@acer 

Thank you for the quick response, and this does exactly what I want, in a very quick way. To answer your inquiry above, yes I am interested in finding the smallest(negative) upon which this inequality is satisfied. 

I knew there was a more efficent way, and what you said about "stopping once found" is what I was interested in doing I was just not sure how to implement in. 

Now for your modification of my procedure, is the time much quicker simply by the way you modified the integral? 

@lemelinm 

This is the downside of my idea.. Interpretation needs to be used. If I find a proper solution I will let you know. 

@acer 

A quick and great response as always. I was unaware of the adaptive option in the plot command, I will keep my eyes out for this in the future.

Much apprecaited.

@ecterrab 

Thank you for this as I now feel embarassed about the simplicity of the problem that I had not considered.

I will forever remember this, and avoid functions with the same names as coordinates if I use this package. 

@Rakshak 

For something to be vacuum means there is zero stress-energy. If the metric itself has unkown function(s) you must solve for those functions before you have the full solution, then the solution will have vanishing Ricci and Einstein tensors. If a metric has unkown functions and is a vacuum solution that is simply stating that you must solve Ricci=0. 

@Jack Zuffante 

If you could upload a worksheet with what you are doing that is the best way to get specific answers to problems you are having. Use the green arrow in top right. 

@acer

A simple fix as I was hoping/assuming, thank you!

@ecterrab 

I am not taking you the wrong way at all. I am aware that my method of comparison is not ideal, I will take fault for that. I was unaware/naive about doing it as simply as you have. But this is where I think the final concern of mine can be seen and probably rectified. 

In your first reponse to my "potential bug" you have in (9) the expression for epsilon=0, a=b=1. Now in your latest post you have either (6) or (7) representing epsilon=0, a=b=1. You can see they do not match, and they should, they are both suppose to be the zeroth order expressions. Now I know we have ignored the physics discussion in this post but I assure you the expression that we obtain will be at most first derivative in both Phi and s. 

If there is something mathematical that I am forgetting and misunderstanding I will admit complete defeat but I am just so surprised that would be the case and that I have not picked that up. I do greatly appreciate your responses and help with post thus far as I have learned way better ways to use the Physics package. 

Edit:  I see your answer with the update you had posted today. Thank you very much for this. 

@ecterrab 

I aggree in that case but the discrepancy in (19) still bothers me. However, consider a different siuation instead. 

Don't let epsilon be zero at the start, instead do subs method for a=b=1 then a series for zeroth order in epsilon they do not agree with what are in the NO_BugMaple_(reviewed).mw file for epslion=0 at the start. Another issue that I think needs to be resolved, because it begs the question with all the options how would one no which is correct? Ofcourse Define is the best approach but requires the most computational time. 

I have attached the issue I described above, sadly I couldn't find a compact way to do it. 

SeriesBug.mw

@ecterrab

Here is attached worked sheet with the suspected bug I mentioned above. 

Curious about your thoughts on this as well.

BugMaple.mw

Thank you for this response. I look forward to seeing what you come up with. 

I also see this as an opportunity to bring to your attention a bug I have found. 

In trying to speed up the process and avoiding the use of Define i tried a method that simply assumed the tensor indices on each term individually and then added said expressions at the end, easy enough. I however noticed there are discrepinces when i condisdered the zeroth order expression in epsilon with this method and using Define. I would be curious to have your thoughts on this problem as well. I will try to make a compact worksheet displaying this issue and reply tho this thread with it. 

@ecterrab 

This is great. The change is self-explanatory. 

Thank you!

@vs140580 

I see what you mean by zeros now. My mistake.

Good question. 

@sursumCorda 

Thank you for your comment, I misread what you had written therefore I retract my previous statement. 

What variable are you trying to solve for? This is one equation with 4 unkowns as I see it. So you would have infinit solutions. Unless I am missing something.