I have a problem that needs to use ode45 in order to find the solution, but one of the state variables is defined using the derivative of another function.
The code I have for this is:
`[~,s]=ode45(#eom,time,IC,options);
function sdot=eom(~,s)
mu=0.01215510;
C=3;
x=s(1);
y=s(2);
gamma=s(3);
O=0/2*(x^2+y^2)+(1-mu)/sqrt((x+mu)^2+y^2)+mu/sqrt((x+mu-1)^2+y^2);
m=sqrt(2*O-C);
sdot=[m*cos(gamma);m*sin(gamma);(Oy*cos(gamma)-Ox*sin(gamma))/sqrt(2*O-C)-2];
sdot=sdot';
end`
Where Ox=dO/dx, and Oy=dO/dy.
However, I have tried putting it into a separate function file and several other methods, but I cannot figure out how to take the derivative of O with respect to x and y, and then transfer those x and y values back into the state defined variables.
In the separate function, I defined Od=O(mu,d), where d would determine what variable the derivative would be taken with respect to. Within this function, I defined it as Od(a,b)=diff(O,d,1). I had thought that I could input this function into my state function where I could define it as Od(s(1),s(2)), but this only returning error messages.
Related
Let the variables A and B be n x 1 doubles and c be a scalar. My goal is to sum the values in A corresponding to indices for which the optimization variable x is greater than B. I have written this up below:
x = optimvar('x',n); % Creates optimization variable
objfun = sum(x); % Creates objective function
constraint = sum(A(x>=B))>=c; % Constraint based on logical indexing
The third line in the code above returns an error message because optimization variables are not compatible with inequality indexing. Specifically, x>=B cannot be input as indexes into A. Is there a way around this? Or am I thinking about this the wrong way?
Thank you!
You need to use function handles for both, the objective function as well as the constraint-function:
objfun = #(var) sum(var);
constraint = #(var) sum(A(var>=B)) >= c;
In fact, for the objective function objfun, you may also use objfun = #sum. This is a function handle. You can think of it as a pointer or reference to a certain function. Function, which work with one input can be used directly (with #). The optimizer calls the function and uses the optimization variable as input.
Now, if you have multiple inputs, you need to create a function, where you define all inputs but one. For this, you create an anonymous function handle, where you tell the handle what variables are placed where: #(var) sum(A(var>=B)) >= c. The variable var is the changing input and the other variables A, B, and c are taken from the workspace at the point of definition (i.e. the function handle is unaffected if you change the variables later or even delete them).
I first defined functions for dy/dt=y and dy/dt=t:
function dy=d(y):
dy=y
end
function ddy=dd(t):
ddy=t
end
And then I used ode45, respectively:
[t,y]=ode45('d',[1 10],1)
[t,y]=ode45('dd',[1 10],1)
which returns the following error: Error using d
Too many input arguments.
My question is:
Where did I go wrong?
How does Matlab know whether y or t is the independent variable? When I define the first function, it could be reasonably interpreted as dt/dy=y instead of dy/dt=y. Is there a built-in convention for defining functions?
First things first: the docs on ode45 are on the mathworks website, or you can get them from the console by entering help ode45.
The function you pass in needs to take two variables, y then t. As you noticed, with just one it would be impossible to distinguish a function of only y from a function of only t. The first argument has to be the independent, the second is the dependent.
Try defining your function as dy = d(t, y) and ddy = dd(t, y) with the same bodies.
one other note, while using a string representing the function name should work, you can use #d and #dd to reference the functions directly.
I have a statement in my MATLAB program:
f = #(A)DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus);
I understood that f is defined as the function handle to the function distancegauss which contains the parameters/arg list present inside the parentheses.
What does the variable A in #(A) do? Does it have any importance? While browsing I found that the variables within parentheses after # would be the input arguments for an anonymous function..
Can anyone explain what does that A do? Will this handle work even without that A after the # symbol ? Because it is already present as an argument to be passed after the function name.
Your code will create an anonymous function f which accepts one input A. In particular f will call the function DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus); where the value of A is whatever you input with f(A) and the other inputs must already exist in your workspace and will be passed to the function. Note: if the other variables don't exist you should get an error when calling f.
Now a reasonable question is why would you want to do this you could just call DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus); directly with whatever values you want, without having to worry about whether some of them exist.
There are two main reasons I can think of why you would do this (I'm sure there are others). Firstly for simplicity where your other inputs don't change and you don't want to have to keep retyping them or have users accidentally change them.
The other reason where you would want this is when optimizing/minimizing a function, for example with fminsearch. The matlab optimization functions will vary all inputs. If you want only vary some of them you can use this sort of syntax to reduce the number of input variables.
As to what A actually is in your case this will depend on what it does in DistanceGauss, which is not a standard MATLAB function so I suggest you look at the code for that.
"f(A)" or "f of A" or "The function of A" here has the handle "f"
DistanceGauss() here is another function that was defined elsewhere in your code.
You would set x_axis, Y_target, Y_initial, numOf, & modus before creating the function f. These arguments would stay the same for Function f, even if you try and set them again later.
'A' though, is different. You don't set it before you make the function. You can later operate on all values of A, such as if you plot the function or get the integral of the function. In that case, it would be performing the DistanceGauss function on every value of 'A', (though we can't see here what DistanceGauss function does. )
Anonymous function should be defined such as:
sqr = #(x) x.^2;
in which x shows the variable of the the function and there is no name for it (it is called anonymous!).
Although you can do something like this:
c = 10;
mygrid = #(x,y) ndgrid((-x:x/c:x),(-y:y/c:y));
[x,y] = mygrid(pi,2*pi);
in which you are modifying an existing function ndgrid to make a new anonymous function.
In your case also:
f = #(A)DistanceGauss(A,x_axis,Y_target,Y_initial,numOf,modus);
This is a new anonymous function by modifying the function DistanceGauss that you may want to have a single variable A.
If you remove (A) from the code, then f would be a handle to the existing function DistanceGauss:
f = #DistanceGauss;
Now you can evaluate the function simply by using the handle:
f(A,x_axis,...)
I'm trying to use the MATLAB function fzero properly but my program keeps returning an error message. This is my code (made up of two m-files):
friction_zero.m
function fric_zero = friction_zero(reynolds)
fric_zero = 0.25*power(log10(5.74/(power(reynolds,0.9))),-2);
flow.m
function f = flow(fric)
f = 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(power(10,4));
z = fzero(#flow,f_initial)
The goal is to return z as the root for the equation specified by f when flow.m is run.
I believe I have the correct syntax as I have spent a couple of hours online looking at examples. What happens is that it returns the following error message:
"Undefined function or variable 'fric'."
(Of course it's undefined, it's the variable I'm trying to solve!)
Can someone point out to me what I've done wrong? Thanks
EDIT
Thanks to all who helped! You have assisted me to eventually figure out my problem.
I had to add another file. Here is a full summary of the completed code with output.
friction_zero.m
function fric_zero = friction_zero(re)
fric_zero = 0.25*power(log10(5.74/(power(re,0.9))),-2); %starting value for fric
flow.m
function z = flow(fric)
re = power(10,4);
z = 1/(sqrt(fric))-1.873*log10(re*sqrt(fric))-233/((re*sqrt(fric))^0.9)-0.2361;
flow2.m
f_initial = friction_zero(re); %arbitrary starting value (Reynolds)
x = #flow;
fric_root = fzero(x,f_initial)
This returns an output of:
fric_root = 0.0235
Which seems to be the correct answer (phew!)
I realised that (1) I didn't define reynolds (which is now just re) in the right place, and (2) I was trying to do too much and thus skipped out on the line x = #flow;, for some reason when I added the extra line in, MATLAB stopped complaining. Not sure why it wouldn't have just taken #flow straight into fzero().
Once again, thanks :)
You need to make sure that f is a function in your code. This is simply an expression with reynolds being a constant when it isn't defined. As such, wrap this as an anonymous function with fric as the input variable. Also, you need to make sure the output variable from your function is z, not f. Since you're solving for fric, you don't need to specify this as the input variable into flow. Also, you need to specify f as the input into fzero, not flow. flow is the name of your main function. In addition, reynolds in flow is not defined, so I'm going to assume that it's the same as what you specified to friction_zero. With these edits, try doing this:
function z = flow()
reynolds = power(10,4);
f = #(fric) 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(reynolds);
z = fzero(#f, f_initial); %// You're solving for `f`, not flow. flow is your function name
The reason that you have a problem is because flow is called without argument I think. You should read a little more about matlab functions. By the way, reynolds is not defined either.
I am afraid I cannot help you completely since I have not been doing fluid mechanics. However, I can tell you about functions.
A matlab function definition looks something like this:
function x0 = f(xGuess)
a = 2;
fcn =#(t) a*t.^3+t; % t must not be an input to f.
disp(fcn);
a = 3;
disp(fcn);
x0 = fsolve(fcn1,xGuess); % x0 is calculated here
The function can then ne called as myX0 = f(myGuess). When you define a matlab function with arguments and return values, you must tell matlab what to do with them. Matlab cannot guess that. In this function you tell matlab to use xGuess as an initial guess to fsolve, when solving the anonymous function fcn. Notice also that matlab does not assume that an undefined variable is an independent variable. You need to tell matlab that now I want to create an anonymous function fcn which have an independent variable t.
Observation 1: I use .^. This is since the function will take an argument an evaluate it and this argument can also be a vector. In this particulat case I want pointwise evaluation. This is not really necessary when using fsolve but it is good practice if f is not a matrix equation, since "vectorization" is often used in matlab.
Observation 2: notice that even if a changes its value the function does not change. This is since matlab passes the value of a variable when defining a function and not the variable itself. A c programmer would say that a variable is passed by its value and not by a pointer. This means that fcn is really defined as fcn = #(x) 2*t.^3+t;. Using the variable a is just a conveniance (constants can may also be complicated to find, but when found they are just a value).
Armed with this knowledge, you should be able to tackle the problem in front of you. Also, the recursive call to flow in your function will eventuallt cause a crash. When you write a function that calls itself like this you must have a stopping criterium, something to tell the program when to stop. As it is now, flow will call ifself in the last row, like z = fzero(#flow,f_initial) for 500 times and then crash. Alos it is possible as well to define functions with zero inputs:
function plancksConstant = h()
plancksConstant = 6.62606957e−34;
Where the call h or h() will return Plancks constant.
Good luck!
I'm writing a program in MATLAB to solve integrals, and I have my function in a .M-file. Now I wonder how I can write a program in the .MAT-file that lets the user set a value that exists in the both files. The .M-file looks like this:
function fh = f(y)
fh = 62.5.*(b-y).*(40-20.*exp(-(0.01.*y).*(0.01.*y)));
and as you can see, the function depends on two variables, y and b. I want the user to set b. I tried putting b = input('Type in the value of b: ') in the .M-file but for some reason the user would then have to put in the same value four times.
Can I ask for the value of b in the .MAT-file?
Firstly, m-files store code (i.e. functions), while MAT-files store data (i.e. variables). You can save workspace variables to a MAT-file using the function SAVE and load them into a workspace from a file using the function LOAD. If you have a user choose a value for b, then save it to a MAT-file ('b_value.mat', for example), you can simply load the value from the MAT-file inside your m-file function like so:
function fh = f(y)
load('b_value.mat','b');
fh = 62.5.*(b-y).*(40-20.*exp(-(0.01.*y).*(0.01.*y)));
However, this is not a very good way to handle the larger problem I think you are having. It requires that you hardcode the name of the MAT-file in your function f, plus it will give you an error if the file doesn't exist or if b isn't present in the file.
Let's address what I think the larger underlying problem is, and how to better approach a solution...
You mention that you are solving integrals, and that probably means you are performing numerical integration using one or more of the various built-in integration functions, such as QUAD. As you've noticed, using these functions requires you to supply a function for the integrand which accepts a single vector argument and returns a single vector argument.
In your case, you have other additional parameters you want to pass to the function, which is complicated by the fact that the integration functions only accept integrand functions with a single input argument. There is actually a link in the documentation for QUAD (and the other integration functions) that shows you a couple of ways you can parameterize the integrand function without adding extra input arguments by using either nested functions or anonymous functions.
As an example, I'll show you how you can do this by writing f as an anonymous function instead of an m-file function. First you would have the user choose the parameter b, then you would construct your anonymous function as follows:
b = input('Type in the value of b: ');
f = #(y) 62.5.*(b-y).*(40-20.*exp(-(0.01.*y).^2));
Note that the value of b used by the anonymous function will be fixed at what it was at the time that the function was created. If b is later changed, you would need to re-make your anonymous function so that it uses the new value.
And here's an example of how to use f in a call to QUAD:
q = quad(f,lowerLimit,upperLimit);
In your m file declare b as a global
function fh = f(y)
global b
fh = 62.5.(b-y).(40-20.*exp(-(0.01.y).(0.01.*y)));
This allows the variable to be accessed from another file without having to create another function to set the value of b. You could also add b to the arguments of your fh function.