Test 1 Review Math 401-501/502 (Spring 2013)
General Information
Test 1 (Thursday, 3/7/13) will have 4 to 6 questions, some with
multiple parts. The problems will be similar to homework problems,
examples done in class and examples worked out in the text. My office
hours for this week will be Monday (3/4/13), 11-3, Tuesday (3/5/13),
2:30-3:30, and Wednesday (3/6/13), 11-3.
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Calculators: You may use scientific calculators to do numerical
calculations logs, exponentials, and so on. You
may not use any calculator that has the capability of doing
algebra or calculus, or of storing course material.
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Other devices: You may not use cell phones, computers, or any
other device capable of storing, sending, or receiving information.
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Formulas: The test will have several formulas that will be
useful in solving the problems, both for differential equations
(e.g. undetermined coefficients), trig identities (e.g., $\sin^3(x)
=(3\sin(x) - \sin(3x))/4$), and a few integrals.
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Coverage: The test will be over material covered from the start
of the semester on through Thursday, 2/28/13.
Topics Covered
- Regular perturbation theory
- Big $\mathcal O$, little $\mathcal o$ notation and applications
- Approximating small roots of polynomials
- Approximating solutions to differential equations on finite intervals. For an example, see assignment 5, problem 3. The answer to this problem is $y(x,\varepsilon) = \frac16 (x^{4/5}-x^{1/5})+\varepsilon\big(\frac{55}{14}\log(2)(x^{1/5}-x^{4/5})+\frac16 (x^{1/5}+4x^{1/5})\log(x)\big)+\mathcal O(\varepsilon^2)$.
- Asymptotic expansions
- Know the difference between approximating with an asymptotic series and approximating with a power series (notes, 1/31/13).
- Be able to do problems similar to problem 3, assignment 2.
- Singular perturbation problems for polynomials
- Scaling in polynomials. Consider assignment 5, problem 2. The perturbation problem, which is singular, is to find the large roots of $\varepsilon x^3 + 4x-12=0$; it requires a balancing technique (Fulling's Notes, p. 16 and class notes, 1/17&1/22): $4|x|\gg 12$ so $\varepsilon x^3 + 4x \sim 0$. Divide by $x$ to get $\varepsilon x^2 + 4\sim 0$. This implies that $x\sim \pm 2i \varepsilon^{-1/2}$. Thus the right scaling is $z=\varepsilon^{1/2}x$. Changing variables turns the problem into one involving a regular perturbation in powers of $\varepsilon^{1/2}$; namely, $z^3 + 4z-12\varepsilon^{1/2}=0$. To finish the problem, use an expansion of the form $z = \pm 2i + a \varepsilon^{1/2} + b\varepsilon+\mathcal O(\varepsilon^{3/2})$ and find $a$ and $b$. The large roots can then be shown to be of the form $x=z\varepsilon^{-1/2}=\pm 2i\varepsilon^{-1/2} -3/2 \pm 27i\varepsilon^{1/2}/16 +\mathcal O(\varepsilon)$.
- Another example. Find the scaling needed for getting a perturbation expansion for the large roots of $\varepsilon x^4 -8x-1=0$. As before, $8|x|\gg 1$, and so $\varepsilon x^4 -8x\sim 0$. Divide by $x$ to get $\varepsilon x^3-8\sim 0$, so $x=\mathcal O(\varepsilon^{-1/3})$. This implies that we should use $z=\varepsilon^{1/3}x$.
- Differential equations
- Nondimensionalizing an equation. See assignments 2 and 3.
- Removing secular terms via the Poincaré-Linsted method. Know what secular terms are and be able to use the Poincaré-Linsted method to avoid them. See assignment 5, problem 4 and assignment 3, problem 4.
- WKBJ method. See assignment 5, problem 1.
- Two-times method. See assignment 5, problem 5. The problems involving two times will not be done in detail, and only the basics of the method will be touched on. The answer to problem 5 is that the order 0 approximation is $y(\sigma,\tau,\varepsilon)=\sin(\sigma-\omega_1\tau)+\mathcal O(\varepsilon)=\sin(t)+\mathcal O(\varepsilon)$, which holds for any value of $\omega_1$.That is, avoiding the secular terms doesn't determine $\omega_1$.
- Boundary-layer problems. We want to look at an example in detail. Consider the perturbation problem $\varepsilon y''+(x+2)y'+y=0$, $y(0)=0$, $y(2)=1$, $0<\varepsilon \ll 1$. You are given that the boundary layer is on the left, where $x=\mathcal O(\varepsilon)$. To solve the problem, do the following.
- In the outer region where $x\approx 2$, start by setting $\varepsilon =0$. Solve the resulting equation, $(x+2)y'+y=0$, to get $y(x) = C(x+2)^{-1}$. Match the right boundary condition, $y(2)=1$. This gives us the outer solution, which is the one valid in the region outside of the boundary layer, $y_{out}=4(x+2)^{-1}$.
- In the boundary layer $x=\mathcal O(\varepsilon)$, scale the equation by making the change of variable $z=x/\varepsilon$. Doing so results in $d^2y/dz^2 + (2+\varepsilon z)dy/dz + \varepsilon y=0$. Keeping terms of order $\varepsilon^0$, we obtain the equation $d^2y_{BL}/dz^2 +2dy_{BL}/dz=0$. Solving this subject to $y_{BL}(0)=0$ yields $y_{BL}(z)=A(1-e^{-2z})$. In terms of $x$, we have $y_{BL}(x)=A(1-\exp(-2x/\varepsilon))$.
- Match in the intermediate region where $x=\mathcal O(\varepsilon^{1/2})$. Let $w=x/\varepsilon^{1/2}$. In terms of $w$, $y_{BL}=A(1-\exp(-2w/\varepsilon^{1/2}))$ and $y_{out}=4(2+\epsilon^{1/2}w)^{-1}$. Then, require $\lim_{\varepsilon\downarrow 0}y_{BL}=\lim_{\varepsilon\downarrow 0}y_{out}$. This gives $A=2$. Also, let $y_{overlap} = 2$.
- Uniform (composite) expansion: $y_{unit}=y_{BL}+y_{out}-y_{overlap}$. For our case, this gives $y_{unif}=4(x+2)^{-1}-2\exp(-2x/\varepsilon)$.
Updated: Mar. 3, 2013 (fjn)