The Computable Differential Equation Lecture ... - Bruce E. Shapiro
The Computable Differential Equation Lecture ... - Bruce E. Shapiro The Computable Differential Equation Lecture ... - Bruce E. Shapiro
78 CHAPTER 4. IMPROVING ON EULER’S METHOD Figure 4.3: Numerical solution of equation 4.80 for h = 0.190 (red), h = 0.205 (blue), and h = 0.23 (green). . Re(hλ) ≤ 0) then the method is said to be A-Stable. A-Stable methods are preferable for stiff problems. A stronger requirement is Stiff Decay: consider the generalized test equation y ′ = λ(y − g(t)) (4.85) where g(t) is any arbitrary, bounded function. Then we say that equation 4.85 has stiff decay if for any fixed t n > 0, lim |y n (t) − g(t n )| = 0. (4.86) hRe(λ)→−∞ To determine the shape of this region we observe that in the complex plane we can write hλ = x + iy (4.87) and therefore 1 ≥ |1 + x + iy| 2 (4.88) = (1 + x + iy)(1 + x − iy) 2 (4.89) = (1 + x) 2 + y 2 (4.90) which is a disk of radius 1 centered at the point (−1, 0). If λ ∈ R then this condition gives −1 ≤ 1 + hλ ≤ 1 (4.91) If λ < 0 then −2 ≤ hλ ≤ 0 (4.92) h ≤ − 2 λ (4.93) Math 582B, Spring 2007 California State University Northridge c○2007, B.E.Shapiro Last revised: May 23, 2007
CHAPTER 4. IMPROVING ON EULER’S METHOD 79 Figure 4.4: The region of absolute stability for Euler’s method is indicated by the gray area. To find the eigenvalue of equation 4.80 we linearize. Since the system is scalar, the Jacobian reduces to a single partial derivative ∂f/∂y. For f(t, y) = −5ty 2 +5/t−1/t 2 we have the linearized equation y ′ ≈ f y (1, 1)y (4.94) ) = (−10(t)(y)| t=1,y=1 y (4.95) = −10y (4.96) Hence λ = −10 and we need h ≤ −2/(−10) = 0.2 to remain in the region of absolute stability. Of course, this eigenvalue will change as the solution progresses, and values of t and y change, so the step size needs to be adjusted dynamically to remain within the desired region. 4.3 Stiffness and the Backward Euler Method An initial value problem is said to be stiff if the absolute stability requirement leads to a much smaller value of h than would otherwise be needed to satisfy the accuracy requirements. Thus stiffness depends on three factors: (1) accuracy; (2) the length of the interval of integration; and (3) the region of absolute stability of the problem. An example is given in figure 4.5. From equation 4.92 this problem, which has λ = −100 requires h < 0.02 for small t. For the test equation the problem is stiff on the interval [0, b] if bRe(λ)
- Page 33 and 34: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 35 and 36: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 37 and 38: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 39 and 40: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 41 and 42: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 43 and 44: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 45 and 46: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 47 and 48: CHAPTER 2. SUCCESSIVE APPROXIMATION
- Page 49 and 50: Chapter 3 Approximate Solutions 3.1
- Page 51 and 52: CHAPTER 3. APPROXIMATE SOLUTIONS 45
- Page 53 and 54: CHAPTER 3. APPROXIMATE SOLUTIONS 47
- Page 55 and 56: CHAPTER 3. APPROXIMATE SOLUTIONS 49
- Page 57 and 58: CHAPTER 3. APPROXIMATE SOLUTIONS 51
- Page 59 and 60: CHAPTER 3. APPROXIMATE SOLUTIONS 53
- Page 61 and 62: CHAPTER 3. APPROXIMATE SOLUTIONS 55
- Page 63 and 64: CHAPTER 3. APPROXIMATE SOLUTIONS 57
- Page 65 and 66: CHAPTER 3. APPROXIMATE SOLUTIONS 59
- Page 67 and 68: CHAPTER 3. APPROXIMATE SOLUTIONS 61
- Page 69 and 70: CHAPTER 3. APPROXIMATE SOLUTIONS 63
- Page 71 and 72: CHAPTER 3. APPROXIMATE SOLUTIONS 65
- Page 73 and 74: CHAPTER 3. APPROXIMATE SOLUTIONS 67
- Page 75 and 76: Chapter 4 Improving on Euler’s Me
- Page 77 and 78: CHAPTER 4. IMPROVING ON EULER’S M
- Page 79 and 80: CHAPTER 4. IMPROVING ON EULER’S M
- Page 81 and 82: CHAPTER 4. IMPROVING ON EULER’S M
- Page 83: CHAPTER 4. IMPROVING ON EULER’S M
- Page 87 and 88: CHAPTER 4. IMPROVING ON EULER’S M
- Page 89 and 90: CHAPTER 4. IMPROVING ON EULER’S M
- Page 91 and 92: CHAPTER 4. IMPROVING ON EULER’S M
- Page 93 and 94: CHAPTER 4. IMPROVING ON EULER’S M
- Page 95 and 96: Chapter 5 Runge-Kutta Methods 5.1 T
- Page 97 and 98: CHAPTER 5. RUNGE-KUTTA METHODS 91 w
- Page 99 and 100: CHAPTER 5. RUNGE-KUTTA METHODS 93 w
- Page 101 and 102: CHAPTER 5. RUNGE-KUTTA METHODS 95 5
- Page 103 and 104: CHAPTER 5. RUNGE-KUTTA METHODS 97 F
- Page 105 and 106: CHAPTER 5. RUNGE-KUTTA METHODS 99 T
- Page 107 and 108: CHAPTER 5. RUNGE-KUTTA METHODS 101
- Page 109 and 110: CHAPTER 5. RUNGE-KUTTA METHODS 103
- Page 111 and 112: CHAPTER 5. RUNGE-KUTTA METHODS 105
- Page 113 and 114: CHAPTER 5. RUNGE-KUTTA METHODS 107
- Page 115 and 116: CHAPTER 5. RUNGE-KUTTA METHODS 109
- Page 117 and 118: CHAPTER 5. RUNGE-KUTTA METHODS 111
- Page 119 and 120: CHAPTER 5. RUNGE-KUTTA METHODS 113
- Page 121 and 122: CHAPTER 5. RUNGE-KUTTA METHODS 115
- Page 123 and 124: Chapter 6 Linear Multistep Methods
- Page 125 and 126: CHAPTER 6. LINEAR MULTISTEP METHODS
- Page 127 and 128: CHAPTER 6. LINEAR MULTISTEP METHODS
- Page 129 and 130: CHAPTER 6. LINEAR MULTISTEP METHODS
- Page 131 and 132: CHAPTER 6. LINEAR MULTISTEP METHODS
- Page 133 and 134: CHAPTER 6. LINEAR MULTISTEP METHODS
CHAPTER 4. IMPROVING ON EULER’S METHOD 79<br />
Figure 4.4: <strong>The</strong> region of absolute stability for Euler’s method is indicated by the<br />
gray area.<br />
To find the eigenvalue of equation 4.80 we linearize. Since the system is scalar, the<br />
Jacobian reduces to a single partial derivative ∂f/∂y. For f(t, y) = −5ty 2 +5/t−1/t 2<br />
we have the linearized equation<br />
y ′ ≈ f y (1, 1)y (4.94)<br />
)<br />
=<br />
(−10(t)(y)| t=1,y=1<br />
y (4.95)<br />
= −10y (4.96)<br />
Hence λ = −10 and we need h ≤ −2/(−10) = 0.2 to remain in the region of absolute<br />
stability. Of course, this eigenvalue will change as the solution progresses, and values<br />
of t and y change, so the step size needs to be adjusted dynamically to remain within<br />
the desired region.<br />
4.3 Stiffness and the Backward Euler Method<br />
An initial value problem is said to be stiff if the absolute stability requirement<br />
leads to a much smaller value of h than would otherwise be needed to satisfy the<br />
accuracy requirements. Thus stiffness depends on three factors: (1) accuracy; (2)<br />
the length of the interval of integration; and (3) the region of absolute stability of<br />
the problem. An example is given in figure 4.5. From equation 4.92 this problem,<br />
which has λ = −100 requires h < 0.02 for small t. For the test equation the problem<br />
is stiff on the interval [0, b] if<br />
bRe(λ)