I get the warning 'Function behaves unexpectedly on array inputs.' when using fplot()
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
LM
am 2 Feb. 2020
Kommentiert: Star Strider
am 2 Feb. 2020
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)
0 Kommentare
Akzeptierte Antwort
Star Strider
am 2 Feb. 2020
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.
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Surface and Mesh Plots finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!