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Edurev123
6. Application of PDE
6.1 A tightly stretched string has its end fixed at ?? =?? and ?? =?? . At time ?? =?? , the
string is given a shape defined by ?? (?? )=???? (?? -?? ) , where ?? is a constant, and then
released. Find the displacement of any point ?? of the string at time ?? >?? .
(2009 : 30 Marks)
Solution:
The one dimensional wave equation is
?
2
?? ??? 2
=
1
?? 2
?
2
?? ??? 2
(1)
Boundary condition (BC) are
?? (0,?? )=?? (??,?? )=0 (2)
Initial condition (IC) are:
?? (?? ,0)=?? (?? )=h?? (?? -?? ) (3)
?? ?? (?? ,0)=0 (4)
Let ?? (?? ,?? )
'
=?? (?? )?? (?? ) be the trial solution where
?? (?? ) is a function of ?? only.
?? (?? ) is a function of ?? only.
Then clearly by (1)
?? '
?? =????
?? '
?? =
?? '
?? =?? (say)
?? -???? =0
and ??? -???? =0
Using (2) and (5), we get
Now,
?? (0,?? )?=0 and ?? (??,?? )=0
?? (?? ,?? )?=?? (?? )?? (?? )
As
Page 2
Edurev123
6. Application of PDE
6.1 A tightly stretched string has its end fixed at ?? =?? and ?? =?? . At time ?? =?? , the
string is given a shape defined by ?? (?? )=???? (?? -?? ) , where ?? is a constant, and then
released. Find the displacement of any point ?? of the string at time ?? >?? .
(2009 : 30 Marks)
Solution:
The one dimensional wave equation is
?
2
?? ??? 2
=
1
?? 2
?
2
?? ??? 2
(1)
Boundary condition (BC) are
?? (0,?? )=?? (??,?? )=0 (2)
Initial condition (IC) are:
?? (?? ,0)=?? (?? )=h?? (?? -?? ) (3)
?? ?? (?? ,0)=0 (4)
Let ?? (?? ,?? )
'
=?? (?? )?? (?? ) be the trial solution where
?? (?? ) is a function of ?? only.
?? (?? ) is a function of ?? only.
Then clearly by (1)
?? '
?? =????
?? '
?? =
?? '
?? =?? (say)
?? -???? =0
and ??? -???? =0
Using (2) and (5), we get
Now,
?? (0,?? )?=0 and ?? (??,?? )=0
?? (?? ,?? )?=?? (?? )?? (?? )
As
Since ?? (?? )=0 leads to ?? =0
So suppose that ?? (?? )?0
Then clearly equation (8) gives us that ?? (0)=0 and ?? (??)=0
Then ( 9 ) is modified boundary condition
Now solving (6) and (9), three cases arise clearly:
Case 1:?? =0 then clearly solution (6) will be ?? '
=0
So, ?? ''
(?? )=??
?????????????????????????????????????? (?? )=???? +??
where ?? and ?? are arbitrary constants
Clearly, ?? (0)=0 and ?? (?? )=0
So, ??? =0 and ???? =0??? =0(?? ?0)
So, ?? (?? )=0
This leads to ?? (?? )=0 which does not satisfy (3) and (4).
So, reject ?? =0 case.
Case 2:?? is positive, i.e., ?? >0
Clearly, ?? =?? 2
:(?? ?0)(?? >0)
So clearly, ?? '
-?? 2
?? =0.
So, ?? 2
-?? 2
=0
? ?? =±??
So, ?? (?? )=?? ?? ????
+?? ?? -????
Using (9), ?? (0)=0 and ?? (??)=0
? ?? +?? =0 and ?? ?? ????
+?? ?? -????
=0
Now, ?? =-?? as -?? (?? ????
-?? -????
)=0
As ?? ????
-?? -????
=0
? ?? =0
So ?? =0
Page 3
Edurev123
6. Application of PDE
6.1 A tightly stretched string has its end fixed at ?? =?? and ?? =?? . At time ?? =?? , the
string is given a shape defined by ?? (?? )=???? (?? -?? ) , where ?? is a constant, and then
released. Find the displacement of any point ?? of the string at time ?? >?? .
(2009 : 30 Marks)
Solution:
The one dimensional wave equation is
?
2
?? ??? 2
=
1
?? 2
?
2
?? ??? 2
(1)
Boundary condition (BC) are
?? (0,?? )=?? (??,?? )=0 (2)
Initial condition (IC) are:
?? (?? ,0)=?? (?? )=h?? (?? -?? ) (3)
?? ?? (?? ,0)=0 (4)
Let ?? (?? ,?? )
'
=?? (?? )?? (?? ) be the trial solution where
?? (?? ) is a function of ?? only.
?? (?? ) is a function of ?? only.
Then clearly by (1)
?? '
?? =????
?? '
?? =
?? '
?? =?? (say)
?? -???? =0
and ??? -???? =0
Using (2) and (5), we get
Now,
?? (0,?? )?=0 and ?? (??,?? )=0
?? (?? ,?? )?=?? (?? )?? (?? )
As
Since ?? (?? )=0 leads to ?? =0
So suppose that ?? (?? )?0
Then clearly equation (8) gives us that ?? (0)=0 and ?? (??)=0
Then ( 9 ) is modified boundary condition
Now solving (6) and (9), three cases arise clearly:
Case 1:?? =0 then clearly solution (6) will be ?? '
=0
So, ?? ''
(?? )=??
?????????????????????????????????????? (?? )=???? +??
where ?? and ?? are arbitrary constants
Clearly, ?? (0)=0 and ?? (?? )=0
So, ??? =0 and ???? =0??? =0(?? ?0)
So, ?? (?? )=0
This leads to ?? (?? )=0 which does not satisfy (3) and (4).
So, reject ?? =0 case.
Case 2:?? is positive, i.e., ?? >0
Clearly, ?? =?? 2
:(?? ?0)(?? >0)
So clearly, ?? '
-?? 2
?? =0.
So, ?? 2
-?? 2
=0
? ?? =±??
So, ?? (?? )=?? ?? ????
+?? ?? -????
Using (9), ?? (0)=0 and ?? (??)=0
? ?? +?? =0 and ?? ?? ????
+?? ?? -????
=0
Now, ?? =-?? as -?? (?? ????
-?? -????
)=0
As ?? ????
-?? -????
=0
? ?? =0
So ?? =0
Reject this case, as again we get a trivial solution.
Case 3:?? <0
So,
?? ?=-?? 2
then
?? (?? )?=?? cos????? +?? sin?????
?? (0)?=0 and ?? (??)=0
So, ?? =0 and ?? cos????? +?? sin????? =0
As clearly ?? ?0
? otherwise ?? =0 which does not satisfy initial condition.
So sin????? =0
????????????????????????????????????sin????? =sin???
? ?? =
????
??
So,
?? (?? )=?? sin?
????
?? ?? ;?? =1,2,…..
Hence, non-zero solution of (6) is clearly given by
?? ?? (?? )=?? ?? sin??????? ??
Then
?? -?? ?? 2
?? =0
??????????????????????????????????????????????????????????????????????? +?? 2
?? 2
?? =0
So ?? '
+
?? 2
?? 2
?? 2
?? 2
?? =0
So,
?? ?? (?? )=?? 0
cos?
???????? ?? +?? 0
sin????????? ??
???????????????????????????????????????????? ?? (?? ,?? )=?? ?? (?? )?? ?? (?? )
=?? ?? sin??? ?? ?? ?? [?? ?? cos????????? ?? +?? ?? sin????????? ?? ]
General solution
Page 4
Edurev123
6. Application of PDE
6.1 A tightly stretched string has its end fixed at ?? =?? and ?? =?? . At time ?? =?? , the
string is given a shape defined by ?? (?? )=???? (?? -?? ) , where ?? is a constant, and then
released. Find the displacement of any point ?? of the string at time ?? >?? .
(2009 : 30 Marks)
Solution:
The one dimensional wave equation is
?
2
?? ??? 2
=
1
?? 2
?
2
?? ??? 2
(1)
Boundary condition (BC) are
?? (0,?? )=?? (??,?? )=0 (2)
Initial condition (IC) are:
?? (?? ,0)=?? (?? )=h?? (?? -?? ) (3)
?? ?? (?? ,0)=0 (4)
Let ?? (?? ,?? )
'
=?? (?? )?? (?? ) be the trial solution where
?? (?? ) is a function of ?? only.
?? (?? ) is a function of ?? only.
Then clearly by (1)
?? '
?? =????
?? '
?? =
?? '
?? =?? (say)
?? -???? =0
and ??? -???? =0
Using (2) and (5), we get
Now,
?? (0,?? )?=0 and ?? (??,?? )=0
?? (?? ,?? )?=?? (?? )?? (?? )
As
Since ?? (?? )=0 leads to ?? =0
So suppose that ?? (?? )?0
Then clearly equation (8) gives us that ?? (0)=0 and ?? (??)=0
Then ( 9 ) is modified boundary condition
Now solving (6) and (9), three cases arise clearly:
Case 1:?? =0 then clearly solution (6) will be ?? '
=0
So, ?? ''
(?? )=??
?????????????????????????????????????? (?? )=???? +??
where ?? and ?? are arbitrary constants
Clearly, ?? (0)=0 and ?? (?? )=0
So, ??? =0 and ???? =0??? =0(?? ?0)
So, ?? (?? )=0
This leads to ?? (?? )=0 which does not satisfy (3) and (4).
So, reject ?? =0 case.
Case 2:?? is positive, i.e., ?? >0
Clearly, ?? =?? 2
:(?? ?0)(?? >0)
So clearly, ?? '
-?? 2
?? =0.
So, ?? 2
-?? 2
=0
? ?? =±??
So, ?? (?? )=?? ?? ????
+?? ?? -????
Using (9), ?? (0)=0 and ?? (??)=0
? ?? +?? =0 and ?? ?? ????
+?? ?? -????
=0
Now, ?? =-?? as -?? (?? ????
-?? -????
)=0
As ?? ????
-?? -????
=0
? ?? =0
So ?? =0
Reject this case, as again we get a trivial solution.
Case 3:?? <0
So,
?? ?=-?? 2
then
?? (?? )?=?? cos????? +?? sin?????
?? (0)?=0 and ?? (??)=0
So, ?? =0 and ?? cos????? +?? sin????? =0
As clearly ?? ?0
? otherwise ?? =0 which does not satisfy initial condition.
So sin????? =0
????????????????????????????????????sin????? =sin???
? ?? =
????
??
So,
?? (?? )=?? sin?
????
?? ?? ;?? =1,2,…..
Hence, non-zero solution of (6) is clearly given by
?? ?? (?? )=?? ?? sin??????? ??
Then
?? -?? ?? 2
?? =0
??????????????????????????????????????????????????????????????????????? +?? 2
?? 2
?? =0
So ?? '
+
?? 2
?? 2
?? 2
?? 2
?? =0
So,
?? ?? (?? )=?? 0
cos?
???????? ?? +?? 0
sin????????? ??
???????????????????????????????????????????? ?? (?? ,?? )=?? ?? (?? )?? ?? (?? )
=?? ?? sin??? ?? ?? ?? [?? ?? cos????????? ?? +?? ?? sin????????? ?? ]
General solution
?? ?? (?? ,?? )=??
8
?? =1
?
sin??????? ?? [?? ?? cos????????? ?? +?? ?? sin????????? ?? ] (6)
Now differentiating (6) w.r.t. ?? , we get
?? ?? (?? ,?? )=??
8
?? =1
?[-?? ?? sin?
???????? ?? (
?????? ?? )+?? ?? cos????????? ?? (
?????? ?? )]
sin??????? ?? Now ??? ?? (?? ,0)=0
???0=??
8
?? =1
??? ?? ?????? ?? sin??????? ?? ????? ?? =0
?? (?? ,?? )=??
8
?? =1
??? ?? cos?2?????? ?? sin??????? ?? ?? (?? ,0)=???? (?? -?? )
?? ?? =
2
?? ? ?
1
0
????? (?? -?? )
sin??????? ?? ????
?=
2?? ?? {(???? -?? 2
)
-
cos??????? ?? ????
?? |
0
1
+
?? ????
? ?
1
0
?(?? -2?? )
cos??????? ?? ????
?=
2?? ????
[(?? -2?? )
sin??????? ?? ????
?? |?+
4?? ?? 2
?? 2
?? 2
·
?? ????
-
cos??????? ?? ]
0
?? ????
?=
-4?? ?? 3
?? 2
?? 3
[cos????? -1]
?={
0 if ?? =2?? 8?? ?? 3
(2?? -1)
3
?? 3
if ?? =2?? -1
So, clearly we have
?? (?? ,?? )= ??
8
?? -1
8?? ?? 3
(2?? -1)
3
?? 3
·
cos?(2?? -1)?? ct
?? ·
sin?(2?? -1)????
??
6.2 Solve the following heat equation
?? ?? -?? ????
?=?? , ???????? ?? <?? <?? ,?? >?? ?? (?? ,?? )?=?? (?? ,?? )=?? ???????? ?? >?? ?? (?? ,?? )?=?? (?? -?? ), ???????? ?? =?? =??
Solution:
Given, the equation is ?? ?? -?? ????
=0 popularly known as "Heat Equation".
Page 5
Edurev123
6. Application of PDE
6.1 A tightly stretched string has its end fixed at ?? =?? and ?? =?? . At time ?? =?? , the
string is given a shape defined by ?? (?? )=???? (?? -?? ) , where ?? is a constant, and then
released. Find the displacement of any point ?? of the string at time ?? >?? .
(2009 : 30 Marks)
Solution:
The one dimensional wave equation is
?
2
?? ??? 2
=
1
?? 2
?
2
?? ??? 2
(1)
Boundary condition (BC) are
?? (0,?? )=?? (??,?? )=0 (2)
Initial condition (IC) are:
?? (?? ,0)=?? (?? )=h?? (?? -?? ) (3)
?? ?? (?? ,0)=0 (4)
Let ?? (?? ,?? )
'
=?? (?? )?? (?? ) be the trial solution where
?? (?? ) is a function of ?? only.
?? (?? ) is a function of ?? only.
Then clearly by (1)
?? '
?? =????
?? '
?? =
?? '
?? =?? (say)
?? -???? =0
and ??? -???? =0
Using (2) and (5), we get
Now,
?? (0,?? )?=0 and ?? (??,?? )=0
?? (?? ,?? )?=?? (?? )?? (?? )
As
Since ?? (?? )=0 leads to ?? =0
So suppose that ?? (?? )?0
Then clearly equation (8) gives us that ?? (0)=0 and ?? (??)=0
Then ( 9 ) is modified boundary condition
Now solving (6) and (9), three cases arise clearly:
Case 1:?? =0 then clearly solution (6) will be ?? '
=0
So, ?? ''
(?? )=??
?????????????????????????????????????? (?? )=???? +??
where ?? and ?? are arbitrary constants
Clearly, ?? (0)=0 and ?? (?? )=0
So, ??? =0 and ???? =0??? =0(?? ?0)
So, ?? (?? )=0
This leads to ?? (?? )=0 which does not satisfy (3) and (4).
So, reject ?? =0 case.
Case 2:?? is positive, i.e., ?? >0
Clearly, ?? =?? 2
:(?? ?0)(?? >0)
So clearly, ?? '
-?? 2
?? =0.
So, ?? 2
-?? 2
=0
? ?? =±??
So, ?? (?? )=?? ?? ????
+?? ?? -????
Using (9), ?? (0)=0 and ?? (??)=0
? ?? +?? =0 and ?? ?? ????
+?? ?? -????
=0
Now, ?? =-?? as -?? (?? ????
-?? -????
)=0
As ?? ????
-?? -????
=0
? ?? =0
So ?? =0
Reject this case, as again we get a trivial solution.
Case 3:?? <0
So,
?? ?=-?? 2
then
?? (?? )?=?? cos????? +?? sin?????
?? (0)?=0 and ?? (??)=0
So, ?? =0 and ?? cos????? +?? sin????? =0
As clearly ?? ?0
? otherwise ?? =0 which does not satisfy initial condition.
So sin????? =0
????????????????????????????????????sin????? =sin???
? ?? =
????
??
So,
?? (?? )=?? sin?
????
?? ?? ;?? =1,2,…..
Hence, non-zero solution of (6) is clearly given by
?? ?? (?? )=?? ?? sin??????? ??
Then
?? -?? ?? 2
?? =0
??????????????????????????????????????????????????????????????????????? +?? 2
?? 2
?? =0
So ?? '
+
?? 2
?? 2
?? 2
?? 2
?? =0
So,
?? ?? (?? )=?? 0
cos?
???????? ?? +?? 0
sin????????? ??
???????????????????????????????????????????? ?? (?? ,?? )=?? ?? (?? )?? ?? (?? )
=?? ?? sin??? ?? ?? ?? [?? ?? cos????????? ?? +?? ?? sin????????? ?? ]
General solution
?? ?? (?? ,?? )=??
8
?? =1
?
sin??????? ?? [?? ?? cos????????? ?? +?? ?? sin????????? ?? ] (6)
Now differentiating (6) w.r.t. ?? , we get
?? ?? (?? ,?? )=??
8
?? =1
?[-?? ?? sin?
???????? ?? (
?????? ?? )+?? ?? cos????????? ?? (
?????? ?? )]
sin??????? ?? Now ??? ?? (?? ,0)=0
???0=??
8
?? =1
??? ?? ?????? ?? sin??????? ?? ????? ?? =0
?? (?? ,?? )=??
8
?? =1
??? ?? cos?2?????? ?? sin??????? ?? ?? (?? ,0)=???? (?? -?? )
?? ?? =
2
?? ? ?
1
0
????? (?? -?? )
sin??????? ?? ????
?=
2?? ?? {(???? -?? 2
)
-
cos??????? ?? ????
?? |
0
1
+
?? ????
? ?
1
0
?(?? -2?? )
cos??????? ?? ????
?=
2?? ????
[(?? -2?? )
sin??????? ?? ????
?? |?+
4?? ?? 2
?? 2
?? 2
·
?? ????
-
cos??????? ?? ]
0
?? ????
?=
-4?? ?? 3
?? 2
?? 3
[cos????? -1]
?={
0 if ?? =2?? 8?? ?? 3
(2?? -1)
3
?? 3
if ?? =2?? -1
So, clearly we have
?? (?? ,?? )= ??
8
?? -1
8?? ?? 3
(2?? -1)
3
?? 3
·
cos?(2?? -1)?? ct
?? ·
sin?(2?? -1)????
??
6.2 Solve the following heat equation
?? ?? -?? ????
?=?? , ???????? ?? <?? <?? ,?? >?? ?? (?? ,?? )?=?? (?? ,?? )=?? ???????? ?? >?? ?? (?? ,?? )?=?? (?? -?? ), ???????? ?? =?? =??
Solution:
Given, the equation is ?? ?? -?? ????
=0 popularly known as "Heat Equation".
???? ?? ?????????????????????????????????????????????????? ?=?? (?? )?? (?? )
?Equation?is???????????? ?? '
-?? ''
?? ?=0
?? '
?? ?=?? ''
?? ?? '
?? ?=
?? ''
?? =??
Case-1:
?? ?=0
?? ''
?? ?=0
?? ?=???? +?? ?? (0,?? ) ?=?? (2,?? )=0
?? ?=0,??? =0???? =0??? =0 which is not possible. ??? ?0.
Case-2:
?? =?? 2
,??? ??? ?
?? ''
?? =?? 2
? ?? ''
-?? 2
?? =0
?
?
2
?? ??? 2
-?? 2
?? =0
? ?? =?? 1
?? ????
+?? 2
Given, ?? (0,?? )=?? (2,?? )=0 ?? 1
+?? 2
=0 and ?? ? ?? 1
=?? 2
=0
? ?? =0??? =0
?? =-?? 2
,??? ??? ?
?? ''
?? =-?? 2
???? ''
+?? 2
?? =0
?
?
2
?? ??? 2
+?? 2
?? =0
???? =?? 3
?? ?????? +?? 4
?? -?????? =?? 5
cos????? +?? 6
sin?????
( ?? 3
,?? 4
are constant)
( ?? 5
,?? 6
are constant)
Given, ?? (0,?? )=?? (2,?? )=0
? ?? 5
=0
and ?? 5
cos?2?? +?? 6
sin?2?? =0
? sin?2?? =0
? 2?? =????
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Nov 23, 2024 Last updated |
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