I want to define a differntiable sym in MatLab so that
sym x y t;
eq = y == x;
diff(eq,'t') = dy/dt == dx/dt
With this Code dy/dt and dx/dt are calculated and 0 is returned. How can I tell matlab that y and x are functions of t?
I do not know the explicit functions, so substituting y and x with their explicit functions is not an option.
What you are looking for is this-
syms x(t) y(t)
Define your functions as-
x(t)='your function';
y(t)='your function';
And then calculate the differential as follows-
diff(x) %since x is only a function of t
diff(y) %since y is only a function of t
Related
This is my code for finding the centered coefficients for lagrange polynomial interpolation:
% INPUT
% f f scalar - valued function
% interval interpolation interval [a, b]
% n interpolation order
%
% OUTPUT
% coeff centered coefficients of Lagrange interpolant
function coeff = lagrangeInterp (f, interval , n)
a = interval(1);
b = interval(2);
x = linspace(a,b,n+1);
y = f(x);
coeff(1,:) = polyfit(x,y,n);
end
Which is called in the following script
%Plot lagrangeInterp and sin(x) together
hold on
x = 0:0.1*pi:2*pi;
for n = 1:1:4
coeff = lagrangeInterp(#(x)sin(x),[0,2*pi],n);
plot(x,polyval(coeff,x,'-'));
end
y = sin(x);
plot(x,y);
legend('1st order','2nd order','3rd order','4th order','sin(x)');
To check for stability I would like to perturb the function (eg g(x) = f(x) + epsilon). How would I go about this?
Well, a little trick for you.
You know randn([m,n]) in matlab generate a m*n random matrix. The point is to generate a random vector, and interp1 to a function of x. Like this:
x = linspace(a,b,n+1); % Your range of input
g = #(ep,xx)f(xx)+interp1(x,ep*randn([length(x),1]),xx);
I have an equation like this:
dy/dx = a(x)*y + b
where a(x) is a non-constant (a=1/x) and b is a vector (10000 rows).
How can I solve this equation?
Let me assume you would like to write a generic numerical solver for dy/dx = a(x)*y + b. Then you can pass the function a(x) as an argument to the right-hand side function of one of the ODE solvers. e.g.
a = #(x) 1/x;
xdomain = [1 10];
b = rand(10000,1);
y0 = ones(10000,1);
[x,y] = ode45(#(x,y,a,b)a(x)*y + b,xdomain,y0,[],a,b);
plot(x,y)
Here, I've specified the domain of x as xdomain, and the value of y at the bottom limit of x as y0.
From my comments, you can solve this without MATLAB. Assuming non-zero x, you can use an integrating factor to get a 10000-by-1 solution y(x)
y_i(x) = b_i*x*ln(x) + c_i*x
with 10000-by-1 vector of constants c, where y_i(x), b_i and c_i are the i-th entries of y(x), b and c respectively. The constant vector c can be determined at some point x0 as
c_i = y_i(x0)/x_0 - b_i*ln(x0)
Consider a function y(x) sampled in an array of values, represented by the arrays x and y. If I have another x value x0, I can evaluate y(x0) using spline
y0 = spline(x,y,x0);
Now, I can also write
pp = spline(x,y);
y0 = ppval(pp,x0);
MY QUESTION: If I already have the coefficient and x matrices, my_coefs (size(my_coefs) = [length(y),4]) and x, how can I create a piecewise polynomial My_pp such that pp.coefs = my_coefs and that y0 = ppval(My_pp,x0)?
OK, There is no "spline object", but rather a piecewise polynomial object. So, if my_coefs was attained by break-points my_x then the code needed is
my_spline = mkpp(my_x,my_coefs);
y0 = ppval(my_spline, x0);
In case that dimensions are dazzeling here, which they are, then
my_coefs is 4*n
my_x is n
y0 is N
x0 is N
Is it possible to have MATLAB do implicit differentiation with its symbols?
I have the following code
syms x;
y = symfun(sym('y(x)'), sym('x'));
yPrime = symfun(sym('y+(2*x)-1'), [sym('x'), sym('y')]);
diff(yPrime, x, 1)
From this I get
ans(x, y) =
2
but what I want to get (in some form) is
ans(x, y) =
dy/dx + 2
Is that possible?
It looks you're trying to use y as both a symbolic variable and an abstract symbolic function (symfun), which isn't possible. A symbolic function can only take symbolic variables as arguments. I think that you can accomplish what you want with (old-style strings not needed)
syms y(x) % Implicitly defines x too
yPrime = y+2*x-1 % Also a symfun because y is a symfun
diff(yPrime,x)
which returns this symfun
ans(x) =
diff(y(x), x) + 2
In order to obtain a graphical representation of the behaviour of a fluid it is common practice to plot its streamlines.
For a given two-dimensional fluid with speed components u = Kx and v = -Ky (where K is a constant, for example: K = 5), the streamline equation can be obtained integrating the flow velocity field components as follows:
Streamline equation: ∫dx/u = ∫dy/v
The solved equation looks like this: A = B + C (where A is the solution of the first integral, B is the solution of the second integral and C is an integration constant).
Once we have achieved this, we can start plotting a streamline by simply assigning a value to C, for example: C = 1, and plotting the resulting equation. That would generate a single streamline, so in order to get more of them you need to iterate this last step assigning a different value of C each time.
I have successfully plotted the streamlines of this particular flow by letting matlab integrate the equation symbolically and using ezplot to produce a graphic as follows:
syms x y
K = 5; %Constant.
u = K*x; %Velocity component in x direction.
v = -K*y; %Velocity component in y direction.
A = int(1/u,x); %First integral.
B = int(1/v,y); %Second integral.
for C = -10:0.1:10; %Loop. C is assigned a different value in each iteration.
eqn = A == B + C; %Solved streamline equation.
ezplot(eqn,[-1,1]); %Plot streamline.
hold on;
end
axis equal;
axis([-1 1 -1 1]);
This is the result:
The problem is that for some particular regions of the flow ezplot is not accurate enough and doesn't handle singularities very well (asymptotes, etc.). That's why a standard "numeric" plot seems desirable, in order to obtain a better visual output.
The challenge here is to convert the symbolic streamline solution into an explicit expression that would be compatible with the standard plot function.
I have tried to do it like this, using subs and solve with no success at all (Matlab throws an error).
syms x y
K = 5; %Constant.
u = K*x; %Velocity component in x direction.
v = -K*y; %Velocity component in y direction.
A = int(1/u,x); %First integral.
B = int(1/v,y); %Second integral.
X = -1:0.1:1; %Array of x values for plotting.
for C = -10:0.1:10; %Loop. C is assigned a different value in each iteration.
eqn = A == B + C; %Solved streamline equation.
Y = subs(solve(eqn,y),x,X); %Explicit streamline expression for Y.
plot(X,Y); %Standard plot call.
hold on;
end
This is the error that is displayed on the command window:
Error using mupadmex
Error in MuPAD command: Division by zero.
[_power]
Evaluating: symobj::trysubs
Error in sym/subs>mupadsubs (line 139)
G =
mupadmex('symobj::fullsubs',F.s,X2,Y2);
Error in sym/subs (line 124)
G = mupadsubs(F,X,Y);
Error in Flow_Streamlines (line 18)
Y = subs(solve(eqn,y),x,X); %Explicit
streamline expression for Y.
So, how should this be done?
Since you are using subs many times, matlabFunction is more efficient. You can use C as a parameter, and solve for y in terms of both x and C. Then the for loop is very much faster:
syms x y
K = 5; %Constant.
u = K*x; %Velocity component in x direction.
v = -K*y; %Velocity component in y direction.
A = int(1/u,x); %First integral.
B = int(1/v,y); %Second integral.
X = -1:0.1:1; %Array of x values for plotting.
syms C % C is treated as a parameter
eqn = A == B + C; %Solved streamline equation.
% Now solve the eqn for y, and make it into a function of `x` and `C`
Y=matlabFunction(solve(eqn,y),'vars',{'x','C'})
for C = -10:0.1:10; %Loop. C is assigned a different value in each iteration.
plot(X,Y(X,C)); %Standard plot call, but using the function for `Y`
hold on;
end