Given real numbers $a, b, c, d$ satisfy $a^{4}+b^{4}=c^{4}+d^{4}=2011, a c+b d=0$. Find the value of $a b+c d$.

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Given $a c+b d=0 \Rightarrow a c=-b d$.
Raising both sides to the fourth power, we get $a^{4} c^{4}=b^{4} d^{4}$.
Since $c^{4}=2011-d^{4}, b^{4}=2011-a^{4}$, we have,
$$
\begin{array}{l}
a^{4}\left(2011-d^{4}\right)=\left(2011-a^{4}\right) d^{4} \\
\Rightarrow 2011 a^{4}-a^{4} d^{4}=2011 d^{4}-a^{4} d^{4} .
\end{array}
$$

Thus, $a^{4}=d^{4} \Rightarrow a= \pm d$.
Combining with the given conditions, we get
$$
b^{4}=c^{4} \Rightarrow b= \pm c \text {. }
$$

From equations (1) and (2), we have
$$
a=d, b=c
$$

or $a=-d, b=-c$
or $a=d, b=-c$
or $a=-d, b=c$.
In the first two cases,
$$
a b+c d= \pm(a c+b d)=0 ;
$$

In the last two cases,
$$
a b=-c d \Rightarrow a b+c d=0 \text {. }
$$

In summary, $a b+c d=0$.
(Chen Kuanhong, Land Reserve Center of Yueyang County, Hunan, 414100)