Lienard ODEs
Description
Examples
The general form of the Lienard ODE is given by the following:
Lienard_ode := diff(y(x),x,x)+f(x)*diff(y(x),x)+y(x)=0;
Lienard_ode≔ⅆ2ⅆx2y⁡x+f⁡x⁢ⅆⅆxy⁡x+y⁡x=0
where f(x) is an arbitrary function of x. See Villari, "Periodic Solutions of Lienard's Equation".
All linear second order homogeneous ODEs can be transformed into first order ODEs of Riccati type. That can be done by giving the symmetry [0,y] to dsolve (all linear homogeneous ODEs have this symmetry) or just calling convert (see convert,ODEs).
with⁡DEtools,odeadvisor:
odeadvisor⁡Lienard_ode
_Lienard
Reduction to Riccati by giving the symmetry to dsolve
ans≔dsolve⁡Lienard_ode,HINT=0,y
ans≔y⁡x=ⅇ∫_b⁡_aⅆ_a+c__1whereⅆⅆ_a_b⁡_a=−_b⁡_a2−_b⁡_a⁢f⁡_a−1,_a=x,_b⁡_a=ⅆⅆxy⁡xy⁡x,x=_a,y⁡x=ⅇ∫_b⁡_aⅆ_a+c__1
The reduced ODE above is of Riccati type
reduced_ode≔op⁡2,2,1,1,ans
reduced_ode≔ⅆⅆ_a_b⁡_a=−_b⁡_a2−_b⁡_a⁢f⁡_a−1
odeadvisor⁡reduced_ode
_Riccati
Converting this ODE into a first order ODE of Riccati type
Riccati_ode_TR≔convert⁡Lienard_ode,Riccati
Riccati_ode_TR≔ⅆⅆx_a⁡x=_F1⁡x⁢_a⁡x2+−f⁡x⁢_F1⁡x−ⅆⅆx_F1⁡x⁢_a⁡x_F1⁡x+1_F1⁡x,y⁡x=ⅇ−∫_a⁡x⁢_F1⁡xⅆx⁢c__1
In the answer returned by convert, there are the Riccati ODE and the transformation of the variable used. Changes of variables in ODEs can be performed using ?PDEtools[dchange]. For example, using the transformation of variables above, we can recover the result returned by convert.
TR≔Riccati_ode_TR2
TR≔y⁡x=ⅇ−∫_a⁡x⁢_F1⁡xⅆx⁢c__1
with⁡PDEtools,dchange
dchange
collect⁡isolate⁡dchange⁡TR,Lienard_ode,_a⁡x,diff⁡_a⁡x,x,_a⁡x,normal
ⅆⅆx_a⁡x=_F1⁡x⁢_a⁡x2−f⁡x⁢_F1⁡x+ⅆⅆx_F1⁡x⁢_a⁡x_F1⁡x+1_F1⁡x
See Also
DEtools
odeadvisor
dsolve
quadrature
missing
reducible
linear_ODEs
exact_linear
exact_nonlinear
sym_Fx
linear_sym
Bessel
Painleve
Halm
Gegenbauer
Duffing
ellipsoidal
elliptic
erf
Emden
Jacobi
Hermite
Lagerstrom
Laguerre
Liouville
Lienard
Van_der_Pol
Titchmarsh
odeadvisor,types
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