Laplace transform

This notebook is a short tutorial of Laplace transform using SymPy.

The main functions to use are laplace_transform and inverse_laplace_transform.



In [1]:

    
from sympy import *



In [2]:

    
init_session()









    



IPython console for SymPy 1.0 (Python 2.7.13-64-bit) (ground types: python)

These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z, t = symbols('x y z t')
>>> k, m, n = symbols('k m n', integer=True)
>>> f, g, h = symbols('f g h', cls=Function)
>>> init_printing()

Documentation can be found at http://docs.sympy.org/1.0/

Let us compute the Laplace transform from variables $t$ to $s$, then, we have the condition that $t>0$ (and real).



In [3]:

    
t = symbols("t", real=True, positive=True)
s = symbols("s")

To calculate the Laplace transform of the expression $t^4$, we enter



In [4]:

    
laplace_transform(t**4, t, s)









    Out[4]:





$$\left ( \frac{24}{s^{5}}, \quad -\infty, \quad 0 < \Re{s}\right )$$

This function returns (F, a, cond) where F is the Laplace transform of f, $\mathcal{R}(s)>a$ is the half-plane of convergence, and cond are auxiliary convergence conditions.

If we are not interested in the conditions for the convergence of this transform, we can use noconds=True



In [5]:

    
laplace_transform(t**4, t, s, noconds=True)









    Out[5]:





$$\frac{24}{s^{5}}$$



In [6]:

    
fun = 1/((s-2)*(s-1)**2)
fun









    Out[6]:





$$\frac{1}{\left(s - 2\right) \left(s - 1\right)^{2}}$$



In [7]:

    
inverse_laplace_transform(fun, s, t)









    Out[7]:





$$\left(- t + e^{t} - 1\right) e^{t}$$

Right now, Sympy does not support the tranformation of derivatives.

If we do



In [8]:

    
laplace_transform(f(t).diff(t), t, s, noconds=True)









    Out[8]:





$$\mathcal{L}_{t}\left[\frac{d}{d t} f{\left (t \right )}\right]\left(s\right)$$

we don't obtain, the expected



In [9]:

    
s*LaplaceTransform(f(t), t, s) - f(0)









    Out[9]:





$$s \mathcal{L}_{t}\left[f{\left (t \right )}\right]\left(s\right) - f{\left (0 \right )}$$

or,

$$\mathcal{L}\lbrace f^{(n)}(t)\rbrace = s^n F(s) - \sum_{k=1}^{n} s^{n - k} f^{(k - 1)}(0)\, ,$$

in general.

We can still, operate with the trasformation of a differential equation.

For example, let us consider the equation

$$\frac{d f(t)}{dt} = 3f(t) + e^{-t}\, ,$$

that has as Laplace transform

$$sF(s) - f(0) = 3F(s) + \frac{1}{s+1}\, .$$



In [10]:

    
eq = Eq(s*LaplaceTransform(f(t), t, s) - f(0),
        3*LaplaceTransform(f(t), t, s) + 1/(s +1))
eq









    Out[10]:





$$s \mathcal{L}_{t}\left[f{\left (t \right )}\right]\left(s\right) - f{\left (0 \right )} = 3 \mathcal{L}_{t}\left[f{\left (t \right )}\right]\left(s\right) + \frac{1}{s + 1}$$

We then solve for $F(s)$



In [11]:

    
sol = solve(eq, LaplaceTransform(f(t), t, s))
sol









    Out[11]:





$$\left [ \frac{s f{\left (0 \right )} + f{\left (0 \right )} + 1}{s^{2} - 2 s - 3}\right ]$$

and compute the inverse Laplace transform



In [12]:

    
inverse_laplace_transform(sol[0], s, t)









    Out[12]:





$$\frac{1}{4 e^{t}} \left(\left(4 f{\left (0 \right )} + 1\right) e^{4 t} - 1\right)$$

and we verify this using dsolve



In [13]:

    
factor(dsolve(f(t).diff(t) - 3*f(t) - exp(-t)))









    Out[13]:





$$f{\left (t \right )} = \frac{1}{4 e^{t}} \left(4 C_{1} e^{4 t} - 1\right)$$

that is equal if $4C_1 = 4f(0) + 1$.

It is common to use practial fraction decomposition when computing inverse Laplace transforms. We can do this using apart, as follows



In [14]:

    
frac = 1/(x**2*(x**2 + 1))
frac









    Out[14]:





$$\frac{1}{x^{2} \left(x^{2} + 1\right)}$$



In [15]:

    
apart(frac)









    Out[15]:





$$- \frac{1}{x^{2} + 1} + \frac{1}{x^{2}}$$

We can also compute the Laplace transform of Heaviside and Dirac's Delta "functions"



In [16]:

    
laplace_transform(Heaviside(t - 3), t, s, noconds=True)









    Out[16]:





$$\frac{1}{s} e^{- 3 s}$$



In [17]:

    
laplace_transform(DiracDelta(t - 2), t, s, noconds=True)









    Out[17]:





$$e^{- 2 s}$$



In [ ]:

The next cell change the format of the notebook.



In [18]:

    
from IPython.core.display import HTML
def css_styling():
    styles = open('./styles/custom_barba.css', 'r').read()
    return HTML(styles)
css_styling()









    Out[18]: