# Modeling and Simulation in Python

Chapter 9



In [1]:

# import everything from SymPy.
from sympy import *

# Set up Jupyter notebook to display math.
init_printing()



The following displays SymPy expressions and provides the option of showing results in LaTeX format.



In [2]:

from sympy.printing import latex

def show(expr, show_latex=False):
"""Display a SymPy expression.

expr: SymPy expression
show_latex: boolean
"""
if show_latex:
print(latex(expr))
return expr



### Analysis with SymPy

Create a symbol for time.



In [3]:

t = symbols('t')
t




Out[3]:

$\displaystyle t$



If you combine symbols and numbers, you get symbolic expressions.



In [4]:

expr = t + 1
expr




Out[4]:

$\displaystyle t + 1$



The result is an Add object, which just represents the sum without trying to compute it.



In [5]:

type(expr)




Out[5]:



subs can be used to replace a symbol with a number, which allows the addition to proceed.



In [6]:

expr.subs(t, 2)




Out[6]:

$\displaystyle 3$



f is a special class of symbol that represents a function.



In [7]:

f = Function('f')
f




Out[7]:

f



The type of f is UndefinedFunction



In [8]:

type(f)




Out[8]:

sympy.core.function.UndefinedFunction



SymPy understands that f(t) means f evaluated at t, but it doesn't try to evaluate it yet.



In [9]:

f(t)




Out[9]:

$\displaystyle f{\left(t \right)}$



diff returns a Derivative object that represents the time derivative of f



In [10]:

dfdt = diff(f(t), t)
dfdt




Out[10]:

$\displaystyle \frac{d}{d t} f{\left(t \right)}$




In [11]:

type(dfdt)




Out[11]:

sympy.core.function.Derivative



We need a symbol for alpha



In [12]:

alpha = symbols('alpha')
alpha




Out[12]:

$\displaystyle \alpha$



Now we can write the differential equation for proportional growth.



In [13]:

eq1 = Eq(dfdt, alpha*f(t))
eq1




Out[13]:

$\displaystyle \frac{d}{d t} f{\left(t \right)} = \alpha f{\left(t \right)}$



And use dsolve to solve it. The result is the general solution.



In [14]:

solution_eq = dsolve(eq1)
solution_eq




Out[14]:

$\displaystyle f{\left(t \right)} = C_{1} e^{\alpha t}$



We can tell it's a general solution because it contains an unspecified constant, C1.

In this example, finding the particular solution is easy: we just replace C1 with p_0



In [15]:

C1, p_0 = symbols('C1 p_0')




In [16]:

particular = solution_eq.subs(C1, p_0)
particular




Out[16]:

$\displaystyle f{\left(t \right)} = p_{0} e^{\alpha t}$



In the next example, we have to work a little harder to find the particular solution.

### Solving the quadratic growth equation

We'll use the (r, K) parameterization, so we'll need two more symbols:



In [17]:

r, K = symbols('r K')



Now we can write the differential equation.



In [18]:

eq2 = Eq(diff(f(t), t), r * f(t) * (1 - f(t)/K))
eq2




Out[18]:

$\displaystyle \frac{d}{d t} f{\left(t \right)} = r \left(1 - \frac{f{\left(t \right)}}{K}\right) f{\left(t \right)}$



And solve it.



In [19]:

solution_eq = dsolve(eq2)
solution_eq




Out[19]:

$\displaystyle f{\left(t \right)} = \frac{K e^{C_{1} K + r t}}{e^{C_{1} K + r t} - 1}$



The result, solution_eq, contains rhs, which is the right-hand side of the solution.



In [20]:

general = solution_eq.rhs
general




Out[20]:

$\displaystyle \frac{K e^{C_{1} K + r t}}{e^{C_{1} K + r t} - 1}$



We can evaluate the right-hand side at $t=0$



In [21]:

at_0 = general.subs(t, 0)
at_0




Out[21]:

$\displaystyle \frac{K e^{C_{1} K}}{e^{C_{1} K} - 1}$



Now we want to find the value of C1 that makes f(0) = p_0.

So we'll create the equation at_0 = p_0 and solve for C1. Because this is just an algebraic identity, not a differential equation, we use solve, not dsolve.

The result from solve is a list of solutions. In this case, we have reason to expect only one solution, but we still get a list, so we have to use the bracket operator, [0], to select the first one.



In [22]:

solutions = solve(Eq(at_0, p_0), C1)
type(solutions), len(solutions)




Out[22]:

(list, 1)




In [23]:

value_of_C1 = solutions[0]
value_of_C1




Out[23]:

$\displaystyle \frac{\log{\left(- \frac{p_{0}}{K - p_{0}} \right)}}{K}$



Now in the general solution, we want to replace C1 with the value of C1 we just figured out.



In [24]:

particular = general.subs(C1, value_of_C1)
particular




Out[24]:

$\displaystyle - \frac{K p_{0} e^{r t}}{\left(K - p_{0}\right) \left(- \frac{p_{0} e^{r t}}{K - p_{0}} - 1\right)}$



The result is complicated, but SymPy provides a method that tries to simplify it.



In [25]:

particular = simplify(particular)
particular




Out[25]:

$\displaystyle \frac{K p_{0} e^{r t}}{K + p_{0} e^{r t} - p_{0}}$



Often simplicity is in the eye of the beholder, but that's about as simple as this expression gets.

Just to double-check, we can evaluate it at t=0 and confirm that we get p_0



In [26]:

particular.subs(t, 0)




Out[26]:

$\displaystyle p_{0}$



This solution is called the logistic function.

In some places you'll see it written in a different form:

$f(t) = \frac{K}{1 + A e^{-rt}}$

where $A = (K - p_0) / p_0$.

We can use SymPy to confirm that these two forms are equivalent. First we represent the alternative version of the logistic function:



In [27]:

A = (K - p_0) / p_0
A




Out[27]:

$\displaystyle \frac{K - p_{0}}{p_{0}}$




In [28]:

logistic = K / (1 + A * exp(-r*t))
logistic




Out[28]:

$\displaystyle \frac{K}{1 + \frac{\left(K - p_{0}\right) e^{- r t}}{p_{0}}}$



To see whether two expressions are equivalent, we can check whether their difference simplifies to 0.



In [29]:

simplify(particular - logistic)




Out[29]:

$\displaystyle 0$



This test only works one way: if SymPy says the difference reduces to 0, the expressions are definitely equivalent (and not just numerically close).

But if SymPy can't find a way to simplify the result to 0, that doesn't necessarily mean there isn't one. Testing whether two expressions are equivalent is a surprisingly hard problem; in fact, there is no algorithm that can solve it in general.

### Exercises

Exercise: Solve the quadratic growth equation using the alternative parameterization

$\frac{df(t)}{dt} = \alpha f(t) + \beta f^2(t)$



In [30]:

# Solution

alpha, beta = symbols('alpha beta')




In [31]:

# Solution

eq3 = Eq(diff(f(t), t), alpha*f(t) + beta*f(t)**2)
eq3




Out[31]:

$\displaystyle \frac{d}{d t} f{\left(t \right)} = \alpha f{\left(t \right)} + \beta f^{2}{\left(t \right)}$




In [32]:

# Solution

solution_eq = dsolve(eq3)
solution_eq




Out[32]:

$\displaystyle f{\left(t \right)} = \frac{\alpha e^{\alpha \left(C_{1} + t\right)}}{\beta \left(1 - e^{\alpha \left(C_{1} + t\right)}\right)}$




In [33]:

# Solution

general = solution_eq.rhs
general




Out[33]:

$\displaystyle \frac{\alpha e^{\alpha \left(C_{1} + t\right)}}{\beta \left(1 - e^{\alpha \left(C_{1} + t\right)}\right)}$




In [34]:

# Solution

at_0 = general.subs(t, 0)




In [35]:

# Solution

solutions = solve(Eq(at_0, p_0), C1)
value_of_C1 = solutions[0]
value_of_C1




Out[35]:

$\displaystyle \frac{\log{\left(\frac{\beta p_{0}}{\alpha + \beta p_{0}} \right)}}{\alpha}$




In [36]:

# Solution

particular = general.subs(C1, value_of_C1)
particular.simplify()




Out[36]:

$\displaystyle \frac{\alpha p_{0} e^{\alpha t}}{\alpha - \beta p_{0} e^{\alpha t} + \beta p_{0}}$



Exercise: Use WolframAlpha to solve the quadratic growth model, using either or both forms of parameterization:

df(t) / dt = alpha f(t) + beta f(t)^2



or

df(t) / dt = r f(t) (1 - f(t)/K)



Find the general solution and also the particular solution where f(0) = p_0.



In [ ]: