I get the warning 'Function behaves unexpectedly on array inputs.' when using fplot()

2 次查看(过去 30 天)
Why do I get a warning when using fplot() in my code:
function solution_heat_equation(t)
f = @(x,y) (exp((-abs(x-y).^2)./(4.*t))).*u_0(y);
a = -10;
b = 10;
z = @(x) integral(@(y) f(x,y) , a,b);
u = @(x) (1/sqrt(4*pi*t))*z(x)
fplot(u)
function val = u_0(x)
if abs(x)<1
val = 1;
else
val = 0;
end
I get the following warning and I don't get a plot of my function:
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly
vectorize your function to return an output with the same size and shape as the input
arguments.
> In matlab.graphics.function.FunctionLine>getFunction
In matlab.graphics.function/FunctionLine/updateFunction
In matlab.graphics.function.FunctionLine.set.Function_I
In matlab.graphics.function.FunctionLine.set.Function
In matlab.graphics.function.FunctionLine
In fplot>singleFplot (line 245)
In fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 200)
In fplot>vectorizeFplot (line 200)
In fplot (line 166)
In solution_heat_equation (line 8)

采纳的回答

Star Strider
Star Strider 2020-2-2
With a scalar ‘t’, there are no problems provided that ‘z’ sets 'ArrayValued' to true in the integral call, and ‘u’ is vectorised:
z = @(x) integral(@(y) f(x,y) , a,b, 'ArrayValued',1);
u = @(x) (1./sqrt(4*pi*t)).*z(x);
The rest of your code is unchanged.

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Argument Definitions 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by