 Author Topic: Reading Week Bonus problem 1  (Read 5077 times)

Victor Ivrii « on: February 16, 2013, 10:27:54 AM »
For "theoretical dudes" I suggest the sequence of the problems devoted comparison theorems and studying consecutive zeroes of  solutions to 2nd order linear homogeneous equations.

Consider two equations:
\begin{gather}
y'' + q(x)y=0,\\
z''+Q(x)z''=0
\end{gather}
with
\begin{equation}
Q(x)\ge q(x).
\end{equation}
Consider Wronskian $W(x):=W[y,z](x)$ and assuming that $y(x)>0$, $z(x)>0$ on interval $[a,b]$ derive differential inequality
\begin{equation}
W'  ?   0.
\end{equation}
« Last Edit: February 16, 2013, 10:55:22 AM by Victor Ivrii »

Brian Bi Re: Reading Week Bonus problem 1
« Reply #1 on: February 16, 2013, 02:41:29 PM »
Is the second equation supposed to read: $z'' + Q(x)z = 0$ ?

Write out the Wronskian explicitly:
\begin{equation}
W[y,z](x) = \begin{vmatrix} y & z \\ y' & z' \end{vmatrix} = yz' - y'z
\end{equation}

Differentiate w.r.t. $x$:
\begin{equation}
W' = \frac{d}{dx} (yz' - y'z) = y'z' + yz'' - y''z - y'z' = yz'' - y''z
\end{equation}

From the equations (1) and (2) we obtain $y'' = -qy$ and $z'' = -Qz$ so
\begin{equation}
W' = y(-Qz) - (-qy)z = (q-Q)yz \leq 0
\end{equation}
where the inequality follows from $Q > q$ and $y, z > 0$.
« Last Edit: February 16, 2013, 02:58:43 PM by Brian Bi »

Victor Ivrii Re: Reading Week Bonus problem 1
« Reply #2 on: February 16, 2013, 05:26:18 PM »
Correct.

Note: You don't need to write $W$ explicitly as for higher orders it only messes things up.

Simple rule how different determinants: determinants are sums (with certain signs $\pm$) of the products; each product includes exactly $1$ element from each row (and from each column, does not matter in formula below); recalling how we differentiate products we see that
\begin{equation*}
\left|\begin{matrix}
a_{11} &a_{12} & \ldots &a_{1m}\\
a_{21} &a_{22} & \ldots &a_{2m}\\
\vdots &\vdots &\ddots &\vdots\\
a_{m1} &a_{m2} & \ldots &a_{mm}
\end{matrix}\right|'=
\underbrace{\left|\begin{matrix}
a'_{11} &a'_{12} & \ldots &a'_{1m}\\
a_{21} &a_{22} & \ldots &a_{2m}\\
\vdots &\vdots &\ddots &\vdots\\
a_{m1} &a_{m2} & \ldots &a_{mm}
\end{matrix}\right|+
\left|\begin{matrix}
a_{11} &a_{12} & \ldots &a_{1m}\\
a'_{21} &a'_{22} & \ldots &a'_{2m}\\
\vdots &\vdots &\ddots &\vdots\\
a_{m1} &a_{m2} & \ldots &a_{mm}
\end{matrix}\right|+\ldots +
\left|\begin{matrix}
a_{11} &a_{12} & \ldots &a_{1m}\\
a_{21} &a_{22} & \ldots &a_{2m}\\
\vdots &\vdots &\ddots &\vdots\\
a'_{m1} &a'_{m2} & \ldots &a'_{mm}
\end{matrix}\right|}_{m \text{ terms}};
\end{equation*}
In particular, when we differentiate Wronskian first $(m-1)$ terms are $0$ as determinant has two equal rows and only the last term survives:
\begin{equation*}
\left|\begin{matrix}
y_1 &y_2 & \ldots &y_m\\
y'_1 &y'_2 & \ldots &y'_m\\
\vdots &\vdots &\ddots &\vdots\\
y^{(m-2)}_1 &y^{(m-2)}_2 & \ldots &y^{(m-2)}_m\\
y^{(m-1)}_1 &y^{(m-1)}_2 & \ldots &y^{(m-1)}_m
\end{matrix}\right|'=
\left|\begin{matrix}
y_1 &y_2 & \ldots &y_m\\
y'_1 &y'_2 & \ldots &y'_m\\
\vdots &\vdots &\ddots &\vdots\\
y^{(m-1)}_1 &y^{(m-1)}_2 & \ldots &y^{(m-1)}_m\\
y^{(m)}_1 &y^{(m)}_2 & \ldots &y^{(m)}_m
\end{matrix}\right|.
\end{equation*}