渐进分析练习题

渐进分析

Autumn, 2023

Consider the eigenvalue problem with \(0<\epsilon\ll1\):

\[ u^{\prime\prime}+(\lambda+\epsilon f(x))u=0, \quad 0<x<1 \]

\[ u(0)=0, \quad u^{\prime}(1)=0. \]

where \(f\) is a given smooth function. Give the asymptotic expansion of \(\lambda\) such that the accuracy is \(O(\epsilon)\).

证明

设 \(\lambda=\lambda_0+\epsilon \lambda_1 + O(\epsilon^2)\)，\(u=u_0+\epsilon u_1 + O(\epsilon^2)\)，代入方程并比较各阶项：

\[ u_0'' + \lambda_0 u_0 = 0, \]

\[ u_1'' + \lambda_0 u_1 = -\lambda_1 u_0 - f(x) u_0. \]

由边界条件 \(u(0)=0\) 和 \(u'(1)=0\) 可得 \(u_0(0)=0\) 和 \(u_0'(1)=0\)。解第一个方程得到 \(\lambda_0 = \left( n - \frac{1}{2} \right)^2 \pi^2\)，其中 \(n=1,2,3,\ldots\)，对应取 \(u_0(x) = \sin\left( \left( n - \frac{1}{2} \right) \pi x \right)\)。

对第二个方程两边乘 \(u_0\) 并积分，得到

\[ \int_0^1 u_0 u_1'' + \lambda_0 u_0 u_1 dx = -\lambda_1 \int_0^1 u_0^2 dx - \int_0^1 f(x) u_0^2 dx. \]

上式左边通过分部积分可化简为

\[ \int_0^1 u_0'' u_1+ \lambda_0 u_0 u_1 dx = 0. \]

故

\[ \lambda_1 = -\frac{\int_0^1 f(x) u_0^2 dx}{\int_0^1 u_0^2 dx}= -2 \int_0^1 f(x) \sin^2\left( \left( n - \frac{1}{2} \right) \pi x \right) dx. \]

因此

\[ \lambda_n=\left( n - \frac{1}{2} \right)^2 \pi^2 - 2 \epsilon \int_0^1 f(x) \sin^2\left( \left( n - \frac{1}{2} \right) \pi x \right) dx + O(\epsilon^2). \]

Autumn, 2024

Consider the following boundary value problem for \(y=y(x)\) on \([0, 1]\) as \(0<\epsilon\ll1\):

\[ \epsilon y^{\prime\prime}+\epsilon(1+x)^{2}y^{\prime}-y=x-1, \quad y(0)=\alpha, \quad y(1)=-1. \]

(i) Suppose \(\alpha=1\). Construct a composite expansion of the above problem and sketch the solution.

(ii) Construct a composite expansion of the above problem for \(\alpha=0\).

(iii) What is the accuracy of your solution in \(\epsilon\)? Formally explain your conclusion. Consider the first case only.

证明

\((i)\)：当 \(\epsilon \to 0\) 时，\(y=1-x\) 不满足 \(y(1)=-1\) 的边界条件，因此在 \(x=1\) 处存在一个边界层。考虑原方程的齐次部分 \(\epsilon \lambda^2 + \epsilon(1+x)^2 \lambda - 1 \approx 0\)。其特征根约为 \(\lambda \approx \pm 1/\sqrt{\epsilon}\)。这说明边界层的厚度尺度为 \(\mathcal{O}(\sqrt{\epsilon})\)。设 \(\eta = \frac{1-x}{\sqrt{\epsilon}}\)，\(y(x)=1-x+v(\eta)\)，则 \(y_x=-\frac{1}{\sqrt{\epsilon}} v_\eta\)，\(y_{xx}=\frac{1}{\epsilon} v_{\eta\eta}\)，代入保留 \(O(1)\) 项得到：

\[ v_{\eta\eta} - v_\eta = 0 \]

解得 \(v(\eta)=A e^{\eta}+B e^{-\eta}\)。由于 \(v(\eta)\) 在 \(\eta \to \infty\) 时必须有界，故 \(A=0\)。边界条件 \(y(1)=-1\) 即 \(v(0)=-1\)，解得 \(B=-1\)。因此

\[ y(x) \approx 1-x - e^{-\frac{1-x}{\sqrt{\epsilon}}}. \]

\((ii)\)：当 \(\epsilon \to 0\) 时，\(y=1-x\) 不满足 \(y(0)=0\) 的边界条件，因此在 \(x=0\) 处也存在一个边界层。同上考虑 \(\xi = \frac{x}{\sqrt{\epsilon}}\)，\(y(x)=1-x+u(\xi)+v(\eta)\)，代入保留 \(O(1)\) 项得到：

\[ u_{\xi\xi} - u = 0 \]

解得 \(u(\xi)=C e^{\xi}+D e^{-\xi}\)。进一步有 \(D=0,C=-1\)，故

\[ y(x) \approx 1-x - e^{-\frac{x}{\sqrt{\epsilon}}}- e^{-\frac{1-x}{\sqrt{\epsilon}}}. \]

\((iii)\)：

\[ \begin{aligned} \text{误差}&=\epsilon (1-x - e^{-\frac{1-x}{\sqrt{\epsilon}}})''+\epsilon(1+x)^2 (1-x - e^{-\frac{1-x}{\sqrt{\epsilon}}})'-(1-x - e^{-\frac{1-x}{\sqrt{\epsilon}}})-(x-1)\\ &=-e^{-\frac{1-x}{\sqrt{\epsilon}}} + \epsilon(1+x)^2(-1-\frac{1}{\sqrt{\epsilon}} e^{-\frac{1-x}{\sqrt{\epsilon}}})+e^{-\frac{1-x}{\sqrt{\epsilon}}} \\ &=O(\sqrt{\epsilon}) \end{aligned} \]

Spring, 2025

Consider a harmonic oscillator with a cubic damping term

\[ y^{\prime\prime}+y+\epsilon(y^{\prime})^{3}=0, \]

where \(y=y(t)\), \(t\ge0\), \(\epsilon>0\), \(y(0)=1\), \(y^{\prime}(0)=0\).

(i) For small \(\epsilon\), use the multiple-scale method to study the behavior of \(y(t)\) for large \(t\), i.e., construct a proper asymptotic solution.

(ii) Make a conclusion on the validity of your asymptotic solution. Briefly justify your conclusion.

证明

\((a)\)：

\[ y(t) \approx \frac{1}{\sqrt{1 - \frac{3}{4} \epsilon t}} \cos t \]

\((b)\)：该近似解在 \(t< \frac{4}{3\epsilon}\) 的时间尺度上保持有效，且误差为 \(O(\epsilon)\)。当 \(t\) 接近 \(\frac{4}{3\epsilon}\) 时，近似解的振幅趋于无穷大，因此该近似解在 \(t \geq \frac{4}{3\epsilon}\) 的时间尺度上失效。

Autumn, 2025

Consider the so-called Rayleigh oscillator

\[ y'' + y + \epsilon \left[ \frac{1}{3}(y')^3 - y' \right] = 0, \quad \epsilon > 0 \]

with initial condition \(y(0)=0, y'(0)=2a\) where \(y=y(t)\), \(a>0\).

(a) For a small \(\epsilon\), construct an approximation of the solution to the above problem which is valid for large \(t\).

(b) What is the accuracy of this approximation? State a conclusion and briefly explain it.

(c) Plot the approximated orbits in the phase plane, i.e., \(y-y'\) plane for several different \(a\). And explain what you observe.

证明

\((a)\)： 设 \(y(t,\epsilon)=y_0(t^+,\tau)+\epsilon y_1(t^+,\tau)+\cdots\)，其中快时间尺度 \(t^+=t\)，慢时间尺度 \(\tau=\epsilon t\)。

关于时间的导数：

\[ \frac{d}{dt}=\frac{\partial}{\partial t^+} + \epsilon \frac{\partial}{\partial \tau} \]

下简记 \(t^+\) 为 \(t\)，则

\[ 0=(\partial_t + \epsilon \partial_\tau)^2 (y_0+\epsilon y_1) + (y_0+\epsilon y_1) + \epsilon \left[ \frac{1}{3}((\partial_t + \epsilon \partial_\tau)(y_0+\epsilon y_1))^3 -(\partial_t + \epsilon \partial_\tau)(y_0+\epsilon y_1) \right]+O(\epsilon^2). \]

由 \(O(1)\) 项可得

\[ \partial_{tt} y_0 + y_0 = 0, \]

解得 \(y_0(t,\tau)=R(\tau)\cos(t+\phi(\tau))\)。

由 \(O(\epsilon)\) 项可得

\[ \begin{aligned} \partial_{tt} y_1 + y_1 &= -2 \partial_{t\tau} y_0- \frac{1}{3} (\partial_t y_0)^3 + \partial_t y_0\\ &= 2 R' \sin\psi + 2 R \phi' \cos\psi +\frac{1}{3} R^3 \sin^3\psi - R \sin\psi\\ &= 2 R' \sin\psi + 2 R \phi' \cos\psi + \frac{1}{4} R^3 \sin\psi - \frac{1}{12} R^3 \sin(3\psi) - R \sin\psi\\ &= (2R' -R + \frac{1}{4} R^3) \sin\psi + 2 R \phi' \cos\psi - \frac{1}{12} R^3 \sin(3\psi), \end{aligned} \]

为了保证 \(y_1\) 的有界性，必须消除 \(\sin\psi\) 和 \(\cos\psi\) 的共振项，即

\[ \begin{aligned} 2R' -R + \frac{1}{4} R^3 &= 0,\\ 2 R \phi' &= 0. \end{aligned} \]

解得 \(R(\tau)=\frac{2}{\sqrt{1-C e^{-\tau}}}\)，\(\phi(\tau)=\phi_0\)，其中 \(C\) 和 \(\phi_0\) 由初值条件确定。最终得到近似解

\[ y(t) \approx \frac{2a}{\sqrt{a^2 + (1-a^2)e^{-\epsilon t}}} \sin t \]

\((b)\)： 该近似解的误差为 \(O(\epsilon)\)，且在 \(t=O(\epsilon^{-1})\) 的时间尺度上保持有效。

\((c)\)：在相平面上，\(t\rightarrow \infty\) 时，振幅 \(R(\tau)\) 收敛到 \(2\)，因此所有轨道最终都会趋近于一个半径为 \(2\) 的圆周。