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regular_parts

  

Find regular parts of a linear ode

 

Calling Sequence

Parameters

Description

Examples

Calling Sequence

regular_parts(L, y, t, [x=x0])

Parameters

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

Description

• 

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.

Examples

> 

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

(1)

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

(2)

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

(3)

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

(4)
> 

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

(5)

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

(6)
> 

newton_polygon⁡ode2,y⁡t,u

0,u,1,u3+6⁢u2+12⁢u+8

(7)

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

(8)
> 

r≔regular_parts⁡ode,y⁡x,t

r≔x⁡t=t,y⁡t=ⅇ13⁢t3⁢t⁢DESol⁡t3⁢ⅆ3ⅆt3y⁡t,y⁡t

(9)

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⁢c⁢x2+b⁢x+a⁢ⅇ13⁢x3⁢x6−6⁢x3+12⁢x3−13x11

(10)

See Also

DEtools

DEtools/formal_sol