Abhinav Baid wrote:
>
> On 02/18/2015 10:57 AM, Waldek Hebisch wrote:
> >
> > Maybe routine for point 5.3.2 (page 93) in the thesis? That
> > is given nontrivial solution to
> >
> > fr + lf =3D 0
> >
> > find a right factor of f.
> >
> Um, I have a doubt here. I feel very silly right now, but I'm not able
> to finish this routine.
Doubts are OK -- just ask here.
> The problem I'm having is this:
> I'm able to find the basis of formal solutions at a regular point of f,=20
> but after that the thesis mentions computing the matrix of map r in this=20
> basis. I think I'm not quite sure about what this means.
> For concreteness, take the example mentioned in the thesis
> f = \partial^4 +
> \frac{6}{x}\partial^3 + \frac{2(x^2-1)}{x^4}\partial^2 -
> \frac{2(3x^2-1)}{x^5}\partial + \frac{1}{x^8}
> r= -x^5\partial^3-x^4\partial^2+2x^3\partial+x\partial
> Using these, I'm able to find the basis of formal solutions at x=1 as
> 1 4 1 3 1 2 1 23 1 4 2 3 2 1 5
> [- -- x + - x - - x + - x + --, - x - - x + x + - x - -,
> 24 6 4 6 24 6 3 3 6
<snip>
> but am not able to figure out what should I do to make operator r act on
> this basis.
> Just applying it to every element in the list and taking the
> coefficients doesn't work as the matrix mentioned in the thesis is 4x4.
> Could you please help me? What should I be do to the basis using r so as
> to get the 4x4 matrix?
The basis got mangled in the mail but it looks wrong. First, you
should have four solutions and it looked as what you posted had
only three. Second, convenient form of basis is such initial
condition is of form (0, ..., 1, 0, ..., 0), that is i-th
element of basis (counting from 0) has i-th derivative at x=1
equal to 1 and other derivatives of order < d (d = 4 in this example)
equal to 0.
I do not know how you produced the basis above, but for this
task convenient routine is 'ode' from UTSODE. You may look
at 'candidates' in ODERAT to see how it is used.
For experimantation 'seriesSolve' may be more convenient:
op := D(y(x), x, 4) + (6/x)*D(y(x), x, 3) + 2*(x^2 - 1)/x^4*D(y(x), x, 2) -
2*(3*x^2 - 1)/x^5*D(y(x), x) + 1/x^8*y(x)
(85)
8 (iv) 7 ,,, 6 4 ,, 5 3 ,
x y (x) + 6x y (x) + (2x - 2x )y (x) + (- 6x + 2x )y (x) + y(x)
-----------------------------------------------------------------------
8
x
Type: Expression(Integer)
b1 := seriesSolve(op, y, x = 1, [1, 0, 0, 0])
Compiling function %CS with type List(UnivariateTaylorSeries(
Expression(Integer),x,1)) -> UnivariateTaylorSeries(Expression(
Integer),x,1)
1 4 7 5 7 6 7
(86) 1 - -- (x - 1) + -- (x - 1) - -- (x - 1) + O((x - 1) )
24 60 30
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
xx := taylor(x, x = 1)
(20) 1 + (x - 1)
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
apr(t) == -xx^5*D(t, x, 3) - xx^4*D(t, x, 2) + (2*xx^3 + x)*D(t, x)
apr(b1)
(88)
3 2 13 3 37 4 171 5
(x - 1) - - (x - 1) + -- (x - 1) - -- (x - 1) + --- (x - 1)
2 6 12 40
+
1393 6 7
- ---- (x - 1) + O((x - 1) )
240
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
As you see first four coefficients (that is 0, 1, -3/2, 13/6)
give you the first column of the matrix. Similarly
b2 := seriesSolve(op, y, x = 1, [0, 1, 0, 0])
1 4 11 5 137 6 7
(89) (x - 1) + - (x - 1) - -- (x - 1) + --- (x - 1) + O((x - 1) )
6 40 360
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
(90) -> apr(b2)
(90)
1 2 5 3 23 4 159 5
3 + 3(x - 1) + - (x - 1) - - (x - 1) + -- (x - 1) - --- (x - 1)
2 3 8 40
+
1243 6 7
---- (x - 1) + O((x - 1) )
240
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
Now coefficient give you second column. Next:
b3 := seriesSolve(op, y, x = 1, [0, 0, 2, 0]);
apr(b3)
(92)
2 1 3 3 4 61 5
- 2 - 2(x - 1) + 2(x - 1) + - (x - 1) - - (x - 1) + -- (x - 1)
3 4 60
+
53 6 7
- -- (x - 1) + O((x - 1) )
45
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
b4 := seriesSolve(op, y, x = 1, [0, 0, 0, 6]);
(94)
2 3 41 4 5 741 6
- 6 - 3(x - 1) + 7(x - 1) - -- (x - 1) + 14(x - 1) - --- (x - 1)
4 40
+
7
O((x - 1) )
Type: UnivariateTaylorSeries(Expression(Integer),x,1)
Note that 'seriesSolve' (and 'ode') takes values of derivatives,
while above I (and van Hoej) used coefficients, so I had to specifiy
factorial(i) as values of i-th derivative. Alternatively, one
can choose basis so that derivatives are 1, then one would have
to divide i-th coefficient by factorial(i). This would lead to
different matrix, but the same eigenvalues.
--
Waldek Hebisch
[email protected]
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