how to integrate a product with function handle

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Abhik Saha
Abhik Saha 2021 年 5 月 7 日
編集済み: Abhik Saha 2021 年 5 月 8 日
I want to perform a numerical integration with function handle. Please find details below
Nz = 256;
L = 1;
dz = L/Nz;
zgrid = dz*(0:Nz-1)'-L/2;
gN = 20000;
U0 = 5*gN;
U1=1.9*gN;
sigma = 5*dz;
xb = L/4;
tau = 0.01;
h=0.2*gN;
n0=mean(dens(abs(zgrid)<L/10,:));
wp=2*pi*sqrt(gN*n0);
Ub = @(t) U1*(t<tau)+(t>=tau)*(U1+h*sin(wp*(t-tau)));
Vext = @(t) (U0*exp(-(zgrid+xb).^2/sigma^2)+Ub(t)*exp(-(zgrid-xb).^2/sigma^2));
Now I want to integrate
integrate (Vext *A, zgrid)
where A is a double array matrix of dimension (zgrid,t)
The problem is Vext is function handle and A is numerical values of the dimension (zgrid,t). please help how to do this integration?

回答 (1 件)

Steven Lord
Steven Lord 2021 年 5 月 7 日
You can't calculate the product of a number and a function handle. What you can do instead is calculate the product of a number and the result of evaluating that function handle.
fh = @(x) x.^2;
This works and returns a number. It evaluates the function handle with x = 5 then multiplies that result by 2.
y1 = 2*fh(5)
y1 = 50
This works and returns another function handle that you can evaluate. Note that inside the function handle fh2 I evaluate fh at the value that was passed into fh2.
fh2 = @(x) 2*fh(x)
fh2 = function_handle with value:
@(x)2*fh(x)
y2 = fh2(7) % Essentially the same as y1 (except with x = 7 instead of x = 5.)
y2 = 98
This does not work which is why I put it last.
y3 = 2*fh
Operator '*' is not supported for operands of type 'function_handle'.
  1 件のコメント
Abhik Saha
Abhik Saha 2021 年 5 月 8 日
編集済み: Abhik Saha 2021 年 5 月 8 日
I understand your point of view. Then one of the possible gateway is to generate the dataset of Vext with dimension zgrid and t. I mean to say Vext(zgrid,t). I am new to matlab so I don't know how to do this can you please tell me how to generate this dataset where lets say zgrid is already defined and t=linspace(0,10,1000).

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