I created a symbolic function in MATLAB R2021b using this script with the goal of solving an ODE.
syms phi(x) lambda L
eqn_x = diff(phi,x,2) == -lambda*phi;
dphi = diff(phi,x);
cond = [phi(0)==0, dphi(1)==0]; % this is the line where the problem starts
disp(cond)
phi = dsolve(eqn_x,x,cond);
This script runs without any errors, but I want to evaluate dphi(L)==0 instead of at x=1. When I try to do this instead, I get the following error.
Obviously, the way MATLAB is interpreting dphi(L)==0 is not correct since it is replacing the dependent parameter x with L which should be an independent parameter. Can someone recommend a fix or workaround?
I've also tried this dphi(x==L)==0 and I got the same error.
Addition following kikon's comment
I tried both options suggested based on the tutorial. These are the results.
dphi | x = L creates a parse error in MATLAB
dphi | x == L gives this result. This seems to indicate x=l OR dphi/dx which is not a condition. I'm tempted to try & in place of | to get an and, but this may not be sufficient.
subs(dphi, x = L) gives the same condition when I use dphi(L)==0 which is incorrect.
Related
This answer explains how to substitute a derivative in a symbolic expression. However, this approach fails when why function is "chained twice", e.g. sin(psi(phi(t))). See the example
clear;
syms t phi(t) psi(phi)
psi = psi(phi)
dphidt = diff(phi, t)
dpsidphi = diff(psi, phi)
x = sin(psi)
dxdt = diff(x, t)
syms phid psip % Substitution variables for derivatives
test1 = subs(dxdt,dphidt,phid) % This works
test2 = subs(dxdt,dpsidphi,psip) % This doesn't work
% Ideal result, without any dependencies
res = psip*phid*cos(psi)
Substituting phid for diff(phi(t), t) works, but substituting psip for D(psi)(phi(t)) does not work. What am I doing wrong? Do I have to specify the chained dependecies in a different way? I also don't quite understand the difference between D and diff.
Second question:
My actual terms are a lot more complicated and I have to take derivatives with respect to phid for example. Therefore, I have to get rid of all the dependencies because I want to generate a function using matlabFunction. I can remove the t dependency using formula, however I can't get rid of the phi(psi) dependency. Is there a command for this or do I have to make another substitution?
I want to solve a system of linear equalities of the type Ax = b+u, where A and b are known. I used a function in MATLAB like this:
x = #(u) gmres(A,b+u);
Then I used fmincon, where a value for u is given to this expression and x is computed. For example
J = #(u) (x(u)' * x(u) - x^*)^2
and
[J^*,u] = fmincon(J,...);
withe the dots as matrices and vectors for the equalities and inequalities.
My problem is, that MATLAB delivers always an output with information about the command gmres. But I have no idea, how I can stop this (it makes the Program much slower).
I hope you know an answer.
Patsch
It's a little hidden in the documentation, but it does say
No messages are displayed if the flag output is specified.
So you need to call gmres with at least two outputs. You can do this by making a wrapper function
function x = gmresnomsg(varargin)
[x,~] = gmres(varargin{:});
end
and use that for your handle creation
x = #(u) gmresnomsg(A,b+u);
I'm using MATLAB 2012b.
I want to get d²/dxdy of a simple function:
f(x,y) = (x-1)² + 2y²
The documentation states that I can use syms and diff as in the following example:
> syms x y
> diff(x*sin(x*y), x, y)
ans =
2*x*cos(x*y) - x^2*y*sin(x*y)
But doing the same I got the wrong answer:
> syms x y
> f = (x-1)^2 + 2*y^2;
> diff(f,x,y)
ans =
4*y
The answer is right if I use diff like this:
diff(diff(f,x),y)
Well, it's not a problem for me to use it this way, but nevertheless why is the first variant not working? Is it a version issue?
The actual documentation from R2010a:
diff(expr) differentiates a symbolic expression expr with respect to its free variable as determined by symvar.
diff(expr, v) and diff(expr, sym('v')) differentiate expr with respect to v.
diff(expr, n) differentiates expr n times. n is a positive integer.
diff(expr, v, n) and diff(expr, n, v) differentiate expr with respect to v n times.
So, the command diff(f,x,y) is the last case. It would be equal to differentiating f w.r.t. x, y times, or w.r.t y, x times.
For some reason I don't quite understand, you don't get a warning or error, but one of the syms variables gets interpreted as n = 1, and then the differentiation is carried out. In this case, what diff seems to do is basically diff(f, y, 1).
In any case, it seems that the behavior changed from version to version, because in the documentation you link to (R2016b), there is an additional case:
diff(F,var1,...,varN) differentiates F with respect to the variables var1,...,varN
So I suspect you're running into a version issue.
If you want to differentiate twice, both w.r.t x and y, your second attempt is indeed the correct and most portable way to do that:
diff( diff(f,x), y )
or equivalently
diff( diff(f,y), x )
NB
I checked the R2010a code for symbolic/symbolic/#sym/diff.m and indeed, n is defaulted to 1 and only changed if one of the input variables is a double, and the variable to differentiate over is set equal to the last syms variable in the argument list. The multiple syms variable call is not supported, nor detected and error-trapped.
Syms is only creating symbolic variables.
The first code you execute is only a single derivative. The second code you provided differentiates two times. So I think you forgot to differentiate a second time in the first piece of code you provided.
I am also wondering what answer you expect? If you want 4*y as answer, than you should use
diff(f,y)
and not
diff(f,x,y)
Performing the second derivative is giving me zero?
diff(diff(f,x),y)
If you want 4 as answer than you have to do following:
diff(diff(f,y),y)
I am trying to find use to Newton-Raphson method to find the roots. It does this by making a guess and then improving the guess after each iteration until you get one of the zeros.
Because the Newton-Raphson method quickly finds the zeros, it gives me a small error immediately and after two or three iterations max it should fail to meet the conditions of the while loop. However, the problem is that when I remove the semi-colon after "error" in my loop, I start getting fractions that should break the while loop, but its like Matlab doesn't know that 123/8328423 is less than 1. It continues to run until I manually force the program to stop running.
How do I fix this? I tried format long, format longe, and using double both in the command window, in the scrip file, and somewhere in the loop.
Thank you in advance for any tips, suggestions, or advice that may help!!
A = [1,2,-4;2,-2,-2;-4,-2,1;];
format longe
% syms x y z
% P = x^4 + 3*x^2*y^2-z^3+y+1;
% feval(symengine,'degree',P,x)
syms x
B = mateigenvalue(A);
f(x) = simplify(matdet(B));
x0 = 1;
error = 10;
while(error > .01)
x1 = x0 - f(x0)/(27*(x0)-3*(x0)^2);
error = abs(((f(x0)-f(x1))/f(x0))*100)
x0 = x1;
end
x0 = double(x0)
I reckon the main problem is with error.
It starts as double but inside the while-loop it turns into a symbolic variable and you can't easily compare symbolic variables with scalar values (the .01 in the while-loop condition).
Check in your workspace if error is symbolic (or type class(error) and check if sym is returned). I guess it is symbolic because a fraction is returned (123/8328423), as instead Matlab treats double values with decimals, not fractions.
If so, try doing (inside the while-loop) a conversion for error that is, under the line
error = abs(((f(x0)-f(x1))/f(x0))*100);
try putting
error=double(error);
So error will be temporarily converted in double and you can easily compare its value with .01 to check the while-loop condition.
Also, it is bad practice to call a variable error since error() is a built-in function in Matlab. By naming a variable error you cannot use the error() function. Same story goes for other built-in functions.
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!