Transcript Superposition Principle & the Method of Undetermined

Reminder
•
•
•
•
Second Hour Exam
Date: July 9 ( Monday).
Room: LC-C3, Time: 10 -11 am
Covers: Sections 5.1, 5.3 and 7.1 7.9 (inclusive).
1
Dirac Delta Function
• Paul A. M. Dirac, one of the great physicists
from England invented the following function:
• Definition: A function (t) having the following
properties: (1)  (t )  0 , for all t  0 , and
(2)


-
 (t ) dt  1,
• is called the Dirac delta function. It follows from
(2) that for any function f(t) continuous in an
open interval containing t = 0, we have



f (t ) (t  a)dt  f (a) .
2
Remarks on Theory of
Distribution.
• Symbolic function, generalized function,
and distribution function.
3
Heuristic argument on the existence of
-function.
• When a hammer strikes an object, it transfer
momentum to the object. If the striking force is
F(t) over a short time interval [t0, t1], then the
total impulse due to F is the integral
t1
Impulse  F(t)dt . By Newton's 2nd law of motion,
t0
dv
t0 F(t) dt  t0 m dt dt  m v(t1 )-m v(t0 ) .
T hismeansthat" Impulse Change in momentum".
t1
t1
4
This heuristic leads to conditions
1 and 2.
5
What is the Laplace Transform of
-function?
• By definition, we have


L{ (t  a)}(s)   e  (t  a)dt   e  st  (t  a)dt  e  as .
 st
0

Note that when
t  a,
t  a,
(*)
t



t
-
 ( x  a)dx  0 , and
 ( x  a)dx  1. It follows that

t
-
δ (x a)dx  u(t  a).
Differentiate both sides of (*), we get (by the F.T .C.):
δ (t  a)  u ' (t  a).
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Application:
• Consider the symbolic Initial Value Problem:
x"9 x  3 (t   ) ; x( 0 )  1 , x'(0 )  0 .
T hisrepresentsa mass at tachedto a springis releasedfrom rest
1 meterbelow theequilibrium positionfor themass - spring
system.It begins to vibrate,but  seconds lat er,themass is struck
by a hammerexertingan impulse on t hemass., where x(t)is the
displacement from theequilibrium positionat timet.
We shall solveit by themethodof LaplaceT ransform.
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Linear Systems can be solved by
Laplace Transform.(7.9)
• For two equations in two unknowns, steps are:
• 1. Take the Laplace Transform of both equations
in x(t) and y(t),
• 2. Solve for X(s) and Y(s), then
• 3. Take the inverse Laplace Transform of X(s)
and Y(s), respectively.
• 4. Work out some examples.
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Examples
• Solve the following IVP:
x' y  1  u (t  2);
x  y'  0;
x(0)  0,
y(0)  0.
• #21, P.438
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Qualitative analysis of Differential
Equations
• Consider three type of differential equations:
(1)
ay"  by'  cy  0 wherea, b, care const ant s,
t hisis a linear equation wit h const antcoefficient s.
(2)
a(t)y"  b(t)y'  c(t)y  0, where a(t), b(t)and c(t) are
functionsof independent variablet . T hisis a linear
equation wit h variable coefficient s.
(3)
y"- 2 y'  4 y 2  0 , or y" (y')2  12 y  0. T heseare
Non - linear equations,(somecomment s).
T hereare no generalsolut ion proceduresfor solvingnonlinear
10
equations.
Reminder
•
•
•
•
Second Hour Exam
Date: July 9 ( Monday).
Room: LC-C3, Time: 10 -11 am
Covers: Sections 5.1, 5.3 and 7.1 7.9 (inclusive).
11
Reminder
•
•
•
•
Final Exam
Date: July 19 ( Thursday).
Room: LC-C3, Time: 10:30 am
Covers: All
12
Generally, we can write a 2nd order
D.E. in the form:
(*)
y"  f (t , y (t ), y ' (t ))  f (t , y, y ' ).
T hisequat ionis nonlinear,if f(t,y,y') is a
nonlinearfunct ionof y and (or) y'. Although
thereare no generalsolutionproceduresfor (*).
W e do havesome qualitative information about
this typeof equat ion, namely
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The Energy Integral Lemma
Lemma. Let y(t) be a solution ot thedifferential equation
(*)
y"  f(y), where f(y) is a continuousfunction
thatdoes not depend on y' or theindependent variablet
explicitly. Let F(y)be an antiderivativeof f , i.e. F'(y)  f(y).
T hen thequantity
1
()
E(t)  (y'(t))2  F(y(t))
2
d
is a constantfor all time t , i.e.
E(t)  0.
dt
14
Sketch of the proof.
1
Since E(t)  (y')2  F(y), it follows t hat
2
1
E'(t)  2(y')(y") F'(y)y'(t) (y')(y") f(y)y'
2
 (y') y"-f(y)  0 .
• Clearly we see that
{E ' (t )  0}  { y"  f ( y )}. T hus
{E(t)  k}  {y"  f(y)}. T heequat ion
E(t)  k is a separableequat ion of first order
dy
in y ;
  2[k  F(y)] t hiscan be solved
dt
implicit ly.
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Examples
• 1. Consider the nonlinear equation:
y"  24 y 1 / 3 . Let F(y)   24 y 1 / 3 dy . T henby the
dy
lemma,we have   2[k  F(y)]. Note that
dt
1
1
3
4
3
4
3
y
3
F(y)  24
 24 y  18y . By takingk  0
1
4
1
3
dy
for simplicity, we see that   36 y 4 / 3  6 y162 / 3 .
dt
There are many other examples
in your book. Select some of
them depends on your field of
interest and go over by yourself.
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Series Solutions of Differential
Equations.
• Recall Taylor polynomials of degree n, at x = a.
• For example solution of the I.V.P.
7
3
y"  3 y ' x y ,
y( 0 )  10,
y'(0 )  5 .
Note : On T aylorRemainder:
Rn,f,a(x)  f(x)  pn,a(x)  En,a(x) 
f (n 1 )( )

(x  a)(n 1 ) ,
(n  1 )!
known as t heLagrangeForm.
18
Cauchy-Euler Equation
Revisited.
Recall t heCauchy - Euler equat ion:
(*) ax2 y"  bxy'  cy  0 , x  0.
W here a  0 , and b, c are real const ant s.If we
rewrit eit in t heSt andard Form :
bx
c
(**) y"  2 y'  2 y  0 , we see t hat
ax
ax
b
c
t hecoefficient s p(x) 
and q(x) 
ax
ax2
havea singular point at x  0.
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Solutions are of the form
r
x
• Consider three possible cases, distinct, double
and complex roots.
• Let w(r,x)=xr , and L[w](x)=0, we obtain
ar2 + (b-a) r + c = 0.
• Note that L commute with w/ r ,
• (i.e. Dr L= L Dr) Thus w/ r is also a
solution.
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Partial Differential Equations
• Definition
• One of the classical partial differential
equation of mathematical physics is the
equation describing the conduction of heat
in a solid body (Originated in the 18th
century). And a modern one is the space
vehicle reentry problem: Analysis of
transfer and dissipation of heat generated by
the friction with earth’s atmosphere.
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For example:
• Consider a straight bar with uniform crosssection and homogeneous material. We wish to
develop a model for heat flow through the bar.
• Let u(x,t) be the temperature on a cross section
located at x and at time t. We shall follow some
basic principles of physics:
• A. The amount of heat per unit time flowing
through a unit of cross-sectional area is
proportional to u / x with constant of
proportionality k(x) called the thermal
conductivity of the material.
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• B. Heat flow is always from points of higher
temperature to points of lower temperature.
• C. The amount of heat necessary to raise the
temperature of an object of mass “m” by an
amount u is a “c(x) m u”, where c(x) is
known as the specific heat capacity of the
material.
• Thus to study the amount of heat H(x) flowing
from left to right through a surface A of a cross
section during the time interval t can then be
given by the formula:
u
H ( x)  k ( x)( area of A) t ( x, t )
x
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Likewise, at the point x + x,
we have
• Heat flowing from left to right across the plane B
during an time interval t is:
u
H ( x  x)  k ( x  x)( area of B)t ( x  x, t ).
t
• If on the interval [x, x+x], during time t ,
additional heat sources were generated by, say,
chemical reactions, heater, or electric currents,
with energy density Q(x,t), then the total change
in the heat E is given by the formula:
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E = Heat entering A - Heat leaving B +
Heat generated .
• And taking into simplification the principle C
above, E = c(x) m u, where m = (x) V .
After dividing by (x)(t), and taking the limits
as x , and t  0, we get:
 
u
u

k ( x)
( x, t )  Q( x, t )  c( x)  ( x)
( x, t )

x 
x
t

• If we assume k, c,  are constants, then the eq.
2
Becomes:
u
2 u

 p( x, t )
2
t
x
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Boundary and Initial conditions
• Remark on boundary conditions and initial
condition on u(x,t).
• We thus obtain the mathematical model for the
heat flow in a uniform rod without internal
sources (p = 0) with homogeneous boundary
conditions and initial temperature distribution
f(x), the follolwing Initial Boundary Value
Problem:
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One Dimensional Heat Equation
u
u
( x, t )   2 ( x, t ) , 0  x  L, t  0 ,
t
x
u (0, t )  u ( L, t )  0 , t  0 ,
2
u ( x,0)  f ( x) , 0  x  L.
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The method of separation of
variables
• Introducing solution of the form
•
u(x,t) = X(x) T(t) .
• Substituting into the I.V.P, we obtain:
X ( x)T ' (t )   X ' ' ( x)T (t ) , 0  x  L, t  0.
t hisleads t o t hefollowingeq.
T ' (t )
X ' ' ( x)

 Const ant s.T hus we have
 T (t )
X ( x)
T ' (t )   kT (t )  0 and X ' ' ( x)  kX ( x)  0.
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Boundary Conditions
• Imply that we are looking for a non-trivial
solution X(x), satisfying:
X ' ' ( x)  kX ( x)  0
X (0)  X ( L)  0
• We shall consider 3 cases:
• k = 0, k > 0 and k < 0.
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• Case (i): k = 0. In this case we have
•
X(x) = 0, trivial solution
• Case (ii): k > 0. Let k = 2, then the D.E gives
X  - 2 X = 0. The fundamental solution set is:
{ e x, e -x }. A general solution is given by:
X(x) = c1 e x + c2 e -x
• X(0) = 0  c1 + c2 = 0, and
• X(L) = 0  c1 e L + c2 e -L = 0 , hence
• c1 (e 2L -1) = 0  c1 = 0 and so is c2 = 0 .
• Again we have trivial solution X(x)  0 .
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Finally Case (iii) when k < 0.
We again let k = - 2 ,  > 0. The D.E. becomes:
X   (x) + 2 X(x) = 0, the auxiliary equation is
r2 + 2 = 0, or r = ±  i . The general solution:
X(x) = c1 e ix + c2 e -ix or we prefer to write:
X(x) = c1 cos  x + c2 sin  x. Now the boundary
conditions X(0) = X(L) = 0 imply:
• c1 = 0 and c2 sin  L= 0, for this to happen, we
need  L = n , i.e.  = n /L or k = - (n /L ) 2.
• We set Xn(x) = an sin (n /L)x, n = 1, 2, 3, ...
•
•
•
•
•
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Finally for T (t) - kT(t) = 0, k = - 2 .
• We rewrite it as: T  +  2 T = 0. Or T 
= -  2 T . We see the solutions are
condition,we try:
theboundary conditions. T o sat isfy the init ial
un ( x, t )  X n ( x)Tn (t ) sat isfies theD.E and
T hus thefunct ion
Tn (t )  bn e   ( n / L ) t , n  1, 2, 3, ...
2
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u(x,t) =  un(x,t), over all n.
• More precisely,

u ( x , t )   cn e
 n 
 
 t
L


1
2
 n
sin 
 L

x .

W e must have:
 n 
u ( x,0)   cn sin 
 x  f ( x) .
 L 
1

• This leads to the question of when it is possible
to represent f(x) by the so called
•
Fourier sine series ??
33
Jean Baptiste Joseph Fourier
(1768 - 1830)
• Developed the equation for heat
transmission and obtained solution under
various boundary conditions (1800 - 1811).
• Under Napoleon he went to Egypt as a
soldier and worked with G. Monge as a
cultural attache for the French army.
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Example
• Solve the following heat flow problem
u
 2u
 7 2 , 0  x   , t  0.
t
x
u (0, t )  u ( , t)  0 , t  0 ,
u ( x,0)  3 sin 2 x  6 sin 5 x , 0  x  π.
• Write 3 sin 2x - 6 sin 5x =  cn sin (n/L)x,
and comparing the coefficients, we see that c2
= 3 , c5 = -6, and cn = 0 for all other n. And
we have u(x,t) = u2(x,t) + u5(x,t) .
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Wave Equation
• In the study of vibrating string such as
piano wire or guitar string.
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