I have a differential equation that looks like this:
dT/dx = (1+alpha\*M1)\*(T^2)\*S(x) - 4\*(2+gamma\*M2)\*x\*T
I want to differentiate this equation let's say over alpha. However, T depends on alpha. How can I get a differential equation that includes the derivative dT/da so I can solve for it (using ode45 or something)?
Note: I can't differentiate the equation by hand as I will need to do this for the Ys as well, where I don't have an easy analytical function.
My code so far:
syms L alpha gamma Y2 Y3 Y4 Y5 Y6 x T X
M1 = 0;
M2 = 7;
Y1 = 1 + M2 / 10;
Y7 = 3 + M1 / 10;
xcp = [0 0.1*L 0.25*L 0.5*L 0.6*L 0.75*L L];
ycp = [Y1 Y2 Y3 Y4 Y5 Y6 Y7];
T0 = 5 + 1/L - 25/(L^2);
S = bezier_syms(); % Assume this returns a function of L,x,Y2,Y3,Y4,Y5,Y6
ode = diff(T,x) == (1+alphaM1)*(T^2)*S(L,x,Y2,Y3,Y4,Y5,Y6) - 4*(2 + gamma*M2)xT;
Given the equation
dT/dx = F(x,T,a)
and U=dT/da, then by the rules of differentiation like the chain rule, you get
dU/dx = dF/da(x,T,a) + dF/dT(x,T,a)*U
where the partial derivatives are easy to determine. For more precise formulas, see https://math.stackexchange.com/a/3699281/115115
Related
I have stumbled upon this matlab code that solves this ODE
y'''(t) + a y(t) = -b y''(t) + u(t)
but I am confused by the ode_system function definition, specifically by the y(2) y(3) part. I would greatly appreciate if someone can shed some light
y(2) y(3) part in the ode_system function confuses me and how it contributes to overaal solution
% Define the parameters a and b
a = 1;
b = 2;
% Define the time horizon [0,1]
time_horizon = [0, 1];
% Define the initial conditions for y, y', and y''
initials = [0; 0; 0];
% Define the function handle for the input function u(t)
%sin(t) is a common example of a time-varying function.
% You can change the definition of u to any other function of time,
% such as a constant, a step function, or a more complex function, depending on your needs
u = #(t) sin(t);
% Define the function handle for the system of ODEs
odefunction = #(t, y) ode_system(t, y, a, b, u);
% Solve the ODEs using ode45
[t, y] = ode45(odefunction, time_horizon, initials);
% Plot the solution
plot(t, y(:,1), '-', 'LineWidth', 2);
xlabel('t');
ylabel('y');
function dydt = ode_system(t, y, a, b, u)
%Define the system of ODEs
dydt = [y(2); y(3); -b*y(3) + u(t) - a*y(1)];
end
This is more of a maths question than a Matlab one.
We would like to rewrite our ODE equation so that there is a single time derivative on the left-hand side and no derivatives on the right.
Currently we have:
y'''(t)+ay(t)=-by''(t)+u(t)
By letting z = y' and x = z' (= y''), we can rewrite this as:
x'(t)+a y(t)=-b x(t)+u(t)
So now we have 3 equations in the form:
y' = z
z' = x
x' = -b * x + u - a *y
We can also think of this as a vector equation where v = (y, z, x).
The right-hand side would then be,
v(1)' = v(2)
v(2)' = v(3)
v(3)' = -b * v(3) + u - a * v(1)
which is what you have in the question.
Could you please help me with the following question:
I want to solve a second order equation with two unknowns and use the results to plot an ellipse.
Here is my function:
fun = #(x) [x(1) x(2)]*V*[x(1) x(2)]'-c
V is 2x2 symmetric matrix, c is a positive constant and there are two unknowns, x1 and x2.
If I solve the equation using fsolve, I notice that the solution is very sensitive to the initial values
fsolve(fun, [1 1])
Is it possible to get the solution to this equation without providing an exact starting value, but rather a range? For example, I would like to see the possible combinations for x1, x2 \in (-4,4)
Using ezplot I obtain the desired graphical output, but not the solution of the equation.
fh= #(x1,x2) [x1 x2]*V*[x1 x2]'-c;
ezplot(fh)
axis equal
Is there a way to have both?
Thanks a lot!
you can take the XData and YData from ezplot:
c = rand;
V = rand(2);
V = V + V';
fh= #(x1,x2) [x1 x2]*V*[x1 x2]'-c;
h = ezplot(fh,[-4,4,-4,4]); % plot in range
axis equal
fun = #(x) [x(1) x(2)]*V*[x(1) x(2)]'-c;
X = fsolve(fun, [1 1]); % specific solution
hold on;
plot(x(1),x(2),'or');
% possible solutions in range
x1 = h.XData;
x2 = h.YData;
or you can use vector input to fsolve:
c = rand;
V = rand(2);
V = V + V';
x1 = linspace(-4,4,100)';
fun2 = #(x2) sum(([x1 x2]*V).*[x1 x2],2)-c;
x2 = fsolve(fun2, ones(size(x1)));
% remove invalid values
tol = 1e-2;
x2(abs(fun2(x2)) > tol) = nan;
plot(x1,x2,'.b')
However, the easiest and most straight forward approach is to rearrange the ellipse matrix form in a quadratic equation form:
k = rand;
V = rand(2);
V = V + V';
a = V(1,1);
b = V(1,2);
c = V(2,2);
% rearange terms in the form of quadratic equation:
% a*x1^2 + (2*b*x2)*x1 + (c*x2^2) = k;
% a*x1^2 + (2*b*x2)*x1 + (c*x2^2 - k) = 0;
x2 = linspace(-4,4,1000);
A = a;
B = (2*b*x2);
C = (c*x2.^2 - k);
% solve regular quadratic equation
dicriminant = B.^2 - 4*A.*C;
x1_1 = (-B - sqrt(dicriminant))./(2*A);
x1_2 = (-B + sqrt(dicriminant))./(2*A);
x1_1(dicriminant < 0) = nan;
x1_2(dicriminant < 0) = nan;
% plot
plot(x1_1,x2,'.b')
hold on
plot(x1_2,x2,'.g')
hold off
I have a set of differential equations of the form:
x1dot = x3;
x2dot = x2;
x3dot = x1;
x4dot = x2 + integral(x1,t,tend)
I have the boundary condition for x1, x2 at tstart and x3, x4 at tend. Without the integral term it's a straight forward implementation using BVP4C.
I am wondering if it is possible to have the previous solution for the states from the BVP solver which can be used for the integral.
One possibility is using ode45 and fsolve in combination for the boundary value problem, where I can have the previous solution, but this approach is not as fast as the BVP setup.
I also feel that there may be some difficulty in convergence when I use the previous solution, x1, for the integral.
Is there a better/quicker or easier way to solve this problem?
Set
x5 = integral(x1,t,tend)
then
x5dot = -x1 with x5(tend) = 0
Since x5dot + x3dot = 0, it follows that x5 + x3 = C = const. Therefore you can use the substitution x5 → C - x3.
The constant C is simply C = x3(tend) + x5(tend) = x3(tend) (because x5(tend) = 0).
I'm working on the numerical integration to a discontinuous ODE in MATLAB, but I'm not getting the results I intended to get.
I have a 2-dof system of two masses connected via springs. (see 2-dof system)
My function in MATLAB is of the form:
function [ xdot ] = two_springs(t, x, m1, m2, c, g, k)
% x(1) = x1
% x(2) = x1'
% x(3) = x2
% x(4) = x2'
delta = max(x(3,1) - x(1,1), 0); % (Discontinuous)
F = k * delta^(3 / 2);
xdot(1,1) = x(2,1);
xdot(2,1) = (-F + m1 * g) / m1;
xdot(3,1) = x(4,1);
xdot(4,1) = (F - c * x(3,1) + m2 * g) / m2;
end
Afterwards I plot force F over time.
But no matter which solver I use ( ode45, ode23, ode113, ode15s, ode23s,...) I get a failure because the step size was reduced below the smallest value allowed.
The topic: MATLAB- ode solver: Unable to meet integration tolerances did unfortunately not help me.
Does anyone have an idea what else I could try?
Thank you very much for your help!
tine-lore
I am trying to use the Matlab "gradient" and "hessian" functions to calculate the derivative of a symbolic vector function with respect to a vector. Below is an example using the sigmoid function 1/(1+e^(-a)) where a is a feature vector multiplied by weights. The versions below all return an error. I am new to Matlab and would very much appreciate any advice. The solution may well be under my nose in the documentation, but I have not been able to solve the issue. Thank you in advance for your help!
%version 1
syms y w x
x = sym('x', [1 3]);
w = sym('w', [1 3]);
f = (y-1)*w.*x + log(1/(1+exp(-w.*x)));
gradient(f, w)
%version 2
syms y w1 w2 w3 x1 x2 x3 x w
x = [x1,x2,x3];
w = [w1,w2,w3];
f = (y-1)*w.*x + log(1/(1+exp(-w.*x)));
%version 3
syms y w1 w2 w3 x1 x2 x3
f = (y-1)*[w1,w2,w3].*[x1,x2,x3] + log(1/(1+exp(-[w1,w2,w3].*[x1,x2,x3])));
Thanks, Daniel - it turns out the problem was in not having used dot() to take the dot product of w and x. Both diff() and gradient() gave the same solution, shown below:
syms y
x = sym('x', [1 3]);
w = sym('w', [1 3]);
f = (y-1)*dot(w,x) + log(1/(1+exp(dot(-w,x))));
diff(f, w(1))
gradient(f, w(1))
%ans =
%x1*(y - 1) + (x1*exp(- x1*conj(w1) - x2*conj(w2) - x3*conj(w3)))/
(exp(-x1*conj(w1) - x2*conj(w2) - x3*conj(w3)) + 1)