Toronto Math Forum

I'm having some trouble getting this problem to work out. There are four critical points: (0,0), (2, 1), (-2, 1), and (-2, -4). At the critical point (-2, -4), the Jacobian is \begin{pmatrix} 10 & -5 \\ 6 & 0 \end{pmatrix} with eigenvalues $5 \pm i\sqrt{5}$. Therefore it looks like it should be an unstable spiral point. However, when I plotted it, it looked like a node. Has anyone else done this problem?

http://www.math.psu.edu/melvin/phase/newphase.html

Quote from: Brian Bi on March 06, 2013, 10:01:33 PM

By inspection, $y = 1/2$ is a solution.

Darn, why did I not see this...
Just goes to show that slowing down during a test and looking at the question with a calm mind can do wonders

I also did not see this during the test. Is there anyone who did?

Assume that $z(x) > 0$ on $(x_0, x_1)$. We will show a contradiction.

Since $y(x) > 0$ and $z(x) > 0$ on $(x_0, x_1)$, recall the result of Bonus Problem 1:
\begin{equation}
W'(x) = (q(x) - Q(x))y(x)z(x)
\end{equation}
We know that $W'(x) \leq 0$ on $(x_0, x_1)$. By the mean value theorem, $W(x_0) \geq W(x_1)$. Consider what happens if we have $W(x_0) = W(x_1)$. Then if for some $x \in (x_0, x_1)$ we had $W(x) \neq W(x_0)$, then by the mean value theorem $W'$ would have to be positive on either $(x_0, x)$ or $(x, x_1)$, a contradiction. Conclude that if $W(x_0) = W(x_1)$, then $W$ is constant on $(x_0, x_1)$, implying that $q - Q$ vanishes identically on $(x_0, x_1)$. We proceed under the assumption that this is not the case, so that $W(x_0) > W(x_1)$.

Writing out the Wronskians explicitly gives
\begin{align}
\begin{vmatrix} y(x_0) & z(x_0) \\ y'(x_0) & z'(x_0) \end{vmatrix} &> \begin{vmatrix} y(x_1) & z(x_1) \\ y'(x_1) & z'(x_1)  \end{vmatrix} \\
\begin{vmatrix} 0 & z(x_0) \\ y'(x_0) & z'(x_0) \end{vmatrix} &> \begin{vmatrix} 0 & z(x_1) \\ y'(x_1) & z'(x_1) \end{vmatrix} \\
-z(x_0) y'(x_0) &> -z(x_1) y'(x_1) \label{eq:foo}
\end{align}
Now, because $y(x_0) = 0$ and $y(x_0) > 0$ on $(x_0, x_1)$, we must have $y'(x_0) \geq 0$. Likewise $y'(x_1) \leq 0$. We thus consider the following:
Case 1: If $z(x_0) > 0$ and $z(x_1) > 0$, then $-z(x_0) y'(x_0) \leq 0$ and $-z(x_1) y'(x_1) \geq 0$; together with $(\ref{eq:foo})$ we obtain the desired contradiction.
Case 2: If either $z(x_0) < 0$ or $z(x_1) < 0$, this contradicts the assumption that $z > 0$ on $(x_0, x_1)$ since $z$ is continuous.
Case 3: If $z(x_0) = 0$, then the RHS of $(\ref{eq:foo})$ must be negative, implying $z(x_1) < 0$. As in Case 2, find a contradiction.
Case 4: If $z(x_1) = 0$, then the LHS of $(\ref{eq:foo})$ must be positive, implying $z(x_0) < 0$. As in Case 2, find a contradiction.

So it is impossible for $z$ to be strictly positive on $(x_0, x_1)$; this implies that $z$ either has a zero, changes sign, or is strictly negative on $(x_0, x_1)$. The latter is impossible by an analysis analogous to the foregoing; and a sign change implies the presence of a zero anyway by the intermediate value theorem, so we are done.

The general solution to a homogeneous linear ODE with constant coefficients is given by
\begin{equation}
y = \sum_{i=1}^n \sum_{j=0}^{p_i - 1} A_{ij} t^j e^{r_i t} \label{eqn:general}
\end{equation}
where $r_1, r_2, ..., r_n$ are the distinct roots of the characteristic polynomial and $p_i$ is the multiplicity of the $i$th distinct root.

We see that a term of the form $Ae^t$ is given by $j = 0$ and $r_i = 1$ in ($\ref{eqn:general}$), and that a term of the form $Bte^{-t}$ is likewise given by $j = 1$ and $r_i = -1$. Since $j = 1$, the multiplicity of the root $r_i = -1$ must be at least 2. So the minimal characteristic polynomial ought to be $$(x - 1)(x + 1)^2 = x^3 + x^2 - x - 1$$Conclude that the desired ODE is $$y''' + y'' - y' - y = 0$$

However I don't think you are correct about Chrome: Preview works for me in Chrome and it is Webkit. I just checked with Opera--Preview works, and with Maxthon--Preview works too!

I didn't say that it doesn't work---I said that it reloads the page when you click "Preview". If everything were working properly then it wouldn't have to reload the page, but would insert the preview directly into the current page. This is evidence that none of the browsers really do what they're supposed to do. Only Firefox has a weird bug that causes the preview not to work at all. Even still, fixing the XML would improve the user experience on all browsers.

I wonder, isn't that when 0<α<1, e^(α-1)t tend to 0 as t tend to infinity, and e^αt tend to infinity? As a result, isn't that all nonzero solutions become unbounded as t tend to zero when α>0 instead of α>1?

The coefficient on $e^{\alpha t}$ might be zero, so you can have solutions that are just $A e^{(\alpha-1)t}$. These will decay to zero as $t \to \infty$.

Add and subtract the first and second equations to obtain:
\begin{align}
(y+z)'' + K(y+z) &= 0 \label{added} \\
(y-z)'' + (K+2L)(y-z) &= 0 \label{subtracted}
\end{align}
Since $K, L > 0$, both equations are of the form $u'' + \omega^2 u = 0$, with general solution $u = A \cos (\omega t) + B \sin (\omega t)$. So the general solution to $(\ref{added})$ is
\begin{equation}
y+z = A \cos (\sqrt{K} t) + B \sin (\sqrt{K} t)
\end{equation}
and the general solution to $(\ref{subtracted})$ is
\begin{equation}
y - z = C \cos (\sqrt{K+2L} t) + D \sin(\sqrt{K+2L}t)
\end{equation}
Using the identities $y = \frac{1}{2}((y+z)+(y-z))$ and $z = \frac{1}{2}((y+z)-(y-z))$ we obtain the general solution to $(\ref{eq-1})$:
\begin{align}
y &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) + C' \cos(\omega_2 t) + D' \sin(\omega_2 t) \\
z &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) - C' \cos(\omega_2 t) - D' \sin(\omega_2 t)
\end{align}
where $A' = A/2$ and so on, and the frequencies are $\omega_1 = \sqrt{K}$, and $\omega_2 = \sqrt{K+2L}$.