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\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Final Examination -- Solutions}
\centerline{December 17, 2004}
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\noindent
I.  A) The equation is separable, so separating and integrating,
$$\eqalign{
\int {dy \over (y-2)^{1/3}} &= \int cos(t)\ dt \cr
{3\over 2}(y-2)^{2/3} &= \sin(t) + c\cr
y &= \left({2\over 3} \sin(t) + c\right)^{3/2} + 2\cr}$$
From the initial condition $y(0) = 3$, we get $c = 1$.

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\noindent
B) In this equation, $f(t,y) = \cos(t)(y-2)^{2/3}$ is continuous
on the whole $t,y$-plane.  But 
${\partial f\over \partial y} = {2\over 3}\cos(t)(y-2)^{-1/3}$
is not continuous at $(0,2)$.  So the hypotheses of the Existence
and Uniqueness Theorem are not satisfied.  $y(t) \equiv 2$ is one
solution since $y'(t) = 0 = \cos(t)(2-2)^{2/3}$.  But there are other
solutions as well:  For instance taking $c = 0$ in solution from A
gives another.

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\noindent
II.  Equilibrium points are $x = \pm\sqrt{a}$ for $a \ge 0$ (none
for $a < 0$).  The phase lines for $a \le 0$ should have
all arrows pointing up, for $a > 0$, the arrows point up 
if $x > \sqrt{a}$ or if $x < -\sqrt{a}$.  They point down 
between $\sqrt{a}$ and $-\sqrt{a}$.  So $x = \sqrt{a}$ is a source,
and $x = -\sqrt{a}$ is a sink.

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\noindent
III.  A)  Taking $t$ in hours, 
$$A(t) = 15\cos\left({\pi t\over 12}\right) + 55$$

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\noindent
B)  The ODE $y' = -1.5y + 1.5A(t)$ is first order linear.  Using entry
\# 9 in the table of integrals we have
$$y(t) = ce^{-1.5t} + 55 + {22.5\over ({\pi/12})^2 + (1.5)^2}\left(
1.5\cos\left({\pi t\over 12}\right) + {\pi\over 12}\sin\left({\pi t\over 12}\right)\right)$$

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\noindent
C) The term containing the $c$ goes to zero as $t\to +\infty$.  So
that term does not affect the long-term behavior of the solution.
(This is like the transient terms in the solutions of our 2nd order
forced oscillator equations).

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\noindent
IV.  A)  For this family of 1st order systems, $T = a$ and $D = -4a$.
The family follows the line $D = -4T$ in the trace-determinant plane.
That line crosses the parabola $D = T^2/4$ at $T = -16$ and $T = 0$.
So, the systems have spiral sinks for $-16 < a < 0$.

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\noindent
B)  With $a = 2$, the characteristic polynomial is $\lambda^2 - 2\lambda + 8$,
so the eigenvalues are $\lambda = 4, -2$.  The origin is a saddle point
in the phase portrait.  The $\lambda = 4$ line is spanned by the 
eigenvector $(4,1)$, and the $\lambda = -2$ line is spanned 
by the eigenvector $(-2,1)$.  

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\noindent
V.  Expanding along row 3, the characteristic polynomial
is $(-3-\lambda)(2-\lambda)^2$.  The dimension of the kernel 
of 
$$A - 2I = \pmatrix{0 & 0 & 1\cr 1 & 0 & 0\cr 0 & 0 & -5\cr}$$
is just 1; the kernel is spanned by $(0,1,0)$.  An eigenvector
for $\lambda = -3$ is $(-5,1,25)$.  So the canonical form is 
$$C = \pmatrix{2 & 1 & 0\cr 0 & 2 & 0\cr 0 & 0 & -3\cr}$$
To find the change of basis matrix here we want a vector 
$V_2$ such that $(A - 2I)V_2 = V_1 = (0,1,0)$.  By inspection 
of the matrix $A - 2I$, we see that $V_2 = e_1$ will work.
So 
$$Q = \pmatrix{0 & 1 & -5\cr 1 & 0 & 1\cr 0 & 0 & 25\cr}$$
Then the solution of the canonical form system $Y' = CY$ is 
$$Y = c_1 e^{2t}\pmatrix{1 \cr 0\cr 0\cr} + c_2 e^{2t} \pmatrix{t \cr 1\cr 0\cr}
+ c_3e^{-3t} \pmatrix{0 \cr 0\cr 1\cr}$$
and the solution of $X' = AX$ is 
$$X = QY = c_1 e^{2t}\pmatrix{0 \cr 1\cr 0\cr} + c_2 e^{2t} \pmatrix{1 \cr t\cr 0\cr}
+ c_3e^{-3t} \pmatrix{-5 \cr 1\cr 25\cr}$$

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\noindent
VI.  A)  Characteristic polynomial is $\lambda^2 + 6\lambda + 9 = 0$,
so $\lambda = -3$ is a repeated root.  
$$y = c_1 e^{-3t} + c_2 t e^{-3 t}$$

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\noindent
B)  Using the particular solution $y_p = A e^{-t} + Bt^2 + Ct + D$
(since no forcing term is a solution of the homogeneous equation),
we get $A = 3/4$, $B = 1/9$, $C = -4/27$, $D = 2/27$.
$$y = c_1 e^{-3t} + c_2 t e^{-3 t} + 3e^{-t}/4 + t^2/9 - 4t/27 + 2/27$$

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\noindent
VII.  A)  From the second equation, at equilibrium either $x = 0$
or $x = 2y - 1$.  In the first case, $y = 1$.  In the second, $y = 0$
and $x = -1$.  So there are two equilibrium points: $(0,1)$ and $(-1,0)$.

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\noindent
B)  The Jacobian of the system is 
$$J(x,y) = \pmatrix{1 & -1\cr 2x - 2y + 1 & -2x\cr}$$
At $(x,y) = (0,1)$, $J(0,1) = \pmatrix{1 & -1\cr -1 & 0\cr}$
which has determinant $-1$.  So $(0,1)$ is a {\it saddle point}.
At $(x,y) = (-1,0)$, $J(-1,0) = \pmatrix{1 & -1\cr -1 & 2\cr}$
which has trace $3$ and determinant $1$.   So $(0,1)$ is a {\it source}
(since $D < T^2/4$ and $T > 0$).  In both cases, the equilbrium
point of the non-linear system has the same type by our general
theorem.

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\noindent
VIII.  A)  Yes, this system is Hamiltonian, with Hamiltonian
function $H = e^{xy}$.  Note $H_y = xe^{xy} = f(x,y)$ and
$-H_x = -ye^{xy} = g(x,y)$.

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\noindent
B)  If a system is Hamiltonian, then the linearization at 
each critical point has coefficient matrix
$$A = \pmatrix{H_{xy} & H_{yy}\cr -H_{xx} & -H_{xy}\cr}$$
(using equality of mixed partials).  The sum of the diagonal
entries is zero, so ${\rm Tr}(A) = 0$.  (Note: it is this fact
that tells us that equilibrium points of Hamiltonian systems
are always either centers or saddles.)

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\noindent
C)  System 1 has a source so it cannot be Hamiltonian.
System 3 has a spiral sink, so it cannot be Hamiltonian either.
Only system 2 is possibly Hamiltonian. 

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\noindent
{\it Extra Credit}:  See Section 5.4 in Blanchard, Devaney, and Hall.
The idea is that if we have a gradient system, then by the 
same chain rule calculation we did in class for Hamiltonian
systems, along a solution,
$${d\over dt}G(x(t),y(t)) = G_x^2 + G_y^2 \ge 0$$
so $G$ increases along solutions.  This means that there cannot be
any closed curves that are solutions.  The linearization of 
a gradient system at an equilibrium has coefficient matrix:
$$A = \pmatrix{G_{xx} & G_{xy} \cr G_{xy} & G_{yy}\cr}$$
This is always a symmetric matrix, so the eigenvalues are real
(by the Spectral Theorem from Linear Algebra).  The equilbria
can be saddles, sinks, or sources, but no ``spiral'' behavior
is possible for gradient systems.
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