evaluating double integral iteratively

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R yan
R yan 2014 年 7 月 14 日
コメント済み: Star Strider 2014 年 7 月 14 日
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

回答 (1 件)

Star Strider
Star Strider 2014 年 7 月 14 日
If K(x,t) is a function, you can use integral2.
  2 件のコメント
R yan
R yan 2014 年 7 月 14 日
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 2014 年 7 月 14 日
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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