Graph plotting of ODE with time variable

Question

Chi Chun Jacky Yeung 2020-10-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/606431-graph-plotting-of-ode-with-time-variable

评论： Star Strider 2020-10-7

采纳的回答： Star Strider

I would like to create a graph with time dependent, I have the below code:

S = 10e-6; %amount of streptavidin (μM)

B = 10e-6; %amount of biotin (μM)

S2 = S+B

S3 = S+2*B

S4 = S+3*B

S5 = S+4*B

syms S(t) S2(t) S3(t) S4(t) S5(t)

ode1 = diff(S,t) == (-4*kon*B*S)+koff*S2;

ode2 = diff(S2,t) == -(3*kon*B+koff)*S2+4*kon*B*S+2*koff*S3;

ode3 = diff(S3,t) == -(2*kon*B+2*koff)*S3+3*kon*B*S2+3*koff*S4;

ode4 = diff(S4,t) == -(kon*B+3*koff)*S4+2*kon*B*S3+4*koff*S5;

ode5 = diff(S5,t) == -4*koff*S5+kon*B*S4;

odes = [ode1; ode2; ode3; ode4; ode5]

Aodes = dsolve(odes)

where I tried to use the ode45 function but it seems doesn't work for me or maybe I misunderstand something. And the graph aims to show the value changes of S(t) to S5(t) from 0 to 200 mins.

The concept will be similar to below graph

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Chi Chun Jacky Yeung 2020-10-6

%% Change in free energy

% assume the temperature is 25 degree

koff = 2.7e-2; %1/s

kon = 1.9e4 % 1/Ms

Kd = kon/koff

Ka = 1/Kd

% Keq = Kd = 1/Ka

R = 8.314472; % J/K*mol universal gas constant

T = 25+273; % K

dG = log(Kd)*R*T % change in free energy

%% number of free and occupied binding site

S = 10e-6; %amount of streptavidin (μM)

B = 10e-6; %amount of biotin (μM)

S2 = S+B

S3 = S+2*B

S4 = S+3*B

S5 = S+4*B

syms S(t) S2(t) S3(t) S4(t) S5(t)

ode1 = diff(S,t) == (-4*kon*B*S)+koff*S2;

ode2 = diff(S2,t) == -(3*kon*B+koff)*S2+4*kon*B*S+2*koff*S3;

ode3 = diff(S3,t) == -(2*kon*B+2*koff)*S3+3*kon*B*S2+3*koff*S4;

ode4 = diff(S4,t) == -(kon*B+3*koff)*S4+2*kon*B*S3+4*koff*S5;

ode5 = diff(S5,t) == -4*koff*S5+kon*B*S4;

odes = [ode1; ode2; ode3; ode4; ode5]

Aodes = dsolve(odes)

Here's the full version of the code.

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Answer 1

Star Strider 2020-10-6

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/606431-graph-plotting-of-ode-with-time-variable#answer_506626

None of the initial conditions are defined (nor are ‘kon’ and ‘koff’), so the initial conditions will be replaced by symbolic constants that do not have numeric values. The result is that none of the equations can be evaluated with fplot or any other plot function.

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Star Strider 2020-10-6

在 MATLAB Online 中打开

My best effort:

% assume the temperature is 25 degree
koff = 2.7e-2; %1/s
kon = 1.9e4; % 1/Ms
Kd = kon/koff;
Ka = 1/Kd;
% Keq = Kd = 1/Ka
R = 8.314472; % J/K*mol universal gas constant
T = 25+273; % K
dG = log(Kd)*R*T; % change in free energy
% %%  number of free and occupied binding site
S = 10e-6; %amount of streptavidin (μM)
B = 10e-6; %amount of biotin (μM)
S2 = S+B;
S3 = S+2*B;
S4 = S+3*B;
S5 = S+4*B;
S00 = 0.65;                                                         % Inferred From Plot
S20 = 0.08;                                                         % Inferred From Plot
S30 = 0.05;                                                         % Inferred From Plot
S40 = 0.05;                                                         % Inferred From Plot
S50 = 0.16;                                                         % Inferred From Plot
syms S(t) S2(t) S3(t) S4(t) S5(t)
ode1 = diff(S,t) == (-4*kon*B*S)+koff*S2;
ode2 = diff(S2,t) == -(3*kon*B+koff)*S2+4*kon*B*S+2*koff*S3;
ode3 = diff(S3,t) == -(2*kon*B+2*koff)*S3+3*kon*B*S2+3*koff*S4;
ode4 = diff(S4,t) == -(kon*B+3*koff)*S4+2*kon*B*S3+4*koff*S5;
ode5 = diff(S5,t) == -4*koff*S5+kon*B*S4;
odes = [ode1; ode2; ode3; ode4; ode5; S(0)==S00; S2(0)==S20; S3(0)==S30; S4(0)==S40; S5(0)==S50];
Aodes = dsolve(odes);
Ss = Aodes.S;
S2s = Aodes.S2;
S3s = Aodes.S3;
S4s = Aodes.S4;
S5s = Aodes.S5;
figure
fplot(Ss,[0 30])
hold on
fplot(S2s,[0 30])
fplot(S3s,[0 30])
fplot(S4s,[0 30])
fplot(S5s,[0 30])
hold off
grid
xlabel('Time')
ylabel('Concentration')
legend('S','S_2','S_3','S_4','S_5', 'Location','NW')

There appear to be problems with the model, since the plot it produces does not match the plot you posted.

I must leave you to sort those, since I have no idea what you are doing.

Chi Chun Jacky Yeung 2020-10-7

Oooops, so is there no anyway to find out the value of S(0) to S5(0) by Matlab itself?

Star Strider 2020-10-7

In certain situations, yes. If you have data for the variables and you want to fit them to a system of differential equations to estimate the parameters, you can certainly estimate the initial conditions as well. (See this Answer where I include them as the last elements of the parameter vector in order to estimate them along with the other parameters.)

Otherwise, MATLAB expects you to supply them, whether you are using the Symbolic Math Toolbox, or one of the numeric ODE solvers (such as ode45 or ode15s).

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