Question 1 · local series and improper analysis
Use one consistent Taylor expansion throughout. For every limit, state whether it is an endpoint limit or an internal limit and identify the expansion or comparison that justifies it. When the quotient is extended at the apparent singular point, define its value there before discussing differentiability. For the improper integral, write it as the sum of separate integrals at every endpoint or internal point where convergence is not automatic; a single unsupported statement that the integral converges or diverges is not sufficient.
Present the solution in the order of the labelled parts. Every final claim must be supported by a theorem, calculation or explicit check.
(a) State Taylor's theorem through order four about x=1, including a valid local remainder. (2 marks)
(b) Expand \(\log x\) about x=1 through fourth order and state the interval or neighbourhood on which the expansion is justified. (4 marks)
(c) On (0,∞), define \(g(x)=(\log x)/(x-1)\) away from x=1. Calculate the three limiting regimes 0^+,1,∞, decide whether the singularity at x=1 is removable, and determine the differentiability of the extension. (7 marks)
(d) Decide whether the integral of \(g(x)/(1+x^2)\) over its full domain converges as an ordinary improper integral. Split at every endpoint or internal point requiring separate analysis and give comparison functions. (7 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** If \(f\) has the required derivatives near \(a\), then
\[
f(x)=\sum_{k=0}^{4}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_4(x),
\]
where, for example under a \(C^5\) hypothesis,
\[
R_4(x)=\frac1{4!}\int_a^x(x-t)^4f^{(5)}(t)\,dt
\]
(or the equivalent Lagrange/Peano local remainder).
**(b)** Put \(h=x-1\). The fourth-order expansion is
\[
\log(1+h)=h-\frac{h^2}{2}+\frac{h^3}{3}-\frac{h^4}{4}+O(h^5),\quad |h|<1.
\]
The stated neighbourhood/radius must be checked before substituting a
new point.
**(c)** In the three regimes listed in the question, the limits are
\[
+\infty,\ 1,\ 0.
\]
The quotient has a removable singularity and is real analytic after setting g(1)=1.
**(d)** Near 0, \(g(x)/(1+x^2)\sim-\log x\), which is integrable; near 1 the point is removable; at infinity it is \(O(\log x/x^3)\). The ordinary improper integral converges absolutely.
Question 2 · conservation, leakage and Fourier data
An open vertical vessel contains an incompressible liquid. Let \(h(t)\) be its depth and let \(V(t)=S h(t)\), where the horizontal cross-sectional area \(S\) is constant. There is no inflow. Liquid leaves simultaneously through a base outlet and through a porous part of the wetted side wall. The two quoted outflow expressions are volume per unit time, are positive while \(h>0\), and vanish when the tank is empty. Adopt a clear sign convention before writing the balance law, retain the initial condition until the constant of integration is fixed, and state which dimensions each coefficient must have. In the limiting comparison, keep the base-loss coefficient fixed while only the porous-loss parameter tends to zero. For the periodic signal, state the Fourier coefficient convention you use and apply the midpoint rule at a discontinuity before extracting the requested numerical series identity.
Present the solution in the order of the labelled parts. Every final claim must be supported by a theorem, calculation or explicit check.
(a) A vertical tank has constant cross-sectional area S=3, initial height \(h(0)=4\), base outflow \(2\sqrt h\), and porous-side outflow \(h\). Derive the conservation equation, verify dimensions, separate it using \(u=\sqrt h\), and express the emptying time in elementary terms. (8 marks)
(b) Introduce a parameter λ multiplying the porous loss. Find the limit as λ→0 and compare the emptying time with the base-only model, including a physical explanation. (4 marks)
(c) A 2π-periodic signal equals 2 on 0<x<π and 0 on −π<x<0. Compute its Fourier coefficients, state convergence at jumps, and evaluate at a suitable point to derive an alternating odd-reciprocal series identity. (8 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** Conservation of volume gives
\[
3\dot h=-2\sqrt h-h.
\]
The coefficient multiplying \(h\) has dimensions area/time. With
\(u=\sqrt h\),
\[
dt=-\frac{6\,du}{2+u},
\]
so the actual two-loss emptying time is
\[
T=6
\log\left(1+\frac{2}{2}\right).
\]
**(b)** After multiplying the porous loss by \(\lambda\),
\[
3\dot h=-2\sqrt h-\lambda h,\qquad
T_\lambda=\frac{6}{\lambda}
\log\left(1+\frac{\lambda\sqrt{4}}{2}\right).
\]
As \(\lambda\to0\),
\[
T_\lambda\longrightarrow T_0
=\frac{6\sqrt{4}}{2}
=6.
\]
Since \(\log(1+x)<x\) for \(x>0\), \(T_\lambda<T_0\): the added
porous loss makes the tank empty sooner.
**(c)** For the signal of height \(A=2\),
\[
a_0=2,\qquad a_n=0,\qquad
b_n=\frac{2}{\pi n}\bigl(1-(-1)^n\bigr).
\]
At a jump the series equals the midpoint \(A/2\). At \(x=\pi/2\),
\[
\frac\pi4=1-\frac13+\frac15-\frac17+\cdots.
\]
Question 3 · linear flow, forcing and exactness
Write the dependent variables as a column vector and show enough work to identify the characteristic polynomial, eigenvalues and a corresponding eigenvector or real fundamental matrix. The phase classification must agree with the large-time behaviour of the solution and with the direction field on a coordinate axis. For the forced system, separate the exponential and constant forcing, test the proposed ansatz for resonance, and then add it to the homogeneous solution. In the differential-form part, identify \(M\) and \(N\), compare the cross-partials on the stated domain, derive rather than guess any integrating factor, and differentiate the recovered potential as a final check.
Present the solution in the order of the labelled parts. Every final claim must be supported by a theorem, calculation or explicit check.
(a) For \(X'=AX\), \(A=\begin{pmatrix}1&-4\\1&-1\end{pmatrix}\), find the spectrum and eigenvectors, write a real general solution, classify the origin and describe large-time behaviour. (8 marks)
(b) Add forcing \(F(t)=(e^t,\,2)^T\). Choose a justified particular-solution ansatz, test resonance and state the complete solution. (5 marks)
(c) Consider the differential form \(2x^{2}y\,dx+(x^{3}+6xy^{2})\,dy=0\) on a domain with x≠0. Test exactness; if needed find a monomial integrating factor and recover a potential function. Verify the result against \(\Phi=x^{2}y+2y^{3}\) only after deriving it. (7 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** For \(A=\begin{pmatrix}1&-4\\1&-1\end{pmatrix}\), direct calculation gives
\[
A^2=-3I,\ \lambda=\pm i\sqrt3,\quad X_h=[I\cos(\sqrt3t)+(A/\sqrt3)\sin(\sqrt3t)]C; the origin is a neutrally stable centre.
\]
**(b)** The forcing is \((e^t,2)^T\).
The scalar rate is not an eigenvalue in this case, so no resonance
multiplier is required. Solving
\[
(I-A)p=\binom10,\qquad Aq=-\binom0{2}
\]
gives
\[
p=\binom{1/2}{1/4},\qquad q=\binom{-8/3}{-2/3}.
\]
Thus \(X=X_h+pe^t+q\). When the exponential rate is zero,
the two constant particular terms may of course be combined.
**(c)** With
\[
M=2x^{2}y,\qquad
N=x^{3}+6xy^{2},
\]
the original cross-partials are unequal. On \(x>0\) or \(x<0\), the
integrating factor \(\mu=x^{-1}\) gives
\[
\widetilde M=2x^{1}y,\qquad
\widetilde N=x^{2}+6y^{2},
\]
whose cross-partials agree. Integrating yields
\[
\Phi(x,y)=x^2y+2y^3=C.
\]
Question 4 · bifurcation, transform and implicit surface
The parameter is allowed to range over all real values for which the vector field is defined. Keep the singular state separate from ordinary equilibria, show the stability of every branch on the parameter intervals between critical values, and label solid, dashed and singular branches on the bifurcation diagram. In the transform problem, state the sign and normalisation convention, transform both differentiation and multiplication by \(x\), and explain why the inverse representation may be written as a real oscillatory integral satisfying the decay requirement. In the implicit problem, evaluate the relevant partial derivative before invoking the implicit-function theorem and then display the full first-order Taylor polynomial at the stated base point.
Use separate labelled panels for the equilibrium branches, transform calculation and local implicit surface.
Present the solution in the order of the labelled parts. Every final claim must be supported by a theorem, calculation or explicit check.
(a) For \(y'=ry-y^2/(1+y)\), find every equilibrium and singular state, determine stability for all parameter ranges, draw the bifurcation diagram on a clearly labelled parameter-state diagram, and classify each exchange or collision. (8 marks)
(b) Treat \(u''-2xu=0\), with u decaying at both infinities, by Fourier transform. State a convention, transform both terms, derive and solve the first-order frequency equation, and explain how an admissible real inverse-integral representation is selected. (6 marks)
(c) Let \(F(x,y,z)=e^{xz}+xy-z-1=0\) define z near (0,1,0). Check the implicit-function hypothesis, compute \(z_x,z_y\) at the base point, and write the first-order Taylor approximation. (6 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** The vector field factors as
\[
F_r(y)=\frac{y(1r+(r-1)y)}{1+y}.
\]
The singular state is \(y=-1\). The equilibria are \(y=0\)
and, for \(r\ne1\),
\[
y_*=\frac{1r}{1-r}.
\]
Their linearisation derivatives are \(F_r'(0)=r\) and
\(F_r'(y_*)=r(r-1)\). Thus \(0\) is stable for \(r<0\) and unstable
for \(r>0\); \(y_*\) is stable for \(0<r<1\) and unstable for
\(r<0\) or \(r>1\). The branches exchange stability in a
transcritical bifurcation at \(r=0\). At \(r=1\) the second branch
escapes through infinity. The singular line must remain a barrier on
the diagram.
**(b)** With
\[
\widehat u(\xi)=\int_{\mathbb R}u(x)e^{-i\xi x}\,dx,
\]
\[
\mathcal F(u'')=-\xi^2\widehat u,\qquad
\mathcal F(2xu)=2i\widehat u',
\]
so
\[
-\xi^2\widehat u-2i\widehat u'=0,\qquad
\widehat u=C\exp\left(\frac{i\xi^3}{6}\right).
\]
Pairing positive and negative frequencies gives the real oscillatory
integral
\[
u(x)=\frac C\pi\int_0^\infty
\cos\left(\xi x+\frac{\xi^3}{6}\right)\,d\xi,
\]
equivalently a scaled Airy \(Ai\) solution selected by decay.
**(c)** At \((0,1,0)\),
\[
F_x=1,\qquad F_y=0,\qquad F_z=-1\ne0.
\]
Hence the implicit-function theorem applies,
\[
z_x=1,\qquad z_y=0,\qquad
z(x,y)=x+
o\!\left(\sqrt{x^2+(y-1)^2}\right).
\]