Getting a system of equations and the using ODE45

Question

Shreshtha Chaturvedi 2022-7-31

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1771440-getting-a-system-of-equations-and-the-using-ode45

评论： Torsten 2022-7-31

Hi, I have a coefficient matrix -

Xi=
   1.0e+03 *
    0.3002    0.3832    0.0577
         0         0         0
         0         0         0
         0         0         0
   -0.1425   -0.1576   -0.0162
         0         0         0
         0         0         0
   -0.0352   -0.0601   -0.0140
         0         0         0
         0         0         0
   -0.0941   -0.1304   -0.0245
    0.1394    0.1897    0.0296
    0.7197    0.8981    0.1662
    0.0058    0.0024    0.0061
   -1.0007   -1.2808   -0.2519
         0         0         0
   -0.0211   -0.0250   -0.0060
    0.3760    0.4933    0.0964
   -0.0242   -0.0253    0.0097
         0         0         0
% 

To get the system of ODEs I first defined the variables, and then multiplied it by transpose(Xi) to get the system-

syms x y z xx xy xz yy yz zz xxx xxy xxz xyy xyz xzz yyy yyz yzz zzz
var = [1;x;y;z;xx;xy;xz;yy;yz;zz;xxx;xxy;xxz;xyy;xyz;xzz;yyy;yyz;yzz;zzz]
A = transpose(Xi)
Xdot = A*var

where Xdot is the column containing the derivatives, ie. Xdot = [x' ; y' ; z']. The code above gives me the output:

Xdot =
 
      (4905784495475543*xxy)/35184372088832 - (6622198076817333*xxx)/70368744177664 - (1253707074885461*xx)/8796093022208 + (6330728071069091*xxz)/8796093022208 + (6526091537672905*xyy)/1125899906842624 - (550152702477171*xyz)/549755813888 - (4953639297776579*yy)/140737488355328 - (743004146196139*yyy)/35184372088832 + (6615295418414269*yyz)/17592186044416 - (3404894971584935*yzz)/140737488355328 + 2640157789935161/8796093022208
   (6675905246377611*xxy)/35184372088832 - (4586552884284477*xxx)/35184372088832 - (2773298922331439*xx)/17592186044416 + (987449040685507*xxz)/1099511627776 + (2666532593051851*xyy)/1125899906842624 - (5632811008410909*xyz)/4398046511104 - (8464188990339351*yy)/140737488355328 - (7048704279715849*yyy)/281474976710656 + (4339015521236507*yyz)/8796093022208 - (7129115399973291*yzz)/281474976710656 + 3370686532933275/8796093022208
(2082177599331991*xxy)/70368744177664 - (863749208859729*xxx)/35184372088832 - (4564685191010399*xx)/281474976710656 + (5847277141854391*xxz)/35184372088832 + (6877937169110265*xyy)/1125899906842624 - (8862390418139439*xyz)/35184372088832 - (491773253308733*yy)/35184372088832 - (3353395911004425*yyy)/562949953421312 + (6782944326528325*yyz)/70368744177664 + (5481194070111587*yzz)/562949953421312 + 2029864570192639/35184372088832
 

My question is

a) Is there a better way to define the system of ODEs?

b) How do I solve this system using ODE45? I do not want to manually input the equations, as my coefficient matrix will vary if I have a different dataset, so I cannot manually input equations everytime.

c) If I have 5 variables, say, x1,x2,x3,x4,x5 can I still use ODE45 to solve a system of ODEs containing 5 equations?

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Torsten 2022-7-31

As far as I can see, you have 3 equations for 19 unknowns. What are the missing 16 equations ?

Shreshtha Chaturvedi 2022-7-31

编辑：Shreshtha Chaturvedi 2022-7-31

It's a non-linear system of 3 equations and 3 variables/unknowns. xx=x^2 and so on.

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Answer 1

Torsten 2022-7-31

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1771440-getting-a-system-of-equations-and-the-using-ode45#answer_1018600

编辑：Torsten 2022-7-31

在 MATLAB Online 中打开

fun = @(x) Xi.'*[1;x(1);x(2);x(3);x(1)^2;x(1)*x(2);x(1)*x(3);x(2)^2;x(2)*x(3);x(3)^2;x(1)^3;x(1)^2*x(2);x(1)^2*x(3);x(1)*x(2)^2;x(1)*x(2)*x(3);x(1)*x(3)^2;x(2)^3;x(2)^2*x(3);x(2)*x(3)^2;x(3)^3];
tspan = [0 1];    % time interval of integration
x0 = [1; 1; 1];   % Initial conditions
[T,X] = ode45(fun,tspan,x0)  % solver call

5 个评论
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Torsten 2022-7-31

在 MATLAB Online 中打开

Xi= 1.0e+03 * ...

[0.3002 0.3832 0.0577

0 0 0

-0.1425 -0.1576 -0.0162

0 0 0

-0.0352 -0.0601 -0.0140

0 0 0

-0.0941 -0.1304 -0.0245

0.1394 0.1897 0.0296

0.7197 0.8981 0.1662

0.0058 0.0024 0.0061

-1.0007 -1.2808 -0.2519

0 0 0

-0.0211 -0.0250 -0.0060

0.3760 0.4933 0.0964

-0.0242 -0.0253 0.0097

0 0 0];

fun = @(t,x) Xi.'*[1;x(1);x(2);x(3);x(1)^2;x(1)*x(2);x(1)*x(3);x(2)^2;x(2)*x(3);x(3)^2;x(1)^3;x(1)^2*x(2);x(1)^2*x(3);x(1)*x(2)^2;x(1)*x(2)*x(3);x(1)*x(3)^2;x(2)^3;x(2)^2*x(3);x(2)*x(3)^2;x(3)^3];

tspan = [0 1]; % time interval of integration

x0 = [1; 1; 1]; % Initial conditions

[T,X] = ode15s(fun,tspan,x0); % solver call

Warning: Failure at t=1.340286e-02. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.775558e-17) at time t.

plot(T,X(:,1))

Walter Roberson 2022-7-31

在 MATLAB Online 中打开

Xi = 1.0e+03 * [

0.3002 0.3832 0.0577

0 0 0

-0.1425 -0.1576 -0.0162

0 0 0

-0.0352 -0.0601 -0.0140

0 0 0

-0.0941 -0.1304 -0.0245

0.1394 0.1897 0.0296

0.7197 0.8981 0.1662

0.0058 0.0024 0.0061

-1.0007 -1.2808 -0.2519

0 0 0

-0.0211 -0.0250 -0.0060

0.3760 0.4933 0.0964

-0.0242 -0.0253 0.0097

0 0 0];

%

fun = @(t, x) Xi.'*[1;x(1);x(2);x(3);x(1)^2;x(1)*x(2);x(1)*x(3);x(2)^2;x(2)*x(3);x(3)^2;x(1)^3;x(1)^2*x(2);x(1)^2*x(3);x(1)*x(2)^2;x(1)*x(2)*x(3);x(1)*x(3)^2;x(2)^3;x(2)^2*x(3);x(2)*x(3)^2;x(3)^3];

tspan = [0 1]; % time interval of integration

x0 = [1; 1; 1]; % Initial conditions

[T,X] = ode45(fun,tspan,x0); % solver call

Warning: Failure at t=1.349254e-02. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.775558e-17) at time t.

for K = 1 : size(X,2)

figure(); plot(T, X(:,K)); title("X" + K);

end

Shreshtha Chaturvedi 2022-7-31

在 MATLAB Online 中打开

I tried doing the same for 5 variables:

fun = @(t,x) Xi.'*[1;x(1);x(2);x(3);x(4);x(5);x(1)^2;x(1)*x(2);x(1)*x(3);x(1)*x(4);x(1)*x(5);x(2)^2;x(2)*x(3);x(2)*x(4);x(2)*x(5);x(3)^2;x(3)*x(4);x(3)*x(5);x(4)^2;x(4)*x(5);x(5)^2;x(1)^3;x(1)^2*x(2);x(1)^2*x(3);x(1)^2*x(4);x(1)^2*x(5);x(1)*x(2)^2;x(1)*x(2)*x(3);x(1)*x(2)*x(4);x(1)*x(2)*x(5);x(1)*x(3)^2;x(1)*x(3)*x(4);x(1)*x(3)*x(5);x(1)*x(4)^2;x(1)*x(4)*x(5);x(1)*x(5)^2;x(2)^3;x(2)^2*x(3);x(2)^3;x(2)^2*x(3);x(2)^2*x(4);x(2)^2*x(5);x(2)*x(3)^2;x(2)*x(3)*x(4);x(2)*x(3)*x(5);x(2)*x(4)^2;x(2)*x(4)*x(5);x(2)*x(5)^2;x(3)^3;x(3)^2*x(4);x(3)^2*x(5);x(3)*x(4)^2;x(3)*x(4)*x(5);x(3)*x(5)^2;x(4)^3;x(4)^2*x(5);x(4)*x(5)^2;x(5)^3];
tspan = [1 10000];    % time interval of integration
x0 = [249.270160982681; 80.9237201495384; 74.2320286869611; 79.7728847180895; 255.583672846571];   % Initial conditions
[T,X] = ode45(fun,tspan,x0)  % solver call

where Xi is a 56*5 matrix. But when I try to run the last line (ode45) it says

Error using  * 
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first
matrix matches the number of rows in the second matrix. To operate on each element of the matrix
individually, use TIMES (.*) for elementwise multiplication.

However, the problem is not with the matrix multiplication as if I comment the last line and run it, my program executes without any error. Any hints as to why is it behaving this way when I change dimensions?

Torsten 2022-7-31

Your vector is 58x1 instead of 56x1.

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Answer 2

Walter Roberson 2022-7-31

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1771440-getting-a-system-of-equations-and-the-using-ode45#answer_1018565

I recommend that you read the first example for odeFunction() to see the flow to turn an array of equations into something that can be evaluated numerically by ode45()