ODE system with variable parameter, Boundary conditions and domain

Question

EldaEbrithil 2020-6-17

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/550032-ode-system-with-variable-parameter-boundary-conditions-and-domain

评论： EldaEbrithil 2020-6-18

采纳的回答： Bjorn Gustavsson

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Hi all

is it possible to solve an ODE system with these characteristics? The problem can be summerized in this way, calling:

Domain=D (integration domain it is a certain set of row vectors. Fo example 0:0.1:1; 0:0.1:2; 0:0.1:5 etc.)

Parameter=P (inside one of the ode equations, it is a column vector; its length is equal to Domain number of set: in D1=> P=2, in D2=> P=4 etc.)

Bc=B (column vectors that rapresent boundary conditions) B=B1,B2,...Bn

at the first step i want to solve the ode system for D1 and P1 with B1. At the second step D2, P2 and B1 and so on...after having completed these steps i want to solve for D1,P1 with B2 etc..

Below there is my code

xRange is the Domain

Tt0,Pt0 and M01 are the IC (collected in Y0). Ch is the variable parameter. The unknowns are Pt, Tt and M

L=0.2:0.1:0.4;
Nc=0.00001;
xRange=[0:Nc:L];
P0=2.200000000000000e+05;
T0=300;
M01=0.0001:0.00001:0.5;
for i=1:length(M01)
Tt0(i)=T0*(1+((gamma-1)/2)*M01(i)^2);
Pt0(i)=P0*(1+((gamma-1)/2)*M01(i)^2)^(gamma/(gamma-1));
end
Y0=[Tt0;Pt0;M01];
[xSol,YSol]=ode23(@chambsinglebobb,xRange,Y0);
function dYdx=tubosinglebobb(x,Y)
gamma=1.667;
hradialchamber=3.5;%mm
Hradialchamber=hradialchamber/1000;%m 
Dexttantalum=0.001500000000000;
Aeffettiva=(Dexttantalum*(Hradialchamber))-(pi*(Dexttantalum^2)/4);%m^2
Perimetro=(Dexttantalum+Hradialchamber+2*pi*(Dexttantalum/2));
Tw=803;
fLc=0.006816427132907;
Tt=Y(1);
Pt=Y(2);
M=Y(3);
Ch=[0.0104838041220848,0.00821863791965557,0.00729962973207637];
dTtdx=Ch*(Tt-Tw)*(Perimetro/Aeffettiva);
dPtdx=Pt*((-gamma*((M^2)/2)*fLc*(Perimetro/Aeffettiva))-...
    (gamma*((M^2)/2)*dTtdx*(1/Tt)));
dMdx=M*(((1+((gamma-1)/2)*M^2)/(1-M^2))*((gamma*M^2*fLc*...
    Perimetro)/(2*Aeffettiva))+...
    (0.5*((gamma*M^2)+1)/(1-M^2))*(Ch*(Tt-Tw)*Perimetro*...
    (1+((gamma-1)/2)*M^2))/(Aeffettiva*Tt));
dYdx=[dTtdx;dPtdx;dMdx];
end

Thank you very much for the help

Regards

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EldaEbrithil 2020-6-18

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Hi all

for what concern varition of the initial conditions i have solved the problem with this simple code:

Nc=0.00001;
xRange=[0:Nc:0.2];
P0=2.200000e+05;
T0=300;
M01=(0.005);
Tt0=T0;
Pt0=P0;
for i=1:20
[xSol,YSol]=ode23(@chambsinglebobb,xRange,[Tt0;Pt0;M01]);
M01=M01+0.01;
YSol_Tt(:,i)=YSol(:,1);
YSol_Pt(:,i)=YSol(:,2);
YSol_M(:,i)=YSol(:,3);
end
figure(1);set(gcf,'Visible', 'on')
plot(xSol,YSol_Tt)
xlabel('S')
ylabel('Tt ')
figure(2);set(gcf,'Visible', 'on')
plot(xSol,YSol_Pt)
xlabel('S')
ylabel('Pt ')
figure(3);set(gcf,'Visible', 'on')
plot(xSol,YSol_M)
xlabel('S')
ylabel('M ')

Now i have to include domain and parameter variations, but actually i don't know how to do it.

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Answer 1

Bjorn Gustavsson 2020-6-18

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/550032-ode-system-with-variable-parameter-boundary-conditions-and-domain#answer_453168

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This could be done something like this:

Dall = {0:0.1:1,0:0.1:2}; % however many domains you need etc
Y0 = {[Tt01;Pt01;M01],[Tt02;Pt02;M02],[Tt03;Pt03;M03]};
for iDom = 1:numel(Dall)
    xRange = Dall{iDom};
    for iInitial = 1:numel(Y0)
        [xSol{iDom,Iinitial},YSol{iDom,Iinitial}]=ode23(@chambsinglebobb,xRange,Y0{iInitial});
    end
end

HTH

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EldaEbrithil 2020-6-18

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I have tried in this way bu with no success

Nc=0.00001;
P0=2.200000e+05;
T0=300;
M01=(0.005);Tt01=T0;Pt01=P0;
M02=(0.006);Tt02=T0;Pt02=P0;
M03=(0.007);Tt03=T0;Pt03=P0;
Dall = {0:Nc:L(1),0:Nc:L(2)}; 
Y0 = {[Tt01;Pt01;M01],[Tt02;Pt02;M02],[Tt03;Pt03;M03]};
Ch_cell={ChL(1),ChL(2)};
for iDom = 1:numel(Dall)
    xRange = Dall{iDom};
    Ch=Ch_cell{iDom};
    for iInitial = 1:numel(Y0)
    [xSol{iDom,iInitial},YSol{iDom,iInitial}]=ode23(@chambsinglebobb,xRange,Y0{iInitial},Ch);
    end
end
function dYdx=tubosinglebobb(x,Y,Ch)
gamma=1.667;
hradialchamber=3.5;%mm
Hradialchamber=hradialchamber/1000;%m 
Dexttantalum=0.001500000000000;
Aeffettiva=(Dexttantalum*(Hradialchamber))-(pi*(Dexttantalum^2)/4);%m^2
Perimetro=(Dexttantalum+Hradialchamber+2*pi*(Dexttantalum/2));
Tw=803;
fLc=0.006816427132907;
Tt=Y(1);
Pt=Y(2);
M=Y(3);
dTtdx=Ch*(Tw-Tt)*(Perimetro/Aeffettiva);
dPtdx=Pt*((-gamma*((M^2)/2)*fLc*(Perimetro/Aeffettiva))-...
    (gamma*((M^2)/2)*dTtdx*(1/Tt)));
dMdx=M*(((1+((gamma-1)/2)*M^2)/(1-M^2))*((gamma*(M^2)*fLc*...
    Perimetro)/(2*Aeffettiva))+...
    ((0.5*((gamma*M^2)+1))*(1+((gamma-1)/2)*M^2)/(1-M^2))*...
    (Ch*(Tw-Tt)*Perimetro)/(Aeffettiva*Tt));
dYdx=[dTtdx;dPtdx;dMdx];
end