Simplify the following expression by matrix calculus and orthonormal properties of eigenfunctions

Question

How can we simplify the following expression? $$\int_0^L\big[(\sum_{k=1}^{K} A_k(x)\textbf{B}_k^T(t))\textbf{C}(\sum_{k=1}^{K} A_k(x)\textbf{B}_k(t))\big]dx$$ In here, corresponding normalized eigenfunction is $$ \{A_k(x)\}_{k=1}^\infty=\Bigg\{\sqrt{\dfrac{2}{L}}\sin(a_kx)\Bigg\}_{k=1}^\infty$$ where $a_k=k\pi/L$

and

$\textbf{B}_k(t), \textbf{C}$ are matrices

and

$\textbf{B}_k^T(t)$ stands for transpose of the matrix.

My effort:

$\int_0^LA_k(x)A_k(x)dx=\sqrt{\dfrac{2}{L}}\sqrt{\dfrac{2}{L}}\int_0^L\sin(k\pi x/L)\sin(k\pi x/L)dx=\dfrac{2}{L}\dfrac{L}{2}=1$

How can we use it?

Answer 1

$$\int_0^L\big[(\sum_{k=1}^{K} A_k(x)\textbf{B}_k^T(t))\textbf{C}(\sum_{k=1}^{K} A_k(x)\textbf{B}_k(t))\big]dx$$

Lets expand the summations, Then each term will be like this $$\int_0^L\big[(A_1(x)\textbf B^T_1+ ... +A_k(x)\textbf B^T_k)\textbf{C}((A_1(x)\textbf B_1+ ... +A_k(x)\textbf B_k)\big]dx$$

$$\sum_{k=1}^K\sum_{j=1}^K\int_0^L A_k A_j\textbf B^T\mathbf C\mathbf B dx$$

Notice that this only changes the basis for $\mathbf C$ which will then stay constant so for $k \ne j$ we get the expression to be zero.

$$\int_0^L \sin (k\pi x/L) \sin(j\pi x/L) = 0$$ So therefore you only evaluate the integral for $k = j$ in which there are $K$ cases.

Therefore you end up with $$\sum_{i=1}^{K} \mathbf B_i^T\mathbf C \mathbf B_i$$

Answer 2

Change the dummy index of one of the sums and you have this.

$$\int_0^L\big[(\sum_{k=1}^{K} A_k(x)\textbf{B}_k^T(t))\textbf{C}(\sum_{k=1}^{K} A_k(x)\textbf{B}_k(t))\big]dx = \sum_{k=1}^{K} \sum_{l=1}^{K} \textbf{B}_k^T(t))\mathbf{C}\textbf{B}_l^T(t)) \int_0^L A_k(x) A_l(x)\,dx. $$

Answer 3

Pulling the sigmas and matrices out of the integral leaves only those orthonormal scalar functions, which are immediately replaced by a Kronecker delta.

In symbols $$\eqalign{ J(t) &= \sum_i\sum_j B_i^TCB_j \int_0^L A_iA_j dx \cr &= \sum_i\sum_j B_i^TCB_j \delta_{ij} \cr &= \sum_i B_i^TCB_i \cr }$$