elliptic PDE with variable coefficient
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Hello,
I am desperately trying to solve the following equation (in 2D) :
-(laplacian)u + u =1
This is basically an elliptic PDE with
c=1 , f=1 and a= B(x,y).
B(x,y) is a function I have created that doesn't have any closed form. Here's the code for B(x,y):
I entered c=1 , f=1 and a= B(x,y) in the pde specification but I cannot solve the equation as I get the following error
Expression evaluates to wrong size. Must be scalar or row vector. In a system case, pass first or second row; for example u(2,:)
Can somebody help me fix this ?
Thank you
2 件のコメント
Geoff Hayes
2014 年 11 月 19 日
Ann - please describe what you mean by I entered c=1 , f=1 and a= V(x,y). in the pde specification but I cannot solve the equation as I get the following error. What line of code are you calling that generates the error? What is the pde specification?
採用された回答
Bill Greene
2014 年 11 月 19 日
The input arguments, x and y are equal length row vectors of x and y coordinates where the a coefficient must be defined. If this length (number of columns) is n, the output argument, v, must be a matrix with dimensions 1 x n (i.e. a row vector). The line
v = M(intx,inty);
is returning an n x n matrix; that is the reason for the error message.
3 件のコメント
Bill Greene
2014 年 11 月 19 日
Yes, x and y are spatial locations in the mesh where the a-coefficient must be defined. The returned value must be a row vector with values at just those points. You are returning a matrix with n x n points where what is required is a row vector with values at the n locations. The simplest way to understand and implement this (though not particularly efficient) is with this snippet of code:
v = zeros(1, length(x))
for i=1:length(x)
v(i) = M(intx(i), inty(i));
end
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