How do I input a lot of values and record each single output in Bvp4c?

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Hussein Al-Tamreh
编辑： Walter Roberson 2024-10-22
Hello everyone,
In my code below, how can I make the code input the values of "Pe" from { 25 to 1025 } and record the generated output values for EACH single input number.
For example; when I input a value of "Pe" of 25, the generated value of "MainpADM" is { 0.0054 }, and another input value of "Pe" of { 26 } will generate { 0.0065 }. and of course I am recording each corresponding output
This old-fashioned way will take A LOT of time to reach the final 1025 input. Therefore, Could you kindly help me in this regard?
The three codes of bvp4c:
MainADM
global Pe Taw Yx B A;
%
Pe = 61;
Taw=4.5; % dimensionless residemnce time
Yx=0.212;
B=0.482;
A=1.01; % A=1+(Xo/So*Yx)
xlow = 0;
xhigh = 1;
y=[];
solinit = bvpinit(linspace(xlow,xhigh,20),[0 1]);
sol = bvp4c(@bvpode5,@bvpbc5,solinit);
plot(sol.x,sol.y(1,:),'r-')
xlabel('Rxtr Length Z/L')
ylabel('Concentration Fraction (S/So)')
axis ([0 1 0 1]);
grid on
%
legend('alpha')
fh=figure(1);
set(fh,'color','white')
title('Concentration Profiles Across the Rxtr')
disp (sol.y(1,end));
bvpode5
function dydx = bvpode5(x,y)
%
global Pe Taw Yx B A;
%
dydx = [y(2)
Pe*y(2)+((Pe*Taw*y(1)*(A-y(1)))/(B*Yx*(A-y(1))+y(1)))];
end
bvpbc5
function res = bvpbc5(ya,yb)
global Pe Taw Yx B;
%
res = [ ya(1)-1-((1/Pe)*ya(2))
yb(2)];
end

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回答（2 个）

Tejas
Tejas 2024-10-22
Hi Hussein,
To record the values of 'MainpADM' for different values of 'Pe', follow these steps:
  • Preallocate the variable 'MainpADM_results' to store the results of 'MainpADM' for each 'Pe' value.
% MainADM
global Pe Taw Yx B A;
Taw = 4.5;
Yx = 0.212;
B = 0.482;
A = 1.01;
% Range of Pe values
Pe_values = 25:1025;
% Initialize storage for results
MainpADM_results = zeros(size(Pe_values));
% Loop over all Pe values
for i = 1:length(Pe_values)
Pe = Pe_values(i);
xlow = 0;
xhigh = 1;
solinit = bvpinit(linspace(xlow, xhigh, 20), [0 1]);
sol = bvp4c(@bvpode5, @bvpbc5, solinit);
% Record the result for the current Pe
MainpADM_results(i) = sol.y(1, end);
end
Walter Roberson
Walter Roberson 2024-10-22
编辑：Walter Roberson 2024-10-22
This old-fashioned way will take A LOT of time to reach the final 1025 input.
Looping is the main practical way to solve this problem. It can take a long time, and that is just how it is.
The less practical way to solve the problem is to transform it into a system of 2 * (1025 - 25 + 1) = 2 * 1001 variables, and vectorize your bvpode5 and bvpbc5. For example,
function dydx = bvpode5(x,y)
%
global Pe Taw Yx B A;
%
dydx = [y(2:2:end),
Pe.*y(2:2:end) + ((Pe.*Taw.*y(1:2:end).*(A-y(1:2:end)))./(B.*Yx.*(A-y(1:2:end))+y(1:2:end)))];
dydx = dydx(:);
end
and then make a single call to bvp4c with Pe set to 25:1025. This will return all the solutions simultaneously

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