Zeroth Order Bessel Equation
Zeroth Order Bessel Equation
Zeroth Order Bessel Equation
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The solution of the zeroth order <strong>Bessel</strong>’s <strong>Equation</strong><br />
Zero <strong>Order</strong> <strong>Bessel</strong> Function<br />
Ang Man Shun<br />
October 16, 2012<br />
Is<br />
t 2 d2 y(t)<br />
dt 2 + t d dt y(t) + t2 y = 0<br />
y(t) =<br />
∞∑<br />
k=0<br />
(−1) k<br />
2 2k (k!) 2 · t2k<br />
1 Review of related mathematics<br />
1.1 Binomial Expansion<br />
A special case ,<br />
(a + b) n =<br />
n∑<br />
Cr n a n−r b r<br />
r=0<br />
(1 + x) n =<br />
n∑<br />
Cr n x r<br />
r=0<br />
The binomial coefficient C n r<br />
For n ∈ R and r > 0<br />
C n r =<br />
n!<br />
(n − r)!r!<br />
n, r ∈ Z + n > r<br />
C n r =<br />
n · (n − 1) · (n − 2) ...(n − r + 1)<br />
r!<br />
1
1.2 Laplace Transfrom<br />
Property 1<br />
Pf.<br />
And thus<br />
{ } d<br />
L<br />
dt x(t) =<br />
{ } d 2 x(t)<br />
L<br />
dt 2<br />
ˆ ∞<br />
0<br />
⎧<br />
L {x(t)} =<br />
( d<br />
dt x(t) )<br />
e −st dt =<br />
ˆ ∞<br />
ˆ ∞<br />
0<br />
e −st dx(t) = e −st x(t)| ∞ 0<br />
= −x(0) + s x(t)e −st dt = sX(s) − x(0)<br />
0<br />
} {{ }<br />
⎫<br />
ˆ ∞<br />
X(s)<br />
⎪⎨ ( )<br />
d dx(t)<br />
⎪⎬ { } dx(t)<br />
= L<br />
= s L<br />
dt dt<br />
dt<br />
⎪⎩ } {{ } ⎪⎭ } {{ }<br />
f(t)<br />
F (s)<br />
0<br />
x(t)e −st dt = X(s)<br />
{ } d<br />
L<br />
dt x(t) = sX(s) − x(0)<br />
−<br />
ˆ ∞<br />
0<br />
x(t)de −st<br />
− dx(t) | x=0 = s 2 X(s) − sx(0) − x ′ (0)<br />
dt<br />
Property 2<br />
L {tx(t)} = − d ds X(s)<br />
Pf.<br />
− d ds X(s) = − d ds L {x(t)} = − d ds<br />
ˆ ∞<br />
0<br />
x(t)e −st dt = −<br />
ˆ ∞<br />
x(t) d ds e−st dt = −<br />
ˆ ∞<br />
0<br />
0<br />
x(t)(−t)e −st dt = L {tx(t)}<br />
Property 3<br />
Pf.<br />
L {t n } = n!<br />
s n+1<br />
L { t 0} =L {1}=<br />
ˆ ∞<br />
0<br />
e −st dt = e−st | ∞ 0<br />
−s<br />
= 1 s = 0!<br />
s 1<br />
L { t 1} = L {t · 1}= − d ds L {1} = − d 1<br />
ds s = 1 s = 1!<br />
2 s 2<br />
L { t 2} = L {t · t}= − d ds L {t} = − d 1<br />
ds s = 2 2 s = 2!<br />
3 s 3<br />
Assume L { t k} = k! , then apply property 2 again ,<br />
sk+1 L { t k+1} = L { t · t k} = − d ds L { t k} = − d k! (k + 1)!<br />
=<br />
ds sk+1 s k+2<br />
So the prove by Mathematical Induction is now complete.<br />
2
2 Solving <strong>Zeroth</strong> Oder <strong>Bessel</strong> Differential <strong>Equation</strong><br />
2.1 General <strong>Bessel</strong> Differential Eqaution<br />
t 2 d2 y(t)<br />
dt 2<br />
+ t dy(t)<br />
dt<br />
p is called the order of <strong>Bessel</strong>’s <strong>Equation</strong><br />
+ ( t 2 − p 2) y = 0 p ≥ 0<br />
2.2 The solution of p = 0 , <strong>Zeroth</strong> <strong>Order</strong><br />
Divide the equation by t<br />
Apply Laplace Transform<br />
t 2 d2 y(t)<br />
dt 2<br />
+ t dy(t)<br />
dt<br />
ty” + y ′ + ty = 0<br />
+ t 2 y(t) = 0<br />
Apply L-property :<br />
L {ty”} + L {y ′ } + L {ty} = 0<br />
ty” ←→ (−1) 1 dF<br />
ds , d 2<br />
dt y(t) ←→ 2 s2 Y (s) − sy(0) − y ′ (0) and d y(t) ←→ sY (s) −<br />
dt<br />
y(0)<br />
⇒<br />
⇒<br />
− d ds L {y”} + L {y′ } − d ds L {y} = 0<br />
− d ds<br />
{<br />
s 2 Y (s) − sy(0) − y ′ (0) } + {sY (s) − y(0)} − d ds Y (s) = 0<br />
⇒<br />
− d ds s2 Y (s) + d<br />
} {{ } ds sy(0) + d<br />
} {{ } ds y′ (0) +sY (s) − y(0) − d<br />
} {{ }<br />
ds Y (s) = 0<br />
2sY +s 2 Y ′ +y(0) 0<br />
⇒ −2sY − s 2 Y ′ + sY − Y ′ = 0 ⇒ ( s 2 + 1 ) dY (s)<br />
ds<br />
+ sY (s) = 0 ⇒<br />
dY (s)<br />
ds<br />
+ s<br />
s 2 + 1 Y (s) = 0<br />
⇒<br />
dY (s)<br />
Y (s) + s<br />
s 2 + 1 ds = 0 ⇒ ln Y (s) + 1 2 ln ( s 2 + 1 ) = C ′ ⇒ ln Y + ln ( s 2 + 1 ) 1 2<br />
= C<br />
⇒<br />
[<br />
ln Y √ ]<br />
s 2 + 1 = C ′ ⇒ Y (s) =<br />
C<br />
√<br />
s2 + 1<br />
Let C = 1 , and rewrite it into the form<br />
Y (s) = ( s 2 + 1 ) − 1 2<br />
= 1 s<br />
(1 + 1 s 2 ) −<br />
1<br />
2<br />
3
Apply Binomial Expansion<br />
Where<br />
Y (s) = 1 s<br />
∞∑<br />
k=0<br />
C − 1 2<br />
k<br />
( 1<br />
s 2 ) k<br />
=<br />
∞∑<br />
k=0<br />
C − 1 2<br />
k<br />
1<br />
s 2k+1<br />
C −1<br />
2<br />
k<br />
=<br />
(<br />
− 1 ) ( −1<br />
2<br />
2 − 1 ) ( −1<br />
2 − 2 )<br />
...<br />
k!<br />
( −1<br />
2 − k + 1 )<br />
=<br />
( ) ( ) ( ( )<br />
1 3 5 2k − 1<br />
(−1) k ...<br />
2 2 2)<br />
2<br />
k!<br />
∴<br />
Since,<br />
(2k)!<br />
{ }} {<br />
= (−1)k 1.3.5... (2k − 1) · 2.4.6...2k<br />
2 k k! · 2.4.6...2k } {{ }<br />
2 k k!<br />
Y (s) =<br />
∞∑<br />
k=0<br />
C − 1 2<br />
k<br />
1<br />
s = ∑ ∞ 2k+1<br />
k=0<br />
= (−1)k (2k)!<br />
2 k k!2 k k!<br />
(−1) k (2k)!<br />
2 2k (k!) 2 1<br />
s 2k+1 = ∞<br />
∑<br />
k=0<br />
= (−1)k (2k)!<br />
2 2k (k!) 2<br />
( )<br />
(−1) k (2k)!<br />
2 2k (k!) 2 s 2k+1<br />
Y (s) = L {y(t)} , which is the solution of t 2 d2 y(t)<br />
+ t dy(t) + t 2 y = 0<br />
dt 2 dt<br />
Denote y(t) = J 0 (t) , take the inverse Laplace Transfrom using the property<br />
∴<br />
J 0 (t) = L −1 { ∞<br />
∑<br />
k=0<br />
}<br />
(−1) k (2k)!<br />
=<br />
2 2k (k!) 2 s 2k+1<br />
L {t n } = n!<br />
s n+1<br />
∞∑<br />
k=0<br />
{ }<br />
(−1) k (2k)!<br />
2 2k (k!) L =<br />
2 s 2k+1<br />
∞∑<br />
k=0<br />
(−1) k<br />
2 2k (k!) 2 t2k<br />
−END−