DEtools
regular_parts
Find regular parts of a linear ode
Calling Sequence
Parameters
Description
Examples
regular_parts(L, y, t, [x=x0])
L
-
linear homogeneous differential equation
y
unknown function to search for
t
name used as parametrization variable
x0
(optional) a rational, an algebraic number or infinity
The regular_parts function computes the minimal generalized exponents of L at the point x0 and the corresponding regular parts. These are operators L_e which result from L by replacing y(x) by exp(int(e, x))*y(x). The Newton polygon of L_e at x_0 has a segment of slope 0 and 0 is a root of the indicial polynomial.
The equation L⁡y=0 must be homogeneous and linear in y and its derivatives, and its coefficients must be rational functions in the variable x.
x0 must be a rational or an algebraic number or the symbol infinity. If x0 is not passed as argument, x0 = 0 is assumed.
The output is a set of solutions which are of the form exp(int(e, x))*y where e is a minimal generalized exponent and y is given as DESol object.
The command with(DEtools,regular_parts) allows the use of the abbreviated form of this command.
with⁡DEtools:
ode≔y⁡x−2⁢x2⁢diff⁡y⁡x,x−6⁢x3⁢diff⁡y⁡x,x,x−2⁢x4⁢diff⁡y⁡x,x,x,x+x6⁢diff⁡y⁡x,`$`⁡x,4
ode≔y⁡x−2⁢x2⁢ⅆⅆxy⁡x−6⁢x3⁢ⅆ2ⅆx2y⁡x−2⁢x4⁢ⅆ3ⅆx3y⁡x+x6⁢ⅆ4ⅆx4y⁡x
Then 0 is a singular point of this equation. Newton polygon is:
newton_polygon⁡ode,y⁡x,u
13,1−2⁢u,1,−2+u
There are slopes > 0 so 0 is an irregular singular point.
r≔regular_parts⁡ode,y⁡x,t
r≔x⁡t=t32,y⁡t=ⅇ−3t⁢t⁢DESol⁡−2027⁢t4+259⁢t3−4627⁢t2+536⁢t−2081⁢t5⁢y⁡t+227⁢t6+2627⁢t5−23⁢t2−43⁢t4−t+227⁢t3⁢ⅆⅆty⁡t+481⁢t7−13⁢t6+16⁢t5−13⁢t3−29⁢t4⁢ⅆ2ⅆt2y⁡t+−281⁢t8+127⁢t7−127⁢t5⁢ⅆ3ⅆt3y⁡t+t9⁢ⅆ4ⅆt4y⁡t324,y⁡t,x⁡t=t,y⁡t=ⅇ−2t⁢t9⁢DESol⁡3024⁢t3+1230⁢t2+141⁢t⁢y⁡t+2016⁢t4+802⁢t3+120⁢t2+8⁢t⁢ⅆⅆty⁡t+432⁢t5+132⁢t4+12⁢t3⁢ⅆ2ⅆt2y⁡t+36⁢t6+6⁢t5⁢ⅆ3ⅆt3y⁡t+t7⁢ⅆ4ⅆt4y⁡t,y⁡t
yields two transformed differential equations:
ode1≔op⁡1,select⁡type,rhs⁡r12,DESol1
ode1≔−2027⁢t4+259⁢t3−4627⁢t2+536⁢t−2081⁢t5⁢y⁡t+227⁢t6+2627⁢t5−23⁢t2−43⁢t4−t+227⁢t3⁢ⅆⅆty⁡t+481⁢t7−13⁢t6+16⁢t5−13⁢t3−29⁢t4⁢ⅆ2ⅆt2y⁡t+−281⁢t8+127⁢t7−127⁢t5⁢ⅆ3ⅆt3y⁡t+t9⁢ⅆ4ⅆt4y⁡t324
ode2≔op⁡1,select⁡type,rhs⁡r22,DESol1
ode2≔3024⁢t3+1230⁢t2+141⁢t⁢y⁡t+2016⁢t4+802⁢t3+120⁢t2+8⁢t⁢ⅆⅆty⁡t+432⁢t5+132⁢t4+12⁢t3⁢ⅆ2ⅆt2y⁡t+36⁢t6+6⁢t5⁢ⅆ3ⅆt3y⁡t+t7⁢ⅆ4ⅆt4y⁡t
These operators have a Newton polygon with slope 0:
newton_polygon⁡ode1,y⁡t,u
0,−u,1,−u2−9⁢u−27,3,−12+u
newton_polygon⁡ode2,y⁡t,u
0,u,1,u3+6⁢u2+12⁢u+8
This can help to find closed-form solutions:
ode≔1x12−15x9+38x6−6x3⁢y⁡x+3x8−18x5+6x2⁢diff⁡y⁡x,x+3x4−3x⁢diff⁡diff⁡y⁡x,x,x+diff⁡diff⁡diff⁡y⁡x,x,x,x
ode≔1x12−15x9+38x6−6x3⁢y⁡x+3x8−18x5+6x2⁢ⅆⅆxy⁡x+3x4−3x⁢ⅆ2ⅆx2y⁡x+ⅆ3ⅆx3y⁡x
r≔x⁡t=t,y⁡t=ⅇ13⁢t3⁢t⁢DESol⁡t3⁢ⅆ3ⅆt3y⁡t,y⁡t
Since the general solution of the regular part is a+b*x+c*x^2 for some constants a,b and c, we obtain the general solution of the original equation by taking into account the exponential transformation:
simplify⁡subs⁡y⁡x=exp⁡13⁢x3⁢x⁢a+b⁢x+c⁢x2,ode
∂3∂x3ⅇ13⁢x3⁢x⁢c⁢x2+b⁢x+a⁢x11+3⁢−x10+x7⁢∂2∂x2ⅇ13⁢x3⁢x⁢c⁢x2+b⁢x+a+3⁢2⁢x9−6⁢x6+x3⁢∂∂xⅇ13⁢x3⁢x⁢c⁢x2+b⁢x+a−6⁢x6−6⁢x3+12⁢x3−13⁢c⁢x2+b⁢x+a⁢ⅇ13⁢x3x11
See Also
DEtools/formal_sol
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