My own notes about the MA269 revision, mainly from the notes and example sheets
- MA269-Asymptotics-and-Integral-Transforms-Revision
- Lecture notes
- Formula sheet
- Example sheets
- Big O notation
- Little o notation
- Big $\Theta$ notation
- Asymptotic Series
- Asymptotic Sequences
- Asymptotics of Algebraic Equations
- Singular perturbations
- The Exponential Integral
- Watson's Lemma
- Laplace's Method
- The Method of Stationary Phase
- The Method of Steepest Descent
- Asymptotics of Differential Equations
- Integral Transforms
- Complex Methods
- Fourier Transforms
- Laplace Transforms
First of all, feel free to download the 2022-2023 lecture notes for MA269 Asymptotics and Integral Transforms.
Second, in this module, we are going to cover a lot of computations, so some expansions and formulas are really useful.
MA269_Asymptotics_Integral_transforms_tutorial_sheet.pdf
- For functions
$f(x)$ and$g(x)$ , we say that$$f(x) = O(g(x))\quad\text{as}\quad x\to a\quad\Leftrightarrow \left|\frac{f(x)}{g(x)}\right|\text{is bounded as}x\to a.$$ - Or it is equivalent to say
$$\exists M,\delta\quad \text{s.t.} \left|\frac{f(x)}{g(x)}\right| < M\quad\forall |x-a| < \delta.$$ - Or it is equivalent to say
$$\lim_{x\to a}\sup\left|\frac{f(x)}{g(x)}\right| < \infty.$$ - Think of this as
$f$ is no bigger than order$g$ . - We can also let
$a = \pm\infty$ .
- For functions
$f(x)$ and$g(x)$ , we say that$$f(x) = o(g(x))\quad\text{as}\quad x\to a\quad\Leftrightarrow \left|\frac{f(x)}{g(x)}\right|\to 0\quad\text{as}\quad x\to a.$$ - Think of this as
$f$ is smaller than order$g$ .
- For functions
$f(x)$ and$g(x)$ , we say that$$f(x) = \Theta(g(x))\quad\text{as}\quad x\to a\Leftrightarrow\quad f(x) = O(g(x))\quad\text{and}\quad g(x) = O(f(x))\quad\text{as}\quad x\to a.$$ - Think of this as
$f$ is equal to order$g$ .
- The function
$f(x)$ is said to have an asymptotic series as$x\to a$ ,$$f(x)\sim\sum_{j=1}^{N}f_{j}(x)\quad\text{as}\quad x\to a\quad\Leftrightarrow\quad\forall\text{finite} M\leq N,\quad f(x) - \displaystyle\sum_{j=1}^{M}f_{j}(x) = o(f_{M}(x)).$$ - This is equivalent as saying
$$\left|\frac{f(x) - \displaystyle\sum_{j=1}^{M}f_{j}(x)}{f_{M}(x)}\right|\to 0\quad\text{as}\quad x\to a.$$ - This means when approximating
$f$ by partial sums$\displaystyle\sum_{j=1}^{M}f_{j}(x)$ , the error is smaller than the last included term provided$|x-a|$ is small enough.
- We often write
$$f(x)\sim\sum_{n=1}^{N} a_{n}\alpha_{n}(x)\quad\text{as}\quad x\to a,$$ where$\alpha_{n}$ is an asymptotic sequence. -
$\alpha_{n}$ is an asymptotic sequence if and only if$\alpha_{n+1} = o(\alpha_{n})\quad\text{as}\quad x\to a\quad\text{for all}\quad n.$
- Given an asymptotic sequence
$\alpha_{n}(x)$ , and an asymptotic series$f(x)\sim\displaystyle\sum_{n=1}^{N} a_{n}\alpha_{n}(x)$ as$x\to a$ , then the coefficients$a_{n}$ are unique since by definition,$$a_{n} = \lim_{x\to a} \frac{f(x)-\displaystyle\sum_{j=1}^{n-1}a_{j}\alpha_{j}(x)}{\alpha_{n}(x)}.$$
- As
$x\to a$ ,$f(x)$ may have different asymptotic series. For example as$x\to 0$ ,$$\tan x\sim x+\frac{1}{3}x^{3}+\frac{2}{15}x^{5}\sim x\cos x+\frac{5}{6}(x\cos x)^{3}+\frac{161}{120}(x\cos x)^{5}\sim \frac{1}{\sqrt{2}}\sinh(\sqrt{2}x)+\frac{1}{10}\left(\frac{1}{\sqrt{2}}\sinh(\sqrt{2}x)\right)^{5}.$$
Consider
-
Rearranging,
$x^{2} = 1-\varepsilon x$ , and so$x_{+} = \sqrt{1-\varepsilon x_{+}}$ . -
Try
$x_{0} = 1$ and$x_{n+1} = \sqrt{1-\varepsilon x_{n}}$ . This gives$$x_{1} = \sqrt{1-\varepsilon x_{0}} = 1-\frac{\varepsilon}{2}- \frac{\varepsilon^{2}}{8} - \frac{\varepsilon^{3}}{16}+...$$ $$x_{2} = \sqrt{1-\varepsilon x_{1}} = 1+\frac{1}{2}\left(-\varepsilon + \frac{\varepsilon^{2}}{2}+\frac{\varepsilon^{3}}{8}+...\right)-\frac{1}{8}\left(-\varepsilon + \frac{\varepsilon^{2}}{2}+\frac{\varepsilon^{3}}{8}+...\right)^{2}-.... = 1-\frac{\varepsilon}{2}+\frac{\varepsilon^{2}}{8}+\frac{\varepsilon^{3}}{4}+...$$ -
Lots of work, so not recommended.
- If
$\varepsilon = 0$ , then$x^{2} - 1 = 0$ and so$x = \pm 1$ and$x_{+} = 1$ . If$\varepsilon \ne 0$ , then try$x = 1+e_{1}$ , where$e_{1}$ is the error in the previous guess, and we hope$e_{1}$ is smaller than the previous guess of$x = 1$ . Then$$0 = (1+e_{1})^{2}+\varepsilon(1+e_{1}) - 1 = 1+2e_{1}+e_{1}^{2} + \varepsilon + \varepsilon e_{1} - 1\approx 2e_{1}+\varepsilon,$$ where we have neglected terms that are smaller than the ones we have included. - This gives
$e_{1}\approx -\frac{\varepsilon}{2}$ , so we set$x = 1-\varepsilon/2 + e_{2}$ , where$e_{2}$ is the next error we hope is smaller than the previous error$e_{1} = \Theta(\varepsilon)$ . Substituting$e_{1} = -\varepsilon/2 + e_{2}$ gives$$0 = 2\left(-\frac{\varepsilon}{2}+e_{2}\right)+\left(-\frac{\varepsilon}{2}+e_{2}\right)^{2}+\varepsilon + \varepsilon\left(-\frac{\varepsilon}{2}+e_{2}\right) = e_{2}^{2}+2e_{2}-\frac{\varepsilon^{2}}{4}\approx 2e_{2}-\frac{\varepsilon^{2}}{4},$$ where we have neglected terms that are smaller than the ones we have included. - This gives
$e_{2}\approx \varepsilon^{2}/8$ , and so$x = 1-\varepsilon/2 + \varepsilon^{2}/8 + o(\varepsilon^{2})$ , provided$e_{3} = o(e_{2}).$
- Guess the series expansion
$x_{+} = a_{0}+\varepsilon a_{1}+\varepsilon^{2} a_{2} + \varepsilon^{3} a_{3} + O(\varepsilon^{4})$ and substitute it in:$$\left(a_{0}+\varepsilon a_{1}+\varepsilon^{2} a_{2}+\varepsilon^{3}a_{3} + O(\varepsilon^{4})\right)^{2} + \varepsilon \left(a_{0}+\varepsilon a_{1}+\varepsilon^{2} a_{2} + \varepsilon^{3} a_{3} + O(\varepsilon^{4})\right) - 1 = 0.$$ - Therefore matching coefficients:
$$O(\varepsilon^{0}): 0 = a_{0}^{2} - 1\implies a_{0} = 1.$$ $$O(\varepsilon^{1}): 0 = 2a_{0}a_{1} + a_{0} = 2a_{1} + 1\implies a_{1} = -\frac{1}{2}.$$ $$O(\varepsilon^{2}): 0 = 2a_{0}a_{2} + a_{1}^{2} + a_{1} = 2a_{2} + \frac{1}{4}-\frac{1}{2}\implies a_{2} = \frac{1}{8}.$$ $$O(\varepsilon^{3}): 0 = 2a_{1}a_{2} + 2a_{0}a_{3} + a_{2} = -\frac{1}{8} + 2a_{3} + \frac{1}{8}\implies a_{3} = 0.$$ - So
$x = 1-\varepsilon/2+\varepsilon^{2}/8+0\varepsilon^{3}+O(\varepsilon^{4})$ .
- The previous problem
$x^{2}+\varepsilon x - 1 = 0$ is called regular; setting$\varepsilon = 0$ gives a solution, and$\varepsilon$ being small but non-zero changes that solution a little. - However consider now
$\varepsilon x^{2} + x - 1 = 0$ . If$\varepsilon = 0$ , we get only one solution. Such problems are singular.
- The exponential integral
$E_{1}(z)$ , is defined for$z > 0$ as$$E_{1}(z) = \int_{z}^{\infty}\frac{e^{-x}}{x}\mathrm{d}x,$$ it turns out that$E_{1}(0)$ is singular, but how singular is it?
- We use the Taylor expansion of
$e^{-x}$ :$$E_{1}(\varepsilon) = \int_{\varepsilon}^{\infty}\frac{e^{-x}}{x}\mathrm{d}x = \int_{\varepsilon}^{\infty}\frac{1}{x} - 1 + \frac{x}{2!}-\frac{x^{2}}{3!}+...\mathrm{d}x = \left[\log x - x + \frac{x^{2}}{2\times 2!}-\frac{x^{3}}{3\times 3!}+...\right]_ {\varepsilon}^{\infty}.$$ - The expansion diverges at infinity. Taylor expansion lost the fact that
$e^{-x}$ decays rapidly as$x\to\infty$ .
- We then split the integral to compare:
$$E_{1}(\varepsilon) = \int_{\varepsilon}^{a}\frac{e^{-x}}{x}\mathrm{d}x + \int_{a}^{\infty}\frac{e^{-x}}{x}\mathrm{d}x = \left[\log x - x + \frac{x^{2}}{2\times 2!} - \frac{x^{3}}{3\times 3!} + ...\right]_ {\varepsilon}^{a} + E_{1}(a) = -\log\varepsilon + \varepsilon - \frac{\varepsilon^{2}}{2\times 2!}+\frac{\varepsilon^{3}}{3\times 3!} + \left[E_{1}(a) + \log a - a + \frac{a^{2}}{2\times 2!}-\frac{a^{3}}{3\times 3!}\right].$$ - The final term is independent of
$\varepsilon$ , as$\varepsilon$ does not occur in the expression and is independent of$a$ , as$a$ dosen't appear in the first term and$E_{1}(\varepsilon)$ should be independent of$a$ . - Therefore, the final term must be a constant.
- Consider
$$I(\lambda) = \int_{0}^{a}f(x)e^{-\lambda x}\mathrm{d}x.$$ If we are interested in$\lambda\to\infty$ , we can rescale using$x = t/\lambda$ , giving$$I(\lambda) = \int_{0}^{a\lambda}f\left(\frac{t}{\lambda}\right)e^{-t}\frac{\mathrm{d}t}{\lambda} = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{\lambda^{n+1}}+O\left(\frac{e^{-\lambda a}}{\lambda}\right).$$
- If
$f(x)\sim\displaystyle\sum_{n=0}^{N}a_{n}x^{\alpha_{n}}$ as$x\to 0$ with$\alpha_{0} > -1$ , and if$f(x)$ is bounded for$x\in[\varepsilon, a]$ for all$\varepsilon > 0$ , then$$\int_{0}^{a}f(x)e^{-\lambda x}\mathrm{d}x\sim\sum_{n=0}^{N}\frac{a_{n}\Gamma(\alpha_{n}+1)}{\lambda^{\alpha_{n}+1}}\quad\text{as}\quad\lambda\to\infty,$$ where$$\Gamma(n+1) = \int_{0}^{\infty}t^{n}e^{-t}\mathrm{d}t = \left[-t^{n}e^{-t}\right]_ {0}^{\infty} + n\int_{0}^{\infty} t^{n-1}e^{-t}\mathrm{d}t = n!.$$
- Consider
$$I(\lambda) = \int_{a}^{b}f(x)e^{\lambda g(x)}\mathrm{d}x\quad\text{as}\quad\lambda\to\infty.$$ - An exponentially large contribution comes from near where
$g(x)$ is largest. If$\max_{x\in[a,b]}g(x)$ is at an endpoint, say at$a$ , and$g'(a)\ne 0$ , then Watson's lemma is applicable:$$I(\lambda) = \int_{a}^{a+\delta}f(x)e^{\lambda g(x)}\mathrm{d}x + \int_{a+\delta}^{b}f(x)e^{\lambda g(x)}\mathrm{d}x.$$ - The second integral will be exponentially small in comparison with the first. So changing variables by setting
$y = g(a) - g(x)$ ,$$I(\lambda) = e^{\lambda g(a)}\int_{0}^{g(a) - g(a + \delta)} f(x(y))e^{-\lambda y}\frac{\mathrm{d}y}{-g'(x(y))}+O\left(e^{\lambda g(a+\delta)}\right) = e^{\lambda g(a)}\int_{0}^{g(a)-g(a+\delta)}F(y)e^{-\lambda y}\mathrm{d}y + O\left(e^{\lambda g(a+\delta)}\right)\sim e^{\lambda g(a)}\sum_{n=0}^{N}\frac{F_{n}\Gamma(\alpha_{n}+1)}{\lambda^{\alpha_{n}+1}},$$ where$F(y) = \frac{f(x(y))}{-g'(x(y))}\sim\displaystyle\sum_{n=0}^{N}F_{n}y^{\alpha_{n}}\quad\text{as}\quad y\to 0.$
- We again consider the integral
$$I(\lambda) = \int_{a}^{b}f(x)e^{\lambda g(x)}\mathrm{d}x\quad\text{as}\quad\lambda\to\infty.$$ - Suppose that
$g(x)$ has a maximum at$x_{0}\in (a,b)$ , where we assume that$g'(x_{0}) = 0$ and that$g''(x_{0}) < 0$ . Then locally near that maximum,$$g(x)\approx g(x_{0}) + (x-x_{0})g'(x_{0}) + \frac{1}{2}(x-x_{0})^{2}g''(x_{0}) + O((x-x_{0})^{3}).$$ - So with the change of variable
$y = (x-x_{0})\sqrt{-\lambda g''(x_{0})}$ ,$$I(\lambda) = e^{\lambda g(x_{0})}\int_{a}^{b}f(x)\exp\set{\frac{1}{2}\lambda g''(x_{0})(x-x_{0})^{2}+O(\lambda(x-x_{0})^{3})}\mathrm{d}x\sim e^{\lambda g(x_{0})}\int_{-\infty}^{\infty}\frac{f(x_{0})e^{-\frac{1}{2}y^{2}}}{\sqrt{-\lambda g''(x_{0})}}\mathrm{d}y\quad\text{as}\quad\lambda\to\infty = \sqrt{\frac{2\pi}{-\lambda g''(x_{0})}}f(x_{0})e^{\lambda g(x_{0})}.$$
- If
$g(x)$ is twice continuously differentiable on$[a,b]$ , and$\exists x_{0}\in (a,b)$ such that$g(x_{0}) = \max_{x\in[a,b]}g(x)$ and$g''(x_{0}) < 0$ , then$$\int_{a}^{b}e^{\lambda g(x)}\mathrm{d}x\sim e^{\lambda g(x_{0})}\sqrt{\frac{2\pi}{-\lambda g''(x_{0})}}\quad\text{as}\quad\lambda\to\infty.$$ -
$a$ and$b$ can be$\pm\infty$ . - Can include
$f(x)$ in the integrand by expanding$f$ in a asymptotic series about$x_{0}$ . - Can get further terms in the asymptotic series of
$I(\lambda)$ by expanding$g(x)$ about$x_{0}$ too. - Can deal with problematic cases (e.g.
$x_{0} = a, g''(x_{0}) = 0, f(x_{0}) = 0$ , etc.) by expanding about$x_{0}$ . - Can deal with multiple maxima by splitting integral so each integral has only one maximum.
- Consider
$$\Gamma(\lambda + 1) = \int_{0}^{\infty}x^{\lambda}e^{-x}\mathrm{d}x.$$ We have already seen that$\Gamma(n+1) = n!$ , but how does$\lambda! = \Gamma(\lambda+1)$ behave as$\lambda\to\infty$ ? - Note that
$$\Gamma(\lambda+1) = \int_{0}^{\infty}e^{-x + \lambda\log x}\mathrm{d}x = \int_{0}^{\infty}e^{-x}e^{\lambda\log x}\mathrm{d}x,$$ so we first try$f(x) = e^{-x}$ and$g(x) = \log x$ . But then$g(x)$ is unbounded, and$g'(x) = \frac{1}{x}\ne 0$ . So our try fails. - The integrand has a maximum when
$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{\lambda}e^{-x}) = x^{\lambda-1}e^{-x}(\lambda - x)\implies x = \lambda.$$ - The motives the change of variable
$x = \lambda t$ , leading to$$\Gamma(\lambda+1) = \int_{0}^{\infty}\lambda^{\lambda}t^{\lambda}e^{-\lambda t}\lambda\mathrm{d}t = \lambda^{\lambda + 1}\int_{0}^{\infty}e^{-\lambda(t-\log t)}\mathrm{d}t,$$ so we take$f(t) = 1$ and$g(t) = -(t-\log t)$ . Hence by differentiation we note that when$t_{0} = 1$ ,$g'(t_{0}) = 0$ and reaches the maximum. Also we calculate that$g''(t_{0}) = -1.$ - Using also that
$g(t_{0}) = -1$ , Laplace's method gives$$\Gamma(\lambda + 1) \sim \lambda^{\lambda + 1}\sqrt{\frac{2\pi}{-\lambda\times-1}}e^{-\lambda} = \sqrt{2\pi\lambda}\lambda^{\lambda}e^{-\lambda},$$ which is the Stirling's approximation.
- Suppose we have a rapidly oscillating integral, say
$$I(\lambda) = \int_{0}^{\infty}f(x)\cos(\lambda g(x))\mathrm{d}x.$$ - Because of cancellation, the integral are hard to compute, as
$\lambda\to\infty$ , the dominant contribution comes from where there is the least cancellation due to oscillation. The Method of Stationary Phase states that$$I(\lambda)\sim\sqrt{\frac{2\pi}{\lambda|g''(x_{0})|}}f(x_{0})\cos(\lambda g(x_{0})\pm\pi/4),$$ where$g'(x_{0}) = 0$ and$\pm 1 = \text{sgn}(g''(x_{0}).$
- Suppose for large
$\lambda$ , we want to evaluate the integral$$I(\lambda) = \int_{\mathcal{C}}f(z)e^{\lambda g(z)}\mathrm{d}z,$$ where$\mathcal{C}$ is some contour in the complex plane (say$\mathcal{C} = \set{q(t):a\leq t\leq b}$ )and$f(z)$ and$g(z)$ are complex differentiable. - The dominant contribution might come from where
$\text{Re}(g(z))$ is maximum. Suppose along$\mathcal{C}$ , this occurs at$z = z_{0} = q(t_{0})$ , where$\frac{\mathrm{d}}{\mathrm{d}t}\text{Re}\left(g(q(t))\right) = 0$ . So near$t_{0}$ ,$$f(q(t)) = f(z_{0})+...\quad g(q(t)) = g(z_{0}) + (t-t_{0})q'(t_{0})g'(z_{0})+...\quad f(z)e^{\lambda g(z)}\cong f(z_{0})e^{\lambda g(z_{0})}e^{\lambda (t-t_{0})q'(t_{0})g'(z_{0})},$$ which is highly oscillatory, so try another approach. - Taking the contour with the lower
$\text{Re}(g(z))$ means passing through the saddle point at$z = z_{* }$ , where$g'(z_{* }) = 0$ . Near$z_{* }$ ,$$f(z) = f(z_{* }) + (z-z_{* })f'(z_{* })+...,\quad g(z) = g(z_{* }) + 0 + \frac{1}{2}(z-z_{* })^{2}g''(z_{* })+...$$ - Steepest descent from
$z_{* }$ means taking the direction where$\frac{1}{2}(z-z_{* })^{2}g''(z_{* }) < 0$ and real. - Parameterising the section of the steepest descent contour
$\mathcal{C}_ {sd}$ near to$z_{* }$ by$t_{* }$ , we therefore set$z-z_{* } = t_{* }/\sqrt{-\lambda g''(z_{* })}$ . -
Changing variables to
$t_{* }$ ,$$I(\lambda) = \int_{-\infty}^{\infty}\left(f(z_{* })+\frac{t_{* }f'(z_{* })}{\sqrt{-\lambda g''(z_{* })}} + O\left(\frac{1}{\lambda}\right)\right)\exp\set{\lambda g(z_{* }) - \frac{1}{2}t_{* }^{2}+O\left(\frac{1}{\sqrt{\lambda}}\right)}\frac{\mathrm{d}t_{* }}{\sqrt{-\lambda g''(z_{* })}} = \sqrt{\frac{2\pi}{-\lambda g''(z_{* })}} f(z_{* })e^{\lambda g(z_{* })}\left(1+O\left(\frac{1}{\lambda}\right)\right). $$ - The dominant contribution comes from a region of size
$z - z_{* } = \Theta(\lambda^{-1/2}),$ so the details of$\mathcal{C}_ {sd}$ aren't important away from the saddle point$z_{* }$ . - Sometimes there are more than one saddle point. The dominant contribution comes from the highest saddle point (with the biggest Re$(g(z))$) that mus tbe deformed through to get from the start point to the end point.
- Sometimes the highest Re$(g(z))$ is at an end point. Then use Watson's Lemma instead for the dominant contribution.
- Consider
$$I(\lambda) = \int_{a}^{b}f(x)e^{i\lambda g(x)}\mathrm{d}x\quad\text{as}\quad\lambda\to\infty,$$ with$f$ and$g$ real. Suppose$g'(x_{0}) = 0$ for some$x_{0}\in(a,b)$ . Then near$x_{0}$ ,$$ig(x) = ig(x_{0}) + i(x-x_{0})g'(x_{0}) + \frac{1}{2}i(x-x_{0})^{2}g''(x_{0})+...$$ - For the steepest descent contour, we want
$$\frac{i}{2}(x-x_{0})^{2}g''(x_{0}) = -\frac{1}{2}t^{2}\implies x = x_{0} + \frac{te^{\pm i\pi/4}}{\sqrt{\pm g''(x_{0})}},$$ where$\pm 1 = \text{sgn}(g''(x_{0})).$ - Applying the steepest descent formula,
$$I(\lambda) = \int_{a}^{b} f(x)e^{i\lambda g(x)}\mathrm{d}x\sim\sqrt{\frac{2\pi}{\lambda|g''(x_{0})|}}f(x_{0})e^{i\lambda g(x_{0})\pm i\pi/4}\quad\text{as}\quad\lambda\to\infty.$$
Consider a damped spring with extension
-
$y^{* }$ = meters$(m)$ , -
$t^{* }$ = seconds$(s)$ , -
$m^{* }$ = kilograms$(kg)$ , -
$K^{* }$ =$kgs^{-2}$ , -
$D^{* }$ =$kgs^{-1}$ .
In the above differential equation, is
To answer this, we nondomensionalise. Pick a timescale
-
$t^{* } = tT^{* } = t\sqrt{\frac{m^{* }}{K^{* }}}$ and$y^{* }(t^{* }) = L^{* }y(t)$ , then -
$$\frac{\mathrm{d}y^{* }}{\mathrm{d}t^{* }} = L^{* }\sqrt{\frac{K^{* }}{m^{* }}}\frac{\mathrm{d}y}{\mathrm{d}t}$$ and$$\frac{\mathrm{d}^{2}y^{* }}{(\mathrm{d}t^{* })^{2}} = L^{* }\frac{K^{* }}{m^{* }}\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}.$$
Note that
Consider
- Using Method 3: series expansion, we guess a series
$$y(t) = y_{0}(t) + \varepsilon y_{1}(t) + \varepsilon^{2}y_{2}(t)+...,$$ - Substituting in,
$$(\ddot{y_{0}} + \varepsilon\ddot{y_{1}}+...) + (\varepsilon\dot{y_{0}}+\varepsilon^{2}\dot{y_{1}}+...) + (y_{0} + \varepsilon y_{1}+...) = 0,\quad y(0) = 0 = y_{0}(0)+\varepsilon y_{1}(0)+...\quad \dot{y}(0) = 1 = \dot{y_{0}}(0) + \varepsilon\dot{y_{1}}(0)+...$$ - Solving at each other,
$$\Theta(\varepsilon^{0}): \ddot{y_{0}}+y_{0} = 0\quad y_{0}(0) = 0\quad \dot{y_{0}}(0) = 1\implies y_{0} = A_{0}\cos t+B_{0}\sin t = \sin t,$$ and$$\Theta(\varepsilon^{1}): \ddot{y_{1}} + \dot{y_{0}} + y_{1} = 0,\quad y_{1}(0) = 0\quad\dot{y_{1}}(0) = 0\implies \ddot{y_{1}}+y_{1} = -\cos t\implies y_{1}(t) = A_{1}\cos t + B_{1}\sin t - \frac{1}{2}t\sin t = -\frac{1}{2}t\sin t.$$ - So the asymptotic solution is
$y\sim\sin t - \frac{1}{2}\varepsilon t\sin t+...$
Sometimes we need to rescale differential equations first. Consider
- Try change of variable
$t = a\tau$ . Then$$\frac{1}{a^{2}}\frac{\mathrm{d}^{2}y}{\mathrm{d}\tau^{2}} + \frac{1}{a\varepsilon}\frac{\mathrm{d}y}{\mathrm{d}\tau} + y = 0\quad, y(0) = 0,\quad \frac{\mathrm{d}y}{\mathrm{d}\tau}(0) = a.$$ - Mathcing terms:
$$\frac{1}{a^{2}} = 1\implies a = 1,$$ $$\frac{1}{a\varepsilon} = 1\implies a = \frac{1}{\varepsilon}$$ and$$\frac{1}{a^{2}} = \frac{1}{a\varepsilon}\implies a = \varepsilon.$$ - For
$a = \varepsilon$ , we get$$\frac{\mathrm{d}^{2}y}{\mathrm{d}\tau^{2}} + \frac{\mathrm{d}y}{\mathrm{d}\tau} + \varepsilon^{2}y = 0\quad y(0) = 0,\quad \frac{\mathrm{d}y}{\mathrm{d}\tau}(0) = \varepsilon.$$ - Use Method 2: successive approximation: for
$\varepsilon = 0$ ,$$y_{0} = A_{0} + B_{0}e^{-\tau} = \varepsilon(1-e^{-\tau}).$$ - So try
$y(\tau) = y_{0}(\tau) + e_{0}(\tau),$ $$\frac{\mathrm{d}^{2}e_{0}}{\mathrm{d}\tau^{2}} + \frac{\mathrm{d}e_{0}}{\mathrm{d}\tau} + \varepsilon^{2}e_{0}+\varepsilon^{3}(1-e^{-\tau}) = 0,\quad e_{0}(0) = 0,\quad \frac{\mathrm{d}y}{\mathrm{d}\tau}(0) = 0,$$ $$\implies e_{0} = -\varepsilon^{3}\tau(1+e^{-\tau}) + 2\varepsilon^{3}(1-e^{-\tau}) + e_{1}(\tau).$$ - Consequently, as
$\varepsilon\to 0$ for fixed$\tau$ ,$$y\sim\varepsilon(1-e^{-\tau}) - \varepsilon^{3}(\tau(1+e^{-\tau}) - 2(1-e^{-\tau})).$$
- Suppose our solution
$f(x)$ has an outer solution which satisfies, as$\varepsilon\to 0$ with$x$ fixed,$$f(x)\sim\sum_{n=0}^{P}\varepsilon^{n} f_{n}(x) = E_{P}f,$$ where$E_{p}$ means taking the first$(P+1)$ terms of the outer asymptotic series. - Suppose also that for
$x = \varepsilon\xi$ , then as$\varepsilon\to 0$ with$\xi$ fixed, we obtain the inner solution$$f(\varepsilon\xi)\sim\sum_{n=0}^{Q}\varepsilon^{n} g_{n}(\xi) = H_{Q}f,$$ where$H_{Q}$ means taking the first$(Q+1)$ terms of the inner asymptotic seires. -
Van Dyke's matching rule states that these two asymptotic series should match according to the matching rule,
$$E_{P}H_{Q}f = H_{Q}E_{P}f,$$ for suitable values of$P$ and$Q$ ; i.e. expanding the inner solution in the outer variable gives the same as expanding the outer solution in the inner variable.
Suppose that, for
- This composite expansion
$C(x;\varepsilon)$ is not a Poincare series, and so is not necessarily unique. For example, a multiplicative composite expansion is given by$$M(x;\varepsilon) = \frac{(E_{P}f)(H_{Q}f)}{E_{P}H_{Q}f}.$$
For nonlinear ODEs for
- To leading order, for
$y = \Theta(1)$ and$t = \Theta(1)$ ,$$0 = y_{0}\dot{y_{0}} - y_{0} = y_{0}(\dot{y_{0}} - 1)\implies y_{0} = 0\quad\text{or}\quad y_{0} = A+t.$$ - In either case,
$y_{0}\approx 0$ for some value of$t$ , and near that value of$t$ we would have$y_{0} = \Theta(\varepsilon)$ , and therefore$\varepsilon\ddot{y_{0}}$ could be important. - So instead, we consider rescaling
$t = \varepsilon^{\alpha}\tau$ and$y(t) = \varepsilon^{\beta}Y(\tau)$ . Then$$\varepsilon^{1+\beta - 2\alpha}Y'' + \varepsilon^{2\beta - \alpha}Y' - \varepsilon^{\beta} Y = 0.$$ - Mathcing the powers of
$\varepsilon$ in these three terms gives$\alpha = \beta = \frac{1}{2}$ , for which the ODE becomes$$\sqrt{\varepsilon}(Y'' + YY' - Y) = 0,$$ which is a full nonlinear problem with no small parameters.
Given a function
Often,
- A useful estimate for the magnitude of a complex integral is that
$$\left|\int_{\mathcal{C}}f(z)\mathrm{d}z\right|\leq\max_{z\in\mathcal{C}}|f(z)|\int_{a}^{b}\left|\frac{\mathrm{d}q}{\mathrm{d}t}\right|\mathrm{d}t = \max_{z\in\mathcal{C}}|f(z)|L,$$ where$L$ is the length of the curve$\mathcal{C}$ .
- If
$f(z)$ is differentiable everywhere in$\mathcal{D}$ and$\mathcal{D}$ is simply connected, which means if any path in$\mathcal{D}$ can be smoothly deformed into any other path in$\mathcal{D}$ between the same ned points without moving the end points, then the integral of$f(z)$ is independent of the path.
First of all, consider an isolated singularity of
- If
$\lim_{z\to z_{0}} f(z)$ exists, then$z_{0}$ is a removable singularity. - If
$\lim_{z\to z_{0}}(f(z) - a_{1}/(z-z_{0}))$ exists for some$a_{1}\ne 0$ , then$f$ has a simple pole at$z_{0}$ with residue$\text{Res}(f,z_{0}) = a_{1}$ . - If the limit
$$\lim_{z\to z_{0}}\left(f(z)-\frac{a_{1}}{z-z_{0}}-\frac{a_{2}}{(z-z_{0})^{2}}-...-\frac{a_{n}}{(z-z_{0})^{n}}\right)$$ exists for some$n$ and some$a_{1},...,a_{n}$ with$a_{n}\ne 0$ , then$f$ has a pole of order$n$ at$z_{0}$ . - Otherwise,
$f$ has an essential singularity at$z_{0}$ , e.g.$$f(z) = e^{1/z}.$$
Consider the function
- Consider the integration contour given by
$\mathcal{C}_ {\pm} = \set{Re^{\pm i\theta}:0\leq\theta\leq\pi}$ . Along$\mathcal{C}_ {+}$ , provided$t > 0$ ,$$\left|\frac{1}{2\pi}\int_{\mathcal{C}_ {+}}\tilde{g}(\omega)e^{i\omega t}\mathrm{d}\omega\right|\leq\frac{1}{2\pi}\frac{2}{R^{2}-2R-5}\cdot 1\cdot \pi R\to 0,$$ as$R\to\infty$ . - The part
$\frac{2}{R^{2}-2R-5}\leq\max|\tilde{g}|$ , the part$1 = \max|e^{i\omega t}|$ and$\pi R = L$ . - Hence if
$\mathcal{C}_ {1} = \set{t:-R\leq t\leq R}$ , then for$t > 0$ ,$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{g}(\omega)e^{i\omega t}\mathrm{d}\omega = \lim_{R\to\infty}\frac{1}{2\pi}\int_{\mathcal{C}_ {1}\cup\mathcal{C}_ {+}}\tilde{g}(\omega)e^{i\omega t}\mathrm{d}\omega,$$ and similarly for$t < 0$ and$\mathcal{C}_ {1}\cup\mathcal{C}_ {-}$ . - We can therefore consider the closed integration contours
$\mathcal{C}_ {1}\cup\mathcal{C}_ {+}$ and$\mathcal{C}_ {1}\cup\mathcal{C}_ {-}$ . - The singularities of
$\tilde{g}(\omega)$ are when$$\omega^{2}-2i\omega - 5 = 0\Leftrightarrow \omega = i\pm 2,$$ so residues are$$\text{Res}(i+2) = \frac{1}{2\pi}\frac{2e^{i(i+2)t}}{2i-2(i+2)} = \frac{1}{4\pi}e^{-t}e^{-2it},$$ and$$\text{Res}(i-2) = \frac{-1}{4\pi}e^{-t}e^{2it}.$$ - For
$t > 0$ , we can use Cauchy's Theorem to deform the contour of integration to around the singularities, and using Residue's Theorem gives for$t > 0$ ,$$g(t) = 2\pi i\left(\frac{1}{4\pi}e^{-t}e^{-2it}-\frac{1}{4\pi}e^{-t}e^{2it}\right) = e^{-t}\sin(2t).$$ - For
$t < 0$ , we get$g(t) = 2\pi i(0) = 0$ . Hence,$$g(t) = 1_{t > 0}e^{-t}\sin(2t).$$
We can check this by using Fourier Transform:
- The Fourier transform of a function
$f(t)$ is$$\tilde{f}(\omega) = (\mathcal{F}f)(\omega) = \int_{-\infty}^{\infty}f(t)e^{-i\omega t}\mathrm{d}t.$$ - The Inverse Fourier transform of a function
$\tilde{f}(\omega)$ is$$f(t) = (\mathcal{F}^{-1}\tilde{f})(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{f}(\omega)e^{i\omega t}\mathrm{d}t.$$
-
Linearity: If
$h(t) = af(t)+bg(t)$ , then$\tilde{h}(\omega) = a\tilde{f}(\omega)+b\tilde{g}(\omega)$ . -
Shift property(I): If
$g(t) = f(t-a)$ , then$\tilde{g}(\omega) = e^{-i\omega a}\tilde{f}(\omega)$ . -
Shift property(II): If
$g(t) = e^{i\Omega t}f(t)$ , then$\tilde{g}(\omega) = \tilde{f}(\omega - \Omega)$ . -
Scaling: If
$g(t) = f(at)$ , then$\tilde{g}(\omega) = \frac{1}{|a|}\tilde{f}(\omega/a)$ . -
Differentiation: If
$g(t) = f'(t)$ , then$\tilde{g}(\omega) = i\omega\tilde{f}(\omega)$ . -
Product: If
$g(t) = tf(t)$ , then$\tilde{g}(\omega) = i\tilde{f}'(\omega)$ .
Consider
- Using Fourier transformation with the properties,
$$(i\omega)^{2}\tilde{y}+i\omega\tilde{y}+i\frac{\mathrm{d}}{\mathrm{d}\omega}\tilde{y} = 0\implies \tilde{y} = Ae^{-\omega^{2}/2-i\omega^{3}/3}.$$ - Hence, by inversing Fourier transform,
$$y(t) = \frac{A}{2\pi}\int_{-\infty}^{\infty}e^{-\omega^{2}/2-i\omega^{3}/3+i\omega t}\mathrm{d}\omega,$$ where we choose$A$ so that$y(0) = 1$ . - In fact, we needed
$y\to 0$ as$t\to-\infty$ for the Fourier transform$\tilde{y}$ to exist.
- Given
$f(t)$ and$g(t)$ , we define the convolution$h = f* g$ by$$h(t) = \int_{-\infty}^{\infty}f(\tau)g(t-\tau)\mathrm{d}\tau = \int_{-\infty}^{\infty}f(t-\tau)g(\tau)\mathrm{d}\tau.$$ - So,
$$\tilde{h}(\omega) = \tilde{f}(\omega)\tilde{g}(\omega).$$
Given two functions
- This is Parseval's theorem. In particular, if
$f(t) = g(t)$ , then$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|\tilde{f}(\omega)|^{2}\mathrm{d}\omega = \int_{-\infty}^{\infty}|f(t)|^{2}\mathrm{d}t,$$ which is known as Plancherel's theorem.
One limitation of Fourier transforms is that they only exist for functions
- Extending
$f:[0,\infty]\to\mathbb{C}$ to$g:\mathbb{R}\to\mathbb{C}$ by setting$$g(t) = f(t)\quad t\geq 0,$$ and$$g(t) = 0\quad t < 0.$$ - This is called zero-extending.
- The Laplace Transform is given by
$$\hat{f}(s) = \int_{0}^{\infty}f(t)e^{-st}\mathrm{d}t.$$ - The Inverse Laplace Transform of a function
$f(t)$ is given by$$f(t) = \frac{1}{2\pi i}\int_{-i\infty+\alpha}^{i\infty+\alpha}\hat{f}(s)e^{st}\mathrm{d}s,$$ with$\alpha > \text{Re}(s_{j})$ for all singularities$s_{j}$ of$\hat{f}(s)$ .
- For
$f(t) = 1$ , provided$\text{Re}(s) > 0$ ,$$\hat{f}(s) = \int_{0}^{\infty}e^{-st}\mathrm{d}t = \frac{1}{s}.$$ - For
$f(t) = e^{at}$ , provided$\text{Re}(s) > \text{Re}(a)$ ,$$\hat{f}(s) = \int_{0}^{\infty}e^{at}e^{-st}\mathrm{d}t = \frac{1}{s-a}.$$ - For
$f(t) = \cos(at)$ , provided$\text{Re}(s) > |\text{Im}(a)|$ ,$$f(t) = \frac{1}{2}(e^{iat}+e^{-iat}\implies\hat{f}(s) = \frac{s}{s^{2}+a^{2}}.$$ - For
$f(t) = \sin(at)$ , provided$\text{Re}(s) > |\text{Im}(a)|$ ,$$f(t) = \frac{1}{2i}(e^{iat}-e^{-iat})\implies\hat{f}(s) = \frac{a}{s^{2}+a^{2}}.$$ - For
$f(t) = t^{n}$ , provided$\text{Re}(s) > 0$ ,$$\hat{f}(s) = \int_{0}^{\infty}t^{n}e^{-st}\mathrm{d}t = \frac{\Gamma(n+1)}{s^{n+1}}.$$ - For
$f(t) = e^{t^{2}}$ , then$\hat{f}(s)$ doesn't exist for any$s$ .
-
Linearity: If
$h(t) = af(t)+bg(t)$ , then$\hat{s} = a\hat{f}(s)+b\hat{g}(s).$ -
Shift property(I): If
$g(t) = f(t-a)$ , then$\hat{g}(s) = e^{-sa}\hat{f}(s)$ . (Have to zero extend first) -
Shift property(II): If
$g(t) = e^{at}f(t)$ , then$\hat{g}(s) = \hat{f}(s-a).$ -
Scaling: If
$g(t) = f(at)$ , then$\hat{g}(s) = \frac{1}{a}\hat{f}(s/a)$ . -
Product: If
$g(t) = tf(t)$ , then$\hat{g}(s) = -\frac{\mathrm{d}\hat{f}}{\mathrm{d}s}(s)$ . -
Differentiation1: If
$g(t) = \frac{\mathrm{d}f}{\mathrm{d}t}(t) = \dot{f}(t)$ , then$\hat{g}(s) = -f(0) + s\hat{f}(s)$ . -
Differentiation2: If
$g(t) = \ddot{f}(t)$ , then$\hat{g}(s) = -\dot{f}(0) - sf(0) + s^{2}\hat{f}(s)$ .
Consider
- Try the Laplace transform:
$$-\dot{y}(0) - sy(0) + s^{2}\hat{y}-y(0) + s\hat{y} - 2\hat{y} = \frac{1}{s+2}\implies\hat{y} = \frac{1}{9}\left(\frac{1}{s-1}-\frac{1}{s+2}-\frac{3}{(s+2)^{2}}\right)+\frac{1}{3}\left(\frac{2}{s-1}+\frac{1}{s+2}\right).$$ - Hence by Inverse Laplace Transform,
$$y = \frac{1}{9}\left(e^{t}-e^{-2t}-3te^{-2t}\right)+\frac{1}{3}\left(2e^{t}+e^{-2t}\right) = \frac{1}{9}\left(7e^{t}+(2-3t)e^{-2t}\right).$$
- If
$f(t) = g(t) = 0$ for$t < 0$ , then$$h(t) = (f* g)(t) = \int_{-\infty}^{\infty}f(\tau)g(t-\tau)\mathrm{d}\tau = \int_{0}^{t}f(\tau)g(t-\tau)\mathrm{d}\tau,$$ and the Laplace transforming gives$$\hat{h}(s) = \hat{f}(s)\hat{g}(s).$$ Note:$f(t)$ here do not have to be$f(t)$ , it can also be$\dot{f}(t)$ . E.g. Suppose we want to find$f(t)$ for$t > 0$ which solves$$f(t) = a\sin t+2\int_{0}^{t}\dot{f}(\tau)\sin(t-\tau)\mathrm{d}\tau,$$ subject to$f(0) = 0$ . - The integral here is a convolution,
$\dot{f}* \sin t$ , and the limits suggest a Laplace transform:$$\hat{f}(s) = \frac{a}{s^{2}+1}+2\left(-f(0)+s\hat{f}\right)\frac{1}{s^{2}+1}\implies\hat{f} = \frac{a}{(s-1)^{2}}\implies f(t) = ate^{t}.$$
For
- Laplace transform
$t\mapsto s$ :$$x(-u(x,0) + s\hat{u}(x,s)) + \frac{\partial\hat{u}}{\partial x} = \frac{x}{s},\quad\text{with}\quad\hat{u}(0,s) = 0.$$
For