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evaluating double integral iteratively

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R yan
R yan el 14 de Jul. de 2014
Comentada: Star Strider el 14 de Jul. de 2014
I am trying to discretize a kernel K(x,t) for solving an integral equation. I need to implement
for i=1...N
for j=1..N
a(i,j) = Int1 Int2 K(x,t)dx dt
where Int1 has limits i/N to (i+1)/N and Int2 has limits j/N to (j+1)/N
I tried the symbolic computation but it takes a lot of time. I want to use the numerical computation and speed it up. thanks

Respuestas (1)

Star Strider
Star Strider el 14 de Jul. de 2014
If K(x,t) is a function, you can use integral2.
  2 comentarios
R yan
R yan el 14 de Jul. de 2014
thanks. Does integral2 implement the iterative version?? I mean in Int1 Int2 K(x,t)dxdt, does it evaluate the integral Int2 K(x,t)dx treating t as constant?? and then evaluate Int1 over t.
Star Strider
Star Strider el 14 de Jul. de 2014
My pleasure!
That’s my understanding of how integral2 works. See the More About in the documentation, and its friends (linked to at the end of the page). Note that only integral can take array-valued functions, so if K(x,t) in one such, you’ll have to nest calls to integral. If it’s scalar-valued, the integral functions will be happy and will return a scalar.

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