The Introduction of boundary conditions in physical problems: A problem with a result in a book

Question

I'm asking this question again, since I incurred in an error while copying the relations from the text.\

Returning to an old book about the foundations of physics by Lindsay and Margenau, I wasn't able to follow a result in a section about the introduction of boundary conditions in physical problems.

The imposed conditions on the wave equation solution in a tube of length $l$ are:

$f_{1}(-ct) + f_{2}(ct)=0$

$f_{1}(l-ct)+f_{2}(l+ct)=0$

from this you`ll get (copying the exact relation from the text)

$f_{2}(ct)=-f_{1}(-ct)$

$f_{2}(ct+l)=-f_{1}(l-ct)= f_{2}(ct-l)$

It is the third term of this equation ($f_{2}(ct-l)$) that I cant figure out how it appears from the others two and the above relations, sadly it is a non-trivial result for me. I`ll appreciate your help.

Answer 1

The first condition $f_1(-ct) + f_2(ct)=0$ basically asserts that $f_1(t) = -f_2(t)$.

To see this, do the substitution $t \to -\dfrac t c$.

Hence we have $-f_1(l-ct) = --f_2(-(l-ct)) = f_2(ct-l)$, via the substitution $t \to l-ct$.

Answer 2

$f_{1}(c(ct+l)) + f_{2}(-c(ct+l))=0$

$f_{1}(l-c(ct+l))+f_{2}(l+c(ct+l))=0$

the same applies to $l-ct$ like above for $ct+l$:

$f_{1}(c(l-ct)) + f_{2}(-c(l-ct))=0$

$f_{1}(l-c(l-ct))+f_{2}(l+c(l-ct))=0$