Ordinary Differential Equations

Question 1

(a) Transfer function (4) Laplace with zero ICs: $\mathcal{L}\{y''\}=s^2Y$, $\mathcal{L}\{y'\}=sY$. So $(s^2+2\zeta\omega_0 s+\omega_0^2)Y(s)=F(s).$ (2 — using derivative rule) $H(s)=\frac{Y}{F}=\frac{1}{s^2+2\zeta\omega_0 s+\omega_0^2}.$ (1) Poles: $s=-\zeta\omega_0\pm i\omega_0\sqrt{1-\zeta^2}=-\zeta\omega_0\pm i\omega_d$. ROC: $\Re(s)>-\zeta\omega_0$ (right of rightmost pole). (1)

(b) Convolution theorem (5) $\mathcal{L}\{(g*h)\}=\int_0^\infty e^{-st}\!\int_0^t g(\tau)h(t-\tau)\,d\tau\,dt.$ (1) Region: $0\le\tau\le t<\infty$. Swap order (Fubini, valid since integrand absolutely integrable when $g,h$ of exponential order and $\Re s$ large enough): (1 for stating condition) $=\int_0^\infty g(\tau)\!\int_\tau^\infty e^{-st}h(t-\tau)\,dt\,d\tau.$ (1) Substitute $u=t-\tau$, $du=dt$: $=\int_0^\infty g(\tau)e^{-s\tau}\!\int_0^\infty e^{-su}h(u)\,du\,d\tau = \Big(\int_0^\infty g(\tau)e^{-s\tau}d\tau\Big)H(s)=G(s)H(s).$ (2)

(c) Impulse response (5) $H(s)=\dfrac{1}{(s+\zeta\omega_0)^2+\omega_d^2}$ (complete the square). (2) Since $\mathcal{L}\{e^{-at}\sin\omega_d t\}=\dfrac{\omega_d}{(s+a)^2+\omega_d^2}$, (1) $h(t)=\frac{1}{\omega_d}e^{-\zeta\omega_0 t}\sin(\omega_d t),\quad t\ge0.$ (2)

(d) Step response two ways (6) Route 1 — convolution: $y=h*f=\int_0^t h(t-\tau)u(\tau-a)d\tau=\int_a^t h(t-\tau)d\tau$ for $t>a$. Let $\sigma=t-\tau$: $y=\int_0^{t-a}h(\sigma)d\sigma=\frac{1}{\omega_d}\int_0^{t-a}e^{-\zeta\omega_0\sigma}\sin\omega_d\sigma\,d\sigma.$ (2) Using $\int_0^T e^{-a\sigma}\sin\omega_d\sigma\,d\sigma=\frac{a\,e^{-aT}(-\sin\omega_d T)-\omega_d e^{-aT}\cos\omega_d T+\omega_d}{a^2+\omega_d^2}$ with $a=\zeta\omega_0$, $a^2+\omega_d^2=\omega_0^2$: $y(t)=\frac{1}{\omega_0^2}\Big[1-e^{-\zeta\omega_0(t-a)}\Big(\cos\omega_d(t-a)+\tfrac{\zeta\omega_0}{\omega_d}\sin\omega_d(t-a)\Big)\Big],\ t>a.$ (2) Route 2 — second shift: For step at $t=0$, $Y(s)=H(s)/s$. Partial fractions / inverse give the bracket above with argument $t$; second-shift theorem $\mathcal{L}\{u(t-a)g(t-a)\}=e^{-as}G(s)$ replaces $t\to t-a$, reproducing the same expression. Both agree. (2)

(e) Euler + steady state (4) State-space $x_1=y,\ x_2=y'$: $\dot x_1=x_2,\ \dot x_2=f-2\zeta\omega_0 x_2-\omega_0^2 x_1$. (1)

h=0.01; x1=0; x2=0; t=0
while t<T:
    f = 1                     # unit step
    dx1 = x2
    dx2 = f - 2*z*w0*x2 - w0**2*x1
    x1 += h*dx1; x2 += h*dx2; t += h
    record(t, x1)

(2) Steady state: set derivatives $=0 \Rightarrow \omega_0^2 y=1$, $y_\infty=1/\omega_0^2=1/4$. (1)

Question 2

(a) Critical points (3) $y=0$ and $\sin x=0 \Rightarrow x=0,\ \pi$. Points: $(0,0)$ and $(\pi,0)$. (3)

(b) Jacobian & classification (9) $J=\begin{pmatrix}0&1\\-\cos x&-\beta\end{pmatrix}$. (2) At $(0,0)$: $J=\begin{pmatrix}0&1\\-1&-\beta\end{pmatrix}$, char eqn $s^2+\beta s+1=0$, $s=\frac{-\beta\pm\sqrt{\beta^2-4}}{2}$. For $0<\beta<2$: complex with negative real part → stable spiral; for $\beta>2$ → stable node. (4) At $(\pi,0)$: $J=\begin{pmatrix}0&1\\1&-\beta\end{pmatrix}$, char eqn $s^2+\beta s-1=0$, $s=\frac{-\beta\pm\sqrt{\beta^2+4}}{2}$: one positive, one negative root → saddle (unstable). (3)

(c) Origin, $\beta=1$ (6) $s^2+s+1=0\Rightarrow s=-\tfrac12\pm i\tfrac{\sqrt3}{2}$. (2) Eigenvector for $s$: $(0-s)v_1+v_2=0\Rightarrow v_2=s v_1$, take $\mathbf v=(1,s)^T$. (2) Real general solution: $\mathbf x(t)=e^{-t/2}\Big[c_1\big(\cos\tfrac{\sqrt3}{2}t\,\mathbf u-\sin\tfrac{\sqrt3}{2}t\,\mathbf w\big)+c_2(\dots)\Big],$ concretely $x(t)=e^{-t/2}\big(A\cos\tfrac{\sqrt3}{2}t+B\sin\tfrac{\sqrt3}{2}t\big)$, $y=\dot x$. (2)

(d) (2) A nonlinear centre has purely imaginary linear eigenvalues; nonlinear terms can push it to a stable/unstable spiral, so linearization is inconclusive (borderline case). Node↔spiral bifurcation at origin occurs at $\beta=2$ (discriminant zero). (2)

Question 3

(a) Recurrence (6) $p(x)=\frac{-2x}{1-x^2}$, $q(x)=\frac{\lambda(\lambda+1)}{1-x^2}$ analytic at $0$ → ordinary point. (1) Sub $y=\sum a_n x^n$: $\sum n(n-1)a_n x^{n-2}-\sum n(n-1)a_n x^n-2\sum n a_n x^n+\lambda(\lambda+1)\sum a_n x^n=0.$ (2) Coefficient of $x^n$: $(n+2)(n+1)a_{n+2}=[n(n-1)+2n-\lambda(\lambda+1)]a_n=[n(n+1)-\lambda(\lambda+1)]a_n,$ $a_{n+2}=\frac{n(n+1)-\lambda(\lambda+1)}{(n+1)(n+2)}a_n.$ (3)

(b) Termination & $P_2$ (5) When $\lambda=n$: numerator vanishes at that $n$, so $a_{n+2}=0$ and the series (same parity) terminates → polynomial. (2) For $\lambda=2$: $a_2=\frac{0-6}{2}a_0=-3a_0$, $a_4=0$. Even series: $y=a_0(1-3x^2)$. Normalize $P_2(1)=1$: $a_0(1-3)=-2a_0=1\Rightarrow a_0=-\tfrac12$: $P_2=\tfrac12(3x^2-1)$. (3)

(c) Orthogonality (5) $P_1=x$, $P_2=\tfrac12(3x^2-1)$. $\int_{-1}^1 x\cdot\tfrac12(3x^2-1)dx=\tfrac12\int_{-1}^1(3x^3-x)dx=0$ (odd integrand). (3) General: $\int_{-1}^1 P_m P_n dx=\frac{2}{2n+1}\delta_{mn}$. (2)

[
{"claim":"H(s) poles have real part -zeta*omega0","code":"s,z,w0=symbols('s zeta omega0',positive=True); roots=solve(s**2+2*z*w0*s+w0**2,s); result = all(simplify(re(r.rewrite(sqrt))+z*w0)==0 for r in [ -z*w0+I*w0*sqrt(1-z**2), -z*w0-I*w0*sqrt(1-z**2) ])"},
{"claim":"impulse response inverse Laplace correct","code":"t,z,w0=symbols('t zeta omega0',positive=True); wd=w0*sqrt(1-z**2); s=symbols('s'); H=1/(s**2+2*z*w0*s+w0**2); h=inverse_laplace_transform(H,s,t); result = simplify(h-exp(-z*w0*t)*sin(wd*t)/wd)==0"},
{"claim":"step steady state = 1/w0^2 with w0=2 gives 1/4","code":"result = Rational(1,4)==Rational(1,2**2)"},
{"claim":"origin beta=1 eigenvalues -1/2 +- i sqrt3/2","code":"s=symbols('s'); r=solve(s**2+s+1,s); result = set(r)=={Rational(-1,2)+I*sqrt(3)/2, Rational(-1,2)-I*sqrt(3)/2}"},
{"claim":"P2 normalized and orthogonal to P1","code":"x=symbols('x'); P1=x; P2=(3*x**2-1)/2; result = (integrate(P1*P2,(x,-1,1))==0) and (P2.subs(x,1)==1) and (integrate(P2*P