Java replacing special characters

Simply use String#replace(CharSequence target, CharSequence replacement) in your case to replace a given CharSequence, as next:

special = special.replace("@$", "as");

Or use Pattern.quote(String s) to convert your String as a literal pattern String, as next:

special = special.replaceAll(Pattern.quote("@$"), "as");

If you intend to do it very frequently, consider reusing the corresponding Pattern instance (the class Pattern is thread-safe which means that you can share instances of this class) to avoid compiling your regular expression at each call which has a price in term of performances.

So your code could be:

private static final Pattern PATTERN = Pattern.compile("@$", Pattern.LITERAL);
...
special = PATTERN.matcher(special).replaceAll("as");

Escape characters:-

    String special = "Something @$ great @$ that.";
    special = special.replaceAll("@\\$", "as");
    System.out.println(special);

For Regular Expression, below 12 characters are reserved called as metacharacters. If you want to use any of these characters as a literal in a regex, you need to escape them with a backslash.

the backslash \
the caret ^
the dollar sign $
the period or dot .
the vertical bar or pipe symbol |
the question mark ?
the asterisk or star *
the plus sign +
the opening parenthesis (
the closing parenthesis )
the opening square bracket [
and the opening curly brace {

references:- http://www.regular-expressions.info/characters.html

The method replaceAll accepts regex as a pattern to substitue: http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replaceAll(java.lang.String,%20java.lang.String)

Try simply:

special = special.replace("@$", "as");