What is the role of [] in ode?

Example1)

[t x] = ode45(@fun, tspan, c0,...)

I'm not sure what exactly is [t x] above. As I understand it, does ode represent t,x in [] by unpacking it? Is it exactly solving for x for t?

Example2)

[t x] = ode45(@fun,tspan,c0,...)

[t y] = ode45(@fun,tspan, c0,...)

Does having two odes as above mean solving for x and y? If a function is a function related to x and y (z=x+y), does ode represent the x and y values ​​for t?

[Stephen23]

"I'm not sure what exactly is [t x] above."

How to call functions with multiple output arguments:

https://www.mathworks.com/help/matlab/getting-started-with-matlab.html

"As I understand it, does ode represent t,x in [] by unpacking it?"

No, MATLAB does not have any concept of output "unpacking". MATLAB functions can return multple outputs: in the example you show, the function ODE45 two outputs. The names of the variables assigned to those outputs are irrelevant.

Answer 1

[Sam Chak]

Hi @채원 이

If the system is

,

then you can solve the ODEs as follows:

kp = 1;
kd = 2;
F = @(t, x) [x(2); ...
             - kp*x(1) - kd*x(2)];
[t, x] = ode45(F, [0 10], [1, 0]);
plot(t, x)

The solver assigns the solution in the time-vector, t and the array.

[채원 이]

Thanks for your reply.

If you have any questions, please leave a reply.

So, if you replace x and y with x(1) and x(2), don't only the x value appear in the workspace? Is it not possible to make x and y appear separately in the workspace?

[Stephen23]

"So, if you replace x and y with x(1) and x(2), don't only the x value appear in the workspace?"

No, that would be assigning the outputs to one array x, where the indexing you use is scalar so your code would throw an error if ODE45 did not return scalar outputs.

"Is it not possible to make x and y appear separately in the workspace?"

These are the two output arguments:

[t, x] = ode45(..)
 ^                 this is the 1st output argument
    ^              this is the 2nd output argument

An output can be:

• assigned to a variables (either creating or replacing any existing variable with that name)
• allocated via indexing to an array.
• implicitly passed to an outer nested function (not relevant here).

It is not possible to "unpack" one single function output argument into multiple variables.

You can do that yourself later from any variable, using basic MATLAB indexing.

[채원 이]

답변주셔서 감사합니다!!

제가 정확히 이해를 못했습니다.

dcdt = 4*dqdt 인 함수가 있다.

1.

위의 함수를 [t c] = ode45(..)이렇게 풀고자 하면

q를 c로 바꿔야만 풀린다는 것으로 이해했습니다.

2.

또한 위의 함수를

[t c] = ode45(..)

[t q] = ode45(..)

이렇게 2번 나눠서 풀면 안된다고 이해했습니다.

제가 이해한 2개가 맞습니까?

3.

ode를 풀때에는 변수 3개(위의 함수에서 c,t,q)를 풀수 없는 것이 맞나요?

4.

답변주신 내용 중

'나중에 기본 MATLAB 인덱싱을 사용하여 모든 변수에서 직접 수행 할 수 있습니다.'

이것을 정확히 이해하지 못했습니다.

[채원 이]

I replied back in English in case the translation was not possible.

Thanks for your reply!!

I didn't understand exactly.

There is a function where dcdt = 4*dqdt.

One.

If we want to solve the above function like [t c] = ode45(..)

I understood it to be solved only by replacing q with c.

2.

Also, the above function

[t c] = ode45(..)

[t q] = ode45(..)

I understood that it shouldn't be solved by dividing it into two.

Is the 2 I understood correct?

3.

Is it true that 3 variables (c, t, q in the above function) cannot be solved when solving ode?

4.

Among your answers

'You can later do it directly on any variable using native MATLAB indexing.'

I didn't understand this correctly.

[Stephen23]

"Is it true that 3 variables (c, t, q in the above function) cannot be solved when solving ode?"

You did not show us an ODE function in any commonly-used mathematical notation. We have no idea what your ODE is.

MATLAB's ODE solvers all solve for vectors of values. You can have as many values in that vector as you wish. If you have three, then it will solve this for you.. If you use indexing inside the objective function to assign them internally to some other variables, then of course you can do that.

As I wrote earlier, you are probably confusing the common use of the word variable with the specific MATLAB meaning. It is easy to use ODE45 to for three input values, you need to think in terms of vectors (not in terms of separately-named variables), for example:

[t,y] = ode45(@myfun, [0,1], [1;2;3])

t = 45×1

0 0.0063 0.0126 0.0188 0.0251 0.0501 0.0751 0.1001 0.1251 0.1501

y = 45×3

1.0000 2.0000 3.0000 1.0507 2.0447 3.0001 1.1022 2.0910 3.0002 1.1547 2.1388 3.0005 1.2082 2.1882 3.0009 1.4308 2.4017 3.0038 1.6707 2.6439 3.0085 1.9299 2.9177 3.0151 2.2105 3.2263 3.0236 2.5151 3.5732 3.0340

function dydt = myfun(t,y)
dydt = zeros(2,1);
dydt(1) = y(1)+2*y(2)+y(3);
dydt(2) = 3*y(1)+2*y(2);
dydt(3) = t*y(3);
end

[Sam Chak]

Thanks @Stephen for the explanations.

@채원 이

Thanks for the translation. It is true that you cannot solve directly because the initial value problem (IVP) has to be expressed as a system of first-order ODEs (according to the way of MATLAB):

.

If you have learned state-space representation previously, then you know that the variables c and q can be assigned as

where is the state vector that @Stephen spoke about. Note that the y is bolded and a small cap; it's a math notation for vector. Naturally, you cannot easily distinguish this in the code if you are new in MATLAB.

Anyhow, the system can be rewritten as a state-space model:

.

If is known, then you write the ODE function as

function dydt = odefcn(t, y)    % Ordinary Differential Equations
    dydt = zeros(2,1);
    dydt(1) = 4*(- y(1) - y(2));
    dydt(2) = - y(1) - y(2);
end

It is preferred if you can reveal the more info about the IVP that you are trying to solve.

[채원 이]

Thanks for your reply. The problem I want to solve is: ∂C/∂t=-u ∂C/∂z+F ∂q/∂z+D (∂^2 C)/(∂z^2 ) I am trying to solve it using the finite difference method.

function dydt = odefun(t,y,D,F,u)
% y(2) = q y(1) = C
dydt = zeros(404,1);
for i = 2:403
       dCdz(i) = 1./(dz).*(C(i)-C(i-1)); 
       d2Cdz2(i) = 1./(dz^2.).*(C(i)-2.*C(i-1)+C(i-2)); 
       dydt(1) = D.*d2Cdz2(i) - u*dCdz(i) + F*(-y(1)-y(2));
end
dydt(2) = -y(1)-y(2);
end

[Walter Roberson]

       dydt(1) = D.*d2Cdz2(i) - u*dCdz(i) + F*(-y(1)-y(2));

why are you always assigning to dydt(1) ? Did you intend to assign to dydt(i) ?

[채원 이]

Ummm...I want the dydt to be all in the i range. But what about dydt(2)?

[Walter Roberson]

Change

       dydt(1) = D.*d2Cdz2(i) - u*dCdz(i) + F*(-y(1)-y(2));

to

       dydt(i) = D.*d2Cdz2(i) - u*dCdz(i) + F*(-y(i-1)-y(i));

[채원 이]

Among the replies above, why is q(y1,y2) = - y(1) - y(2);