M3M6: Methods of Mathematical Physics 2018

$$ \def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\E{{\rm e}} \def\dx{\D x} \def\dt{\D t} \def\dz{\D z} \def\C{{\mathbb C}} \def\R{{\mathbb R}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\I{{\rm i}} \def\qqqquad{\qquad\qquad} \def\qqfor{\qquad\hbox{for}\qquad} \def\qqwhere{\qquad\hbox{where}\qquad} \def\Res_#1{\underset{#1}{\rm Res}}\, \def\sech{{\rm sech}\,} \def\vc#1{{\mathbf #1}} $$

Dr. Sheehan Olver
s.olver@imperial.ac.uk

Website: https://github.com/dlfivefifty/M3M6LectureNotes

Solution Sheet 1

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Problem 1.1

1.

Use fundamental theorem of algebra: a polynomial is a constant times a product of terms like $z - \lambda_k$, where $\lambda_k$ are the roots. In this case, the roots are $a$ times the quartic-root of $-1$, hence this gives us: \begin{align*} z^4 + a^4 = (z-a \E^{\I \pi /4})(z-a \E^{3\I \pi /4})(z-a \E^{5\I \pi /4})(z-a \E^{7\I \pi /4}) \end{align*} We are only interested in the root $a \E^{\I \pi /4}$, thus we simplify the expression Therefore $$ {z^3 \sin z \over z^4 + a^4} = {z^3 \sin z \over (z-a \E^{3\I \pi /4})(z-a \E^{5\I \pi /4})(z-a \E^{7\I \pi /4})} {1 \over z-a \E^{\I \pi /4}} = {a^3\E^{3\I \pi /4} \sin(a \E^{\I \pi /4}) \over a^3 (\E^{\I \pi / 4} - \E^{3\I \pi /4})(\E^{\I \pi / 4} - \E^{5\I \pi /4})(\E^{\I \pi / 4} - \E^{7\I \pi /4})} {1 \over z-a \E^{\I \pi /4}} + O(1) $$ Therefore, $$ \Res_{z = a \E^{\I \pi /4}} {z^3 \sin z \over z^4 + a^4} = { \E^{3\I \pi /4} \sin(a \E^{\I \pi /4}) \over \E^{3\I \pi /4} (1 - \E^{\I \pi \over 2})(1 - \E^{\I \pi })(1 - \E^{3\I \pi/2 })}= { \sin(a \E^{\I \pi /4}) \over (1 - \I) (2) (1+ \I)}= { \sin(a \E^{\I \pi /4}) \over 4} $$

Let's check our work: we compare the numerically calculated residue to the formula we have derived:

2.

We have $$ (z^2-1)^2 = (z-1)^2(z+1)^2 $$ Thus this is a slightly more challenging since it has a double pole. But we can expand using Geometric series: $$ {z+1 \over (z^2-1)^2} = {1 \over (z-1)^2} {1 \over 2 - (1-z)} = {1 \over (z-1)^2} {1 \over 2} (1 + (1-z)/2 +O(1-z)^2) = {1 \over 2(z-1)^2} - {1 \over 4 (z-1)} + O(1) $$ Thus the residue is the negative-first Laurent coefficient, namely $-{1 \over 4}$.

We again check our work:

3.

$$ {z^2 \E^z \over z^3 - a^3} = {z^2 \E^z \over (z-a) (z^2 + az + a^2)} $$

We thus need only evaluate the extra term at $z=a$: $$ \Res_{z = a} {z^2 \E^z \over z^3 - a^3} = {\E^a \over 3} $$

Let's check:

Problem 1.2

1.

Change of variables $z = \E^{\I \theta}$, $\dz = \I \E^{\I \theta} \D\theta = \I z \D\theta$, $\cos \theta = {z + z^{-1} \over 2}$ gives $$ \int_0^{2 \pi} {\D\theta \over 5-4 \cos \theta} = - \I \oint {\dz \over 5z - 2z^2 - 2} = \I \oint {\dz \over (z-2)(2z-1)} = -\pi \Res_{z=1/2} {1 \over (z-2)(z-1/2)} = {2 \over 3} \pi $$

2.

Use $\cos 2 \theta = { \E^{2 \I \theta} + \E^{-2 \I \theta} \over 2} = {z^2 + z^{-2} \over 2}$ to get $$ \int_0^{2 \pi} {\cos 2 \theta \D\theta \over 5+4 \cos \theta} = - {\I \over 4} \oint { (z^4 + 1)\dz \over z^2 (z+1/2)(z+2)} = {\pi \over 2} \left( \Res_{z=-1/2} + \Res_{z=0}\right) {z^4+1 \over z^2(z+2)(z+1/2)} = {\pi \over 6} $$

3.

Because the integrand is analytic and $O(z^{-2})$ in the upper half plane, we can use the residue theorem in the upper half plane using $$ {1 \over (z^2+1) (z^2 + 4)} = {1 \over (z+\I)(z-\I) (z+2\I)(z-2\I)} $$

This has two poles in the upper half plane:

\begin{align*} \int_{-\infty}^\infty {1 \over (x^2+1)(x^2+4)} \dx &= 2 \pi \I \left( \Res_{z = \I} + \Res_{z = 2\I}\right){1 \over (z^2+1)(z^2+4)} \\ & = 2 \pi \I \left( {1 \over 2 \I 3 \I (-\I)} + {1 \over 3 \I \I 4\I} \right) = \pi/ 6 \end{align*}

We can check the result numerically:

4.

Again, decays like $O(z^{-2})$ in upper half plane so we can use residue calculus. This integrand has poles at $z = \I$ and $z = 3 \I$:

The residues are $(-1-\I)/16$ and $(3-7\I)/48$ giving the answer $$ 5\pi \over 12 $$

5.

$$ \int_{-\infty}^\infty {1 \over x + \I } \dx $$

Trick question: it's undefined because the integral doesn't decay fast enough. But what if I had asked for $$ \dashint_{-\infty}^\infty {1 \over x + \I } \dx? $$

We can't use residue theorem since it doesn't decay fast enough, but we can use, with a contour $C_R = \{ R \E^{\I theta} : 0 \leq \theta \leq \pi \}$ $$ \oint_{[-M, M] \cup C_R} {1 \over z + \I} = 0 $$ Further, by direct substitution, we have $$ \int_{C_R} {1 \over z + \I}\dz = \I \int_0^\pi R {\E^{\I \theta} \over R \E^{\I \theta} + \I} \D \theta $$ Letting $R \rightarrow \infty$, the integrand tends to one uniformly hence
$$ \int_{C_R} {1 \over z + \I}\dz \rightarrow \I \int_0^\pi \D \theta = \I \pi. $$ Therefore, we have $$ \dashint_{-\infty}^\infty {1 \over x + \I } \dx = \lim_{M\rightarrow \infty} \int_{-M}^M {1 \over x + \I} \dx = - \I \pi . $$

Indeed:

6.

$$ \int_{-\infty}^\infty {\sin 2x \over x^2 + x + 1} \dx = \Im \int_{-\infty}^\infty {\E^{ 2 \I x} \over x^2 + x + 1} \dx $$

This can be deformed in the upper half plane with a pole at ${-1 + \I \sqrt{3} \over 2}$, using residue calculus gives us $$ -{2 \pi \over \sqrt 3} {\sin 1 \over \E^\sqrt{3}} $$

7.

$$ \int_{-\infty}^\infty {\cos x \over x^2 + 4} \dx = \Re \int_{-\infty}^\infty {\E^{\I x}\over x^2 + 4} \dx $$

and residue calculus gives ${\pi \over 2 \E^2}$

8.

$$ \int_{-\infty}^\infty {x \sin x \over x^2 + 1} \dx = { \pi \over \E} $$

using Residue calculus. You need to appeal to Jordan's lemma to argue that it can still be done even with only $O(x^{-1})$ decay.

9.

$$ \int_{-\infty}^\infty {\cos a x - \cos b x \over x^2} \dx \qqwhere a,b > 0 $$

We have for $f(x) = {\E^{\I a x} - \E^{\I b x} \over x^2}$ $$ \Re f(x) = {\cos a x - \cos b x \over x^2} $$

Note that, since $\cos x = 1 + x^2/2 + O(x^4)$, the integrand is fine near zero: $$ {\cos a x - \cos b x \over x^2} = {(a -b) \over 2} + O(x^2) $$ But $f(x)$ has a pole: $$ {\E^{\I a x} - \E^{\I b x} \over x^2} = {\I (a-b) \over x} + O(1) $$ To rectify this, we need to be a bit more careful. First note that $$ \int_{\infty}^\infty {\cos a x - \cos b x \over x^2} \dx = \lim_{\epsilon \rightarrow 0} \left(\int_{-\infty}^{-\epsilon} + \int_\epsilon^\infty \right){\cos a x - \cos b x \over x^2} \dx = \Re \dashint_{-\infty}^\infty f(x) \dx $$ Then we construct a contour avoiding zero as follows:

Note that $\oint_\gamma f(z) \dz = 0$, $$\int_{C_\epsilon} {\E^{\I a z } - \E^{\I b z} \over z^2} \dz = \int_\pi^0 { (b-a) \E^{\I \theta} + O(\epsilon) \over \E^{ \I \theta}} \D\theta \rightarrow (a-b) \pi $$ Also, as the integrand is $O(z^{-2})$ the integral over $C_M$ vanishes as $M \rightarrow \infty$. We therefore get $$ \dashint f(x)\dx = (b-a)\pi $$

10.

Use binomial formula \begin{align*} \int_0^{2 \pi} (\cos \theta)^n \D \theta &= {1 \over 2^n \I} \oint (z+z^{-1})^n {\dz \over z} \\ &= {1 \over 2^n \I} \sum_{k=0}^n {n! \over k! (n-k)!} \oint z^k z^{k-n} {\dz \over z} \\ &= {1 \over 2^n \I} \sum_{k=0}^n {n! \over k! (n-k)!} \oint z^{2k-n-1}\dz \end{align*}

We only have a residue of $2k-n-1 = -1$, that is, if $2k = n$. If $n$ is odd, this can't happen (duh! the integral is symmetric with respect to $\theta$). If it's even, then we have $$ \int_0^{2 \pi} (\cos \theta)^n \D \theta = {\pi \over 2^{n-1} } {n! \over 2 (n/2)!} $$

Problem 2.1

By integrating around a rectangular contour with vertices at $\pm R$ and $\pi \I \pm R$ and letting $R \rightarrow \infty$, show that: $$\int_0^\infty \sech x \dx = {\pi \over 2}$$ where $\sech x = {2 \over \E^{-x} + \E^x}$.

Recall $\sech x = {2 \over \E^{-x} + \E^x} $. This shows that $\sech(-x) = \sech x$ But we also have $ \sech(x + \I \pi) = {2 \over \E^{-x-\I \pi} + \E^{x+ \I \pi}} = {2 \over -\E^{-x} - \E^{x}} = -\sech x $ Thus we have $$ 4\int_0^\infty \sech x \dx = \left[ \int_{-\infty}^\infty + \int_{\infty+\I\pi}^{-\infty + \I \pi} \right] \sech z \dz $$

We can approximate this using $$ \left[ \int_{-R}^R + \int_R^{R+\I \pi} + \int_{R+\I \pi}^{-R+\I\pi} + \int_{-R+\I \pi} \right ] \sech z \dz = 2 \pi \I \Res_{z = {\I \pi \over 2}} \sech z = 2 \pi $$ since, for $z_0 = {\I \pi \over 2}$, we have $$ \sech z = {1 \over \cos \I z} = {1 \over - \I \sin \I z_0 (z- z_0) + O(z-z_0)^2} = -{\I \over (z- z_0)} + O(1) $$

Finally, we need to show that the limit as $R \rightarrow \infty$ tends to the right value. In this case, it follows since $$ \left|\int_R^{R+\I \pi} \sech z \dz \right| \leq {2 \pi \E^{-R} \over 1 - \E^{-2R}} \rightarrow 0 $$ (and by symmetry for $\int_{-R+\I \pi}^{-R}$.)

Problem 2.2

Show that the Fourier transform of $\sech x$ satisfies $$ \int_{-\infty}^\infty \E^{\I k x } \sech x \dx = \pi \sech {\pi k \over 2} $$

Define $$f(z) = \E^{\I k z } \sech z = {2 \E^{(1+\I k) z} \over \E^{2 z } + 1}$$ In this case, we have the symmetry $$ f(x + \I \pi) = - \E^{- k \pi} \E^{\I k x} \sech x = - \E^{-k \pi} f(x) $$

In other words, we have $$ (1 + \E^{-k \pi}) \int_{-\infty}^\infty f(x) \dx =\left(\int_{-\infty}^\infty + \int_{\infty+\I \pi}^{-\infty + \I \pi}\right) f(z) \dz $$ By similar logic as above, we can show that the integral over the rectangular contour converges to this.

Again, the only pole inside is at $z = {\I \pi \over 2}$, where the residue is $-\I \E^{-\pi k \over 2}$. Thus we have $$ \int_{-\infty}^\infty f(x) \dx = {2 \pi \E^{-\pi k \over 2} \over 1 + \E^{-k \pi}} = \pi \sech{\pi k \over 2} $$

Problem 3.1

We have $$ \rho(A) \subseteq B(1,3) \cup B(2,3) \cup B(4,1) $$ where $B(z_0,r)$ is the ball of radius $r$ around $z_0$.

Here's a depiction:

Problem 3.2

We get $$ \rho(A) \subseteq B(1,2) \cup B(2,3) \cup B(4,2) $$

Problem 3.3

Because the spectrum live in the intersection of the two estimates, the sharpest bound is $$ \rho(A) \subseteq B(2,3) $$

Thus we can take $2 + 3 \E^{\I \theta}$ as the contour.

Problem 4.1

Note that in the scalar case $u'' = a u$ we have the solution $$ u(t) = u_0 \cosh \sqrt{a} t + v_0 {\sinh \sqrt{a} t \over \sqrt a} $$ Write $$A = Q \begin{pmatrix} \lambda_1 \\ & \ddots \\ && \lambda_n \end{pmatrix} Q^\top $$ where $\lambda_k > 0$ and then the solution has the form, where $\gamma$ is a contour surrounding the eigenvalues and to the right of zero: \begin{align*} \vc u(t) = Q \begin{pmatrix} \cosh \sqrt{\lambda_1} t \\ & \ddots \\ && \cosh \sqrt{\lambda_n} t \end{pmatrix} Q^\top \vc u_0 + Q \begin{pmatrix} {\sinh \sqrt{\lambda_1} t \over \sqrt{\lambda_1}}\\ & \ddots \\ && {\sinh \sqrt{\lambda_n} t \over \sqrt{\lambda_n}} \end{pmatrix} Q^\top \vc v_0 \\ = {1 \over 2 \pi \I} Q \begin{pmatrix} \oint_\gamma {\cosh \sqrt{z} t \dz \over z - \lambda_1}\\ & \ddots \\ && \oint_\gamma {\cosh \sqrt{z} t \dz \over z - \lambda_n}\end{pmatrix} Q^\top \vc u_0 + {1 \over 2 \pi \I} Q \begin{pmatrix} \oint_\gamma {\sinh \sqrt{z} t \over \sqrt z} {\dz \over z - \lambda_1} \\ & \ddots \\ && \oint_\gamma {\sinh \sqrt{z} t \over \sqrt z} {\dz \over z - \lambda_n} \end{pmatrix} Q^\top \vc v_0 \\ = {1 \over 2 \pi \I} \oint_\gamma \cosh \sqrt{z} t Q \begin{pmatrix} (z - \lambda_1)^{-1}\\ & \ddots \\ && (z - \lambda_n)^{-1}\end{pmatrix} Q^\top \vc u_0 \dz + {1 \over 2 \pi \I} \oint_\gamma {\sinh \sqrt{z} t \over \sqrt z} Q \begin{pmatrix} (z - \lambda_1)^{-1}\\ & \ddots \\ && (z - \lambda_n)^{-1}\end{pmatrix} Q^\top \vc v_0 \dz \\ = {1 \over 2 \pi \I} \oint_\gamma \cosh \sqrt{z} t (z I - A)^{-1} \vc u_0 \dz + {1 \over 2 \pi \I} \oint_\gamma {\sinh \sqrt{z} t \over \sqrt z} (z I - A)^{-1} \vc v_0 \dz \end{align*}

Here we verify the formulae numerically:

Time-stepping solution:

Solution via diagonalization:

Solution via elliptic integrals. We chose the ellipse to surround all the spectrum of our particular $A$ with eigenvalues:

Problem 4.2

I put the restriction in because of the $\sqrt z$ term, which look like it is not analytic on $(-\infty,0]$. However, this restriction was NOT necessary, since in fact $\cosh \sqrt z t$ and ${\sinh \sqrt z t \over \sqrt z}$ are entire:

This follows from Taylor series, though I prefer the following argument: we have $$ \cosh z = {\E^{z} + \E^{-z} \over 2} $$ hence $\cosh{\I t } = \cos t = \cosh{-\I t}$. Therefore, on the possible branch cut $\cosh \sqrt z$ is continuous (hence analytic): $$ \cosh \sqrt{x}_+ = \cosh \I \sqrt{|x|} = \cosh -\I \sqrt{|x|} = \cosh\sqrt{x}_- $$ Similarly, $$ \sinh z = {\E^z - \E^{-z} \over 2} $$ implies $\sinh{\I t} = \I \sin t = - \I \sin(-t) = -\sinh{-\I t}$, which gives us continuity: $$ {\sinh \sqrt{x}_+ \over \sqrt{x}_+} = {\sinh \I \sqrt{|x|} \over \I \sqrt{|x|}} = {\sinh(-\I \sqrt{|x|)} \over -\I \sqrt{|x|}} ={\sinh \sqrt{x}_- \over \sqrt{x}_-} $$ furthermore, they are both bounded at zero, hence analytic there too.

Here's a numerical example:

Here's the solution using an elliptic integral:

Problem 5.1

We have $$ {1 \over n} \sum_{j=0}^{n-1} g(\theta_j) = \sum_{k=0}^\infty g_k {1 \over n} \sum_{j=0}^{n-1} \E^{\I k \theta_j} $$ Define the $n$-th root of unity as $\omega = e^{2 \pi \I \over n}$ (that is $\omega^n = 1$), and simplify $$ \sum_{j=0}^{n-1} \E^{\I k \theta_j} =\sum_{j=0}^{n-1} e^{{2 \pi j \I k \over n}} =\sum_{j=0}^{n-1} \omega^{kj}= \sum_{j=0}^{n-1} (\omega^k)^j $$ If $k$ is a multiple of $n$, then $\omega^k = 1$, and this sum is equal to $n$. If $k$ is not a multiple of $n$, use Geometric series: $$ \sum_{j=0}^{n-1} (\omega^k)^j = {\omega^{nk} - 1 \over \omega^k -1} = {1^k - 1 \over \omega^k-1} = 0 $$

Problem 5.2

From lecture 4, we have $|f_k| \leq M_r r^{-|k|}$ for any $1 \leq r < R$, where $M_r$ is the supremum of $f$ in an annulus $\{ z : r^{-1} < |z| < r \}$. Thus from the previous part we have (using geometric series) $$ \left| {1 \over n} \sum_{j=0}^{n-1} g(\theta_j) - { 1\over 2 \pi} \int_0^{2\pi} g(\theta) \D \theta \right| \leq \sum_{K=1}^\infty |f_{Kn}| + |f_{-Kn}| \leq 2M \sum_{K=1}^\infty r^{-Kn} = 2M_r {r^{-n} \over 1 - r^{-n}}. $$ This is an upper bound that decays exponentially fast.

Problem 5.3

Note that $f(z) = 2 z/(4z - z^2 - 1)$ satisfies $f(\E^{\I \theta}) = g(\theta)$. This has two poles at $2 \pm \sqrt 3$:

Here's a phase plot showing the location of poles:

Note that $2+\sqrt 3 = 1/(2-\sqrt 3)$ so in the previous result, we take $R = 2+\sqrt 3$. For any $1 \leq r < 2+\sqrt 3$ we have $$ M_r = {2 \over 4 - r - r^{-1}} $$

Thus we get the upper bounds $$ {4 \over 4 - r - r^{-1}} {r^{-n} \over 1 - r^{-n}} $$

Let's see how sharp it is: