Thanks ! I have been looking for such a clear and detailed explanation for a long time. Either too simplified in 10' visuals or too obscure 2h lectures. This is just the golden mean. I will now dig into more détails with this good overview.
Danke dir! Das ist das erste Video auf UA-cam, das mir wirklich hilft meine Vorlesung zu verstehen :) Toll aufgearbeitet und jeder Schritt gut und verständlich erklärt
Excellent, thank you for the time to put this video together. Also thank you for additional information on the lecture notes. Looking forward to see more on this.
Yes, you are correct the RHS should be integrated over the master element K(1). This is a mistake in the video but at least it is later correct for the Laplace/stiffness term. Thank you for pointing this out 🤗
This is an excelente video. I really strong recommend it. I fond of the way that you explained and the notation that you use it, very formal, but easy to understand.
In 14:54, why is this step valid? Why is decomposing Фh into it's basis functions Фi valid with the integral when one of the requirements for this weak form formulation is that the test function you integrate with satisfies the boundary conditions. Фh satisfies them but not every Фi. Let's say in 9:43, you write that the weak formulation is valid with Ф(x) being in the function space V, and earlier you defined V as a function space where all functions satisfy the B.C. For example, what if you had a B.C. like f(0) = 100. Then one of the linear basis functions, the one that ramps up from 0, won't satisfy it.
Yes, I agree that your question is valid. Most FEM tutorials focus on homogeneous Dirichlet BCs and then don't really talk about inhomogeneous Dirichlet BCs like f(0) = 100. Your questions has already been answered at the following link: scicomp.stackexchange.com/questions/5072/how-to-properly-apply-non-homogeneous-dirichlet-boundary-conditions-with-fem The short answer is that for inhomogeneous Dirichlet BCs the solution u_h and the test functioj Φ_h belong to different function spaces. u_h needs to specify the inhomogeneous Dirichlet BCs, e.g. u_h(0) = 100, but the test function Φ_h needs to satisfy the homogenous Dirichlet BCs, e.g. Φ_h(0) = 0.
Do you mean that the matrix A should be transposed? If this is what you meant, it does not make any difference here since the bilinear form is symmetric and the linear system is also symmetric. In general, one needs to be cautious though with the assembly of the linear system, since for more complicated problems the order of i and j matters. That's why if you derive the weak form by multiplying from the right with a test function, like in this video, you need to switch i and j in your code in the assembly of the linear system. Thus, many engineers multiply with a test function from the left, i.e. (φ, Δu), to later avoid any confusions in the assembly of the linear system.
Concise and clear, thank you very much. Do you know of a nice python library i could use to solve pdes with the finite element method ? The only ones I've found seem to be aimed at mechanical engineering simulations only...
It depends on which PDE you want to solve. For the Poisson equation there is plenty of stuff on GitHub. If you want something that is more versatile, I would go for FEniCS/FEniCSx, Ngsolve, Dune, MFEM or deal.II, but some of the latter might not have Python bindings yet. Personally, I have mostly worked with FEniCS in Python and deal.II in C++. If you can implement your PDE in FEniCS on your first try, then FEniCS is my favorite, but sometimes I needed low-level access to the library and found deal.II to be more convenient. Under the hood both libraries use C++ anyway.
@@numericalanalysisbyjulianr4242 i thought I had replied to you but I can't see my comment when checking again : just wanted to thank you one more time for the helpful answer, you rock !
Hi Sir, phi(0) = phi(1) = u(0) = u(1) = 0, phi and u share the same function space and boundary conditions, is this called Galerkin method to choose the testing function?
Yes, if you solve the weak form of a differential equation with the same trial and test space, i.e. the same function space for u and φ, this is called Galerkin method. Some people also use different function spaces for u and φ, this is then called Petrov-Galerkin method.
To recap the first part: we must find the solution u to the bvp in C². But since it is not possible to do it directly in C² we move to the subspace of test functions V (in which the test functions respect the boundary conditions) and here we reformulate the bvp in the weak form. Right? And what role does the minimization of potential energy play in all this?
Yes, this is correct. For the finite element method we cannot directly work with the strong form in C^2(Ω) \cap C^1(\bar{Ω}) but instead use the weak form and solve it in H^1_0(Ω). (Note: There are also methods that work directly with the strong form, e.g. the Finite Difference method.) So far we have not used the energy minimzation anywhere and could ignore that. Nevertheless, especially in engineering, many people start off with some functional that needs to be minimized and derive the weak form from this energy minimization. Furthermore, sometimes one needs to use the energy minimization to prove theoretical results, e.g. the existence of solutions.
So i have an engineering background. Could you explain to me from a mathematicians perspective, what the difference is between using finite difference and finite element method? Both solve PDE with a mesh. Both methods construct an eigenvalue equation, where the matrices are sparse. But with FDM the PDE is approximated with derivatives and a the grid is structured the sparse matrices is populated relatively straight forward. With FEM polynomials here the basis functions are tuned to approximate solutions of each element. Which then together forms the big solution over the hole mesh. Is this somehow correct understanding?
Yes, your summary is correct. With FDM we just approximate the derivatives in the strong form of the PDE and with FEM we work with element-wise polynomial approximations. We don't construct an eigenvalue problem though but a linear equation system. A big difference is that FDM uses the strong formulation of the PDE, e.g. -u" = f, whereas FEM uses the weak/variational formulation of the PDE, e.g. (u', φ') = (f, φ). Like you mentioned, imho FEM is more flexible when it comes to meshes, since FDM requires "structured" grids. With FEM it is easier to deal with more complicated geometries, e.g. plate with hole. This has then also many engineering applications, e.g. Isogemetric Analysis (IGA) uses spline basis functions to work on CAD generated domains directly.
You are almost right. Solutions of the strong formulation need to be twice continuously dofferentiable and solutions of the weak formulation need to be once weakly differentiable. So automatically each strong solution is also a weak solution. But a weak solution is only a strong solution if it is also twice continuously differentiable. Otherwise you can't perform integration by parts and recover the Δu term. TL;DR: For u in C^2(Ω) and C^1(\bar{Ω}) the strong and weak form are equivalent.
For future references, please provide concrete examples to make answering your question easier. Nevertheless, here is a very general answer: The choice of the finite element basis depends on your function space, which in turn depends on your PDE. At the example of the Poisson equation, here we have a second order PDE, but for the weak formulation we apply partial integration and thus have at most only first derivatives. This needs to be reflected by the function space and we choose the space H^1 (or H^1_0 if we have homogeneous Dirichlet BC) which contains all functions whose first (weak) derivative is L^2 integrable. We can thus choose the Lagrange FE basis of piecewise polynomials of up to order r which is a conforming subspace of H^1, i.e. all functions from our FE space are also contained in H^1. If we have a fourth order PDE like the biharmonic problem, we apply integration by parts and arrive at a weak formulation which has up to second order derivatives. By the same argumentation as above, we thus use H^2 as our function space, since the second order derivatives need to be L^2-integrable. Now we can't use the same FE basis anymore as for the Poisson problem, since between elements we only ensure continuity of the FE basis (C^0), but we would need also the continuity of the derivatives of the FE basis (C^1). Therefore you need to use C^1 conforming FE, like the Hermite basis, or split the problem into a mixed system similar to fenicsproject.org/olddocs/dolfin/1.6.0/python/demo/documented/biharmonic/python/documentation.html. The problem of choosing an appropriate more complicated finite element is not only specific to fourth order PDEs, but also applies to other problems like Maxwell's equations from electromagnetism, where for the simplest problem (Eddy-Current) one solves - curl(curl u) = f and after integration by parts we need a function space where curl u is L^2-integrable (H^curl). There you can't use the standard Langrange basis either, but need Nedelec elements to be H^curl conforming.
@@numericalanalysisbyjulianr4242 thank you so much... I used biharmonic equation. But i cant create a stiffness matrix. Because i don't know the basis when i used discrete finite element method
My example is in hilbert space of fourth order The problem is biharmonic plus poisson =f I used green theorem to obtain weak form twice for biharmonic and once for poission Then i have galerkin equation B(y, v) =F(v) Vn is a subspace(??? ) of H2 Yn is a basis of linear combination What is a basis of finite element method? To creat stiffness matrix Is it yn=sum Cj Yj, and j =1,..., n??
@@emanalquraishi9279 Yes, in your case your FE space needs to be a subspace of H^2. For this you could use the cubic Hermite basis: hplgit.github.io/INF5620/doc/pub/H14/fem/html/._main_fem007.html. If you don't want to use C^1 FE, then you could also transform the bihramonic into two second order equations and then use the regular Lagrange basis: fenicsproject.org/olddocs/dolfin/1.6.0/python/demo/documented/biharmonic/python/documentation.html.
Dear Julian, I always want to make a video with nice text animation like this video. May I ask how you could do that?( what software, library , etc). Thank you.
Dear Tien, thank you very much for the kind words. I am using the community version of manim (docs.manim.community/en/stable/), 3blue1brown's animation library, to create the animations.
Thank you for the very informative but brief video. This is really helpful for me to learn what FEM is and prepares me to simulate some models using FEM. However, there is one thing that I wonder about. @17:43 Why do you put (1/h) for the basis function under integral? I am confusing where that (1/h) came from.
You have to look at the picture on the left hand side and at the basis function ϕ_i in the interval [x_i, x_{i+1}]. In this interval the basis function starts at 1 for the left boundary x_i and goes to 0 at the right boundary x_{i+1}. Additionally, the basis function is linear in this interval. Hence, its derivative is constant in this interval. We observe that in this interval the derivative of the basis function is simply (0 - 1) / (x_{i+1} - x_i). Now we said that the distance between two nodes is fixed and we set it to be h := x_{i+1} - x_i. Overall we have thus shown that the derivative of ϕ_i equals to -1/h in the interval [x_i, x_{i+1}].
How do I deal with inhomogeneous Dirichlet boundary conditons, e.g. u(0)=a and u(1)=b? Like the homogeneous Dirichlet boundary conditions this needs to be included in the function space. Before we had that u is in H^1_0 to satisfy u(0) = u(1) = 0. Now we need to find u in H^1 such that u(0) = a and u(1) = b. But be careful one still uses homogenous boundary conditions for the test functions, i.e. φ in H^1_0. You don't change the weak formulation but only need to enforce u(0) = a and u(1) = b on the linear system. TL;DR: For the boundary conditions u(0) = a, u(1) = b, you have the probelm statement: Find u € {g € H^1((0,1)) | g(0) = a, g(1) = b} such that (u', φ') = (f, φ) for all φ € H^1_0((0,1))
Almost correct. The weak form is defined in H^1(Ω) resp. H^1_0(Ω) when we include the boundary conditions. As you correctly observed, for the weak form we only need to use first derivatives of the solution. But since these derivatives only need to make sense when we integrate over Ω, e.g. (u', Φ') needs to make sense, it is sufficient if the solution u is once *weakly* differentiable. We denote this then by H^1 for the space of function that are once weakly differentiable. If you want more details on weak derivatives and the space H^1, you can take a look at Sections 7.1.4 and 7.1.5 in the lecture notes from the video description.
@@nick45be Yes, H^1 is a Sobolev space. Sometimes people also write Sobolev spaces as W^{k,p} with k denoting how often it is differentiable and p denoting that these derivatives need to be in L^p. For many PDEs, we just need p = 2 and hence the short hand notation H^k = W^{k,2} is used.
Sir can you pl tell me What are functional spaces here Pl help me to understand it I am not from maths background Pl sir And why these are to be finite here
In this video we discussed how the Finite Element Method can be used to solve a Partial Dofferential Equation. Here, we used the Poisson problem as an example of such a problem. It is hard (or sometimes even impossible) to write down the exact solution to a differential equation. That's why one tries to find a way how to get a good approximation of the solution by using a computer. For this we try to transform our problem into a linear equation system, because this is what computers are good at solving. In the first step, we don't solve the strong formulation of the Poisson problem -u''=f anymore but choose some other function φ with which we multiply the PDE and then we integrate by parts to get (u',φ') = (f, φ). And now we compute to the function spaces. What kind of properties do u and φ need to have for all this to make sense? Going back to -u''=f we need to be able to take second derivatives of u, so we usually require u to be twice continuously differentiable. When we now look at (u',φ') = (f, φ), we see that u and φ need to only be once differentiable. Furthermore, since these derivatives only show up in the integrals we can use a bigger class of differentiable functions, called weakly differentiable functions. That means u and φ only need to be once weakly differentiable. If you want to learn more about weak differentiability, you can read soke books on functional analysis and PDE theory and look for Sobolev spaces. Now, that we have a rough idea which kind of function space u comes from we want to solve it with a computer, but the basis of the function space is infinite dimensional, so you would need an infinite amount of numbers to represent the solution u. But your computer has only a finite memory... That's why we now take our function space and replace it by another space which has a finite basis, e.g. the space of piecewise linear functions. With this we can finally solve a linear system to get an approximate solution for u. Moreover, the more elements you have in your mesh, the better your piecewise linear approximation and the more accurate your solution u
@@numericalanalysisbyjulianr4242 sir how we choose phi functions and on which basis How we choose basis function And in fem what are these function in function spaces Are these functions related to to the function on each node pl explain
The general version of Poisson's problem is given at 3:04 with -u'' = f. Chossing f = -1 means that we have a uniform force acting in the entire domain, like gravity.
Yes, if (w,φ) = 0, then by definition w and φ are orthogonal in the inner product (•,•). In the video I used the L2-inner product, so then w and φ would be orthogonal in L2.
It would be helpful to have some knowledge of calculus, linear algebra and possibly even PDE theory. Hopefully this video is self-contained enough though that you don't need any prerequisites beyond introductory math courses.
@@mediwise2474 I created a few videos on the Finite Element Method, including also this one. This should cover a few important aspects of FEM. If you want more details, please also checkout the video lectures by Wolfgang Bangerth www.math.colostate.edu/~bangerth/videos.html that also partially give an introduction to the FEM software library deal.II. If you prefer a PDF, take a look at the lecture notes of Thomas Wick: doi.org/10.15488/11709
Jaein. Ich gebe dir Recht, dass nicht jede schwache Lösung auch eine starke Lösung der PDE ist. Unter hinreichenden Annahmen an die Regularität der Lösung, hier u ist 2 mal stetig differenzierbar, sind schwache Lösungen auch starke Lösungen. Ich habe die hinreichende Regularität der Lösung im Video aber wahrscheinlich nicht explizit erwähnt...
@@hakuhakuji No, I still don't see any mistake around 1:56. Everything looks correct to me. There are however different conventions in the FEM literature, how the Laplace operator is written. Some people prefer the notation ∇^2 , whereas I use the convention Δ for the Laplacian. These different notation conventions can also be found in fenicsproject.org/pub/tutorial/html/._ftut1004.html in the text below equation (2.2).
@@numericalanalysisbyjulianr4242 Ah.. I see it now. I would say it's not the best convention since it is greek Delta, which means just a difference. Just puts more confusion. Why not to use the most widespread convention?
@@hakuhakuji The most widespread convention in the mathematical literature is using the greek Delta for the Laplace operator. Hence, I also use this convention.
OMG! I really cannot express how thankful I am 🙏. This is a fantastic contribution to the community.
Thanks ! I have been looking for such a clear and detailed explanation for a long time. Either too simplified in 10' visuals or too obscure 2h lectures. This is just the golden mean. I will now dig into more détails with this good overview.
Danke dir! Das ist das erste Video auf UA-cam, das mir wirklich hilft meine Vorlesung zu verstehen :) Toll aufgearbeitet und jeder Schritt gut und verständlich erklärt
This is very useful. The explaination is easy to follow and the animations are beautiful! Amazing job! Thank you so much!
Thank You, visualization is the best way to understand new stuff.
Excellent, thank you for the time to put this video together. Also thank you for additional information on the lecture notes. Looking forward to see more on this.
I had watch lot of videos and I can say you are the only one who I understood with
First complete demonstration, thanks man
This is the best intro video to FEM ever!!! Huge thanks!
thank you so much for the explaining it! Best resource I have seen on the material so far!!
The way I see it in 25:17, RHS of the bottom integration's interval should be K(1) not Ki(hi).
Yes, you are correct the RHS should be integrated over the master element K(1). This is a mistake in the video but at least it is later correct for the Laplace/stiffness term. Thank you for pointing this out 🤗
Thank you very much! This is a very well made video! Big ups!
I would love some more videos on numerical methods with examples! Thank you very much
Thanks for describing slow. It's easier to understand.
Excellent!! simple, clear. Thank you!
Großen Dank Julian!
That was an amazing lecture thank you
Thank you very much for this perfect tutorial 💐
Such an amazing explanation! Thanks.
Thank you. Would like to know more of such insights in advance topics of FEM
Awesome videos, Julian! 🥰 You inspired me to try Manim as well. Hope to see some followup videos on your channel. Keep it up!
Great Explanation! Thank you!
Nice and clear explanation!
Which type of maths is required to learn to understand this
Hi Sir, 19:37 is it supposed to be F = -h... instead of F = h...?
Yes, the minus sign was missing there in the video.
This is a very great intro to FEM!!
This is an excelente video. I really strong recommend it. I fond of the way that you explained and the notation that you use it, very formal, but easy to understand.
Which maths to learn to study fem
At 19:40 the vector F should be -h \dot (0,1,1,1,1,1,1...,0) right?
Yes, you are correct. I forgot the minus sign in the vector F at 19:40
Very good introduction ! Even if the part on "master element" is a a little fast for me...
Excellent video! Thanks for sharing!
In 14:54, why is this step valid? Why is decomposing Фh into it's basis functions Фi valid with the integral when one of the requirements for this weak form formulation is that the test function you integrate with satisfies the boundary conditions. Фh satisfies them but not every Фi.
Let's say in 9:43, you write that the weak formulation is valid with Ф(x) being in the function space V, and earlier you defined V as a function space where all functions satisfy the B.C.
For example, what if you had a B.C. like f(0) = 100. Then one of the linear basis functions, the one that ramps up from 0, won't satisfy it.
Yes, I agree that your question is valid. Most FEM tutorials focus on homogeneous Dirichlet BCs and then don't really talk about inhomogeneous Dirichlet BCs like f(0) = 100. Your questions has already been answered at the following link: scicomp.stackexchange.com/questions/5072/how-to-properly-apply-non-homogeneous-dirichlet-boundary-conditions-with-fem
The short answer is that for inhomogeneous Dirichlet BCs the solution u_h and the test functioj Φ_h belong to different function spaces. u_h needs to specify the inhomogeneous Dirichlet BCs, e.g. u_h(0) = 100, but the test function Φ_h needs to satisfy the homogenous Dirichlet BCs, e.g. Φ_h(0) = 0.
an amazing video, but in minute 15:28 should not phi (i) be constant per row?
Do you mean that the matrix A should be transposed? If this is what you meant, it does not make any difference here since the bilinear form is symmetric and the linear system is also symmetric. In general, one needs to be cautious though with the assembly of the linear system, since for more complicated problems the order of i and j matters. That's why if you derive the weak form by multiplying from the right with a test function, like in this video, you need to switch i and j in your code in the assembly of the linear system. Thus, many engineers multiply with a test function from the left, i.e. (φ, Δu), to later avoid any confusions in the assembly of the linear system.
@@numericalanalysisbyjulianr4242 Thank you so much!
This is so good.
thank you for the video, i need more videos like this, please
Concise and clear, thank you very much. Do you know of a nice python library i could use to solve pdes with the finite element method ? The only ones I've found seem to be aimed at mechanical engineering simulations only...
It depends on which PDE you want to solve. For the Poisson equation there is plenty of stuff on GitHub. If you want something that is more versatile, I would go for FEniCS/FEniCSx, Ngsolve, Dune, MFEM or deal.II, but some of the latter might not have Python bindings yet. Personally, I have mostly worked with FEniCS in Python and deal.II in C++. If you can implement your PDE in FEniCS on your first try, then FEniCS is my favorite, but sometimes I needed low-level access to the library and found deal.II to be more convenient. Under the hood both libraries use C++ anyway.
@@numericalanalysisbyjulianr4242 perfect, i will look into these. Thank you so much for the awesome video and your quick reply !
@@numericalanalysisbyjulianr4242 i thought I had replied to you but I can't see my comment when checking again : just wanted to thank you one more time for the helpful answer, you rock !
Excelente video!! Muchas Gracias!
Nice tutorial! very clear and thanks
Thank you for this video. Very helpful!
Please do the equivalent formulation for Helmholtz PDE, i would love u if u do it
Hi Sir, phi(0) = phi(1) = u(0) = u(1) = 0, phi and u share the same function space and boundary conditions, is this called Galerkin method to choose the testing function?
Yes, if you solve the weak form of a differential equation with the same trial and test space, i.e. the same function space for u and φ, this is called Galerkin method. Some people also use different function spaces for u and φ, this is then called Petrov-Galerkin method.
To recap the first part: we must find the solution u to the bvp in C². But since it is not possible to do it directly in C² we move to the subspace of test functions V (in which the test functions respect the boundary conditions) and here we reformulate the bvp in the weak form. Right? And what role does the minimization of potential energy play in all this?
Yes, this is correct. For the finite element method we cannot directly work with the strong form in C^2(Ω) \cap C^1(\bar{Ω}) but instead use the weak form and solve it in H^1_0(Ω). (Note: There are also methods that work directly with the strong form, e.g. the Finite Difference method.) So far we have not used the energy minimzation anywhere and could ignore that. Nevertheless, especially in engineering, many people start off with some functional that needs to be minimized and derive the weak form from this energy minimization. Furthermore, sometimes one needs to use the energy minimization to prove theoretical results, e.g. the existence of solutions.
@@numericalanalysisbyjulianr4242 thank you very much your answer and video are awesome!
Great video, thanks!
Thanks alot. Subscribed very fast
Thanks so much;
It was awesome
I am curious to know the software you used to make the animations
Would you please tell me the name?
I used the community version of manim, see e.g. docs.manim.community/en/stable/
@@numericalanalysisbyjulianr4242 I really appreciate your help
So i have an engineering background. Could you explain to me from a mathematicians perspective, what the difference is between using finite difference and finite element method? Both solve PDE with a mesh. Both methods construct an eigenvalue equation, where the matrices are sparse. But with FDM the PDE is approximated with derivatives and a the grid is structured the sparse matrices is populated relatively straight forward. With FEM polynomials here the basis functions are tuned to approximate solutions of each element. Which then together forms the big solution over the hole mesh. Is this somehow correct understanding?
Yes, your summary is correct. With FDM we just approximate the derivatives in the strong form of the PDE and with FEM we work with element-wise polynomial approximations. We don't construct an eigenvalue problem though but a linear equation system. A big difference is that FDM uses the strong formulation of the PDE, e.g. -u" = f, whereas FEM uses the weak/variational formulation of the PDE, e.g. (u', φ') = (f, φ). Like you mentioned, imho FEM is more flexible when it comes to meshes, since FDM requires "structured" grids. With FEM it is easier to deal with more complicated geometries, e.g. plate with hole. This has then also many engineering applications, e.g. Isogemetric Analysis (IGA) uses spline basis functions to work on CAD generated domains directly.
It seems the strong and weak formulation are only equivalent for u in V, right?
You are almost right. Solutions of the strong formulation need to be twice continuously dofferentiable and solutions of the weak formulation need to be once weakly differentiable. So automatically each strong solution is also a weak solution. But a weak solution is only a strong solution if it is also twice continuously differentiable. Otherwise you can't perform integration by parts and recover the Δu term. TL;DR: For u in C^2(Ω) and C^1(\bar{Ω}) the strong and weak form are equivalent.
Great video!!!
Best best best lecture sir.... please, I need your Google scholar link so that I can follow u...my research is in the same area...
Bro I need some help
What about the basis of fourth order partial differential equations?
For future references, please provide concrete examples to make answering your question easier.
Nevertheless, here is a very general answer: The choice of the finite element basis depends on your function space, which in turn depends on your PDE. At the example of the Poisson equation, here we have a second order PDE, but for the weak formulation we apply partial integration and thus have at most only first derivatives. This needs to be reflected by the function space and we choose the space H^1 (or H^1_0 if we have homogeneous Dirichlet BC) which contains all functions whose first (weak) derivative is L^2 integrable. We can thus choose the Lagrange FE basis of piecewise polynomials of up to order r which is a conforming subspace of H^1, i.e. all functions from our FE space are also contained in H^1. If we have a fourth order PDE like the biharmonic problem, we apply integration by parts and arrive at a weak formulation which has up to second order derivatives. By the same argumentation as above, we thus use H^2 as our function space, since the second order derivatives need to be L^2-integrable. Now we can't use the same FE basis anymore as for the Poisson problem, since between elements we only ensure continuity of the FE basis (C^0), but we would need also the continuity of the derivatives of the FE basis (C^1). Therefore you need to use C^1 conforming FE, like the Hermite basis, or split the problem into a mixed system similar to fenicsproject.org/olddocs/dolfin/1.6.0/python/demo/documented/biharmonic/python/documentation.html.
The problem of choosing an appropriate more complicated finite element is not only specific to fourth order PDEs, but also applies to other problems like Maxwell's equations from electromagnetism, where for the simplest problem (Eddy-Current) one solves - curl(curl u) = f and after integration by parts we need a function space where curl u is L^2-integrable (H^curl). There you can't use the standard Langrange basis either, but need Nedelec elements to be H^curl conforming.
@@numericalanalysisbyjulianr4242 thank you so much... I used biharmonic equation. But i cant create a stiffness matrix. Because i don't know the basis when i used discrete finite element method
My example is in hilbert space of fourth order
The problem is biharmonic plus poisson =f
I used green theorem to obtain weak form twice for biharmonic and once for poission
Then i have galerkin equation
B(y, v) =F(v)
Vn is a subspace(??? ) of H2
Yn is a basis of linear combination
What is a basis of finite element method? To creat stiffness matrix
Is it yn=sum Cj Yj, and j =1,..., n??
@@emanalquraishi9279 Yes, in your case your FE space needs to be a subspace of H^2. For this you could use the cubic Hermite basis: hplgit.github.io/INF5620/doc/pub/H14/fem/html/._main_fem007.html.
If you don't want to use C^1 FE, then you could also transform the bihramonic into two second order equations and then use the regular Lagrange basis: fenicsproject.org/olddocs/dolfin/1.6.0/python/demo/documented/biharmonic/python/documentation.html.
Thanks. Great explanation. Gerne noch ein anderes Beispiel zur FEM.
Dear Julian, I always want to make a video with nice text animation like this video. May I ask how you could do that?( what software, library , etc). Thank you.
Dear Tien, thank you very much for the kind words. I am using the community version of manim (docs.manim.community/en/stable/), 3blue1brown's animation library, to create the animations.
@@numericalanalysisbyjulianr4242 awesome !thank you very much. I will have a look at it.
Great explanation! Is this made using manim or similar technology?
Yes, I used the community version of manim (docs.manim.community/en/stable/index.html) for this video.
Thank you for the very informative but brief video. This is really helpful for me to learn what FEM is and prepares me to simulate some models using FEM. However, there is one thing that I wonder about. @17:43 Why do you put (1/h) for the basis function under integral? I am confusing where that (1/h) came from.
You have to look at the picture on the left hand side and at the basis function ϕ_i in the interval [x_i, x_{i+1}]. In this interval the basis function starts at 1 for the left boundary x_i and goes to 0 at the right boundary x_{i+1}. Additionally, the basis function is linear in this interval. Hence, its derivative is constant in this interval. We observe that in this interval the derivative of the basis function is simply (0 - 1) / (x_{i+1} - x_i). Now we said that the distance between two nodes is fixed and we set it to be h := x_{i+1} - x_i. Overall we have thus shown that the derivative of ϕ_i equals to -1/h in the interval [x_i, x_{i+1}].
How would I change the weak form if I have a different set of initial conditions? say u(0)=x and u(1)=y, x=/=y=/=0
How do I deal with inhomogeneous Dirichlet boundary conditons, e.g. u(0)=a and u(1)=b?
Like the homogeneous Dirichlet boundary conditions this needs to be included in the function space. Before we had that u is in H^1_0 to satisfy u(0) = u(1) = 0. Now we need to find u in H^1 such that u(0) = a and u(1) = b. But be careful one still uses homogenous boundary conditions for the test functions, i.e. φ in H^1_0. You don't change the weak formulation but only need to enforce u(0) = a and u(1) = b on the linear system.
TL;DR: For the boundary conditions u(0) = a, u(1) = b, you have the probelm statement:
Find u € {g € H^1((0,1)) | g(0) = a, g(1) = b} such that
(u', φ') = (f, φ)
for all φ € H^1_0((0,1))
So the weak form is defined in C¹?
Almost correct. The weak form is defined in H^1(Ω) resp. H^1_0(Ω) when we include the boundary conditions. As you correctly observed, for the weak form we only need to use first derivatives of the solution. But since these derivatives only need to make sense when we integrate over Ω, e.g. (u', Φ') needs to make sense, it is sufficient if the solution u is once *weakly* differentiable. We denote this then by H^1 for the space of function that are once weakly differentiable. If you want more details on weak derivatives and the space H^1, you can take a look at Sections 7.1.4 and 7.1.5 in the lecture notes from the video description.
Thank you a lot. H¹ is Sobolev space, right? I ask you this because I use another letter in my book
@@nick45be Yes, H^1 is a Sobolev space. Sometimes people also write Sobolev spaces as W^{k,p} with k denoting how often it is differentiable and p denoting that these derivatives need to be in L^p. For many PDEs, we just need p = 2 and hence the short hand notation H^k = W^{k,2} is used.
@@numericalanalysisbyjulianr4242 🤩🤩🤩
Sir can you pl tell me
What are functional spaces here
Pl help me to understand it
I am not from maths background
Pl sir
And why these are to be finite here
In this video we discussed how the Finite Element Method can be used to solve a Partial Dofferential Equation. Here, we used the Poisson problem as an example of such a problem. It is hard (or sometimes even impossible) to write down the exact solution to a differential equation. That's why one tries to find a way how to get a good approximation of the solution by using a computer. For this we try to transform our problem into a linear equation system, because this is what computers are good at solving. In the first step, we don't solve the strong formulation of the Poisson problem -u''=f anymore but choose some other function φ with which we multiply the PDE and then we integrate by parts to get (u',φ') = (f, φ). And now we compute to the function spaces. What kind of properties do u and φ need to have for all this to make sense? Going back to -u''=f we need to be able to take second derivatives of u, so we usually require u to be twice continuously differentiable. When we now look at (u',φ') = (f, φ), we see that u and φ need to only be once differentiable. Furthermore, since these derivatives only show up in the integrals we can use a bigger class of differentiable functions, called weakly differentiable functions. That means u and φ only need to be once weakly differentiable. If you want to learn more about weak differentiability, you can read soke books on functional analysis and PDE theory and look for Sobolev spaces. Now, that we have a rough idea which kind of function space u comes from we want to solve it with a computer, but the basis of the function space is infinite dimensional, so you would need an infinite amount of numbers to represent the solution u. But your computer has only a finite memory... That's why we now take our function space and replace it by another space which has a finite basis, e.g. the space of piecewise linear functions. With this we can finally solve a linear system to get an approximate solution for u. Moreover, the more elements you have in your mesh, the better your piecewise linear approximation and the more accurate your solution u
@@numericalanalysisbyjulianr4242 sir how we choose phi functions and on which basis
How we choose basis function
And in fem what are these function in function spaces
Are these functions related to to the function on each node pl explain
@@mediwise2474 Please watch the video again from 11:10 until 15:52 for an explanation of basis functions
Nice information
One takes the second derivative of a function wrt space and equates it to -1. How on earth gravity comes into this story? Thx for the nice video.
The general version of Poisson's problem is given at 3:04 with -u'' = f. Chossing f = -1 means that we have a uniform force acting in the entire domain, like gravity.
if integral (w,phi)=0, it could also mean that w and phi are orthogoanl right?
Yes, if (w,φ) = 0, then by definition w and φ are orthogonal in the inner product (•,•). In the video I used the L2-inner product, so then w and φ would be orthogonal in L2.
Which maths is required to learn finite element
It would be helpful to have some knowledge of calculus, linear algebra and possibly even PDE theory. Hopefully this video is self-contained enough though that you don't need any prerequisites beyond introductory math courses.
@@numericalanalysisbyjulianr4242 sir can you recommend me any good UA-cam detailed lectures to master fem
@@mediwise2474 I created a few videos on the Finite Element Method, including also this one. This should cover a few important aspects of FEM. If you want more details, please also checkout the video lectures by Wolfgang Bangerth www.math.colostate.edu/~bangerth/videos.html that also partially give an introduction to the FEM software library deal.II. If you prefer a PDF, take a look at the lecture notes of Thomas Wick: doi.org/10.15488/11709
¡Great! Thank you :)))))
this is beautiful:)
What are functional spaces here
!! thanks !!
1.25 x speed is perfect.
I see he do just fine, this kind of topic doesn’t need marathon speeche
Die schwache und die Starke Form sind nicht äquivalent, da der Lösungsraum der starken Form größer ist, trotzdem super Video :)
Jaein. Ich gebe dir Recht, dass nicht jede schwache Lösung auch eine starke Lösung der PDE ist. Unter hinreichenden Annahmen an die Regularität der Lösung, hier u ist 2 mal stetig differenzierbar, sind schwache Lösungen auch starke Lösungen. Ich habe die hinreichende Regularität der Lösung im Video aber wahrscheinlich nicht explizit erwähnt...
The concept of node becomes predominant, a concept of no relevance in the classical methods.
1:56 Laplacian is written with a mistake.
@hakuhakuji I don't see a mistake. Could you please point out what is wrong?
A sign (triangle) is upside-down
@@hakuhakuji No, I still don't see any mistake around 1:56. Everything looks correct to me. There are however different conventions in the FEM literature, how the Laplace operator is written. Some people prefer the notation ∇^2 , whereas I use the convention Δ for the Laplacian. These different notation conventions can also be found in fenicsproject.org/pub/tutorial/html/._ftut1004.html in the text below equation (2.2).
@@numericalanalysisbyjulianr4242 Ah.. I see it now. I would say it's not the best convention since it is greek Delta, which means just a difference. Just puts more confusion. Why not to use the most widespread convention?
@@hakuhakuji The most widespread convention in the mathematical literature is using the greek Delta for the Laplace operator. Hence, I also use this convention.
very bad explanation
you are like reading from a textbook
This is really beautiful. Thank you very much!