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I am studying some theories around FEM method in 2D, and I am trying to solve this problem from Ciarlet's book (the proof was not provided): Consider a simplex $T$ in $R^d$ with $N_1(T) = \left\{N_i\right\}_{i=0}^{d}\subset P_1^{*}(T)$ be the Lagrange nodal variables (or nodal evaluation). By the Riesz representation theorem, there exist functions $\lambda_{i}^{*}\in P_1(T)$ (which is the representation of $N_i\in P_1(T)$) such that $N_j(\lambda_{i}) = \int_{T} \lambda_{i}\lambda_{j}^{*} = \delta_{ij}$ for each $0\leq i\leq d$. Show that $\lambda_{i}^{*} = \frac{(1+d)^2}{|T|}\lambda_{i} - \frac{1+d}{|T|}\sum_{j\neq i} \lambda_j$

My attempt: I was thinking that $\lambda_i^{*}$ plays a role like a Fourier coefficient, so I multiply both sides by $\lambda_j$ and integrate over $T$. Then LHS would be $\delta_{ji}$, while RHS $= \frac{(1+d)^2}{|T|}\int_{T} \lambda_{i}\lambda_{j} - \frac{1+d}{|T|}\sum_{j\neq i}\int_{T} \lambda_{i}\lambda_{j}$. Now, for $i=j$, then $LHS=1$, and I was thinking of $\int_{T} \lambda_{i}\lambda_{i} = $ the area of region $T$, but then it would result in $(1+d)^2,$ which is certainly not equal to $1$ (contradiction). I got stuck here completely....

My question: Could anyone please help me with this problem? It seems to be such a beautiful identity, yet quite hard to prove. Any thoughts would really be appreciated.

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  • $\begingroup$ Nobody wants to help me with this difficult problem? $\endgroup$
    – user177196
    Commented Dec 9, 2016 at 5:39
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    $\begingroup$ You can form a system for combination of integrals. E.g., the first system is for integrals of the products of lambda_1 and others lambda(_1,2,3) with righthand side (1,0,0) in case d=2. Then you can solve the system and find what are the values of integral lambda_i* lambda_j in terms of area of T. Then you need only to check that this is true by direct computations maybe. At least you can give it a try I guess. $\endgroup$
    – VorKir
    Commented Dec 10, 2016 at 0:06
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    $\begingroup$ Or, you can google it as me and find, e.g., the computed integrals ljll.math.upmc.fr/~ledret/M1English/M1ApproxPDE_Chapter5-2.pdf, page 158, $\endgroup$
    – VorKir
    Commented Dec 10, 2016 at 0:09
  • $\begingroup$ @VorKir: thank you so much for your help. But why for $d=2$, you got $RHS = (1,0,0)$? So for the general case with arbitrary $d$, is the area of a $d$-dimensional reference triangle $T$ constructed in a similar fashion as $T^{hat}_k$ on page 158 equal to $\frac{J}{2}$ still? I am asking if I could compute $\int_{T_k} \lambda_{1}(x) dx = \int_{T^{hat}_k} \lambda^{hat}_{1}(x^{hat}) J dx = 2area|T|\int_{0}^{1} \lambda^{hat}_{1} \int_{0}^{1}....\int_{0}^{1-x^{hat}_{1}-x^{hat}_2-...-x^{hat}_{d-1}} dx^{hat}_2dx^{hat}_3...dx^{hat}_{d-1}$? $\endgroup$
    – user177196
    Commented Dec 10, 2016 at 6:11
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    $\begingroup$ Actually, after change of variables you get the Jacobian $J$, which in 2D is equal to $2 area|T|$ for the reference triangle as described in the reference. In d-case, for an affine transform you will get area $T = J * $ volume of reference d-simplex. $\endgroup$
    – VorKir
    Commented Dec 10, 2016 at 23:57

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(failed to edit the comment-answer in 5 minutes fully)

In fact, no, I am sure there is a nice way to compute all this but now can think only of a brute force one. $\int_{T^{hat}} \lambda_1^2 = \int_{T^{hat}} x_1^2 dx = \int_0^1 x_1^2 V_{d-1}(1-x_1) dx_1$ where $V_k(r)$ is the volume of $k$-dimensional canonical simplex with side lentghs equal to $r$. So, I guess $V_k(r) = r^k V_k(1) = r^k / k!$. So I got something like $\int_{T^{hat}} \lambda_1^2 dx = 1/(d-1)! \cdot 2 / (d(d+1)(d+2)) = \frac{2}{ (d+2)!}$. Which differs from your value. The integral for a product $\lambda_i \lambda_j$ is more tedious, since leads to something like 2d-integral $\int_0^1 x_1 \int_0^1 x_2 (1-x_1-x_2)^{d-2} dx_1 dx_2 \cdot \frac{1}{(d-2)!}$ which is doable. Hope someone would come up with a nice and easy answer to overall question.

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    $\begingroup$ check,e.g. the related question which has the answer math.stackexchange.com/questions/1074675/… $\endgroup$
    – VorKir
    Commented Dec 12, 2016 at 5:03
  • $\begingroup$ thank you so much for your help. I got exactly the same result as you got for both integrals yesterday:) $\endgroup$
    – user177196
    Commented Dec 13, 2016 at 0:09
  • $\begingroup$ You are welcome, I am glad to be able to help $\endgroup$
    – VorKir
    Commented Dec 13, 2016 at 7:09

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