SKKUicc.skku.ac.kr/~yeonlee/Eng_Math/Kreyszig_01.pdf · 2012. 10. 23. · Kreyszig by YHLee;100302;...
Transcript of SKKUicc.skku.ac.kr/~yeonlee/Eng_Math/Kreyszig_01.pdf · 2012. 10. 23. · Kreyszig by YHLee;100302;...
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Kreyszig by YHLee;100302; 1-1 Chapter 1. First‐Order Ordinary Differential Equations (ODE) 1.1 Basic Concepts. Modeling
Mathematical modeling : A problem is formulated by variables, functions and equations. An ordinary differential equation contains derivatives of a function ( )y x of one variable. ′ = cosy x ′′ + =9 0y y ( )′′′ ′ ′′+ = +2 2 22 2xx y y e y x y A partial differential equation contains derivatives of a function ( ),u x y of more than one variable.
∂ ∂+ =∂ ∂
2 2
2 20
u ux y
The order of an ODC is equal to the order of the highest derivative. A first‐order ODE contains only ( )′y x ( )′ =, , 0F x y y : Implicit form ( )′ = ,y f x y : Explicit form Concept of Solution = ( )y h x is a solution if it satisfies the ODE in a given interval <
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Kreyszig by YHLee;100302; 1-2 Example 2 Solution Curves Solve =' cosy x By integrating both sides
= = +∫cos siny x dx x c , c is an arbitrary constant The equation represents a family of curves, Fig. 2.
If c=2, the equation represents a specific curve in Fig. 2. Example 3 Exponential Growth or Decay Differentiate = 3ty ce and obtain = 3' 3 ty ce , which is equal to =' 3y y .
→ = 3ty ce is a solution of =' 3y y Similarly, −= 0.2ty ce is a solution of = −' 0.2y y Example 4 Initial Value Problem Solve =' 3y y with ( )0 5.7y = The general solution is = 3ty ce Apply the initial condition → ( ) 00 5.7y ce c= = = The particular solution is 35.7 ty e=
Example 5 Radioactivity
Experiments show that a radioactive material decomposes at a rate proportional to the present amount.
Step 1 Setting up a model
dy
kydt
= , k: a constant measured by experiments
11 11.4 10 [sec ]k X − −= − for radium. When the initial amount is given by 0.5g, the initial condition is ( )0 0.5y = . Step 2 Solving the eq. ( ) kty t ce= Applying the initial condition ( ) 0.5 kty t e= Check the solution ' 0.5 kty ke ky= =
0(0) 0.5 0.5y e= =
Step 3 Interpretation of the result The radioactive material decays exponentially from the initial amount of 0.5g because k is negative.
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Kreyszig by YHLee;100302; 1-3 Example 6 A geometric application
Find the curve passing through the point (1, 1) and having the slope ‐y/x at each of its points. The mathematical formulation
y
yx
′ = −
The general solution
c
yx
= : A family of hyperbolas
Apply the initial condition
11c
=
→ 1 /y x= : A particular solution shown by the thick curve in Fig. 5.
1.2 Geometric Meaning of ( )' ,y f x y= . Direction Fields • A first‐order ODE ( )' ,y f x y= represents the slope of ( )y x .
→ Short straight‐line segments, lineal elements, can be drawn in xy‐plane. → An approximate solution by connecting lineal elements, Fig. 7(a).
CAS means Computer Algebra System
• ( )' ,y f x y k= = represents curves of equal inclination, isoclines, Fig. 7(b).
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Kreyszig by YHLee;100302; 1-4 1.3 Separable ODEs. In most cases ODEs can be written as ( ) ( )g y y f x′ = : Separable equation Integration with respect to x
( ) ( )g y y dx f x dx c′ = +∫ ∫
Using /y dy dx′ = , the general solution is obtained as ( ) ( )g y dy f x dx c= +∫ ∫ : Method of separating variables Example 1 A separable ODE Solve 21y y′ = +
By separating variables, 21
dydx
y=
+
By integration, arctany x c= + or tan( )y x c= +
The constant of integration should follows just after the integration. Study the following, Integration → arctany x= → tany x= → tany x c= + (NOT a solution)
Example 2 Radioactivity Facts: The iceman, Oetzi, found in Southern Tyrolia in 1991. The ratio of radioactive carbon 146C
to ordinary carbon 126C was 52.5% of that of a living organism.
Physics : Ordinary carbon becomes radioactive carbon by cosmic rays in the atmosphere. The ratio 146C /
126C in a living organism is constant due to breathing and eating.
146C decays to 12
6C in a dead organism with the half‐life of 5715 years.
A radioactive material decomposes at a rate proportional to the present amount Problem : Find when Oetzi died. Solution: A radioactive material decomposes at a rate proportional to the present amount
→ dy
kydt
=
Separating variables
dy
kdty=
→ ln y kt c= +
→ ktoy y e=
Determine k using the half‐life H=5715 years.
0.5kHo oy e y=
→ ln0.5
0.0001213kH
= = −
Using 52.5% ratio
12 12
6 6
0.525kt
o oy e y
C C=
→
ln0.5255312
0.0001213t years= =
−
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Kreyszig by YHLee;100302; 1-5 Example 3 Mixing Problem A tank contains 1000 gal of water with 100 lb of salt dissolved. Brine runs into the tank at a rate of 10 gal/min and each gallon contains 5 lb of dissolved salt. Brine runs out of the tank at a rate of 10 gal/min. Find the amount of slat in the tank at any time t. Solution: Let y(t) be the amount of salt in the tank. Then, y’(t) represents the rate of change of y in time.
10
501000
y y′ = −
Separating variables
0.015000dy
dty
= −−
→ − = − + 1ln 5000 0.01y t c → 0.01 5000ty ce−= +
The initial condition 0100 5000ce= + → c=‐4900 The solution 0.014900 5000ty e−= − +
Example 4 Newton's law of cooling(SKIP)
Facts: The daytime temperature in an office building is maintained at 70oF. The heating is on from 6 AM to 10 PM. The outside temperature was 50 oF at 10 PM and dropped to 40 oF at 6 AM. The inside temperature was 65 oF at 2 AM. What is the inside temperature at 6 AM. Physics : The time rate of change of the temperature is proportional to the temperature difference. Modeling : Let T(t) be the inside temperature and TA the outside temperature. From Newton’s law
( )AdT
k T Tdt
= −
General solution: TA is a function of time and we do not know. So we take the average value of 45oF as TA. Then, separating variables,
45
dTkdt
T=
− → 1ln 45T kt c− = + → 145
kt cT e += + → ( ) 45 ktT t ce= +
Particular solution: We choose 10 PM to be t=0. ( ) 00 45 70T ce= + = → ( ) 45 25 ktT t e= + The inside temperature was 65 oF at 2 AM. ( ) 44 45 25 65kT e= + = → 0.056k = − → ( ) 0.05645 25 tT t e−= + 6 AM is t=8, ( ) 0.056 88 45 25 61oT e F−= + =i
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Kreyszig by YHLee;100302; 1-6 Example 5 Leaking tank. Torricelli’s Law
The tank has diameter 2m and the hole has diameter 1cm. The initial height of the water is 2.25. When will the tank be empty? Physics: Due to the gravity the outflowing water velocity is
( ) ( )0.6 2v t gh t= : Torricelli’s Law where the acceleration of gravity 2980 / secg cm= . Modeling: The volume of the outflowing water in a short time tΔ , V Av tΔ = Δ : A is area of hole Using the change of water height in the tank, B h Av t− Δ = Δ : B is area of tank By letting tΔ approach to zero,
26.56dh A A
v hdt B B
= − = −
General solution: By separating variables
26.56dh A
dtBh
= − → 26.56dh A
dtBh
= − → 12 26.56A
h t cB
= − + → ( ) ( )20.000332h t c t= −
Particular solution: Using ( )0 225h cm= , we obtain c=15.0, ( ) ( )215.0 0.000332h t t= − Tank empty:
( )20 15.0 0.000332t= − → t=45,181 seconds=12.6 hours Reduction to Separable Form A first‐order ODE can be made separable if it is of the form of
y
y fx
⎛ ⎞′ = ⎜ ⎟⎝ ⎠
Using change of variable
y
ux
= → y ux= → ' 'y u x u= +
The ODE becomes ( )'u x u f u+ = By separating variables
( )du dx
f u u x=
−
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Kreyszig by YHLee;100302; 1-7 Example 6 Reduction to separable form
Solve 2 22xyy y x′ = − Rewrite it as
2 2
2y x
yxy−′ =
By setting u=y/x
1
2 2u
u x uu
′ + = −
By separating variables
221udu dx
xu= −
+
By integrating
( )2 1ln 1 lnu x c+ = − + → 21 cu x+ = → 2 2x y cx+ = or
2 22
2 4c c
x y⎛ ⎞− + =⎜ ⎟⎝ ⎠
The general solution represent a family of circles passing through the origin 1.4 Exact ODEs. Integrating Factors The differential of a function ( , )u x y is given as,
u u
du dx dyx y∂ ∂
= +∂ ∂
.
If ( , ) constant, then =0u x y du= • A first‐order ODE is given, ( , ) ( , ) 0M x y dx N x y dy+ =
It is an exact differential equation if the left side is of the form u udx dy
x y∂ ∂
+∂ ∂
,
In this case,
,
u uM N
x y∂ ∂
= =∂ ∂
(4)
Then the solution is ( , )u x y c= • The necessary and sufficient condition for an exact differential equation By differentiating M and N
2 2
, M u N uy y x x x y
∂ ∂ ∂ ∂= =
∂ ∂ ∂ ∂ ∂ ∂
By assuming the two second partial derivatives to be continuous
M Ny x
∂ ∂=
∂ ∂
This is a necessary and sufficient condition for (Mdx+Ndy) to be exact differential.
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Kreyszig by YHLee;100302; 1-8 • The solution From (4a)
u
Mx∂
=∂
→ ( )u Mdx k y= +∫
By differentiating with respect to y,
/ / /( ) ( )
u y N M y N xu dk y dk y
Mdx N Mdxy y dy dy y
∂ ∂ = ∂ ∂ =∂ ∂∂ ∂ ∂⎡ ⎤ ⎡ ⎤= + ⎯⎯⎯→ = − ⎯⎯⎯⎯⎯→⎣ ⎦ ⎣ ⎦∂ ∂ ∂∫ ∫ ( )dk y N
N dxdy x
∂⎡ ⎤= − ⎢ ⎥∂⎣ ⎦∫ Then,
{ }Part of depending only on u Mdx N y dy= +∫ ∫
• The solution in another method From (4b)
uN
y∂
=∂
→ ( )u Ndy l x= +∫
By differentiating with respect to x,
/ / /( ) ( )
u x M M y N xu dl x dl x
Ndy M Ndyx x dx dx x
∂ ∂ = ∂ ∂ =∂ ∂∂ ∂ ∂⎡ ⎤ ⎡ ⎤= + ⎯⎯⎯→ = − ⎯⎯⎯⎯⎯→⎣ ⎦ ⎣ ⎦∂ ∂ ∂∫ ∫ ( )dl x M
M dydx y
∂⎡ ⎤= − ⎢ ⎥∂⎣ ⎦
∫ Then,
{ }Part of depending only on u Ndy M x dx= +∫ ∫
Example 1 An exact ODE
( ) ( )2cos 3 2 cos 0x y dx y y x y dy⎡ ⎤+ + + + + =⎣ ⎦
(7)
1. Test for exactness
( )cosM x y= + , ( )23 2 cosN y y x y= + + +
( )sinM x yy
∂= − +
∂
( )sinN x yx
∂= − +
∂
The differential equation (7) is exact!
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Kreyszig by YHLee;100302; 1-9 2. The general solution
( ) ( )( ) sinu Mdx k y x y k y= + ⇒ + +∫
( )∂ ⇒ + + =∂
cosu dk
x y Ny dy
→ ( ) 2cos 3 2dk N x y y ydy
= − + ⇒ +
→ 3 2 1k y y c= + + Therefore ( ) 3 2sinu x y y y c= + + + =
• In another method
{ }Part of depending only on u Mdx N y dy= +∫ ∫
( ) 2cos (3 2 )u x y dx y y dy= + + +∫ ∫ or
{ }Part of depending only on u Ndy M x dx= +∫ ∫
( ) { }23 2 cos 0u y y x y dy dx⎡ ⎤= + + + +⎣ ⎦∫ ∫
3. Check the solution
u u
du dx dyx y∂ ∂
= +∂ ∂
= ( ) ( )2cos 3 2 cos 0x y dx y y x y dy⎡ ⎤+ + + + + =⎣ ⎦
OK! The same as (7). Example 2 An initial value problem
Solve ( ) ( )sinh cos 1 cosh sin 0x y dx x y dy+ − = , ( )1 2y = It can be shown that the equation is exact. Find u using
{ }Part of depending only on u Mdx N y dy= +∫ ∫
( ) { } 1sinh cos 1 0 cosh cos constantu x y dx dy x y x c= + + ⇒ + + =∫ ∫
cosh cos cu x y x= + =
The initial condition ( ) ( )cosh 1 cos 2 1 c+ = → c=0.358 The particular solution cosh cos 0.358x y x+ =
Example 3 Non‐exact equation Consider 0ydx xdy− + = This is not exact. Try to solve it using the above method anyhow,
( )( )u Mdx k y xy k y= + ⇒ − +∫
→ u dk
xy dy∂
= − +∂
→ dk
x xdy
= + (??)
↑ k should depend only on y It should be solved by another method than this.
0, 0.358,1, 2, 3c =
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Kreyszig by YHLee;100302; 1-10 Reduction to Exact Form. Integrating Factors A first‐order ODE can be made to be exact by multiplying an integrating factor, F ( , ) ( , ) 0P x y dx Q x y dy+ = : Not exact 0FPdx FQdy+ = : Exact The exactness condition
( ) ( )FP FQy x∂ ∂
=∂ ∂
→ y y x xF P FP F Q FQ+ = + : Too complicated to be solved
Try ( )F F x= , then 'y xFP F Q FQ= +
→ ( )1 1 y xdF P Q RF dx Q= − ≡ If R is a function of only x,
( ) RdxF x e∫= • Similarly, Try ( )F F y= , then ' y xF P FP FQ+ =
→ ( )1 1 x ydF Q P RF dy P= − ≡ If R is a function of only y,
( ) RdyF y e∫= Example 5 Initial Value Problem Solve
( ) ( )1 0x y y ye ye dx xe dy+ + + − = , ( )0 1y = − Check the exactness
Find the integrating factor
R = Using R , ( ) yF y e−= The differential equation becomes
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Kreyszig by YHLee;100302; 1-11 The general solution
{ }Part of depending only on u Mdx N y dy= +∫ ∫ → ( ) { }‐x yu e y dx e dy−= + +∫ ∫
→ x yu e xy e c−= + + = The particular solution 0 10( 1) 3.72e e c+ − + = ⇒
→ 3.72x ye xy e−+ + = 1.5 Linear ODEs. Bernoulli Equation. A first‐order ODE is linear if it is of the form of ( ) ( )y p x y r x′ + = , : ( ) ( )is input and is outputr x y x • Homogeneous Linear ODE ( ) 0y p x y′ + = By separating variables
( )dy
p x dxy= −
→ ( )
( )p x dx
y x ce−∫=
Another solution ( ) 0y x = : Trivial solution • The general solution of nonhomogeneous linear ODE ( ) 0py r dx dy− + = An integrating factor
( )1 P Q p xQ y x
∂ ∂⎛ ⎞− ⇒⎜ ⎟∂ ∂⎝ ⎠
( )pdx
F x e∫→ = Then
( )pdx pdx
e y py e r∫ ∫′ + =
→ ( )'pdx pdxe y e r∫ ∫=
→ ( ) h hy x e e rdx c− ⎡ ⎤= +⎣ ⎦∫ , ( )h p x dx= ∫ Rewrite this,
( ) h h hy x e e rdx ce− −= +∫ ↑ ↑ Response to the input r. Response to the initial value.
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Kreyszig by YHLee;100302; 1-12 Example 1 First‐order ODE, general solution Solve 2xy y e′ − = .
The general solution
Example 2 First‐order ODE, initial value problem Solve tan sin2y y x x′ + = , ( )0 1y =
tan ln sech xdx x= ⇒∫ → sec , cosh he x e x−= = The general solution
( ) ( ) 2cos sec sin2 cos 2cosy x x x xdx c c x x= + ⇒ −∫ i The initial condition 1=c cos(0)‐2 cos2(0) → c=3 The particular solution ( ) 23cos 2cosy x x x= − Example 3 Hormone Level Reduction to Linear Form. Bernoulli Equation Certain nonlinear ODE can be transformed to linear ODE. Bernoulli equation,
( ) ( ) ay p x y g x y′ + = , : a is any real number. a=0 or 1, the equation is linear Otherwise, nonlinear Let
[ ]1( ) ( ) au x y x −= . → ( )1(1 ) (1 ) a au a y y a g py− −′ ′= − ⇒ − −
→ (1 ) (1 )u a pu a g′ + − = − : Linear ODE Example 4 Logistic equation 2y Ay By′ = − : y is a function of t Since a=2, let 1u y−= .
( )2 2 2 u y y y By Ay B Au− −′ ′= − ⇒ − − + ⇒ − It becomes an linear ODE, whose general solution is given by
/Atu ce B A−= + or 1
/Aty
ce B A−=
+
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Kreyszig by YHLee;100302; 1-13 Population Dynamics(SKIP) The logistic equation is important in population dynamics.
[ ]1 ( / )y Ay B A y′ = − 0B=⎯⎯→ 1 Aty ec
= , exponential growth
↑ Braking term, y increases when it is small but y decreases when large. The independent variable t does not appear explicitly, ( )'y f y= : Autonomous ODE When ( ) 0f y = , its solution is y=constant. ↑ ↑ Critical points. Equilibrium solution. Stable if its neighboring solutions remain close to it for all t, y=4 in Fig. 18. Unstable if otherwise, y=0 in Fig. 18. Example 5 Stable and unstable equilibrium solutions(SKIP) ( )( )' 1 2y y y= − − Fig. 19(a), direction field Fig. 19(b), the direction field is squeezed horizontally. Fig. 19(c), plot of f(y)
From (a) y=1, stable equilibrium solution. y=2, unstable equilibrium solution.
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Kreyszig by YHLee;100302; 1-14 1.6 Orthogonal Trajectories Orthogonal trajectories intersect a given family of curves at right angles. An example:
A family of curves is given in xy‐plane with a parameter c, ( ), , 0G x y c = The ODE of the given family of curves, ( )' ,y f x y= The ODE of the orthogonal trajectories,
( )1
',
yf x y
= −
Proof: ( ) ( )1 1 1 1' , ' , 1y x y y x y = − , two curves are perpendicular. Example Parabola A family of curves, ( ) 2G , , 0x y c y cx= − = The ODE of the family of curves,
2 2 32
' 2 0 'y
yx c y x yx yx
− − −= → − = → =
The ODE of the orthogonal trajectories
'2x
yy
= −
By separating variables
→ 2 2 112
y x c+ =
( ) 2
2 2 3
An example: G , , 0
The corresponding ODE can be obtained as
2 ' 2 0 '
x y c y cx
yyx c y x yx y
x− − −
= − =
= → − = → =
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Kreyszig by YHLee;100302; 1-15 1.7 Existence and Uniqueness of Solutions An initial value problem may have no solution, only one solution, or many solutions. ( ) ( ), , o oy f x y y x y′ = = Theorem 1 Existence Theorem
If ( ),f x y is continuous and bounded, ( ),f x y K≤ , in some region R, and o ox x a y y b− < − < ,
the initial value problem has at least one solution in the interval ox x− < α ,
where α is the smaller of a and b/K. Example ( ) 2 2,f x y x y= + is bounded with K=2 in the square, 1, 1x y< < . ( ) ( ), tanf x y x y= + in not bounded for / 2x y+ < π . Theorem 2 Uniqueness Theorem
If ( ),f x y and /f y∂ ∂ are continuous and bounded in the region R,
( ),f x y K≤ and f My∂
≤∂
,
the initial value problem has at most one solution in the interval ox x− < α .
If the two theorems are combined, there should exist only one solution in the subinterval ox x− < α .
Understanding These Theorems ( ),f x y K≤ means 'K y K− ≤ ≤ . → A solution passing a point ( ),o ox y should be between 1 2and l l in Fig. 27(a) and (b).
(a) Blue curve is a solution satisfying ( ),f x y K≤ . A solution exists in ox x a− < Dark curve is not a solution because ( ),f x y K≤ is not satisfied. At some points the slope is smaller than –K.
(b) Blue curve is a solution satisfying ( ),f x y K≤ . A solution exists in − < 'ox x a
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Kreyszig by YHLee;100302; 1-16 Example 1 Choice of a Rectangle An initial value problem 2' 1y y= + , ( )0 0y = in a rectangle, 5 and 3x y< < .
Then
( ) 2, 1 10f x y y K= + ≤ =
2 6df
y Mdy
= ≤ =
0.3b
aK= <
A solution exists in 0.3ox x− < .
• Lipschitz condition From the mean value theorem,
( ) ( ) ( )2 1 2 1, ,y y
ff x y f x y y y
y =
∂− = −
∂, y is a value between 1 2and y y
f
My∂
≤∂
can be replaced by
( ) ( )2 1 2 1, ,f x y f x y M y y− ≤ − Example 2 Nonuniqueness The initial value problem
( )' , 0 0y y y= = Two solutions exist,
y=0 and 2
2
/ 4 if 0
/ 4 if 0
y x x
y x x
⎧ = ≥⎪⎨
= −