Past exam of the differential equations and applications course of the University of Oxford / 2007 / A

Use dimensionless variables $X=X_{\ast }x$, $Y=Y_{\ast }y$, $Z=Z_{\ast }z$, $t=T\tau$, withSubstitution gives the requested system, with dots now meaning differentiation with respect to $\tau$, andIn particular the same scale for $Y$ makes the first two coefficients of the cubic equal in magnitude.

Its derivative is $f^{\prime }(y)=1-2y+\gamma y^2$. For $0<\gamma <1$ its two simple zeros areThe cubic increases below $y_{-}$, decreases between the two fold points of a critical manifold, and increases above $y_{+}$. It has a local maximum at $y_{-}$ and a local minimum at $y_{+}$.

After a rapidly decaying transient response, $z\simeq f(y)$ when its relaxation is faster than the other variables. The reduced autonomous system isFor $ϵ\ll 1$, the fast $y$ dynamics at fixed $x$ attracts to the branches where $f^{\prime }(y)>0$ and repels from the middle branch. On the low branch $y<y_{-}<\alpha$, $x$ increases slowly. At its fold point of a critical manifold $x=f(y_{-})$, the attracting low branch disappears and $y$ jumps upward, with nearly constant $x$, to the high branch. On that branch $y>y_{+}>\alpha$, $x$ decreases slowly. At $x=f(y_{+})$ it jumps back to the low branch. These slow arcs and fast jumps form the singular relaxation cycle. For small positive $ϵ$, and parameters bounded away from the fold point of a critical manifold-endpoint cases, the nearby attracting periodic motion has the same pattern. The equilibrium $(f(\alpha ),\alpha )$ is on the repelling branch: its Jacobian matrix has determinant $1/ϵ>0$ and trace $-f^{\prime }(\alpha )/ϵ>0$.

The limiting landing values are $3/\gamma -2y_{-}$ after the upward jump and $3/\gamma -2y_{+}$ after the downward jump, obtained from the third root of $f(y)=f(y_{\pm })$. Thus the two vertical segments in the phase plane plot really connect roots of the same fast equation.

If $\gamma >1$, $f^{\prime }$ is strictly positive everywhere, so the cubic is monotone and has no fold points of a critical manifold. The reduced equilibrium is asymptotically stable. In factThe only invariant trajectory in the zero-derivative set is the equilibrium, so bounded reduced solutions converge to it; relaxation oscillations disappear.

The stronger condition $\boxed{\delta \ll ϵ}$ matters during a fast jump: $y$ then varies on the $ϵ$ time scale, not on the slow unit scale. Only if $z$ adjusts faster still may one keep $z\simeq f(y)$ through that jump. Merely imposing $\delta \ll 1$ would not justify this reduction on all pieces of the cycle. This is a three-time-scale cubic relaxation oscillator.

For zero damping the conserved mechanical energy isFor $\gamma <0$, the potential grows to infinity in both directions, so every nonstationary finite-energy trajectory is trapped between two turning points and is periodic; the zero-energy solution is stationary. The case $\gamma =0$ is the harmonic oscillator. The negative-$\gamma$ comparison extends the parameter range initially stated in the source.

For $\gamma >0$, the origin lies in a local potential well, with barriers at $x=\pm {\gamma }^{-1/2}$ of height $1/(4\gamma )$. A trajectory launched from the origin oscillates precisely when its initial energy is below that barrier:At equality the trajectory approaches a barrier equilibrium asymptotically, rather than being periodic; for example the positive branch is $x={\gamma }^{-1/2}\tanh (t/\sqrt{2})$. Above the barrier the trajectory escapes the well. The source's inequality without the absolute value tacitly takes $v\ge 0$; for signed initial velocity the energy criterion requires $|v|$.

Next set $\beta =\gamma =0$ and assume $\alpha >0$. The energy $H=({\dot{x}}^2+x^2)/2$ satisfies $\dot{H}=-\alpha x^2{\dot{x}}^2\le 0$. This bounds the complete trajectory. The zero-dissipation set is the union of $x=0$ and $\dot{x}=0$, and its only invariant trajectory is the origin: a nonzero point on either axis immediately leaves it under the differential equation. The LaSalle invariance principle therefore gives $\boxed{x(t)\to 0,\dot{x}(t)\to 0}$. If $\alpha =0$ as well, damping is absent and that conclusion fails for nonzero $v$; the positive order-one damping assumption is essential.

For the weakly nonlinear periodic motion, write $\gamma =\alpha \kappa$, set $\tau =\omega t$, $\omega =1+\alpha {\omega }_1+\cdots$, and expand $x=A\cos \tau +\alpha x_1+\cdots$. The first correction obeysResonant forcing at the fundamental frequency would generate secular terms. Eliminating both coefficients yieldsConsequently the maximum displacement is $\boxed{2\sqrt{\beta }+O(\alpha )}$ for fixed positive $\beta$. A possible third-harmonic correction is $x_1=-A^3\sin 3\tau /32-\kappa A^3\cos 3\tau /32$, up to homogeneous amplitude and phase choices. The leading amplitude also balances averaged damping: its slow evolution is $\dot{A}=\alpha A(\beta /2-A^2/8)$, whose nonzero fixed point is attracting. The case $\beta =0$ is a degenerate zero-amplitude limit, rather than a nonzero limit cycle.

Multiplication by an integrating factor givessince $p^{\prime }=ap$ and hence $(p{w}^{\prime }{)}^{\prime }=p(w^{\prime \prime }+a{w}^{\prime })$.

For the singular-endpoint equation, take $p=x^{-1/2}$, $r=x^{1/2}/2$, $q=0$. The indicial equation is $\alpha (2\alpha -3)=0$. The constant branch violates the zero endpoint value unless its coefficient vanishes; the nontrivial admissible branch therefore has $\boxed{\alpha =3/2}$. More precisely, an eigenfunction has $w=C{x}^{3/2}+O(x^{9/2})$ and $w^{\prime }=(3C/2)x^{1/2}+O(x^{7/2})$ near zero.

Integrate the self-adjoint equation after multiplying by $w$, initially on $[a,1]$ with $a>0$:At zero the boundary expression is $O(x^{3/2})$, the first integrand is $O(x^{1/2})$, and the second is $O(x^{7/2})$. Both integrals converge and the boundary term vanishes. Division by the positive weighted norm gives $\boxed{\lambda =I(w)}$.

The change of coordinate $z=x^{3/2}$, $w(x)=W(z)$ gives exactlyThus ${\lambda }_n=9n^2{\pi }^2/2$ and $w_n=\sin (n\pi x^{3/2})$. To prove a minimum over trial functions, rather than just identify the least eigenvalue, transform the quotient itself:The Dirichlet Poincaré inequality gives the last step, with equality precisely for a nonzero multiple of $W=\sin \pi z$. The trial class is the nonzero finite-energy class with zero Sobolev traces; endpoint values alone without an integrability requirement do not define a finite quotient. This is the Power-coordinate reduction of a singular Sturm-Liouville quotient.

With diffusion absent, the characteristic equations for a first-order quasilinear equation are $du/dt=0$, $dx/dt=q^{\prime }(u)$. A characteristic curve labelled by its starting point $s$ isThis determines a smooth single-valued solution while $s\mapsto x$ is one-to-one. Its derivative isAll characteristic curves move right because $q^{\prime }>0$. Because $q^{\prime \prime }<0$, increasing portions of the initial profile have slower characteristic curves ahead than behind, so they steepen. The first positive gradient catastrophe occurs atThis is a minimum when attained. The numerator at such a point is positive and the denominator decreases to zero from above, so $u_x\to +\infty$.

Resolve the thin transition with $X=(x-x_s(t))/ϵ$ and $u\simeq U(X)$. The leading balance is $(-c+q^{\prime }(U))U^{\prime }=(D(U)U^{\prime }{)}^{\prime }$, where $c={\dot{x}}_s$. Integrating once and matching to the state behind givesMatching the state ahead requires $F(u_{+})=0$ and gives the Rankine-Hugoniot speedThe strictly concave graph of $F$ has zeros at the two states and lies above its chord, hence is strictly positive between them. Since $D>0$, $U^{\prime }>0$ through the layer. A heteroclinic profile increasing with $X$ can therefore connect the rear state at $X=-\infty$ to the front state at $X=+\infty$ only when $\boxed{u_{-}<u_{+}}$. The concavity also gives $q^{\prime }(u_{-})>c>q^{\prime }(u_{+})$: characteristic curves enter the viscous shock from both sides. Reversing the states would demand a decreasing profile, contradicting the sign of $F/D$.