I get this error The number of derivatives returned by func() (1) must equal the length of the initial conditions vector (3)
Can some kindly suggest how resolve ?
Thanks
# Define the ODE system
diffeq <- function(t, y, params) {
with(as.list(c(y, params)), {
dCc <- -k12 * c_central + k21 * c_peripheral - k10 * c_central
dCp <- k12 * c_central - k21 * c_peripheral
dy <- -alpha * exp(-beta * y) * y + (c_central - gamma) * (c_central - gamma)
dw <- g * (t - delay) - kw * w
dg <- ks * kw * w0 * (w/W0)^(-param) * exp(-u * y) - (ks + v * y) * g
return(list(dCc, dCp,dy, dw, dg))
})
}
# Set the initial conditions and parameter values
y0 <- 0
g0 <- 0
w0 <- 100
c_central0 <- 500/5
c_peripheral0 <- 0
params <- c(alpha = 1, beta = 2, gamma = 3, kw = 0.1, k12 = 1.2,
k21 = 1.5, k10 = 0.04,ks = 0.01, delay = 0.5,
W0 = 100, param = 2, u = 1, v = 2,c_central = c_central0, c_peripheral = c_peripheral0)
# Solve the differential equations using an ODE solver
result <- ode(y = c(y = y0, w = w0, g = g0), times = seq(0, 40, by = 0.1), func = diffeq, parms = params)
# Plot the solution
plot(result[, "time"], result[, "y"], type = "l", xlab = "Time (t)", ylab = "y(t)")
lines(result[, "time"], result[, "w"], col = "red")
lines(result[, "time"], result[, "g"], col = "blue")```
Related
I want to get extract intermediate parameter values from below ODE function. Can someone figure out how to extract those values from the ode solver.
I want to get values of "a, b,s,& w" apart from the main outputs of the ode solver. I tried to modify return option in the function, but that doesn't work.
Be kind to explain by providing sample codes as I am bit new to python.
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt
# parameters
S = 0.0001
M = 30.03
K = 113.6561
Vr = 58
R = 8.3145
T = 298.15
Q = 0.000133
Vp = 0.000022
Mr = 36
Pvap = 1400
wf = 0.001
tr = 1200
mass = 40000
# define t
time = 14400
t = np.arange(0, time + 1, 1)
# define initial state
Cv0 = (mass / Vp) * wf # Cv(0)
Cr0 = (mass / Vp) * (1 - wf)
Cair0 = 0 # Cair(0)
# define function and solve ode
def model(x, t):
C = x[0] # C is Cair(t)
c = x[1] # c is Cv(t)
a = Q + (K * S / Vr)
b = (K * S * M) / (Vr * R * T)
s = (K * S * M) / (Vp * R * T)
w = (1 - wf) * 1000
Peq = (c * Pvap) / (c + w * c * M / Mr)
Pair = (C * R * T) / M
dcdt = -s * (Peq - Pair)
if t <= tr:
dCdt = -a * C + b * Peq
else:
dCdt = -a * C
return [dCdt, dcdt]
x = odeint(model, [Cair0, Cv0], t)
C = x[:, 0]
c = x[:, 1]
For the purpose of generalization, I hope Matlab can automatically compute the 1st & 2nd derivatives of the associated function f(x). (in case I change f(x) = sin(6x) to f(x) = sin(8x))
I know there exists built-in commands called diff() and syms, but I cannot figure out how to deal with them with the index i in the for-loop. This is the key problem I am struggling with.
How do I make changes to the following set of codes? I am using MATLAB R2019b.
n = 10;
h = (2.0 * pi) / (n - 1);
for i = 1 : n
x(i) = 0.0 + (i - 1) * h;
f(i) = sin(6 * x(i));
dfe(i) = 6 * cos(6 * x(i)); % first derivative
ddfe(i) = -36 * sin(6 * x(i)); % second derivative
end
You can simply use subs and double to do that. For your case:
% x is given here
n = 10;
h = (2.0 * pi) / (n - 1);
syms 'y';
g = sin(6 * y);
for i = 1 : n
x(i) = 0.0 + (i - 1) * h;
f(i) = double(subs(g,y,x(i)));
dfe(i) = double(subs(diff(g),y,x(i))); % first derivative
ddfe(i) = double(subs(diff(g,2),y,x(i))); % second derivative
end
By #Daivd comment, you can vectorize the loop as well:
% x is given here
n = 10;
h = (2.0 * pi) / (n - 1);
syms 'y';
g = sin(6 * y);
x = 0.0 + ((1:n) - 1) * h;
f = double(subs(g,y,x));
dfe = double(subs(diff(g),y,x)); % first derivative
ddfe = double(subs(diff(g,2),y,x)); % second derivative
i recently made a simple neural network, and i found that the error is a matrix when its supposed to be a single number, which causes the output to change from a simple 4*1 matrix to a 4*20 matrix, can someone pleas help me figure out how i have to redefine the error to change the l5_error into a 4*1 matrix while preserving the accuracy of the network
<import numpy as np
def nonlin(x, deriv=False):
if (deriv == True):
return (x * (1 - x))
return 1 / (1 + np.exp(-x))
X = np.array([[1,1,0],
[0,1,1],
[0,0,1],
[1, 0, 0]])
y = np.array([[0],
[1],
[0],
[1]])
np.random.seed(1)
syn0 = 2 * np.random.random((len(X[1]), 100)) - 1
syn1 = 2 * np.random.random((100, 80)) - 1
syn2 = 2 * np.random.random((80, 60)) - 1
syn3 = 2 * np.random.random((60, 40)) - 1
syn4 = 2 * np.random.random((40, 20)) - 1
syn5 = 2 * np.random.random((20, 1)) - 1
#the layers are only defined here so i can see the dimensions of the error
l0 = X
l1 = nonlin(np.dot(l0, syn0))
l2 = nonlin(np.dot(l1, syn1))
l3 = nonlin(np.dot(l2, syn2))
l4 = nonlin(np.dot(l3, syn3))
l5 = nonlin(np.dot(l4, syn4))
l5_error = y - l5
print('beggining', l5_error, 'ending')
for i in range(1000):
l0 = X
l1 = nonlin(np.dot(l0, syn0))
l2 = nonlin(np.dot(l1, syn1))
l3 = nonlin(np.dot(l2, syn2))
l4 = nonlin(np.dot(l3, syn3))
l5 = nonlin(np.dot(l4, syn4))
l5_error = y - l5
if (i % 10) == 0:
print( "Error: " + str(np.mean(np.abs(l5_error))))
print(l5_error, nonlin(l5, deriv=True))
l5_delta = l5_error * nonlin(l5, deriv=True)
l4_error = l5_delta.dot(syn4.T)
l4_delta = l4_error * nonlin(l4, deriv=True)
l3_error = l4_delta.dot(syn3.T)
l3_delta = l3_error * nonlin(l3, deriv=True)
l2_error = l3_delta.dot(syn2.T)
l2_delta = l2_error * nonlin(l2, deriv=True)
l1_error = l2_delta.dot(syn1.T)
l1_delta = l1_error * nonlin(l1, deriv=True)
#syn5 += l5.T.dot(l6_delta)
syn4 += l4.T.dot(l5_delta)
syn3 += l3.T.dot(l4_delta)
syn2 += l2.T.dot(l3_delta)
syn1 += l1.T.dot(l2_delta)
syn0 += l0.T.dot(l1_delta)
print ("Output after training")
print (l5)
You've got a mistake in the definition of l5_error, it should be
l5_error = np.linalg.norm(y-l5)
This same issue occurs some of the other error variables; simply add a call np.linalg.norm to each.
I have an equation which goes like this:
Here, I_L(lambdap) is the modified bessel function. This and product with exponential function can be written in matlab as besseli(L,lambdap,1). "i" stands for square root of -1. I want to solve:
1+pt+it=0
where I have to vary 'k' and find values of 'w'. I had posted similar problem at mathematica stack exchange, but I couldn't solve the problem fully, though i have got a clue (please go through the comments at mathematica stack exchange site). I could not convert my equation to the code that has been posted in clue. Any help in this regards will be highly appreciated.
Thanks in advance...
I never attempted this before, but... is this returning a suitable result?
syms w k;
fun = 1 + pt(w,k) + it(w,k);
sol = vpasolve(fun == 0,w,k);
disp(sol.w);
disp(sol.k);
function res = pt(w,k)
eps_l0 = w / (1.22 * k);
lam_k = 0.25 * k^2;
res = sym('res',[5 1]);
res_off = 1;
for L = -2:2
gam = besseli(L,lam_k) * exp(-lam_k);
eps_z = (w - L) / (1.22 * k);
zeta = 1i * sqrt(pi()) * exp(-eps_z^2) * (1 + erfc(1i * eps_z));
res(res_off,:) = ((25000 * gam) / k^2) * (1 + (eps_l0 * zeta));
res_off = res_off + 1;
end
res = sum(res);
end
function res = it(w,k)
eps_l0 = (w - (0.86 * k)) / (3.46 * k);
lam_k = 0.03 * k^2;
res = sym('res',[5 1]);
res_off = 1;
for L = -2:2
gam = besseli(L,lam_k) * exp(-lam_k);
eps_z = (w - (8 * L) - (0.86 * k)) / (3.46 * k);
zeta = 1i * sqrt(pi()) * exp(-eps_z^2) * (1 + erfc(1i * eps_z));
res(res_off,:) = ((2000000 * gam) / k^2) * (1 + (eps_l0 * zeta));
res_off = res_off + 1;
end
res = sum(res);
end
EDIT
For numeric k and symbolic w:
syms w;
for k = -3:3
fun = 1 + pt(w,k) + it(w,k);
sol = vpasolve(fun == 0,w);
disp(sol.w);
end
I am right now stuck on a problem in matlab. What I have done is that I have an equation that is passed on into another function which works by the bisection-method.
But I have a multiplier that I am trying to implement which somehow leads to the function crashing.
Before I introduced the multiplier it all worked, I tried breaking it down by entering the multiplier value manually and it didn't work
P_{1} = 0.6;
P_{2} = 0.2;
P_{3} = 0.2;
a_1 = 4/3;
a_2 = -7/3;
b_1 = -1/3;
b_2 = 4/3;
persistent multiplier
multiplier = exp(a_1 * 44 + a_2 * 14 + 0);
eqn = #(x) ((a_1 * x + b_1)^a_1) * ((a_2 * x + b_2)^a_2) * x ...
-(P_{1}^a_1) * (P_{2}^a_2) * P_{3} * multiplier;
Q_{3} = Bisectionmethod(a_1, a_2, b_1, b_2, eqn);
Here is the calculating part of the bisection method.
x_lower = max(0, -b_1 / a_1);
x_upper = -b_2 / a_2;
x_mid = (x_lower + x_upper)/2;
Conditional statement encompassing the method of bisection
while abs(eqn(x_mid)) > 10^(-10)
if (eqn(x_mid) * eqn(x_upper)) < 0
x_lower = x_mid;
else
x_upper = x_mid;
end
x_mid = (x_lower + x_upper)/2;
end
Based on the information you provided this is what I came up with
function Q = Stackoverflow
persistent multiplier
P{1} = 0.6;
P{2} = 0.2;
P{3} = 0.2;
a1 = 4/3;
a2 = -7/3;
b1 = -1/3;
b2 = 4/3;
multiplier = exp(a1 * 44 + a2 * 14 + 0);
eqn = #(x) ((a1 .* x + b1).^a1) .* ((a2 .* x + b2).^a2) .* x -(P{1}.^a1) .* (P{2}.^a2) .* P{3} .* multiplier;
Q{3} = Bisectionmethod(eqn, max([0, -b1/a1]), -b2/a2, 1E-10);
end
function XOut = Bisectionmethod(f, xL, xH, EPS)
if sign(f(xL)) == sign(f(xH))
XOut = [];
error('Cannot bisect interval because can''t ensure the function crosses 0.')
end
x = [xL, xH];
while abs(diff(x)) > EPS
x(sign(f(mean(x))) == sign(f(x))) = mean(x);
end
XOut = mean(x);
end