How can I run an app if another app is running?

If you know the name of the application and you are sure that only instantiation of it can be running at a given time, then the solution is pretty simple:

running="$(ps -aux | grep yourappname)"

if [ -n "$running" ] ; then
   launch your secondary application
fi

Obviously, you'd need to change 'yourappname' to the appropriate name of the app you seek, but that's all you need to do.

If you could have more than one version of the process running or otherwise don't have an easy way to identify it uniquely, then you have a greater challenge. For example, if you wanted to check for the Apache Web server (process name "httpd") but wanted to make sure that you weren't seeing a child process but the parent, then you'd also need to take into account the parent process ID (ppid) because the ppid of child processes is the parent process (which makes sense!): in the case of child Apache processes, their ppid's are the process ID (pid) of the parent Apache process.

And that's exactly how you do it. Use a flag like '-j' to get the pid and ppid, use grep to pull out the matching processes by name, then check the ppid to identify which has the odd ppid out. Let me show you. Here's a set of processes shown in this format:

$ ps -jax | grep http
root    6706     1     0 Ss    ??    0:03.43 /usr/local/apache/bin/httpsd
www     6707  6706     0 I     ??    0:11.91 /usr/local/apache/bin/httpsd
www     6708  6706     0 S     ??    0:13.22 /usr/local/apache/bin/httpsd
www     6709  6706     0 S     ??    0:13.46 /usr/local/apache/bin/httpsd
www     6710  6706     0 I     ??    0:13.82 /usr/local/apache/bin/httpsd
www     6711  6706     0 S     ??    0:12.36 /usr/local/apache/bin/httpsd
www    19003  6706     0 S     ??    0:00.09 /usr/local/apache/bin/httpsd
www    19046  6706     0 I     ??    0:00.01 /usr/local/apache/bin/httpsd
www    19429  6706     0 S     ??    0:00.02 /usr/local/apache/bin/httpsd
www    19432  6706     0 S     ??    0:00.04 /usr/local/apache/bin/httpsd
www    19531  6706     0 I     ??    0:00.00 /usr/local/apache/bin/httpsd
www    19612  6706     0 S     ??    0:00.01 /usr/local/apache/bin/httpsd
www    19615  6706     0 S     ??    0:00.00 /usr/local/apache/bin/httpsd
www    19616  6706     0 S     ??    0:00.00 /usr/local/apache/bin/httpsd
www    19622  6706     0 S     ??    0:00.01 /usr/local/apache/bin/httpsd
www    19624  6706     0 Z     ??    0:00.00  (httpsd)

Notice that the second column is the process ID and the third is the parent process id: as you can see, the parent process is 6706 in this instance.

You can identify this in your script (I've written about it here before) and then screen for a process by ID rather than by name (use the '-p process' flag to ps) then use the same basic test shown earlier to launch your secondary application, as needed.

Hope that helps you out!