Variable not getting picked up inside EOF - shell

Question

I'm trying to run a commands remotely by doing SSH but variable which is defined inside EOF is not getting picked up.
Variable $BASE_PATH is not getting called inside another variable which i'm defining by name FOLDER_NAME . Mentioned inside script too.
I couldn't use EOF in single quotes ('EOF') because i have to use variable from exiting shell too.
#!/usr/bin/ksh
FILE_NAME=$2
jvm_list=$1
for jvm in `echo $jvm_list`
do
ssh $jvm << EOF
echo FILE_NAME=${FILE_NAME}
export BASE_PATH="\${WEB_DOMAIN_HOME}/Server";
echo BASE_PATH=\${BASE_PATH}; ##Value of BASE_PATH is getting picked up##
export FOLDER_NAME=\`ls -1d $"{BASE_PATH}"/properties* | grep -i -v old\` ##Value of BASE_PATH is coming blank here##
echo $FOLDER_NAME
EOF
done

Answer 1

The construct started with << EOF acts like double-quoted string, so dollar signs are special within it, unless escaped.
Here, you escaped the $, so the shell eventually started by ssh evaluates that variable
echo BASE_PATH=\${BASE_PATH}; ##Value of BASE_PATH is getting picked up##
Here, you didn't escape it, so the evaluation happens in the outer shell:
export FOLDER_NAME=\`ls -1d $"{BASE_PATH}"/properties* ...\`
I do also suspect that the quotes there are misplaced. In Bash $".." is a locale-specific translation, and it seems to be the same in ksh. You probably don't have a translation for that, so the string should come back as is: {BASE_PATH}.
Somewhat related to this may be the backticks, since they need to be quoted too. You could use the $( ... ) form of command substitution, so you'd again, only need to think about the $.
I think this may be what you want:
ssh "$jvm" << EOF
export FOLDER_NAME=\$(ls -1d "\${BASE_PATH}"/properties* | grep -i -v old)
EOF
Sanity check:
$ foo=out; bash <<EOF
> foo=in; echo $foo
> EOF
out
$ foo=out; bash <<EOF
> foo=in; echo \$foo
> EOF
in