Function may not have a root

Question

Matthew Palermo 2023-11-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2049457-function-may-not-have-a-root

评论： Walter Roberson 2023-11-20

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

I am getting this eror with this code and associated functions:

Exiting fzero: aborting search for an interval containing a sign change

because no sign change is detected during search.

Function may not have a root.

Here is the code and functions:

close all

clear all

clc

% %FOR A = 3

%

x_til = linspace(0,1,10);

dt = 0.1;

tmax = 1;

t_til = linspace(dt,tmax,10);

nt = length(t_til);

nx = length(x_til);

x0 = [0.0000001 1];

tp1 = ones(nt);

tp2 = zeros(nt);

for A = 3 %finding tp actual

for jx = 1:nx

Tfun1 = @(tp) fX10B1T0(x_til(jx),tp);

Tfun1A = @(tp) (1*(10^(-A))) - Tfun1(tp);

tp1(jx) = fzero(Tfun1A,x0(1));

end

for jx = 1:nx

Tfun2 = @(tp) fX11B10T0(x_til(jx),tp,A);

Tfun2A = @(tp) (1*(10^(-A))) - Tfun2(tp);

tp2(jx) = fzero(Tfun2A,x0(1));

end

Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root. Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.

Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.

for ix = 1:nx

for it = 1:nt

TX10B1T0 = fX10B1T0(x_til,t_til);

end

plot (x_til, TX10B1T0(ix,:));

end

function [T_til1] = fX10B1T0 (x_til, t_til)

lengthx=length(x_til);

lengtht=length(t_til);

T_til1=zeros(lengthx,lengtht); % Preallocating Arrays for speed

xd = zeros(lengthx);

td = zeros(lengtht);

for ix=1:lengthx % Begin time loop

xd_ix=xd(ix); % Set current time

for it=1:lengtht % Begin space loop

td_it=td(it); % Set current space

if td_it == 0 % For time t=0 condition

T_til1(it,ix)=0; % Set inital temperature

else

% Solution at any time

T_til1(ix,it)=erfc(xd_ix/sqrt(4*td_it));

end % if td_it

end % for ix

end % for it

end % function

function [T_til2] = fX11B10T0 (t_til, x_til,A)

lengthx=length(x_til);

lengtht=length(t_til);

T_til2=zeros(lengtht,lengthx); % Preallocating Arrays for speed

for it=1:lengtht % Begin time loop

td_it=t_til(it); % Set current time

for ix=1:lengthx % Begin space loop

xd_ix=x_til(ix); % Set current space

td_dev1=(1/(10*A))*(2-xd_ix)^2; % First deviation time

td_dev2=(1/(10*A))*(2+xd_ix)^2; % Second deviation time

if td_it == 0 % For time t=0 condition

T_til2(it,ix)=0; % Set inital temperature

elseif td_it <= td_dev1 % Solution for first small times:

T_til2(it,ix)=erfc(xd_ix/sqrt(4*td_it));

elseif td_it > td_dev1 && td_it <= td_dev2 % Solution for second small times:

T_til2(it,ix)=erfc(xd_ix/sqrt(4*td_it))-erfc((2-xd_ix)/sqrt(4*td_it));

else

m_max=ceil(sqrt(A*log(10)/(td_it*pi^2))); % Set max. No. terms

% Start X11B10T0 case for large times:

T_til2(it,ix)=1-xd_ix; % steady-state temperature solution

for m=1:m_max % Continue X11B10T0 case for large times

% Series solutions:

betam=m*pi; % Define eigenvalues

T_til2(it,ix)=T_til2(it,ix)-2*exp(-betam^2*td_it)*sin(betam*xd_ix)/betam;

end % for m

end % if td_it

if xd_ix == 1 % For location x=1 condition

T_til2(it,ix)=0; % Set x=1 boundary temperature

end % if xd_ix

end % for ix

end % for it

end % function

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Answer 1

Walter Roberson 2023-11-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2049457-function-may-not-have-a-root#answer_1356162

在 MATLAB Online 中打开

td = zeros(lengtht);

You initialize td to all zeros

for ix=1:lengthx % Begin time loop
    xd_ix=xd(ix); % Set current time
    for it=1:lengtht % Begin space loop
        td_it=td(it); % Set current space

You extract a value from td. But td is all zero. And you never change td anywhere in the function after you initialize it to zero.

        if td_it == 0 % For time t=0 condition
            T_til1(it,ix)=0; % Set inital temperature

But it does indeed equal 0 because td remains 0. So you are always setting all of T_til1 to 0.

Your fX10B1T0 function returns constant values. There is no possibility of finding a sign change in a function that is constant.

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Matthew Palermo 2023-11-20

I pulled the two functions from the textbook and copied them as they are presented. My understanding of how fX10B1T0 is supposed to run is that

if td_it == 0 % For time t=0 condition

T_til1(it,ix)=0; % Set inital temperature

is supposed to be set to 0 since I am working with non-dimensional figures here.

Could the issue be in the script that I wrote?

Walter Roberson 2023-11-20

在 MATLAB Online 中打开

In your code, your time is always zero. You set td = zeros(lengtht) and you never change it. So the td_it == 0 is always true, and none of the elseif/else are executed.

You do get

        if xd_ix == 1 % For location x=1 condition
            T_til2(it,ix)=0; % Set x=1 boundary temperature
        end % if xd_ix

executed in some meaningful way... but since at time 0 T_til2 entries are set to 0, setting one or more entries to 0 at this point makes no difference to the outcome.

Therefore, no matter what your inputs are, your T_til2 array is going to be all 0. And a function that is all zero does not have a zero crossing.

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