# What is the error I am making on getting general solutions for this Sturm-Liouville problem?

Given this Sturm-Liouville problem: $$X'' + \lambda X = 0$$

There are general solutions (Eigenfunctions) for three cases on $$\lambda$$: $$\lambda > 0$$ Has the characteristic equation: $$r^2+\lambda = 0$$ with roots $$r_1 = i \sqrt{\lambda}, r_2 = -i \sqrt{\lambda}$$, inserting them into the derivative eigenfunction $$e^{rx}$$ results into: $$X(x) = c_1 e^{i \sqrt{\lambda}x} = c_1cos\sqrt{\lambda}x + c_2 sin \sqrt{\lambda}x$$

For the case: $$\lambda < 0$$ $$X(x) = c_1 e^{\sqrt{\lambda}x} + c_2 e^{-\sqrt{\lambda}x}$$

Finally, for case $$\lambda = 0$$ the characteristic equation is $$r^2 = 0$$, thus $$r=0$$ and: $$X(x) = ce^{r x} = c$$

However the solution for this case is known to be: $$X(x) = c_1 x + c_2$$

What is the error I am making?

The special cases go away if you normalize the solutions at one endpoint, such as at $$x=0$$. For example, start by solving $$X''+\lambda X = 0 \\ X(0)=A,\;\; X'(0) = B.$$ The solutions then depend on $$\lambda$$ through a power series expansion. So special cases at, say $$\lambda=0$$, become limiting cases as $$\lambda\rightarrow 0$$, which allows you to ignore special cases during the process of finding the general solution. In this case, for $$\lambda \ne 0$$, the solutions are $$X_\lambda(x)=A\cos(\sqrt{\lambda}x)+B\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}.$$ The solution when $$\lambda=0$$ is the limit of the above as $$\lambda\rightarrow 0$$: $$X_{\lambda=0}(x)=A+Bx.$$ But, if you do not normalize the endpoint conditions with constants, then the special case solutions are not necessarily limits of a family of solutions that you can find with arbitrary constants, which is something you have already discovered.

• So where in my steps would I need to normalize the solutions and how would that look like? Oct 25 '18 at 14:39
• @hirschme : Compare your general solution with mine. I have solutions that have constant values of $X(0), X'(0)$. As your $\lambda\rightarrow 0$, you lose a solution. Oct 25 '18 at 18:34

$$X^{''} + \lambda X\tag{1}$$

In addition, it is simpler to just call the second one hyperbolics. If $$\lambda < 0$$

$$r = \pm \sqrt{ -\lambda} \\ \lambda = -s \tag{2}$$

$$X^{''} = sX \tag{3}$$ $$X(x) = c_{1}e^{\sqrt{s}x} + c_{2}e^{-\sqrt{s}x} \tag{4}$$

which gives us hyperbolics

$$X(x) = c_{3}\cosh(\sqrt{s}x) +c_{4} \sinh(\sqrt{s}x) \tag{5}$$

in case $$3$$ we have to integrate

$$X^{''} = 0\cdot X \\ X^{''} = 0 \tag{6}$$

integrate as usual

$$X^{'} = \int 0 dx = c_{1} \tag{7}$$

now integrate again

$$\int X^{'} = \int c_{1} dx \implies X(x) = c_{1}x + c_{2} \tag{8}$$

You don't need to use a characteristic equation. Or also that there are two roots that are $$0$$ so you need to make one have a polynomial power. You can't have two equal solutions. In the general problem is

$$ay^{''} + by^{'} +cy =0 \tag{9}$$

we have the characteristic equation $$ar^{2} +br + c= 0\tag{10}$$

if we have double roots $$r_{1} = r_{2} =r$$

$$y_{1}(t) =e^{r_{1}t} = e^{rt} ,y_{2}(t) =e^{r_{2}t} = e^{rt} \tag{11}$$

$$r_{1,2} = \frac{-b \pm \sqrt{b^{2} -4ac }}{2a} \tag{12}$$

double roots come from $$r_{1,2} = \frac{-b}{2a} \tag{13}$$

$$y_{1}(t) = e^{\frac{-bt}{2a}} \tag{14}$$

to get a second solution you have multiply by $$v(t)$$

$$y_{2}(t) = v(t)y_{1}(t) \tag{15}$$

you eventually figure out that $$v(t) = t$$

• Integrating the equation is definitely simpler and the solution its very intuitive. But I still do not understand why I do not arrive to the same solution with the characteristic equation. You say I don't 'need' to use it. But can one use it? How would the steps look using it and 'adding a polynomial power' as you say? Oct 25 '18 at 14:38
• tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx I don't really want to write all of this.. but it is the repeated roots section. if you would like me to I can.
– user3417
Oct 25 '18 at 17:40
• Ok so I need to know that in general, we avoid having two equal solutions and whenever this is the case, we apply another general solution of the form $y(t)=c_1 e^{rt}+c_2 t e^{rt}$ Oct 25 '18 at 18:06
• Yes, I'll just simply note that.
– user3417
Oct 25 '18 at 18:11