Problem source

The functions $\mathrm{x}$ and $\mathrm{y}$ are related by 
\[
\mathrm{x}(t)=\int_{0}^{t}\mathrm{y}(u)\,\mathrm{d}u,
\]
so that $\mathrm{x}'(t)=\mathrm{y}(t)$. Show that 
\[
\int_{0}^{1}\mathrm{x}(t)\mathrm{y}(t)\,\mathrm{d}t=\tfrac{1}{2}\left[\mathrm{x}(1)\right]^{2}.
\]
In addition, it is given that $\mbox{y}(t)$ satisfies 
\[
\mathrm{y}''+(\mathrm{y}^{2}-1)\mathrm{y}'+\mathrm{y}=0,\mbox{ }(*)
\]
with $\mathrm{y}(0)=\mathrm{y}(1)$ and $\mathrm{y}'(0)=\mathrm{y}'(1)$.
By integrating $(*)$, prove that $\mathrm{x}(1)=0.$ 
By multiplying $(*)$ by $\mathrm{x}(t)$ and integrating by parts, prove the relation 
\[
\int_{0}^{1}\left[\mathrm{y}(t)\right]^{2}\,\mathrm{d}t=\tfrac{1}{3}\int_{0}^{1}\left[\mathrm{y}(t)\right]^{4}\,\mathrm{d}t.
\]
Prove also the relation 
\[
\int_{0}^{1}\left[\mathrm{y}'(t)\right]^{2}\,\mathrm{d}t=\int_{0}^{1}\left[\mathrm{y}(t)\right]^{2}\,\mathrm{d}t.
\]

Solution source

Consider $\frac12 x(t)^2$ then differentiating we obtain $x(t)x'(t) = x(t)y(t)$. Also note that $x(0) = \int_0^0 y(u) \d u = 0$

Therefore,

\begin{align*}
\int_0^1 x(t)y(t) \d t &= \left [  \frac12 x(t)^2 \right]_0^1 \\
&= \frac12[x(1)]^2
\end{align*}

\begin{align*}
&& 0 &= y'' + (y^2-1)y' + y  \\
\Rightarrow && 0 &= \int_0^1 \l y'' + (y^2-1)y' + y \r \d t \\
&&&= \left [y'(t) + \frac13y^3-y+x \right]_0^1 \\
&&&= x(1)
\end{align*}

Therefore $x(1) = 0$.

\begin{align*}
&& 0 &= xy'' + (y^2-1)y' x+ yx  \\
\Rightarrow && 0 &= \int_0^1 \l xy'' + (y^2-1)y'x + xy \r \d t \\
&&&= \left [ x y' +(\frac13 y^3-y)x \right]_0^1 - \int_0^1 yy'+\frac13y^4-y^2 \d t \\
&&&= 0 - \frac13 \int_0^1 [y(t)]^4 \d t - \int_0^1 [y(t)]^2 \d t \\
\Rightarrow && \int_0^1 [y(t)]^2 \d t &=  \frac13 \int_0^1 [y(t)]^4 \d t
\end{align*}

\begin{align*}
&& 0 &= yy'' + (y^2-1)y' y+ y^2  \\
\Rightarrow && 0 &= \int_0^1 \l yy'' + (y^2-1)y'y + y^2 \r \d t \\
&&&= \left [ y y' +(\frac14 y^4-\frac12y^2) \right]_0^1 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\
&&&= 0 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\
\Rightarrow && \int_0^1 [y'(t)]^2 \d t &=  \int_0^1 [y(t)]^2 \d t
\end{align*}