Clarification of Qualitative Behaviour of BVP Solutions Example I have the following example from my PDE textbook (Essential Partial Differential Equations by Griffiths, Dold, and Silvester):



*

*I don't understand how the authors found the solution $u(x)$. The exercise (5.2) they reference is as follows:



Verify that (5.6) and (5.8) are solutions to the BVP in Example 5.1.

But how does verifying this find us $u(x)$?


*Why does $|u(\frac{1}{2})| \to \infty$ as $b \to -(2n - 1)^2 \pi^2$ indicate that the BVP is not well posed for these values of $b$?

*For the graph (figure 5.1), in a), the authors state that, when $\epsilon > 0$, the solution is negative throughout the interval $0 < x < 1$. But we can see from the graph (figure 5.1) that, when $\epsilon = 1 > 0$, the solution is not always negative throughout the interval $0 < x < 1$. Specifically, the solution becomes positive for $b = 100$ and $b = -100$.
I would greatly appreciate it if someone could please take the time to clarify the author's work here.
 A: *

*Exercise (5.2) does not ask you to find the solution, and the
authors do not say how they found it. The exercise only says
"verify" which means just plug in and check.
If you do want to derive the solution, which you are not asked to do,
it seems that the ordinary methods of ODE suffice, but are rather
tedious; you know when $b\ne0$ the solution is in the form
$$
 u(x) = -\frac{\epsilon}{b} +c_1 e^{\sqrt{b}x}+c_2 e^{-\sqrt{b}x},
$$
where $\sqrt{b}$ is one of the square roots of $b$,
and you only need to determine $c_1$ and $c_2$. The algebra is
awful. It works out better to recognize the left--right symmetry of the problem
across the point $x=\frac{1}{2}$, and use instead
$$
 u(x) = -\frac{\epsilon}{b} +a_1 e^{\sqrt{b}(x-1/2)}+a_2 e^{-\sqrt{b}(x-1/2)}.
$$
You quickly come to the boundary conditions giving
$$
 a_1 = a_2 = \frac{\epsilon/b}{e^{-\sqrt{b}/2}+e^{\sqrt{b}/2}}
$$
provided that the denominator is not zero, and no solution
in the case that the denominator is zero. This gives the answers stated.

*"indicates" does not mean proves, it means suggests. The proof is 
that this is the zero denominator case above, where there is no
solution to the BC equations for $a_1$ and $a_2$.

*I think this is a mis-reading of the authors sentence, which is
slightly ambiguous. They meant that $b>0$ and $\epsilon > 0$. 
A: So there is a proposal:
$$u(x) = -\dfrac{\epsilon}{b} +c_1 e^{\sqrt{b}\ x}+c_2 e^{-\sqrt{b}\ x},$$
but, incredibly, there is another proposal:
$$u(x) = -\dfrac{\epsilon}{b} +c_1 \cosh(\sqrt{b}\ x)+c_2 \sinh(\sqrt{b}\ x).$$
With $u(0)=0$ and $u(1)=0$, you have:
$$0=-\dfrac{\epsilon}{b} + c_1, $$
$$0=-\dfrac{\epsilon}{b} + c_1 \cosh(\sqrt{b}/2)+c_2 \sinh(\sqrt{b}/2).$$
Solve for $c_1$ and $c_2$:
$$c_1 = \dfrac{\epsilon }{b},\quad c_2= \dfrac{\epsilon\ \text{csch}\left(\sqrt{b}\right) - \epsilon  \coth \left(\sqrt{b}\right)}{b}.$$
So the solution is (after simplifications):
$$u(x) = -\dfrac{\epsilon}{b} + \dfrac{\epsilon}{b} \cdot \dfrac{\cosh \left(\dfrac{1}{2} \sqrt{b}\ (1-2 x)\right)}{\cosh\left(\dfrac{\sqrt{b}}{2}\right)}. $$
As you can see, the denominator has problems with $b=0$, that is why you solve $u''(x) =0$, and you get that solution:
$$u(x) = -\dfrac{1}{2} \epsilon x (1-x).$$
Now for the case $b < 0$. Well, $\cosh(\sqrt{- |b|}\ x) =\cosh(i\sqrt{|b|}\ x)$, with $i=\sqrt{-1}$, which means:
$$\cosh(i\sqrt{|b|}\ x) = \cos(\sqrt{|b|}\ x).$$
You and I conclude that for $b<0$:
$$u(x) = -\dfrac{\epsilon}{b} + \dfrac{\epsilon}{b} \cdot \dfrac{\cos \left(\dfrac{1}{2} \sqrt{|b|}\ (1-2 x)\right)}{\cos\left(\dfrac{\sqrt{|b|}}{2}\right)}. $$
Wait! The denominator still has problems (again) when $\cos\left(\dfrac{\sqrt{|b|}}{2}\right)=0$. Of course you can find those values of $b$:
$$b=-(2n-1)^2 \pi^2.$$
Choose $n$, say $n=2$, now find $u(1/2)$ for $b=-9\pi^2 \pm 0.000001$, you will be surprised.
If you have doubts about figure 5.1, plot. For $\epsilon = 1, b=100$:

For $\epsilon = 1, b=-100$:

Figure with the three solutions:

