# Modeling and Simulation in Python

Chapter 9



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# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'

# 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.



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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.



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t = symbols('t')



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



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expr = t + 1



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



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type(expr)



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



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expr.subs(t, 2)



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



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f = Function('f')



The type of f is UndefinedFunction



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type(f)



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



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f(t)



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



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dfdt = diff(f(t), t)




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type(dfdt)



We need a symbol for alpha



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alpha = symbols('alpha')



Now we can write the differential equation for proportional growth.



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



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



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solution_eq = dsolve(eq1)



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



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C1, p_0 = symbols('C1 p_0')




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particular = solution_eq.subs(C1, p_0)



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:



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r, K = symbols('r K')



Now we can write the differential equation.



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



And solve it.



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solution_eq = dsolve(eq2)



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



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general = solution_eq.rhs



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



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at_0 = general.subs(t, 0)



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, , to select the first one.



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




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value_of_C1 = solutions



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



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particular = general.subs(C1, value_of_C1)



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



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particular = simplify(particular)



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



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particular.subs(t, 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:



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A = (K - p_0) / p_0




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



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



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simplify(particular - logistic)



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

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



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# Solution goes here




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# Solution goes here




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# Solution goes here




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# Solution goes here




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# Solution goes here




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# Solution goes here




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# Solution goes here



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.



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