Past exam of the perturbation methods and applied complex variables course of the University of Oxford

Source: released examination bundle, Part C 2007 11.pdf, pages 345–350. The complete scanned paper, including both halves, is covered. No mark allocations are printed.

Put $\delta =\sqrt{ϵ}$ and write the outer asymptotic expansion as $y_0+\delta y_{1/2}+{\delta }^2{y}_1+\cdots$. Equating powers in the differential equation givesThe first two equations give $y_0=A{e}^{x^2/2}$ and $y_{1/2}=B{e}^{x^2/2}$. Since $y_0^{(3)}=A(x^3+3x)e^{x^2/2}$, an integrating factor yieldsHere $A,B,C$ are matching constants; boundary conditions have not yet been imposed.

In a layer of thickness $\ell \ll 1$, the third-derivative and first-derivative terms have sizes $ϵ{\ell }^{-3}y$ and ${\ell }^{-1}y$. Their dominant balance requires ${\ell }^2\sim ϵ$, soThe local fast modes are exponentials with rates $\pm 1/\sqrt{ϵ}$. A mode decaying into the interval can occur at either endpoint: $e^{-x/\sqrt{ϵ}}$ at zero and $e^{-(1-x)/\sqrt{ϵ}}$ at one. The zero fast exponent represents the matching outer component. The term $xy$ is lower order on either fast scale, and even smaller near zero; it creates no further independent layer thickness.

For the left inner coordinate $X=x/\delta$, the equation becomes $Y_{XXX}-Y_X+{\delta }^{2}XY=0$. At leading order the solution that matches a bounded outer expansion is $Y_0=A+D{e}^{-X}$; the growing exponential is excluded. The boundary value gives $A+D=1$. Since $y^{\prime }(0)=1$ is finite, the possible $O({\delta }^{-1})$ derivative must vanish, so $D=0$ and $A=1$.

At the next order, $Y_{1/2}=B+E{e}^{-X}$. Its boundary value is zero, giving $E=-B$, while its physical derivative at zero is $-E=1$. Hence $B=1$, and the left inner expansion is $1+\delta (1-e^{-X})+O({\delta }^2)$.

At the right endpoint put $\xi =(1-x)/\delta$. The equation is $Y_{\xi \xi \xi }-Y_{\xi }=\delta (1-\delta \xi )Y$. Matching the outer expansion at one and imposing $Y(0)=1$ giveAt the next order, $Y_{1/2}^{(3)}-Y_{1/2}^{\prime }=Y_0$. The solution with no growing exponential, the correct matching polynomial, and $Y_{1/2}(0)=0$ isIndeed the operator $D^3-D$ sends $-\sqrt{e}\xi$ to $\sqrt{e}$ and sends $\xi e^{-\xi }$ to $2e^{-\xi }$.

Add the two decaying inner corrections to the outer solution, subtracting their common matching parts. Since $\delta \xi =1-x$, the composite asymptotic expansion isIt reproduces all three boundary conditions to the retained algebraic orders, with exponentially small interactions between endpoints. The right endpoint has an $O(1)$ boundary layer in $y$. The left correction has only amplitude $O(\sqrt{ϵ})$ in $y$, but its derivative is $O(1)$ and is necessary for the derivative boundary condition. Thus the left layer is a weak derivative boundary layer; if “genuine” refers specifically to an order-one change in the function itself, only the right layer is genuine in that sense. Both nonzero corrections belong in the expansion through $O(\sqrt{ϵ})$.

For an asymptotic sequence $a_{n+1}(ϵ)=o(a_n(ϵ))$, the asymptotic expansion condition is that every fixed truncation has remainder smaller than its last retained term:This limit is taken as $ϵ\to 0$ in the specified direction or sector. The number $N$ stays fixed while taking that limit. The infinite series need not converge for any nonzero parameter.

Let $g_x(t)=(1+xt{)}^{-1}$. Repeated integration by parts using the factor $e^{-t}$ givesBoundary terms at infinity vanish, and $g_x^{(n)}(t)=(-1{)}^{n}n!x^n(1+xt{)}^{-n-1}$. ThereforeFor every fixed $N$, this is $o(x^N)$ as $x\downarrow 0$, provingFor each fixed $x>0$ the displayed infinite series diverges, since successive term magnitudes have ratio $(n+1)x\to \infty$. Its usefulness is precisely that of an asymptotic expansion, not a convergent power series.

Fix $0<ϵ<\pi /2$. On $[ϵ,\pi /2]$, ${\sin }^{2}t\ge {\sin }^2ϵ>0$, so that part of the integral is bounded by $(\pi /2)e^{-x{\sin }^2ϵ}$, exponentially smaller than $x^{-1/2}$.

By the assumed local equivalence, the remaining part is asymptotic to ${\int }_0^{ϵ}{e}^{-x{t}^2}dt$. Substitute $s=\sqrt{x}t$:The last step uses the Gaussian integral and monotone convergence of the upper limit to infinity. Hence

A fixed small-angle approximation alone is insufficient because its error is multiplied by $x$. Introduce the shrinking cutoff $\eta =x^{-1/3}$, which satisfies $\eta \ll ϵ$ but $\sqrt{x}\eta \to \infty$.

On $[0,\eta ]$, the Taylor expansion ${\sin }^{2}t=t^2+O(t^4)$ gives $x({\sin }^{2}t-t^2)=O(x{\eta }^4)=O(x^{-1/3})$ uniformly. Thusuniformly there, and the two central integrals are asymptotically equal. On $[\eta ,ϵ]$, choose a fixed $c>0$ with $\sin t\ge ct$ for $0\le t\le ϵ$. The sine-integral tail is at most $ϵe^{-c^{2}x{\eta }^2}=ϵe^{-c^2{x}^{1/3}}$; the Gaussian tail has the same exponential bound with $c=1$. Both are $o(x^{-1/2})$, while the central Gaussian integral is asymptotic to $\sqrt{\pi }/(2\sqrt{x})$. HenceThis justifies the localized Laplace method rather than interchanging a nonuniform approximation with integration.

Matching powers of $ϵ$ gives $u_0^{\prime \prime }+u_0=0$, $u_1^{\prime \prime }+u_1=u_0^2$, and $u_2^{\prime \prime }+u_2=2u_0{u}_1$, with $u_0(0)=2$ and homogeneous initial data for every higher correction. ThusThere is no resonant fundamental forcing at first order. At second order,The first term resonates with the homogeneous oscillator. Solving with the initial data givesConsequently the regular perturbation approximation is $u=u_0+ϵu_1+{ϵ}^2{u}_2+O({ϵ}^3)$ for fixed $t$. The first secular term is $(10/3){ϵ}^{2}t\sin t$, at order ${ϵ}^2$. It makes the regular expansion nonuniform on times of order ${ϵ}^{-2}$.

Use the fast time $T_0=t$ and slow time $T={ϵ}^{2}t$. A possible intermediate scale $ϵt$ has no leading modulation: the absence of first-order resonance imposes independence of that scale. WriteAt first order choose the nonresonant particular solutionwith any needed fundamental homogeneous correction assigned consistently when matching the initial data. At second order,The coefficient of $e^{i{T}_0}$ on the right is $(10/3)|A{|}^{2}A-2i{A}_T$. A bounded oscillatory correction requires this resonance coefficient, and its conjugate, to vanish. The quadratic oscillator phase-modulation condition is thereforeIt resums the secular phase accumulation into a slow modulation instead of leaving it as a growing term in the correction.

Write $A=a{e}^{i\varphi }$. The modulation equation gives $a_T=0$ and ${\varphi }_T=-5a^2/3$. Matching the leading initial conditions gives $a=1$, $\varphi (0)=0$. ThereforeThe leading amplitude remains two and has no slow growth or decay. The frequency changes from one to $1-5{ϵ}^2/3+O({ϵ}^3)$; expanding this shifted cosine recovers the second-order secular term from part (i).

There are bounded first-order waveform changes as well. If they are retained while matching $u(0)=2$, one may write $2\cos \tau +ϵ[2-(4/3)\cos \tau -(2/3)\cos 2\tau ]+\cdots$, where $\tau =(1-5{ϵ}^2/3)t$. Thus the first harmonic's coefficient is $2-4ϵ/3$, with a mean shift and a second harmonic. These bounded shape corrections are distinct from the constant leading amplitude in the zeroth-order multiple-scales solution.

At a vertex with exterior turning angle ${\beta }_j\pi$, the interior angle is $(1-{\beta }_j)\pi$. A local conformal map from a half-plane therefore behaves as $f(Z)-z_j\sim c_j(Z-X_j{)}^{1-{\beta }_j}$, so its derivative has local exponent $-{\beta }_j$.

On each real boundary interval between prevertices, $\arg f^{\prime }$ is constant because the image is a straight side. Across a prevertex its change is the exterior turning angle. Taking $\arg (Z-X_j)\in (0,\pi )$, the product ${\prod }_j(Z-X_j{)}^{-{\beta }_j}$ has exactly these changes. Dividing $f^{\prime }$ by that product removes the corner singularities and leaves a holomorphic nonzero function with constant boundary argument. Schwarz reflection and the regular behavior at infinity make it constant. This gives the Schwarz-Christoffel formulaThere are three freely normalizable real prevertices, because an automorphism of the upper half-plane has three real parameters. After those are fixed, the remaining $n-3$ prevertex parameters are determined by the polygon's shape through the side-length conditions. They are not further independent normalization choices.

If the last prevertex is at infinity, absorb its constant scale into $C$ and omit its finite factor:For a bounded polygon, ${\sum }_j{\beta }_j=2$, so this derivative behaves like $Z^{{\beta }_n-2}$, consistent with the corner described in the local coordinate $1/Z$.

Use the branch of $\sqrt{1-Z^2}$ that is positive for $Z=iY$, $Y>0$. The right-half-plane slit mapis holomorphic in the upper half-plane, with derivative $Z(1-Z^2{)}^{-3/2}$, the required Schwarz-Christoffel formula form. Its derivative has no interior zero.

To verify the image and global injectivity, factor the construction. Squaring maps the upper half-plane bijectively onto $\mathbb{C}\setminus [0,\infty )$. Thus $1-Z^2$ ranges over $\mathbb{C}\setminus (-\infty ,1]$. The positive-real-part square root maps this onto the right half-plane with the interval $(0,1]$ removed; reciprocation maps it onto the right half-plane slit along $[1,\infty )$. This also identifies the slit tip as a boundary point; the strict inequality in the printed slit description must be understood with that boundary convention.

Approaching the real axis from above givesIn particular $Z=-1$ maps to the lower infinite prime end $A$, $Z=0$ to the tip $B=1$, and $Z=1$ to the upper infinite prime end $C$. The real-axis point at infinity maps to zero on the vertical wall.

Let $g=f^{-1}$, so $g(z)=\sqrt{1-z^{-2}}$, choosing the analytic branch with $\mathrm{Im}g>0$; for example $g(1/2)=i\sqrt{3}$. Let $\zeta \in \Omega$ be the source point. For the convention $-\Delta G={\delta }_{\zeta }$, the Dirichlet Green function in the upper half-plane is obtained by an opposite-sign reflected source:It vanishes for real $Z$, has the logarithmic singularity $-(2\pi {)}^{-1}\log |Z-W|$, and is harmonic elsewhere. Conformal invariance of two-dimensional harmonic functions now givesThe branch specified above maps every boundary bank to the real axis, so the Green function vanishes on both slit banks and the vertical wall. Local conformality preserves the coefficient of the logarithmic singularity. If the convention is instead $\Delta G={\delta }_{\zeta }$, negate this expression.

Orient the smooth contour so the plus side is to its left. Assume the density is Hölder continuous near the observation point and the integrals converge at any other endpoints. The Cauchy principal value isSubtract $f(t)$ from the numerator. The resulting quotient is locally integrable, so its limits from the two sides coincide. For the remaining constant density, straightening the contour gives a half-circle contribution $\pm i\pi f(t)$. Division by $2\pi i$ yields the Sokhotski–Plemelj formulaFor the requested real integral assume $0<\alpha <1$ and $-1<x<1$, ensuring integrable endpoints and an interior principal-value pole. Put $W(t)=(1-t{)}^{\alpha -1}/(1+t{)}^{\alpha }$. Choose $w(z)=(z-1{)}^{\alpha -1}/(z+1{)}^{\alpha }$ analytic off $[-1,1]$, positive for $z>1$, and $w(z)\sim 1/z$ at infinity. Its two boundary values are $w_{\pm }(t)=-e^{\pm i\pi \alpha }W(t)$.

The jump is $-2i\sin (\pi \alpha )W(t)$. The Cauchy representation for this decaying analytic function is therefore $w(z)=-\sin (\pi \alpha ){\int }_{-1}^{1}W(t)/(t-z)dt/\pi$. Taking the mean of its two boundary values removes the half-residue terms and gives

The complex potential of a two-dimensional potential flow maps the fluid region to the strip $0<\mathrm{Im}w<1$. Its lower boundary is the wall, with $B$ at $w=0$; upstream is $\mathrm{Re}w\to -\infty$ and downstream is $\mathrm{Re}w\to +\infty$. The free streamline is the upper strip boundary.

Write $q=w^{\prime }=u-iv$. The horizontal wall gives positive real velocities running from $q=1$ to the stagnation value $q=0$ at $B$. The vertical wall gives negative imaginary velocities from zero to $-i$. On the free surface $|q|=1$, with the angle turning continuously from zero to $-\pi /2$. Thus the hodograph plane domain is the fourth-quadrant quarter of the unit disk, bounded by those two radii and the unit-circle arc.

The following successive conformal maps straighten it. First $r=q^2$ maps to the lower half unit disk. Next $s=(r-1)/(r+1)$ maps to the third quadrant, since $\mathrm{Re}s=(|r{|}^2-1)/|r+1{|}^2<0$ and $\mathrm{Im}s=2\mathrm{Im}r/|r+1{|}^2<0$. Finally $s^2$ maps onto the upper half-plane. The three wall landmarks $q=1,0,-i$ go to $0,1,\infty$ respectively.

The strip map $e^{\pi w}$ also maps onto the upper half-plane, with these same landmarks $0,1,\infty$. The unique half-plane automorphism fixing all three is the identity. ThereforeThe free surface maps to the negative real axis in either construction, as required by its streamfunction value one. This completes the right-angle free-streamline hodograph map construction.

Glue the two analytic functions on their overlap, where their values agree. Since $\alpha <\beta$, the two half-planes cover the whole complex plane. The gluing therefore defines an entire function. It is bounded outside a sufficiently large disk by the two given infinity bounds, and bounded on the remaining compact disk by continuity. The Liouville theorem then gives one common constant.

For the standard canonical Wiener-Hopf factorization, first assume the nonzero kernel has Index of a Wiener-Hopf kernel zero and a continuous analytic logarithm $\ell =\log F$ tending to zero at both ends of the strip, with sufficient decay for the Cauchy projections. Choose $\alpha <a<b<\beta$. Let ${\Gamma }_{+}$ be the line $\mathrm{Im}\zeta =a$ oriented left to right, with its factor used above that line; let ${\Gamma }_{-}$ be the line $\mathrm{Im}\zeta =b$ oriented left to right, with its factor used below that line.

For $a<\mathrm{Im}z<b$, closing the rectangle between the two lines and using the Cauchy integral formula givesThus $F=F_{+}/F_{-}$. Multiplying $F{w}_{+}=w_{-}$ gives $F_{+}{w}_{+}=F_{-}{w}_{-}$ in the overlap. The canonical factors are analytic, nonzero, and normalized at infinity in their respective half-planes, so part (a) yieldsContours may be shifted within the strip without changing their factors on the appropriate side. When necessary, a subtracted Cauchy kernel fixes convergence and the common normalization constant.

The printed hypotheses omit the index and factor-growth qualifications; $F\to 1$ alone does not ensure them. An explicit counterexample is $F(z)=(z+i)/(z-i)$ on $-1/2<\mathrm{Im}z<1/2$, with $w_{+}=(z-i)/(z+i)$ and $w_{-}=1$. Both $w$ functions are analytic and bounded at infinity in their prescribed half-planes, and the relation holds. If admissible analytic factors made their common product a nonzero constant, $w_{-}=1$ would force $F_{-}$ constant and hence $F_{+}$ proportional to $F$, impossible because $F$ has a pole at $i$ in the plus half-plane. The natural factors $F_{+}=z+i$, $F_{-}=z-i$ instead give the nonconstant common polynomial $z-i$. Thus the boxed constant conclusion is the intended index-zero result, not a consequence of the literal assumptions alone.

Let $k=s+i\eta$. On the negative half-line, the integrand magnitude is bounded at infinity by a constant times $e^{(\alpha -\eta )x}$, which is integrable as $x\to -\infty$ when $\eta <\alpha$. On the positive half-line the corresponding bound is $e^{(\beta -\eta )x}$, integrable when $\eta >\beta$. ThusOn compact subsets strictly inside these half-planes, the same exponential domination absorbs every power of $|x|$. Differentiation under the integral is therefore valid and proves holomorphy, rather than only absolute convergence. The bars here denote the printed one-sided Fourier transforms, not complex conjugation.

Use the Fourier convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$. Transforming the differential equation givesDefine $k_{+}(k)=\sqrt{k}$ analytic above the downward cut from zero, and $k_{-}(k)=\sqrt{k-i}$ analytic below the upward cut from $i$. More explicitly take $\arg k\in (-\pi /2,3\pi /2)$ and $\arg (k-i)\in (-3\pi /2,\pi /2)$. In the strip $0<\mathrm{Im}k<1$, put $q=k_{+}{k}_{-}$. Then $q^2=k^2-ik$ and $\mathrm{Re}q>0$; at $k=i/2$, $q=1/2$, fixing the choice throughout the strip. The decaying $y$ mode is $e^{-qy}$.

The positive-half-line boundary value has transform ${\int }_0^{\infty }{e}^{ikx}dx=i/k$ for $\mathrm{Im}k>0$. Thus $\hat{u}(k,0)=\overline{f}+i/k$ and ${\hat{u}}_y(k,0)=\overline{g}$. The transformed boundary relation isLet $c=k_{-}(0)=e^{-i\pi /4}$. Splitting off the pole at zero givesThe left side is analytic in the lower domain up to $\mathrm{Im}k<1$: its apparent pole at zero is removable. The right side is analytic for $\mathrm{Im}k>0$. By the supplied boundedness of $k\overline{f}$ and $\overline{g}$, both sides tend to zero at infinity in their half-planes. The gluing and Liouville theorem argument therefore makes their common value zero. ConsequentlyChoose the inversion contour $\Gamma =\{s+i\eta :s\in \mathbb{R}\}$ oriented left to right, with any fixed $0<\eta <1$. It passes above the pole at zero and below the branch point at $i$. Fourier inversion gives exactlyThus the appropriate value is $\sqrt{-i}=e^{-i\pi /4}$. The other proposed value reverses the residue of the boundary step and violates its prescribed positive value.

There is also a far-field qualification in the source. The formula is the intended bounded mixed-boundary solution, but it does not tend to zero along every path to infinity. This can be checked explicitly: in parabolic coordinates $x=a^2-b^2$, $y=2ab$, with $a,b>0$, the mixed-boundary convection-diffusion solution in parabolic coordinates isIndeed $\Delta u=u_{bb}/[4(a^2+b^2)]$ and $u_x=-b{u}_b/[2(a^2+b^2)]$; the identity $u_{bb}=-2b{u}_b$ proves the differential equation. At $b=0$ the boundary value is one, and at $a=0$ its normal derivative is zero. Its traces are $f(x)=\mathrm{erfc}\sqrt{-x}$ for $x<0$ and $g(x)=-1/\sqrt{\pi x}$ for $x>0$, whose one-sided transforms are the expressions derived above.

At a fixed angle away from the positive $x$ axis, $b\to \infty$ and $u\to 0$. But at fixed $y>0$ and $x\to +\infty$, $b\sim y/(2\sqrt{x})$ and $u\to 1$. Hence the radial decay statement cannot hold uniformly as printed; the intended decay is away from the downstream boundary layer. The exact expression also identifies its thickness as $O(\sqrt{x})$.

For $x>0$, the real and imaginary parts of the exponent divided by $x$ areThus constant-phase contours satisfy $u^3-3u{v}^2=c$. Through zero, $c=0$ gives the six rays $\theta =\pi /6+j\pi /3$. The three rays $\theta =\pi /6,5\pi /6,3\pi /2$ have decreasing real part of the exponent and are steepest descent contours; the alternating three are steepest ascent rays.

Through one, the phase is $c=1$. Its branch passing through that point can be writtenwith the cube root continuous from $t=1$. The exponent is $ix-xs$, so increasing $s$ from zero gives descent and decreasing $s$ gives ascent.

Deform the finite interval into the ray $t=e^{i\pi /6}r$, $0\le r<\infty$, followed by the reverse of the descent branch from one. The integrand is entire, so Cauchy's integral theorem permits the deformation.

For a precise connector at infinity, use $t=(a+iR{)}^{1/3}$, $0\le a\le 1$. Its length is $O(R^{-2/3})$ and the exponential has modulus $e^{-xR}$. Its contribution is therefore $O(R^{-2/3}{e}^{-xR})\to 0$. This avoids relying on an undifferentiated claim that every large circular arc decays.

A saddle point of the exponential phase occurs where $(i{t}^3{)}^{\prime }=3i{t}^2=0$. Hence the only saddle isIt is degenerate: the second phase derivative vanishes there, while the third is $6i\ne 0$. This cubic degeneracy produces the $x^{-1/3}$ scale. The other endpoint $t=1$ is not a saddle, since the phase derivative there is $3i$.

On the selected steepest descent contour put $t=e^{i\pi /6}r$, $r\ge 0$. Then $i{t}^3=-r^3$ and $dt=e^{i\pi /6}dr$. Its exact contribution isUse the Gamma function integral and its recurrence $\Gamma (4/3)=\Gamma (1/3)/3$ to obtainThere is no remainder here: the complete infinite descent ray was integrated exactly.

The endpoint branch has $t=(1+is{)}^{1/3}$ and $dt=(i/3)(1+is{)}^{-2/3}ds$. Its reversed orientation in the deformed contour givesAt $s=0$,Watson's lemma, or elementary Laplace's method with a Taylor remainder, now yieldsAdding the exact cubic-saddle contribution gives the two leading termsIf the next term is retained, it is $-2e^{ix}/(9x^2)$. The different powers reflect a degenerate saddle at zero and an ordinary endpoint at one.