130_notes.dvi

Take the differential equation
d^2 u
dy^2

• (ǫ−y^2 )u= 0

and plug
u(y) =h(y)e−y

(^2) / 2
into it to get
d^2
dy^2
h(y)e−y
(^2) / 2
+ǫh(y)e−y
(^2) / 2
−y^2 h(y)e−y
(^2) / 2
= 0
d^2 h(y)
dy^2
e−y
(^2) / 2
− 2
dh(y)
dy
ye−y
(^2) / 2
−h(y)e−y
(^2) / 2
+h(y)y^2 e−y
(^2) / 2
+ǫh(y)e−y
(^2) / 2
−y^2 h(y)e−y
(^2) / 2
= 0
d^2 h(y)
dy^2
− 2 y
dh(y)
dy
−h(y) +y^2 h(y) +ǫh(y)−y^2 h(y) = 0
d^2 h(y)
dy^2
− 2 y
dh(y)
dy

• (ǫ−1)h(y) = 0
  This is our differential equation for the polynomialh(y).
  Writeh(y) as a sum of terms.
  h(y) =

∑∞

m=0

amym

Plug it into the differential equation.

∑∞

m=0

[am(m)(m−1)ym−^2 − 2 am(m)ym+ (ǫ−1)amym] = 0

We now want ot shift terms in the sum so that we see the coefficient ofym. To do this, we will shift the
termam(m)(m−1)ym−^2 down two steps in the sum. It will now show up asam+2(m+2)(m+1)ym.

∑∞

m=0

[am+2(m+ 2)(m+ 1)− 2 am(m) + (ǫ−1)am]ym= 0

(Note that in doing this shift the first term form= 0and form= 1get shifted out of the sum. This
is OK sinceam(m)(m−1)ym−^2 is zero form= 0orm= 1.)

For the sum to be zero for ally, each coefficient ofymmust be zero.

am+2(m+ 2)(m+ 1) + (ǫ− 1 − 2 m)am= 0

Solve foram+2

am+2=

2 m+ 1−ǫ

(m+ 1)(m+ 2)

am

and we have a recursion relation giving us our polynomial.

But, lets see what we have. For largem,

am+2=

2 m+ 1−ǫ

(m+ 1)(m+ 2)

am→

2

m

am