# Modeling and Simulation in Python

Chapter 9

Copyright 2017 Allen Downey

License: [Creative Commons Attribution 4.0 International](https://creativecommons.org/licenses/by/4.0)



```python
# 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.


```python
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.


```python
t = symbols('t')
```

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


```python
expr = t + 1
```

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


```python
type(expr)
```




    sympy.core.add.Add



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


```python
expr.subs(t, 2)
```

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


```python
f = Function('f')
```




    f



The type of `f` is `UndefinedFunction`


```python
type(f)
```




    sympy.core.function.UndefinedFunction



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


```python
f(t)
```

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


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


```python
type(dfdt)
```




    sympy.core.function.Derivative



We need a symbol for `alpha`


```python
alpha = symbols('alpha')
```

Now we can write the differential equation for proportional growth.


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

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


```python
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`


```python
C1, p_0 = symbols('C1 p_0')
```


```python
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:


```python
r, K = symbols('r K')
```

Now we can write the differential equation.


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

And solve it.


```python
solution_eq = dsolve(eq2)
```

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


```python
general = solution_eq.rhs
```

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


```python
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](https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem), but we still get a list, so we have to use the bracket operator, `[0]`, to select the first one.


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




    (list, 1)




```python
value_of_C1 = solutions[0]
```

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


```python
particular = general.subs(C1, value_of_C1)
```

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


```python
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`


```python
particular.subs(t, 0)
```

This solution is called the [logistic function](https://en.wikipedia.org/wiki/Population_growth#Logistic_equation).

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:


```python
A = (K - p_0) / p_0
```


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

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


```python
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) $


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


```python
alpha,beta = symbols('alpha beta')

eq1 = Eq(dfdt,(alpha * f(t)) + (beta * f(t)**2))
```


```python
solution_eqn = dsolve(eq1)
```


```python
general = solution_eqn.rhs
```


```python
at_0 = general.subs(t, 0)
```


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


```python
particular = general.subs(C1, value_of_C1)
```


```python
particular = simplify(particular)
```

**Exercise:**  Use [WolframAlpha](https://www.wolframalpha.com/) 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`.


```python
general = solution_eqn.rhs
```


```python
particular.subs(t,0)
```

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