Recent content by kuan

Yes. This is an open question. The integration of the polynomial f(x^2) can be computed accurately up to machine precision using the quadrature rule. For example, using the Gauss-Legendre rule, we need approximately N sampling points. My question is that if we can reduce the number of the...

Let f(x^2) be a polynomial. I would like to carry out the integration \int_a^b f(x^2) dx using quadrature rule. Suppose a and b > 0 and are arbitrary and the degree of f(x^2) is 2N. I would like to know if there is a possibility to reduce the sampling points down to N/2?

hmmm, i might see your question now. Can you set up 2 pseudo-grids at the boundary? One for the diffusion in domain 1 and the other one for domain 2.

Hmmm, I am not sure I understand you properly? Do you ask if these 2 systems are ought be solved at the same time or not? 2 systems are coupled, thus generally they need to be solved at the same time.

Yes, JJacquelin. You are right, I totally agree with you!

JJacquelin, I see what you mean! For me, I simply ignored these 2 points at (0 , L) and (L , L). No matter what conditions they are, the Fourier series is going to diverge pointwisely at at (0 , L) and (L , L).

I am confused! It seem for me this problem can be split in 2 diffusion problems with the coupling at the boundary. For the stability, the simplest one is dt / dx^2 <=1. I cannot remember what it is called. You find it in the Numerical Recipes.

I think the sinuous functions in x direction are the basis you choose. However, for the y direction, it must be both sinuous and co-sinuous functions. When you get the matrix representation in y direction, I think we need to impost the boundary condition on both sides. The simple way would be...

For example we define a 3D array, like f(9, 9, 9). I think transpose f with the second index fixed would be: transpose( f ( :, 1, : ) ). Does this solve your problem?

Dear All, I am right now facing a problem of the adjoint operator. From the mathematical point of view, if T is a linear operator mapping from Hilbert space H to another Hilbert space H' , there exist an unique adjoint operator T^{\dagger} mapping from H' back to H . However I am...