# How are $\sinh \beta x$ and $\cosh \beta x$ related to $\sinh \beta (L-x)$ and $\cosh \beta (L-x)$?

I'm reading Section 2.3 of DuChateau and Zachmann's Applied Partial Differential Equations. The section deals with Sturm-Liouville Eigenvalue Problems, and I'm currently looking at Example 2.3.2 on page 68.

The problem in the example is as follows:

$\begin{matrix} -u^{\prime\prime}(x) = \mu u(x) & 0<x<L \\ u^{\prime}(0)=0, & u^{\prime}(L)=0.\end{matrix}$

So, I am now looking at how they find the negative eigenvalues (i.e., $\mu < 0$).

For this case, what it says to do is to write $\mu = -\beta^{2}$ for $beta$ real and not zero, and then the general solution to the differential equation can be written as $u(x) = A \cosh \beta x + B \cosh \beta (L - x)$.

It said earlier in the book that the general solution to an eigenvalue problem can be written as a linear combination of any two of the following six linearly independent functions: $e^{\beta x}$, $e^{-\beta x}$, $\sinh \beta x$, $\cosh \beta x$, $\sinh \beta (L-x)$, $\cosh \beta (L-x)$.

Now, my previous experience with ODEs is limited, so much so in fact, that whenever I saw an equation like $-u^{\prime\prime}(x)=-\beta^{2}u(x)$, after forming the characteristic equation $-r^{2} + \beta^{2} = 0 \, \implies \, r^{2} - \beta^{2} = 0$, since we have distinct real roots $r = \pm \beta$, I would just say that the solutions are $u_{1} = e^{-\beta x}$ and $u_{2} = e^{\beta x}$.

I've never even seen solutions containing the other linearly independent functions for problems with characteristic equations with distinct real roots.

So, I've been sitting here trying to convince myself that they really are all equivalent.

I'm assuming that since $e^{-\beta x} = \cosh(-\beta x)+ \sinh (-\beta x) = \cosh (\beta x) - \sinh (\beta x)$ is how we would be justified in writing our solution $u_{1}$ in terms of $\cosh (\beta x)$ or $\sinh (\beta x)$, but I can't figure out where the $\cosh \beta (L-x)$ and $\sinh \beta (L-x)$ comes from.

I tried writing $\sinh \beta (L-x)$ as $\sinh (\beta L - \beta x)$ and then used the $\sinh$ difference identity to get $\sinh (\beta L) \cosh (\beta x) - \cosh (\beta L) \sinh (\beta x)$.

I then used the fact that this is equal to $\displaystyle \left( \frac{e^{\beta L} - e^{-\beta L}}{2}\right) \left(\frac{e^{\beta x}-e^{-\beta x}}{2} \right) - \left(\frac{e^{\beta L} + e^{-\beta L}}{2} \right) \left(\frac{e^{\beta x} -e^{-\beta x}}{2} \right)$, but after multiplying everything out and simplifying, I ended up right beck where I started with $\sinh \beta (L-x)$.

So, for my first question: could somebody please explain in thorough detail how these six linearly independent functions are related as things we can choose to write our solution in terms of?

For my second question, why is it more beneficial to choose some solution forms over others? Feel free to provide examples to illustrate. What would you choose, for instance, in the case that the left boundary condition was a derivative and the right boundary condition just $u$, or vice-versa?

Finally, the textbook says that the general solution can be written in the form $u(x) = A \cosh \beta x + B \cosh \beta (L - x)$. Now, I'm assuming that they chose $\cosh \beta (L-x)$ for $u_{2}(x)$ since they already chose $\cosh \beta x$ for $u_{1}(x)$ and we want the two solutions to be linearly independent. But, for my third and last question, could someone please explain to me the rationale behind choosing these solutions in this particular problem? (I suppose this also kind of goes hand in hand with what I asked on my first and second questions, but don't just ignore them in favor of answering this one. I need to know how to do this in general!)

Thank you for your time and patience, just please make your answer as thorough and detailed as possible, as though you were trying to explain this to the most absolutely clueless person you ever met in your life (because right now, I feel like that person!).

$e^{\beta x}$, $e^{-\beta x}$, $\sinh(\beta x)$, $\cosh(\beta x)$, $\sinh(\beta(L-x))$ and $\cosh(\beta(L-x))$ are not "six linearly independent functions". The vector space they span is just two-dimensional. For example, the first two can be used as a basis.
\eqalign{\sinh(\beta x) &= \dfrac{1}{2} e^{\beta x} - \dfrac{1}{2} e^{-\beta x} \cr \cosh(\beta x) &= \dfrac{1}{2} e^{\beta x} + \dfrac{1}{2} e^{-\beta x}\cr \sinh(\beta(L-x)) &= -\frac{e^{-\beta L}}{2} e^{\beta x} + \frac{e^{\beta L}}{2} e^{-\beta x} \cr \cosh(\beta(L-x)) &= \frac{e^{-\beta L}}{2} e^{\beta x} + \frac{e^{\beta L}}{2} e^{-\beta x}\cr}
Which basis you choose is a matter of choice. An advantage of $\sinh(\beta x)$ and $\cosh(\beta x)$ is that they are convenient for initial conditions at $x=0$, since the solution with initial conditions $y(0) = a$, $y'(0)=b$ is $a \cosh(\beta x) + (b/\beta) \sinh(\beta x)$. Similarly, $\sinh(\beta(L-x))$ and $\cosh(\beta(L-x))$ are convenient for conditions at $x=L$.
• how does the coefficient $b/\beta$ come about? – user100463 Nov 16 '16 at 18:32
• $\dfrac{d}{dx} (b/\beta) \sinh(\beta x) = b \cosh(\beta x)$, and $\cosh(0) = 1$. – Robert Israel Nov 16 '16 at 18:33