import sympy
from sympy import (laplace_transform, inverse_laplace_transform,
laplace_correspondence, laplace_initial_conds, pi)
sympy.init_printing()
from IPython.display import display
print("sympy: ", sympy.__version__)sympy: 1.14.0In recent versions of sympy (1.13 and later), \(~\)sympy.laplace_transform can transform derivatives of unknown functions, \(~\)apply both translation theorems, \(~\)the scaling rule \(f(at)\), \(~\)and the rule for \(t^n f(t)\). \(~\)Two companion functions finish the job:
laplace_correspondence(expr, {f: F}) \(~\)replaces \(\mathcal{L}\{f(t)\}\) by \(F(s)\)
laplace_initial_conds(expr, t, {y: [y0, y1, ...]}) \(~\)substitutes the initial values \(y(0)\), \(y'(0)\), \(\dots\)
A few rules are still missing in sympy 1.14: \(~\)integrals \(\int_0^t\), \(~\)convolution integrals, \(~\)and products such as \(e^{ct}\, t^n f(t)\). \(~\)For these we use a small wrapper _laplace_transform_expansion.py, \(~\)whose function laplace_transform_() adds only those rules and otherwise calls sympy.laplace_transform
a, b = sympy.symbols("a, b", positive=True)
n, m = sympy.symbols("n, m", positive=True, integer=True)
t, tau, tau1, tau2 = sympy.symbols("t, tau, tau1, tau2", positive=True)
s = sympy.symbols("s")
y, f, g = sympy.symbols("y, f, g", cls=sympy.Function)
Y, F, G = sympy.symbols("Y, F, G", cls=sympy.Function)def Lt(expr):
return laplace_correspondence(laplace_transform_(expr, t, s), {y: Y, f: F, g: G})laplace_transform(1, t, s, noconds=True)\(\displaystyle \frac{1}{s}\)
laplace_transform(t, t, s, noconds=True)\(\displaystyle \frac{1}{s^{2}}\)
laplace_transform(sympy.exp(-3 *t), t, s, noconds=True)\(\displaystyle \frac{1}{s + 3}\)
laplace_transform(2 *sympy.Heaviside(t -3), t, s, noconds=True)\(\displaystyle \frac{2 e^{- 3 s}}{s}\)
laplace_transform(sympy.sin(2 *t)**2, t, s, noconds=True)\(\displaystyle \frac{8}{s \left(s^{2} + 16\right)}\)
inverse_laplace_transform(1 / s**3, s, t)\(\displaystyle \frac{t^{2}}{2}\)
inverse_laplace_transform((-2 *s +6) / (s**2 + 4), s, t)\(\displaystyle 3 \sin{\left(2 t \right)} - 2 \cos{\left(2 t \right)}\)
eq = y(t).diff(t, 2) +2 *y(t).diff(t) +10 *y(t)
eq\(\displaystyle 10 y{\left(t \right)} + 2 \frac{d}{d t} y{\left(t \right)} + \frac{d^{2}}{d t^{2}} y{\left(t \right)}\)
Leq = laplace_correspondence(laplace_transform(eq, t, s, noconds=True), {y: Y})
Leq\(\displaystyle s^{2} Y{\left(s \right)} + 2 s Y{\left(s \right)} - s y{\left(0 \right)} + 10 Y{\left(s \right)} - 2 y{\left(0 \right)} - \left. \frac{d}{d t} y{\left(t \right)} \right|_{\substack{ t=0 }}\)
Leq = laplace_initial_conds(Leq, t, {y: [1, 1]}) # y(0) = 1, y'(0) = 1
Leq\(\displaystyle s^{2} Y{\left(s \right)} + 2 s Y{\left(s \right)} - s + 10 Y{\left(s \right)} - 3\)
sol = sympy.solve(Leq, Y(s))[0]
sol\(\displaystyle \frac{s + 3}{s^{2} + 2 s + 10}\)
eq = sympy.integrate(g(tau1), (tau1, 0, t)) +sympy.integrate(sympy.integrate(f(tau2), (tau2, 0, tau1)), (tau1, 0, t))
eq\(\displaystyle \int\limits_{0}^{t} g{\left(\tau_{1} \right)}\, d\tau_{1} + \int\limits_{0}^{t}\int\limits_{0}^{\tau_{1}} f{\left(\tau_{2} \right)}\, d\tau_{2}\, d\tau_{1}\)
sympy.laplace_transform leaves these integrals unevaluated:laplace_transform(eq, t, s, noconds=True)\(\displaystyle \mathcal{L}_{t}\left[\int\limits_{0}^{t} g{\left(\tau_{1} \right)}\, d\tau_{1}\right]\left(s\right) + \mathcal{L}_{t}\left[\int\limits_{0}^{t}\int\limits_{0}^{\tau_{1}} f{\left(\tau_{2} \right)}\, d\tau_{2}\, d\tau_{1}\right]\left(s\right)\)
Lt(eq)\(\displaystyle \frac{G{\left(s \right)}}{s} + \frac{F{\left(s \right)}}{s^{2}}\)
eq = sympy.integrate(f(tau)*g(t -tau), (tau, 0, t)) +sympy.integrate(f(t -tau)*g(tau), (tau, 0, t))
eq\(\displaystyle \int\limits_{0}^{t} f{\left(\tau \right)} g{\left(t - \tau \right)}\, d\tau + \int\limits_{0}^{t} f{\left(t - \tau \right)} g{\left(\tau \right)}\, d\tau\)
Lt(eq)\(\displaystyle 2 F{\left(s \right)} G{\left(s \right)}\)
Lt(sympy.integrate(sympy.exp(tau) *sympy.sin(t -tau), (tau, 0, t))) # e^t * sin t\(\displaystyle - \frac{s}{2 \left(s^{2} + 1\right)} - \frac{1}{2 \left(s^{2} + 1\right)} + \frac{1}{2 \left(s - 1\right)}\)
eq = y(t) *sympy.exp(-a *t)
eq\(\displaystyle y{\left(t \right)} e^{- a t}\)
Lt(eq)\(\displaystyle Y{\left(a + s \right)}\)
eq = f(t -a) *sympy.Heaviside(t -a)
eq\(\displaystyle f{\left(- a + t \right)} \theta\left(- a + t\right)\)
Lt(eq)\(\displaystyle F{\left(s \right)} e^{- a s}\)
eq = g(t) *sympy.Heaviside(t -a)
eq\(\displaystyle g{\left(t \right)} \theta\left(- a + t\right)\)
Lt(eq) # alternative form: exp(-a s) L{g(t + a)}\(\displaystyle \mathcal{L}_{t}\left[g{\left(a + t \right)}\right]\left(s\right) e^{- a s}\)
eq = sympy.cos(t) *sympy.Heaviside(t -pi)
eq\(\displaystyle \cos{\left(t \right)} \theta\left(t - \pi\right)\)
Lt(eq)\(\displaystyle - \frac{s e^{- \pi s}}{s^{2} + 1}\)
eq = 3 *f(t) *t**m *t**n
eq\(\displaystyle 3 t^{m} t^{n} f{\left(t \right)}\)
Lt(eq)\(\displaystyle 3 \left(-1\right)^{m + n} \frac{d^{m + n}}{d s^{m + n}} F{\left(s \right)}\)
eq = t**n *t**m *f(t)*sympy.exp(-2*t)
eq\(\displaystyle t^{m} t^{n} f{\left(t \right)} e^{- 2 t}\)
Lt(eq)\(\displaystyle \left(-1\right)^{m + n} \left. \frac{d^{m + n}}{d \sigma^{m + n}} F{\left(\sigma \right)} \right|_{\substack{ \sigma=s + 2 }}\)
eq = t**n *sympy.diff(f(t), t, t)
eq\(\displaystyle t^{n} \frac{d^{2}}{d t^{2}} f{\left(t \right)}\)
Lt(eq)\(\displaystyle \left(-1\right)^{n} \frac{d^{n}}{d s^{n}} \left(s^{2} F{\left(s \right)} - s f{\left(0 \right)} - \left. \frac{d}{d t} f{\left(t \right)} \right|_{\substack{ t=0 }}\right)\)
eq = t *sympy.integrate(f(tau), (tau, 0, t))
eq\(\displaystyle t \int\limits_{0}^{t} f{\left(\tau \right)}\, d\tau\)
Lt(eq)\(\displaystyle - \frac{\frac{d}{d s} F{\left(s \right)}}{s} + \frac{F{\left(s \right)}}{s^{2}}\)
eq = t *f(t) *sympy.exp(-t)
eq\(\displaystyle t f{\left(t \right)} e^{- t}\)
Lt(eq)\(\displaystyle - \left. \frac{d}{d \sigma} F{\left(\sigma \right)} \right|_{\substack{ \sigma=s + 1 }}\)
sympy.laplace_transform itself (1.14) raises an error:try:
laplace_transform(eq, t, s)
except ValueError as err:
print("ValueError:", err)ValueError:
Can't calculate derivative wrt s + 1.eq = t *sympy.diff(f(t), t) *sympy.exp(-4*t)
eq\(\displaystyle t e^{- 4 t} \frac{d}{d t} f{\left(t \right)}\)
Lt(eq)\(\displaystyle - \left(s + 4\right) \left. \frac{d}{d \sigma} F{\left(\sigma \right)} \right|_{\substack{ \sigma=s + 4 }} - F{\left(s + 4 \right)}\)
eq = sympy.exp(-a*t) *sympy.diff(f(t), t, t)
eq\(\displaystyle e^{- a t} \frac{d^{2}}{d t^{2}} f{\left(t \right)}\)
Lt(eq)\(\displaystyle \left(a + s\right)^{2} F{\left(a + s \right)} - \left(a + s\right) f{\left(0 \right)} - \left. \frac{d}{d t} f{\left(t \right)} \right|_{\substack{ t=0 }}\)
eq = sympy.exp(-4*t) *sympy.integrate(f(tau), (tau, 0, t))
eq\(\displaystyle e^{- 4 t} \int\limits_{0}^{t} f{\left(\tau \right)}\, d\tau\)
Lt(eq)\(\displaystyle \frac{F{\left(s + 4 \right)}}{s + 4}\)
eq = f(3*a*t)
eq\(\displaystyle f{\left(3 a t \right)}\)
Lt(eq) # scaling: F(s/k)/k with k = 3a\(\displaystyle \frac{F{\left(\frac{s}{3 a} \right)}}{3 a}\)
\(~\)
A technique that can be used to solve certain ODE problems is to Laplace transform the ODE, which for many problems results in an algebraic equation that is easier to solve. The solution to the algebraic equation can then be transformed back to the original domain with an inverse Laplace transform, to obtain the solution to the original problem. For example, consider the following differential equation for a driven harmonic oscillator:
\[ \frac{d^2 y}{dt^2} + 2\frac{dy}{dt} +10 y = 2\sin 3t \]
\(~\)
t = sympy.symbols('t', positive=True)
s = sympy.symbols('s')
y, Y = sympy.symbols('y, Y', cls=sympy.Function)ode = y(t).diff(t, 2) +2 *y(t).diff(t) +10 *y(t) -2 *sympy.sin(3*t)
ode\(\displaystyle 10 y{\left(t \right)} - 2 \sin{\left(3 t \right)} + 2 \frac{d}{d t} y{\left(t \right)} + \frac{d^{2}}{d t^{2}} y{\left(t \right)}\)
Lode = laplace_correspondence(laplace_transform(ode, t, s, noconds=True), {y: Y})
Lode\(\displaystyle s^{2} Y{\left(s \right)} + 2 s Y{\left(s \right)} - s y{\left(0 \right)} + 10 Y{\left(s \right)} - 2 y{\left(0 \right)} - \left. \frac{d}{d t} y{\left(t \right)} \right|_{\substack{ t=0 }} - \frac{6}{s^{2} + 9}\)
laplace_initial_conds:Lode = laplace_initial_conds(Lode, t, {y: [1, 0]})
Lode\(\displaystyle s^{2} Y{\left(s \right)} + 2 s Y{\left(s \right)} - s + 10 Y{\left(s \right)} - 2 - \frac{6}{s^{2} + 9}\)
Ysol = sympy.solve(Lode, Y(s))
Ysol[0]\(\displaystyle \frac{s^{3} + 2 s^{2} + 9 s + 24}{s^{4} + 2 s^{3} + 19 s^{2} + 18 s + 90}\)
ysol = inverse_laplace_transform(Ysol[0], s, t)
ysol.expand().collect([sympy.exp(-t)])\(\displaystyle \left(\frac{43 \sin{\left(3 t \right)}}{111} + \frac{49 \cos{\left(3 t \right)}}{37}\right) e^{- t} + \frac{2 \sin{\left(3 t \right)}}{37} - \frac{12 \cos{\left(3 t \right)}}{37}\)
dsolve:sympy.dsolve(ode, y(t), ics={y(0): 1, y(t).diff(t).subs(t, 0): 0})\(\displaystyle y{\left(t \right)} = \left(\frac{43 \sin{\left(3 t \right)}}{111} + \frac{49 \cos{\left(3 t \right)}}{37}\right) e^{- t} + \frac{2 \sin{\left(3 t \right)}}{37} - \frac{12 \cos{\left(3 t \right)}}{37}\)
In some instances, the Laplace transform can be used to solve linear differential equations with variable monomial coefficients
\[ty'' + y' +ty =0, \;\;y(0)=1, \;y'(0)=0\]
\(~\)
s, t = sympy.symbols('s, t', positive=True)
y, Y = sympy.symbols('y, Y', cls=sympy.Function)
ode = t *y(t).diff(t, 2) +y(t).diff(t) +t *y(t)
ode\(\displaystyle t y{\left(t \right)} + t \frac{d^{2}}{d t^{2}} y{\left(t \right)} + \frac{d}{d t} y{\left(t \right)}\)
\(~\)
Lode = laplace_transform(ode, t, s, noconds=True)
Lode = laplace_initial_conds(laplace_correspondence(Lode, {y: Y}), t, {y: [1, 0]})
Lode = Lode.doit().expand().collect(Y(s).diff(s))
Lode\(\displaystyle - s Y{\left(s \right)} + \left(- s^{2} - 1\right) \frac{d}{d s} Y{\left(s \right)}\)
sol = sympy.dsolve(Lode, Y(s))
sol\(\displaystyle Y{\left(s \right)} = \frac{C_{1}}{\sqrt{s^{2} + 1}}\)
y = inverse_laplace_transform(sol.rhs, s, t)
y\(\displaystyle C_{1} J_{0}\left(t\right)\)
C1 = sympy.solve(sympy.Eq(y.subs(t, 0), 1))[0]
y.subs(sympy.Symbol('C1'), C1)\(\displaystyle J_{0}\left(t\right)\)