How to handle a missing environment variable set in `crontab`

I have a bash script that I need to be able to start from either: 1. cron 2. an interactive (logon) shell This script needs to know whether it was started from cron, or started from an interactive shell. I thought I had resolved this issue when I declared an environment variable in the root crontab:

RUN_BY_CRON="TRUE"

In the script, I test RUN_BY_CRON and use the result to set another variable:

if [ "$RUN_BY_CRON" = "TRUE" ]; then            
    ((wait_time=DELAY_HALT*60))
fi

This worked until I added set -u to my script (as a *"common defensive programming strategy"*). Since then, when I run the script from the command line, set -u flags RUN_BY_CRON as an "unbound variable" error:

$ sudo ./skedrtc.sh
./skedrtc.sh: line 24: RUN_BY_CRON: unbound variable

FWIW, I ran shellcheck on this script, and got no warning or error. I tried adding a test for RUN_BY_CRON, but got the same error. I tried testing for an interactive shell, but testing from within the script itself isn't helpful:

...
if [ -z "$RUN_BY_CRON" ]; then        # test for null string
    RUN_BY_CRON="FALSE"
fi
...

if [[ $- == *i* ]]; then              # test for interactive shell
    RUN_BY_CRON="FALSE"
fi

This feels like a [*"catch 22"*](https://www.merriam-webster.com/dictionary/catch-22) situation. I've looked for ways to create a **try-catch** block, but AIUI there's nothing like that in bash. ### Q: How can I avoid this "unbound variable" error without removing the set -u ?