Let
\[
I = \int_0^1 \bigl((y')^2 -y^2\bigr)\d x \qquad\text{and}\qquad
I_1=\int_0^1 (y'+y\tan x)^2 \d x \,,
\]
where \(y\) is a given function of \(x\) satisfying \(y=0\) at \(x=1\). Show that \(I-I_1=0\) and deduce that \(I\ge0\). Show further that \(I=0\) only if \(y=0\) for all \(x\) (\(0\le x \le 1\)).
Let
\[
J = \int_0^1 \bigl((y')^2 -a^2y^2\bigr)\d x
\,,
\]
where \(a\) is a given positive constant and \(y\) is a given function of \(x\), not identically zero, satisfying \(y=0\) at \(x=1\). By considering an integral of the form
\[
\int_0^1 (y'+ay\tan bx)^2 \d x \,,
\]
where \(b\) is suitably chosen, show that \(J\ge0\). You should state the range of values of \(a\), in the form \(a < k\), for which your proof is valid. In the case \(a=k\), find a function \(y\) (not everywhere zero)
such that \(J=0\).
Solution:
\begin{align*}
&& I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\
&&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\
&&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\
&&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\
&&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\
&&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\
&&&= \left [-y^2 \tan x \right]_0^1 \\
&&&= 0 \\
\\
\Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0
\end{align*}
The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so
\begin{align*}
&& \frac{\d y}{\d x} &= - y \tan x \\
\Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\
\Rightarrow && \ln |y| &= \ln |\cos x| + C \\
\Rightarrow && y &= A \cos x \\
\Rightarrow && A &= 0 \Rightarrow y = 0
\end{align*}
Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then
\begin{align*}
&& J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\
&&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\
&&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\
&&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\
&&&= \left [ - a y^2 \tan a x \right]_0^1 = 0
\end{align*}
This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\).
Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and
\begin{align*}
&& \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\
&&&= 0
\end{align*}