# Problem about mixture of ode-pde

12 visualizaciones (últimos 30 días)
ZIYI LIU el 23 de Jul. de 2020
Comentada: Paul Maurerlehner el 24 de En. de 2022
Hello,
I have a system that I want to solve numerically (attached is the pde-ode file). I know a little about how to solve ode and pde separately, but I don't know how to combine ode and pde parts in MATLAB.
Here is my code: (I am sorry for my rough code)
T=120;
c10 = 1.3;
c20 = 0.03;
g10 = 0.07;
g20 = 1.37;
[t, state_variable]=ode45(@LV,[0 T],[c10,c20,g10,g20]);
c1 = state_variable(:,1);
c2 = state_variable(:,2);
g1 = state_variable(:,3);
g2 = state_variable(:,4);
%[c1,t]
x = linspace(0,L,20);
%t = [linspace(0,0.05,20), linspace(0.5,5,10)];
m = 0;
sol = pdepe(m,@heatpde,@heatic,@heatbc,x,t);
hold on
plot(t,c1,'LineWidth',2)
plot(t,c2,'LineWidth',2)
% Various commands to make graph clearer
h=legend('c1','c2','g1','g2');
xlabel('Time','Fontsize',16)
ylabel('concentration','Fontsize',16)
set(gca,'XTick')
set(h,'Fontsize',20)
function f=LV(t,state_variable)
c1=state_variable(1);
c2=state_variable(2);
g1=state_variable(3);
g2=state_variable(4);
%here i deleted many parameters
% Equations
dc1 =(k0+kcat*c1^n)*g1*C(0,t)-kbar*c1;
dc2 =(k0+kcat*c2^n)*g2*C(L,t)-kbar*c2;
dg1=kon/(1+K*c1^m)*G(0,t)-koff*g1;
dg2=kon/(1+K*c2^m)*G(L,t)-koff*g2;
%f=[dx;dy;];
f=[dc1;dc2;dg1;dg2;];
end
function [c,f,s] = heatpde(x,t,C,dCdx)
c = 1/Dc;
f = dCdx;
s = 0;
end
function C0 = heatic(x)
C0 = 0.17;
end
function [pl,ql,pr,qr] = heatbc(xl,Cl,xr,Cr,t)
pl = -Dc*Cl+dc1;
ql = 0;
pr = -Dc*Cr-dc2;
qr = 0;
end
Thank you very much!
Best,
Ziyi
##### 4 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
Bill Greene el 24 de Jul. de 2020
Can you please provide values for all the parameters in the model?
ZIYI LIU el 24 de Jul. de 2020
Editada: ZIYI LIU el 25 de Jul. de 2020
Below are my values in the code I wrote by using MOL just now。 But I cannot let C,C,g,G change with time and position, my result is just the initial result. Maybe I didn't make correct connections between every part. Thanks!
%main program
clear all
clc
close all
% Data
L=1;
Nz=100; %n=Nz+1
Dc=3;
Dg=3;
dz=L/Nz;
Ctot=1.58;
Gtot=1.5;
c10=1.3;
c20=0.03;
g10=0.07;
g20=1.37;
k0 =0.1;
kcat= 40;
kbar = 1;
m =2;
n=2;
kon=1;
K=8;
koff=0.9;
%g1=1;%
%g2=0.1;%
t = [0:0.01:50];
z= [0:0.01:L];%?I am not sure about linspace of t and (0,L,Nz)?
%initial condition at t=0
IC=zeros(Nz+1,1);%or zeros(Nz+1?1)?zeros(1,Nz)? and should if be (Nz+1,1)like the vedio
%IC(1:Nz+1)=(Ctot-c10-c20)/Nz;% i assume.or(1:Nz+1)
IC(1:Nz+1)=0.25;% i assume.or(1:Nz+1)
IG=zeros(Nz+1,1);
%IG(1:Nz+1)=(Gtot-g10-g20)/Nz;
IG(1:Nz+1)=0.06;% i assume.or(1:Nz+1)
Ic=zeros(Nz+1,1);%or(Nz+1,1).
Ic(1)=c10;
Ic(Nz+1)=c20;
Ig=zeros(Nz+1,1);%or(Nz+1,1).
Ig(1)=g10;
Ig(Nz+1)=g20;
Iy=[IC;Ic;IG;Ig];
%Solver
%[t, state_variable]=ode45(@LV,tspan,[c10,c20]);
%c1 = state_variable(:,1);
%c2 = state_variable(:,2);
[T ,Y]=ode15s(@(t,y) fun(t,y,z,Nz,Dc,Dg,dz,c10,c20,g10,g20,L,k0,kcat,kbar,kon,K,koff,n,m),t,Iy);%IC,[],Nz,Dc,dc1dt,dc2dt,c1,c2);
%plot 2D
hold on
plot(T, Y)%c1
hold on
%plot(T, Y(:,2*Nz+2))%c2
%plot(T, Y)%why only two lines?
ylabel('c1');
xlabel('time span');
%bar image 3D
%imagesc(z,t, Y(1:Nz+1))
%imagesc(z,t, Y(Nz+2:2*Nz+2))% z-xaxis,t-yaxis, and i only want c
%imagesc(t, Y(Nz+2:2*Nz+2)), i cannot get result(error)
%grid on
%xlabel('z position');
%ylabel('time span');
%colormap jet
%colorbar
%dispersion
function DyDt=fun(t,y,z,Nz,Dc,Dg,dz,c10,c20,g10,g20,L,k0,kcat,kbar,kon,K,koff,n,m)
C=zeros(Nz+1,1);
c=zeros(Nz+1,1);
G=zeros(Nz+1,1);
g=zeros(Nz+1,1);
%C=zeros(1,Nz+1);
%c=zeros(1,Nz+1);
DCDt=zeros(Nz+1,1);
DcDt=zeros(Nz+1,1);
DGDt=zeros(Nz+1,1);
DgDt=zeros(Nz+1,1);
DyDt=zeros(4*(Nz+1),1);
%zhalf=zeros(Nz,1);% WHY NEED THIS?
%D2CDz2=zeros(Nz-1,1);
C=y(1:Nz+1);
c=y(Nz+2:2*(Nz+1));
G=y(2*(Nz+1)+1:3*(Nz+1));
g=y(3*(Nz+1)+1:4*(Nz+1));
%zhalf(1:Nz)=(z(1:Nz)+z(2:Nz+1))/2;
%boundary conditions
C(1)=1/3*(4*C(2)-C(3)-2*dz/Dc*DcDt(1));
C(Nz+1)=1/3*(-2*dz/Dc*DcDt(Nz+1)+4*C(Nz)-C(Nz-1));
G(1)=1/3*(4*G(2)-G(3)-2*dz/Dg*DgDt(1));
G(Nz+1)=1/3*(-2*dz/Dg*DgDt(Nz+1)+4*G(Nz)-G(Nz-1));
for i=2:Nz
D2CDz2(i)=1/(dz^2)*(C(i+1)-2*C(i)+C(i-1));
D2GDz2(i)=1/(dz^2)*(G(i+1)-2*G(i)+G(i-1));
DcDt(i)=0;%as there is no c between 2:Nz, c is only existing in two poles
DgDt(i)=0;
end
D2CDz2(Nz+1)=1/(dz^2)*(-C(Nz-1)+2*C(Nz)-C(Nz+1));
D2CDz2(1)=1/(dz^2)*(C(3)-2*C(2)+C(1));
D2GDz2(Nz+1)=1/(dz^2)*(-G(Nz-1)+2*G(Nz)-G(Nz+1));
D2GDz2(1)=1/(dz^2)*(G(3)-2*G(2)+G(1));
%for time, (1:Nz+1)
%the rate of translation between C and c(G and g) at two poles.
DcDt(1) =(k0+kcat*c(1)^n)*g(1)*C(1)-kbar*c(1);
DcDt(Nz+1) =(k0+kcat*c(Nz+1)^n)*g(Nz+1)*C(Nz+1)-kbar*c(Nz+1);
DgDt(1)=kon/(1+K*c(1)^m)*G(1)-koff*g(1);
DgDt(Nz+1)=kon/(1+K*c(Nz+1)^m)*G(Nz+1)-koff*g(Nz+1);
for i=1:Nz+1
DCDt(i)=Dc*D2CDz2(i);
DGDt(i)=Dg*D2GDz2(i);
end
DyDt=[DCDt;DcDt;DGDt;DgDt];
%plot(t,c(1))
end
%problem is:
%I got from the workspace that my result is almost the initial conditions.
%C G c g aren't change.

Iniciar sesión para comentar.

Bill Greene el 29 de Jul. de 2020
Editada: Bill Greene el 8 de Oct. de 2020
I have developed a PDE solver, pde1dM, that Ibelieve can solve your coupled PDE/ODE system.
The solver runs in MATLAB and is similar to the standard pdepe solver but it allows a set of ODE to
be coupled to the set of PDE.
I have used your pdf document and example code to solve a problem which I think
is close to what you want to solve. I have attached this MATLAB script below.
Unfortunately, I don't understand your problem well enough to know if I have
If you want to try or modify this example yourself, you can easily download pde1dM at this location and
run the example code shown below:
% Data
L=1;
Nz=100; %n=Nz+1
Dc=3;
Dg=3;
dz=L/Nz;
Ctot=1.58;
Gtot=1.5;
c10=1.3;
c20=0.03;
g10=0.07;
g20=1.37;
k0 =0.1;
kcat= 40;
kbar = 1;
m =2;
n=2;
kon=1;
K=8;
koff=0.9;
%g1=1;%
%g2=0.1;%
tFinal=15;
t=linspace(0,tFinal,30);
z=linspace(0,L,20);
xOde = [0 L]'; % couple ODE at the two ends
mGeom = 0;
%% pde1dM solver
pdef = @(z,t,u,DuDx) pdefun(z,t,u,DuDx,Dc,Dg);
ic = @(x) icfun(x);
odeF = @(t,v,vdot,x,u,DuDx,f, dudt, du2dxdt) ...
odeFunc(t,v,vdot,x,u,DuDx,f, dudt, du2dxdt, ...
k0, kcat, n, kbar, kon, K, m, koff);
odeIcF = @() odeIcFunc(c10,c20,g10,g20);
%figure; plot(x, ic(x)); grid; return;
[sol, odeSol] = pde1dM(mGeom,pdef,ic,@bcfun,z,t,odeF, odeIcF,xOde);
C=sol(:,:,1);
G=sol(:,:,2);
figure; plot(z, C(end,:)); grid;
title 'C at final time';
figure; plot(z, G(end,:)); grid;
title 'G at final time';
figure; plot(t, C(:,1)); grid;
title 'C at left end as a function of time';
figure; plot(t, C(:,end)); grid;
title 'C at right end as a function of time';
figure; plot(t, G(:,1)); grid;
title 'G at left end as a function of time';
figure; plot(t, G(:,end)); grid;
title 'G at right end as a function of time';
% plot ode variables as a function of time
figure;
hold on;
for i=1:4
plot(t, odeSol(:,i));
end
legend('c1', 'c2', 'g1', 'g2');
grid; xlabel('Time');
title 'ODE Variables As Functions of Time'
end
function [c,f,s] = pdefun(z,t,u,DuDx,Dc,Dg)
c = [1 1];
f = [Dc Dg]'.*DuDx;
s = [0 0]';
end
function c0 = icfun(x)
c0 = [.25 .06]';
end
function [pl,ql,pr,qr] = bcfun(xl,cl,xr,cr,t,v,vdot)
dc1dt = vdot(1);
dc2dt = vdot(2);
dg1dt = vdot(3);
dg2dt = vdot(4);
pl = [-dc1dt -dg1dt]';
ql = [1 1]';
pr = [dc2dt dg2dt]';
qr = [1 1]';
end
function R=odeFunc(t,v,vdot,x,u,DuDx,f, dudt, du2dxdt, ...
k0, kcat, n, kbar, kon, K, m, koff)
C1 = u(1,1);
C2 = u(1,2);
G1 = u(2,1);
G2 = u(2,2);
c1 = v(1);
c2 = v(2);
g1 = v(3);
g2 = v(4);
Dc1Dt =(k0+kcat*c1^n)*g1*C1-kbar*c1;
Dc2Dt =(k0+kcat*c2^n)*g2*C2-kbar*c2;
Dg1Dt=kon/(1+K*c1^m)*G1-koff*g1;
Dg2Dt=kon/(1+K*c2^m)*G2-koff*g2;
R=vdot-[Dc1Dt, Dc2Dt, Dg1Dt, Dg2Dt]';
end
function v0=odeIcFunc(c10,c20,g10,g20)
v0=[c10,c20,g10,g20]';
end
##### 11 comentariosMostrar 9 comentarios más antiguosOcultar 9 comentarios más antiguos
Bill Greene el 24 de En. de 2022
The ODE variables are available to the PDE definition function. Look at section 3.2.3 in this pde1dM user manual that shows the function signature for the PDE function when there are ODE variables.
But looking at equations (5) in your doc, they look like PDE to me because they contain the u_2 and u_1 which are functions of x.
Why don't you post a new question here concerning your problem?
Paul Maurerlehner el 24 de En. de 2022
You are right, at least the second equation at equations (5) is a PDE since it contains a spatial deriavative (of u_1). I could define all equations as PDEs, but I have the problem that I don't have boundary conditions for the auxiliary variables..
I postet it here, because this specific issue concerns the pde1dM solver which is not a standard Matlab function and the probleme here is very similar to mine.

Iniciar sesión para comentar.

### Categorías

Más información sobre Boundary Conditions en Help Center y File Exchange.

### Community Treasure Hunt

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

Start Hunting!

Translated by