I am trying to solve those four equations for "p, x, y, z," , which contains two parameters "b", "a", and I have used solve command as shown in the code:
clc,clear;
syms p x y z b a
eqn1 = (1)*p + (exp(0.5*b^(1/3)))*x + (exp(-0.25*b^(1/3)) * cos(0.25*sqrt(3)*b^(1/3)))*y + (exp(-0.25*b^(1/3)) * sin(0.25*sqrt(3)*b^(1/3)))*z == 0;
eqn2 = (1)*p + (exp(-0.5*b^(1/3)))*x + (exp(0.25*b^(1/3)) * cos(0.25*sqrt(3)*b^(1/3)))*y - (exp(0.25*b^(1/3)) * sin(0.25*sqrt(3)*b^(1/3)))*z == 0;
eqn3 = (b^(1/3)*exp(0.5*b^(1/3))+a*b^(2/3)*exp(0.5*b^(1/3)))*x + (-.866*b^.333*sin(.433*b^.333)*exp(-.25*b^.333)-.5*cos(.433*b^.333)*b^.333*exp(-.25*b^.333) + a(-.5*b^.666*cos(.433*b^.333)*exp(-.25*b^.333)+.866*b^.666*sin(.433*b^.333)*exp(-.25*b^.333)))*y + (.866*cos(.433*b^.333)*b^.333*exp(-.25*b^.333)-.5*b^.333*sin(.433*b^.333)*exp(-.25*b^.333)+a(-.5*b^.666*sin(.433*b^.333)*exp(-.25*b^.333)-.866*b^.666*cos(.433*b^.333)*exp(-.25*b^.333)))*z == 0;
eqn4 = (b^(1/3)*exp(-0.5*b^(1/3))-a*b^(2/3)*exp(-0.5*b^(1/3)))*x + (.866*b^.333*sin(.433*b^.333)*exp(.25*b^.333)-.5*cos(.433*b^.333)*b^.333*exp(.25*b^.333) - a(-.5*b^.666*cos(.433*b^.333)*exp(.25*b^.333)-.866*b^.666*sin(.433*b^.333)*exp(.25*b^.333)))*y + (.866*cos(.433*b^.333)*b^.333*exp(.25*b^.333)+.5*b^.333*sin(.433*b^.333)*exp(.25*b^.333)-a(.5*b^.666*sin(.433*b^.333)*exp(.25*b^.333)-.866*b^.666*cos(.433*b^.333)*exp(.25*b^.333) ))*z == 0;
sol = solve([eqn1, eqn2, eqn3, eqn4 ], [p, x, y, z]);
pSol = sol.p
xSol = sol.x
ySol = sol.y
zSol = sol.z
, but it keeps showing errors, would you please suggest any solution to this issue.
Let's look at this :
eqn3 = (b^(1/3)*exp(0.5*b^(1/3))+a*b^(2/3)*exp(0.5*b^(1/3)))*x + (-.866*b^.333*sin(.433*b^.333)*exp(-.25*b^.333)-.5*cos(.433*b^.333)*b^.333*exp(-.25*b^.333) + a(-.5*b^.666*cos(.433*b^.333)*exp(-.25*b^.333)+.866*b^.666*sin(.433*b^.333)*exp(-.25*b^.333)))*y + (.866*cos(.433*b^.333)*b^.333*exp(-.25*b^.333)-.5*b^.333*sin(.433*b^.333)*exp(-.25*b^.333)+a(-.5*b^.666*sin(.433*b^.333)*exp(-.25*b^.333)-.866*b^.666*cos(.433*b^.333)*exp(-.25*b^.333)))*z == 0;
I pick one of the wrong parts out, 4 in total.
a(-.5*b^.666
You see, if a(index) exists, index must be a positive integer. However, in your code the value of -.5*b^.666*cos(.433*b^.333)*exp(-.25*b^.333)+.866*b^.666*sin(.433*b^.333)*exp(-.25*b^.333) might not be a positive integer. And A may not even be a matrix, let alone an index.
Plus, before you post your question, you' d better check the basic grammar issues to save each other's time.
Related
I am solving a fourth order equation in matlab using the solve function.My script looks like this:
syms m M I L Bp Bc g x
m = 0.127
M = 1.206
I = 0.001
L = 0.178
Bc = 5.4
Bp = 0.002
g = 9.8
eqn = ((m + M)*(I + m*L^2) - m^2*L^2)*x^4 + ((m + M)*Bp + (I + m*L^2)*Bc)*x^3 + ((m + M)*m*g*L + Bc*Bp)*x^2 + m*g*L*Bc*x == 0
S = solve(eqn, x)
In the answer, I should get 4 roots, but instead I get such strange expressions:
S =
0
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 1)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 2)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 3)
The first root, which is 0, is displayed clearly. Is it possible to make the other three roots appear as numbers as well? I looked for something about this in the documentation for the solve function, but did not find it.
So I'm trying to figure out how to solve a given equation y'=-y+2te^(-t+2) for t in [0,10], step of 0.01 and y(0)=0.
I am supposed to solve it using the Lobatto IIIA method following a Butcher tableau:
Coefficients table
So far, this is what I got:
function lob = Lobatto_solver()
h = 0.01;
t = 0:h:10;
y = zeros(size(t));
y(1) = 0;
f = #(t,y) -y + (2*t*(exp(-t+2)));
% Lobatto IIIA Method
for i=1:numel(y)-1
f1 = f(t(i), y(i));
f2 = f(t(i)+(1/2)*h, y(i) + (5/24)*h*f1 + (1/3)*h*f2 + (-1/24)*h*f3);
f3 = f(t(i)+h, y(i) + (1/6)*h*f1 + (2/3)*h*f2 + (1/6)*h*f3);
y(x) = y(i) + h*((-1/2)*f1 + (2)*f2 + (-1/2)*f3);
end
end
It obviously makes no sense from the point when I equal f2 to itself, when the variable is still undefined.
Any help would be much appreciated :)
Cheers
You will need a predictor-corrector loop, in the simple case the corrector uses the slope-equations as basis of a fixed-point iteration. In the code below I use also the value of an explicit Euler step, in principle you could initialize all slopes with f1.
function lob = Lobatto_solver()
h = 0.01;
t = 0:h:10;
y = zeros(size(t));
y(1) = 0;
f = #(t,y) -y + (2*t*(exp(-t+2)));
% Lobatto IIIA Method
for i=1:numel(y)-1
f1 = f(t(i), y(i));
f3 = f(t(i)+h, y(i)+h*f1)
f2 = (f1+f3)/2;
for k=1:3
f2 = f(t(i)+(1/2)*h, y(i) + (5/24)*h*f1 + (1/3)*h*f2 + (-1/24)*h*f3);
f3 = f(t(i)+h, y(i) + (1/6)*h*f1 + (2/3)*h*f2 + (1/6)*h*f3);
end;
y(i+1) = y(i) + h*((-1/2)*f1 + (2)*f2 + (-1/2)*f3);
end
plot(t,y,t,0.05+t.^2.*exp(-t+2))
end
The graph shows that the result (blue) is qualitatively correct, the exact solution curve (green) is shifted so that two distinct curves can be seen.
Given a system of differential equations such as:
dy/dt = f(t)
dx/dt = g(t)
A solution can be found using dsolve, such as:
dsolve(diff(y) == f(t), diff(x) == g(t), y(0) == 1, x(0) == 1);
But what about a system where all the variables depend on each other:
dy/dt = f(y,z)
dx/dt = g(x,y)
dz/dt = h(z,x)
When approached in the same way, with initial conditions, for a system which does have a solution, I cannot find a solution.
I know the system I have tried can produce solutions as I have used a stochastic/deterministic simulator - think there's probably some strange syntax to use.
I'm specifically looking for the solution where the derivatives are all zero, if that helps.
EDIT:
Here is an example:
PX/dt = (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PZ^n))))/kd_mRNA)-kd_prot*PX;
PY/dt = (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PX^n))))/kd_mRNA)-kd_prot*PY;
PZ/dt = (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PY^n))))/kd_mRNA)-kd_prot*PZ;
with the coefficients:
eff = 20;
KM = 40;
tau_mRNA=2.0;
tau_prot=10;
ps_a=0.5;
ps_0=5.0E-4;
t_ave = tau_mRNA/log(2);
k_tl=eff/t_ave;
a_tr=(ps_a-ps_0)*60;
a0_tr=ps_0*60;
kd_mRNA = log(2)/tau_mRNA;
kd_prot = log(2)/tau_prot;
beta = tau_mRNA/tau_prot;
alpha = a_tr*eff*tau_prot/(log(2)*KM);
alpha0 = a0_tr*eff*tau_prot/(log(2)*KM);
n=2;
And the initial conditions:
PX0 = 20;
PY0 = 0;
PZ0 = 0;
This produces a response:
This clearly has a steady state solution (all derivatives 0).
In MATLAB I have tried:
%%
syms PX(t) PY(t) PZ(t);
z = dsolve(diff(PX) == (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PZ^n))))/kd_mRNA)-kd_prot*PX, diff(PY) == (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PX^n))))/kd_mRNA)-kd_prot*PY, diff(PZ)==(k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PY^n))))/kd_mRNA)-kd_prot*PZ,PX(0)==20)
and:
%%
eq1 = (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PZ^n))))/kd_mRNA)-kd_prot*PX;
eq2 = (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PX^n))))/kd_mRNA)-kd_prot*PY;
eq3 = (k_tl*(a0_tr + ((a_tr*KM^n)/((KM^n) + (PY^n))))/kd_mRNA)-kd_prot*PZ;
dsolve(diff(PX)==eq1,PX(0)==20,diff(PY)==eq2,PY(0)==0,diff(PZ)==eq3,PZ(0)==0)
Both produce no errors but return an empty sym.
Your numeric solution appears to have an oscillatory component. The "steady state" may be a zero amplitude limit cycle, which is a non-trivial solution. You definitely shouldn't expect a system like this to have an easy-to-find analytic solution. The cyclic relations between your three variables also doesn't help. For what it's worth, Mathematica 10's DSolve also is unable to find a solution.
Though it won't get you to a solution, the way you're using symbolic math is less than optimal. When you use something like log(2) in a symbolic math equation, 2 should be converted to a symbolic value first. For example, sym(log(2)) yields the approximation 6243314768165359/9007199254740992, whereas log(sym(2)) returns the exact log(2). This latter form is much more likely to lead to solutions if they exist. Here's a modified version of your code, which unfortunately still returns "Warning: Explicit solution could not be found":
eff = 20;
KM = 40;
tau_mRNA=2;
tau_prot=10;
ps_a=1/sym(2);
ps_0=5/sym(10000);
ln2 = log(sym(2));
t_ave = tau_mRNA/ln2;
k_tl=eff/t_ave;
a_tr=(ps_a-ps_0)*60;
a0_tr=ps_0*60;
kd_mRNA = ln2/tau_mRNA;
kd_prot = ln2/tau_prot;
beta = tau_mRNA/tau_prot;
alpha = a_tr*eff*tau_prot/(ln2*KM);
alpha0 = a0_tr*eff*tau_prot/(ln2*KM);
n=2;
PX0 = 20;
PY0 = 0;
PZ0 = 0;
syms PX(t) PY(t) PZ(t);
eq1 = (k_tl*(a0_tr + a_tr*KM^n/(KM^n + PZ^n))/kd_mRNA)-kd_prot*PX;
eq2 = (k_tl*(a0_tr + a_tr*KM^n/(KM^n + PX^n))/kd_mRNA)-kd_prot*PY;
eq3 = (k_tl*(a0_tr + a_tr*KM^n/(KM^n + PY^n))/kd_mRNA)-kd_prot*PZ;
s = dsolve(diff(PX,t)==eq1,diff(PY,t)==eq2,diff(PZ,t)==eq3,PX(0)==20,PY(0)==0,PZ(0)==0)
I have written the following code in order to try to plot the following integral against values of r, but MATLAB gives me an error -- is fun too long? am I going wrong somewhere else?
figure(1); %f_1
r = 0:0.001:50;
q = zeros(1, size(r));
for m = 1:numel(r)
fun = #(t) ((-3*(r(m).^3)*sin(3*t) + 2*(r(m)^2)*cos(2*t) + 7*r(m)*cos(t) -2*sin(t))*(-6*(r(m)^3)*sin(3*t) + 2*(r(m)^3)*cos(3*t) - 3*(r(m)^4)*cos(4*t) - 2*(r(m)^3)*sin(3*t) + 2*(r(m)^2)*cos(2*t) + 7*(r(m)^2)*sin(2*t))) - ((3*(r(m).^3).*cos(3*t) + 2*(r(m).^2).*sin(2*t) + 7*r(m).*sin(t) - 2*r(m).*cos(t))*(-6*(r(m).^3).*cos(3*t) + 2*(r(m).^3).*sin(3*t) + 3*(r(m).^4).*sin(4*t) - 2* (r(m).^3).*cos(3*t) - 2*(r(m).^2).*sin(2*t) + 7*(r(m).^2).*cos(2*t)))./((-3*(r(m).^3).*sin(3.*t) + 2*(r(m).^2)*cos(2.*t) + 7*r(m).*cos(t) - 2*sin(t)).^2 + (3*(r(m).^3).*cos(3*t) + 2*(r(m).^2).*sin(2*t) + 7*r(m).*sin(t) - 2*r(m).*cos(t)).^2);
q(m) = quad(fun, 0, 2*pi);
end
The error I get is
Error using * Inner matrix dimensions must agree.
Error in #(t)......
Error in quad (line 76) y = f(x, varargin{:});
I'll show you a way you may proceed, based on a retained r and fun (I did not pick all the terms of the native function):
r = 0:0.1:50;
q = zeros(size(r));
for ii = 1:numel(r)
fun = #(t) (-3.*(r(ii).^3).*sin(3.*t) + 2.*(r(ii).^2).*cos(2.*t) + 7.*r(ii).*cos(t) -2.*sin(t));
q(ii) = quad(fun,0,2.*pi);
end
Since your r is quite huge (50001 elements if I remember right), I'd go for parfor insted of simple for loop, too.
EDIT
As alternative solution, you could achieve the same results without any anonymous function, by following this way:
r = 0:.01:50;
fun1 = zeros(size(r));
t = 0:.001:(2*pi);
for ii = 1:numel(r)
fun1(ii) = trapz(t,(-3.*(r(ii).^3).*sin(3.*t) + 2.*(r(ii).^2).*cos(2.*t) + 7.*r(ii).*cos(t) -2.*sin(t)));
end
smys x y A;
L = A*(-2*x -y -3) + 2*x^2 - 2*x*y + 3*y^2;
Lx = diff(L, x);
Ly = diff(L, y);
LA = diff(L, A);
a = solve(Lx, Ly, LA);
a = [a.x, a.y, a.A]
a =
[ -7/6, -2/3, -5/3]
Now this piece looks ok and doing
subs(Lx, {x,y,A}, a)
Gives the expected 0, however, if I were to take the values and manually calculate the Lx/Ly/LA I get anything but 0. Why is it so?
It is correct. If I didn't break the calculations it looks like
Lx = -2*A+4*x-2*y
and thus
Lx = -2*(-5/3)+4*(-7/6)-2*(-2/3) = 10/3 - 14/3 + 4/3 = 0
The same holds for Ly and LA.