Operator Methods
description
Transcript of Operator Methods
MATH F211:MATHEMATICS–III
Presented by
Dr. M.S. Radhakrishnan
Email: [email protected]
BITS-PILANI HYDERABAD CAMPUS
9
Apr 22, 2023
2
Operator Methods for finding Particular Solutions of LDE’s with constant coefficients
Ch. 3 Section 23George F. Simmons, Differential Equations with Applications and Historical notes, Tata McGraw-Hill, 2nd Ed, 2003 (Twelfth reprint, 2008)
Lecture 9
Presented by Dr. M.S. Radhakrishnan BITS, Pilani
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 3
You don’t have to be famous. You just have to make your mother and father proud of you.
Operator Methods for finding Particular Solutions of Linear Differential Equations with constant coefficients
Apr 22, 2023
4Presented by Dr. M.S. Radhakrishnan BITS, Pilani
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 5
In this section we give a very brief sketch of the use of differential operators for finding particular solutions of linear differential equations with constant coefficients.
These methods are more efficient than the previous methods and were mainly due to the English applied mathematician Oliver Heaviside.
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 6
Heaviside’s methods seemed so strange to the scientists of his time that he was widely regarded as a crackpot, which unfortunately is a common fate for thinkers of unusual originality.
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 7
Consider the th order linear differential equation with constant cofficients:
𝑎0𝑑𝑛 𝑦𝑑𝑥𝑛 +𝑎1
𝑑𝑛−1 𝑦𝑑𝑥𝑛− 1 +…+𝑎𝑛−1
𝑑𝑦𝑑𝑥
+𝑎𝑛 𝑦
Using the symbols etc.
we can write the above equation as
¿h (𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 8
(𝑎¿¿ 0𝐷𝑛+𝑎1 𝐷𝑛− 1+…+𝑎𝑛− 1𝐷+𝑎𝑛)𝑦 ¿
Or more compactly as 𝑝 (𝐷 ) 𝑦=h(𝑥)where is the th order linear differential operator
𝑝 (𝐷 )≡𝑎0 𝐷𝑛+𝑎1 𝐷
𝑛−1+…+𝑎𝑛−1 𝐷+𝑎𝑛
¿h (𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 9
A particular solution of the given d.e. is got by “solving” for
𝑦=1
𝑝 (𝐷 )h(𝑥 )
Thus 𝑦=1
𝑝 (𝐷 )h(𝑥 ) if and only if
𝑝 (𝐷 ) 𝑦=h(𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 10
For example, a particular solution of
𝐷𝑦=cos 𝑥
is 𝑦=1𝐷
cos 𝑥¿ sin 𝑥
A particular solution of (𝐷2+4 )𝑦=𝑒2𝑥
is 𝑦=1
𝐷2+4𝑒2𝑥¿
18𝑒2𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 11
Commutativity of the differential operators
Suppose we can factorize as
𝑝 (𝐷 )= 𝑓 (𝐷 )𝑔 (𝐷)
Then we easily verify that
𝑝 (𝐷 ) 𝑦=¿ 𝑓 (𝐷 ) [𝑔 (𝐷 ) ] 𝑦¿𝑔 (𝐷 ) [ 𝑓 (𝐷 ) ] 𝑦
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 12
Hence a particular solution of
𝑝 (𝐷 ) 𝑦=h(𝑥)
is 𝑦=1
𝑝 (𝐷 )h(𝑥 )
¿1
𝑓 (𝐷 )1
𝑔 (𝐷 )h(𝑥)
Also ¿1
𝑔 (𝐷 )1
𝑓 (𝐷 )h(𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 13
We also note that from the principle of superposition, we get
1𝑝 (𝐷 )
[𝑅 (𝑥 )+𝑆 (𝑥 ) ]
¿1
𝑝 (𝐷 )𝑅 (𝑥 )+ 1
𝑝 (𝐷 )𝑆 (𝑥 )
Also1
𝑝 (𝐷 ) [𝑐 {h (𝑥 ) }]=¿𝑐1
𝑝 (𝐷 )h (𝑥)
a constant.
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 14
We also note that from definition, we get
𝑝 (𝐷 ) 1𝑝 (𝐷 )
h (𝑥)=¿
i.e. and 1
𝑝 (𝐷 )are inverses to each other.
h (𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 15
𝐷𝑒𝑎𝑥=𝑎𝑒𝑎𝑥
𝐷2𝑒𝑎𝑥=𝑎2𝑒𝑎𝑥
…𝐷𝑛𝑒𝑎𝑥=𝑎𝑛𝑒𝑎𝑥
Hence 𝑝 (𝐷 )𝑒𝑎𝑥=𝑝(𝑎)𝑒𝑎𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 16
Or
And so 1
𝑝 (𝐷 )𝑒𝑎𝑥=¿
1𝑝 (𝑎 )
𝑒𝑎𝑥if
1𝑝 (𝐷 )
𝑝 (𝑎)𝑒𝑎𝑥=¿𝑒𝑎𝑥
i.e. 𝑝 (𝑎)1
𝑝 (𝐷 )𝑒𝑎𝑥=¿𝑒𝑎𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 17
We easily verify that
1𝐷−𝑎
𝑒𝑎𝑥=¿
𝑥2
2!𝑒𝑎𝑥1
(𝐷−𝑎 )2𝑒𝑎𝑥=¿
𝑥𝑒𝑎𝑥
…𝑥𝑚
𝑚!𝑒𝑎𝑥1
(𝐷−𝑎 )𝑚𝑒𝑎𝑥=¿
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 18
𝐷2sin𝑎𝑥=¿ −𝑎2 sin𝑎𝑥
𝐷4 sin𝑎𝑥=¿ (−𝑎2)2 sin𝑎𝑥
…Thus 𝑝 (𝐷¿¿2)sin 𝑎𝑥=¿¿𝑝 (−𝑎2 ) sin𝑎𝑥
Hence1
𝑝 (𝐷¿¿ 2)sin𝑎𝑥=¿¿1
𝑝 (−𝑎¿¿2)sin𝑎𝑥 ¿
if
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 19
Similarly1
𝑝 (𝐷¿¿ 2)cos𝑎𝑥=¿¿1
𝑝 (−𝑎¿¿2)cos𝑎𝑥 ¿
if
We easily see that1
𝐷2+𝑎2 sin 𝑎𝑥=¿− 𝑥2𝑎
cos𝑎𝑥
1
𝐷2+𝑎2 cos𝑎𝑥=¿ 𝑥2𝑎
sin𝑎𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 20
If is a polynomial in we can find a particular solution of
𝑝 (𝐷 ) 𝑦=h(𝑥)
as 𝑦=1
𝑝 (𝐷 )h(𝑥 )
¿ [1+𝑏1𝐷+𝑏2 𝐷2+…]h(𝑥)
where is the power1
𝑝 (𝐷 ).series expansion of
Series Expansion of Operators
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 21
The Exponential Shift Rule
𝐷 [𝑒¿¿ 𝑎𝑥𝑔(𝑥 )]=¿¿
𝐷2[𝑒𝑎𝑥𝑔 (𝑥 )]=𝑒𝑎𝑥 (𝐷+𝑎)2𝑔 (𝑥)
…𝐷𝑛 [𝑒𝑎𝑥𝑔 (𝑥 )]=𝑒𝑎𝑥 (𝐷+𝑎 )𝑛𝑔(𝑥 )
Hence
¿𝑒𝑎𝑥 (𝐷+𝑎)𝑔(𝑥)
𝑝 (𝐷 )[𝑒¿¿𝑎𝑥𝑔 (𝑥)]=𝑒𝑎𝑥𝑝 (𝐷+𝑎)𝑔(𝑥 )¿
𝑒𝑎𝑥𝐷𝑔(𝑥 )+𝑎𝑒𝑎𝑥𝑔(𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 22
Hence
𝑝 (𝐷){𝑒𝑎𝑥 1𝑝 (𝐷+𝑎)
𝑔(𝑥)}¿𝑒𝑎𝑥𝑝 (𝐷+𝑎)
1𝑝 (𝐷+𝑎)
𝑔 (𝑥)
¿𝑒𝑎𝑥𝑔(𝑥)
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 23
Thus we get the
1𝑝 (𝐷 )
[𝑒¿¿𝑎𝑥𝑔 (𝑥)]=¿¿𝑒𝑎𝑥 1𝑝 (𝐷+𝑎 )
𝑔 (𝑥 )
The Exponential Shift Rule
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 24
An Application
1𝐷−𝑎
𝑒𝑎𝑥=¿
¿ 𝑥2
2 !𝑒𝑎𝑥1
(𝐷−𝑎 )2𝑒𝑎𝑥=¿
¿ 𝑥𝑒𝑎𝑥
…
𝑒𝑎𝑥 1
𝐷𝑚 (1)1
(𝐷−𝑎 )𝑚𝑒𝑎𝑥=¿
𝑒𝑎𝑥 1𝐷
(1)
𝑒𝑎𝑥 1
𝐷2 (1)
¿ 𝑥𝑚
𝑚 !𝑒𝑎𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 25
PROBLEMS
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 26
1. Find a particular solution of
𝑦 ′ ′+4 𝑦=𝑒2𝑥
i.e. (𝐷¿¿2+4)𝑦=𝑒2𝑥¿
A particular solution is
𝑦=1
𝐷2+4𝑒2𝑥
¿18𝑒2𝑥
¿1
22+4𝑒2 𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 27
2. Find a particular solution of
𝑦 ′ ′−5 𝑦 ′+6 𝑦=𝑒2𝑥
i.e. (𝐷¿¿2−5𝐷+6)𝑦=𝑒2 𝑥¿
A particular solution is𝑦=1
𝐷2−5𝐷+6𝑒2𝑥
¿−𝑥 𝑒2 𝑥
¿1
(𝐷−2)(𝐷−3)𝑒2𝑥
¿ 1(𝐷−2)
1(𝐷−3)
𝑒2 𝑥
¿1
(𝐷−2)(−1 )𝑒2𝑥¿−
1(𝐷−2)
𝑒2𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 28
A particular solution is𝑦=1
𝐷2−5𝐷+6𝑒2𝑥
−𝑥𝑒2𝑥
¿1
(𝐷−2)(𝐷−3)𝑒2𝑥
¿ [ 1(𝐷−3)
−1
(𝐷−2) ]𝑒2𝑥
¿ (−1 )𝑒2𝑥
We can also do the above problem by “Partial Fractions”:
¿− (𝑥+1 )𝑒2𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 29
3. Find a particular solution of
𝑦 ′ ′+𝑦 ′+𝑦=sin 2𝑥
i.e. (𝐷¿¿2+𝐷+1)𝑦=sin 2 𝑥¿
A particular solution is𝑦=1
𝐷2+𝐷+1sin 2 𝑥
¿1
−22+𝐷+1sin 2 𝑥¿
1𝐷−3
sin 2𝑥
¿ (𝐷+3) [ 1
𝐷2−9sin 2 𝑥]
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 30
¿ (𝐷+3) [ 1
−22−9sin 2 𝑥 ]
¿ (𝐷+3) [− 11 3
sin 2𝑥 ]¿−
113
(𝐷+3 )sin 2 𝑥
¿−113
(2cos2𝑥+3sin 2𝑥 )
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 31
4. Find a particular solution of
4 𝑦 ′ ′+𝑦=𝑥4
i.e. (4𝐷¿¿2+1)𝑦=𝑥4 ¿
A particular solution is 𝑦=1
4 𝐷2+1𝑥4
¿ 𝑥4−4 8 𝑥2+384
¿ [1−4𝐷2+16𝐷4−64𝐷6+… ]𝑥4
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 32
5. Find a particular solution of
i.e. (𝐷¿¿2−2𝐷+2)𝑦=𝑒𝑥 sin𝑥 ¿
A particular solution is
𝑦=1
𝐷2−2𝐷+2𝑒𝑥sin 𝑥
𝑦 ′ ′−2 𝑦 ′+2 𝑦=𝑒𝑥 sin 𝑥
¿𝑒𝑥 1
(𝐷+1)2−2(𝐷+1)+2sin 𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 33
¿𝑒𝑥 1
𝐷2+1sin 𝑥
¿𝑒𝑥 ¿¿−
𝑥2𝑒𝑥 cos 𝑥
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 34
6. Find a particular solution of
i.e. (𝐷¿¿2−4𝐷+3)𝑦=𝑥3𝑒2𝑥 ¿
A particular solution is
𝑦=1
𝐷2−4𝐷+3𝑥3𝑒2𝑥
𝑦 ′ ′−4 𝑦 ′+3 𝑦=𝑥3𝑒2 𝑥
¿𝑒2𝑥 1
(𝐷+2)2−4(𝐷+2)+3𝑥3
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 35
¿𝑒2𝑥 1
𝐷2−1𝑥3
¿−𝑒2 𝑥 [1+𝐷2+𝐷4+… ]𝑥3
¿−𝑒2 𝑥 [𝑥3+6 𝑥 ]
¿−𝑒2 𝑥 1
1−𝐷2 𝑥3
Apr 22, 2023
Presented by Dr. M.S. Radhakrishnan BITS, Pilani 36
7. Find a particular solution of
(𝐷−2)3 𝑦=𝑒2𝑥
A particular solution is
𝑦=1
(𝐷−2)3 𝑒2𝑥
¿𝑒2𝑥 1
𝐷3 (1)¿𝑒2𝑥 1
𝐷2
1𝐷
(1)¿𝑒2𝑥 1
𝐷2 𝑥
¿𝑒2𝑥 1𝐷𝑥2
2¿𝑒2𝑥 𝑥3
3 !¿ 𝑥
3
3 !𝑒2 𝑥