What is the significance of -Mu in equation equality and boundedness?

[ssky]

http://dc191.4shared.com/img/6pyFHiMb/s7/0.8584304740152441/706869958.jpg

I need more explanation for the first equation in the picture above:

Why we said that each side in the equation is equal (-Mu) ?

 

[tiny-tim]

hi ssky!

(have a mu: µ )

the LHS is the same for all x, and the RHS is the same for all t,

and since they're equal, they must be the same for all x and t

 

[ssky]

:shy:

I am sorry,
can you explain more.
and why we took the negative value?

 

[tiny-tim]

the equation has to be true for all values of x and t

the LHS is a function of t, but for a fixed value of t it is the same for all values of x …

if you were to plot it on a 3D graph, with horizontal x and t axes, it would be a hillside on which all the contour lines (lines of level height) were parallel to the t axis…

but the RHS would be a hillside on which all the contour lines (lines of level height) were parallel to the x axis …

but the LHS and the RHS have to be the same hillside …

the only way that can happen is if the hillside is completely flat! ​

and why we took the negative value?

without seeing the previous page (which i don't want to ), I've no idea

 

[ssky]

Actually,

i didn't understand



can you give me a clear example about this problem?


​

 

[Office_Shredder]

The equation looks like this:

f(t)=g(x)

The left side is a function of t only, and the right side a function of x only. Now, pick t=0.

f(0)=g(x)

This is true for ANY value of x, so it must be that g(x) is a constant function. Different values of x don't change g(x), because you always get f(0).

Similarly, if we set x=0 we get
f(t)=g(0)

This is true for every value of t, so f(t) is a constant function as well

 

[HallsofIvy]

:shy:

I am sorry,
can you explain more.
and why we took the negative value?

It says in what you posted "where the negative value was forced to warrent the boundedness of the function $\Gamma(t)$ as $t\to \infty$".

I suspect that if you set it equal to "a" where a could be any constant, you would eventually get a function involving $e^{at}$ which will go to infinity if a is positive. Writing $a= -\mu^2$ where $\mu$ can be any real number forces a to be negative so that $e^{at}$ does NOT go to infinity and is bounded.