution? Is accuracy dependent of the test functions chosen? Computing speed definitely is but the result should be the same and if that's the case, adding more equations should add no new information as previously mentioned.

@@5eurosenelsuelo I think, as the discretization of u has an influence on the accuracy, the choice of v will also have an influence on the accuracy. For example, if you consider a two-dimensional domain, I am sure that choosing many test functions that are zero at the interesting parts of the domain (e.g., at edges or holes) is not benificial. Unfortunately, I don't have any experience with this. It would be interesting to dig in the literature and see if there is any research on this.. 🤔

Ah I see a little further in the video you mean with gradient what I would call the total derivative. Never mind :)

@@aqibrasheed4874 Thank you!! :)

@@5eurosenelsuelo Thanks a lot! Sharing the video with your friends would help a lot 😁🤗

@@krokodilvomnil5327 Tough question. FEM is related to math, CM is related to physics. So it depends if you have more a math or physics background.. 🤔

@@soumyadas9896 You can take other functions than N_i, but they should be linearly independent. If you take linearly dependent functions then the system is not uniquely solvable.

Can you be more specific about the problem you are interested in? Is your solution function u also complex?

@@DrSimulate I am sorry, I did it a little bit wrong. It turns out to be a little bit more complicated. Edit: The system looks like: 1 / (i * k) * u'' - u = f; where only u is complex. The physical system contains a time-alternating flux density (field source = forcing term, homogeneously distributed along x) that penetrates an electrically conductive material and therefore induces a voltage in the material that causes eddy currents and damping reaction fields (Lenz's rule). f is the flux density (e. g. 0.1 T) and k can be e. g. 2 * pi * frequency * 4 * pi * 10^-7 * 625000 u = 0 at the left and right boundary.

@@maxhullmann5660 Mhh. I have never worked with such a system. Did you already derive a weak form? Maybe you can discretize both the real and imaginary part of u and substitute this into the weak form (just a guess).

Is f periodic in time (e.g., sin or cos)? If yes, you may assume a periodic ansatz for u. If not, you may have to discretize in time (e.g., Euler discetization in time). Is the problem in 1D?

@@DrSimulate 1 / (i * k) * u''(x) - u(x) = f is 1-Dimensional and f is a constant (= flux density amplitude, e.g.). I could figure out a solution: weak form: 1 / (i * k) * Integral u''(x) * v(x) dx - Integral u(x) * v(x) dx = Integral f(x) * v(x) dx. The only difference to your example is the complex factor of u''(x) * v(x) and the additional term - u(x) * v(x). The final solution turns out to be ( i / k * K - K' ) * U = F where K = Integral N' N'T dx and K' = Integral N NT dx = IdentityMatrix * ElementLength with u0 = 0 and uend = 0. The analytical solution can be computed by u_an(x) = f * cosh( sqrt( i * k ) * x) / cosh( sqrt( i * k) * Interval_Length / 2) - f (symetrical interval, e.g. -0.015 <= x <= 0.015)

Nice work ❤❤

I use Manim www.manim.community/

Ist erstmal nicht geplant ... sorry

Thanks Angu 😁

That is a very specific request 😅

THANKS!

Thank you so much!! 😃

The purpose of the weak form is to reduce computational effort. Linear functions require less computational power compared to quadratic or higher-order functions. Additionally, the strong form requires both essential and natural boundary conditions, while the weak form only requires essential boundary conditions, thereby reducing the continuity requirements. In conclusion, using the weak form reduces computational power and lowers continuity requirements. If I am mistaken, please correct me.

@@hamzazaheer3783 thanks. I do agree that linear shape functions and the weak form are computationally less intensive. But my question was more about the way it was framed, i.e. "let's use linear shape functions", and then "oh, these have 2nd derivative = 0", therefore "we must solve the weak form". It seems that this motivation would've been avoided if the shape functions were 2nd order or higher. So I wonder if there would be more hurdles if one would proceed with this approach (use higher order polynomials to avoid the weak form).

@@LucasVieira-ob6fx Yes, you can solve the strong form without using the weak form, but it requires higher-order shape functions compared to the weak form. Additionally, it must satisfy both essential and natural boundary conditions. You might wonder why we use the weak form if higher-order shape functions are needed. The reason is that, in the weak form, when using higher-order shape functions, you only need to satisfy the essential boundary conditions. The primary advantage of using the weak form is that it lowers the continuity requirements.

I would like to add another point. Even if the second derivative of the ansatz would not be zero, there may be a problem: It is very likely that it is not possible to tweak the parameters in the ansatz in a way that the strong form is exactly fulfilled at all points x. In other words: the "exact" or "true" solution of the differential equation cannot be expressed by the ansatz. In this case, what people do is to minimize the norm of the residuals of the strong form at some chosen points x. This is called collocation and there is still active research on that. But collocation methods are by far not as successful as methods based on the weak form.

@@DrSimulate you are talking about galerkin and rayleigh ritz method ?

I don't know about a video explaining this, but you may check out these lecture notes, which are very didactic and have one section about this: ethz.ch/content/dam/ethz/special-interest/mavt/mechanical-systems/mm-dam/documents/Notes/IntroToFEA_red.pdf

@@DrSimulate Thanks

That's an interesting suggestion. I definitely want to show more applications in future videos. The nice thing about the Poisson problem is that it appears in many different disciplines.

Yes, this is definitely planned in the future. But it will take some time, thanks for your patience :)

Keep grinding 💪

Thanks a lot! If you find this too easy, wait until we get to nonlinear continuum mechanics 🤯🤯

@@DrSimulate heheheh :) i meant easy in the sense that i somehow felt that understood the basics of it very quickly, like dunning-kruger effect, feels like i "know" so much already. Maybe because in the video i didnt get too much info about what i dont know, you know? 😁 im just spitballing

@@DrSimulate do you have a discord or a server? i would love to chat with you occasionally!

@@utof Are you by any chance on the Summer of Math Exposition discord server? We can have a chat there if you like :) discord.com/invite/WZvZMVsXXR

@@DrSimulate yeah, im there! im @utof