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.

  • $\begingroup$ So where in my steps would I need to normalize the solutions and how would that look like? $\endgroup$
    – hirschme
    Oct 25 '18 at 14:39
  • $\begingroup$ @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. $\endgroup$ 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 $v(t) = t$

  • $\begingroup$ 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? $\endgroup$
    – hirschme
    Oct 25 '18 at 14:38
  • 1
    $\begingroup$ 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. $\endgroup$
    – user3417
    Oct 25 '18 at 17:40
  • $\begingroup$ 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 $y(t)=c_1 e^{rt}+c_2 t e^{rt}$ $\endgroup$
    – hirschme
    Oct 25 '18 at 18:06
  • $\begingroup$ Yes, I'll just simply note that. $\endgroup$
    – user3417
    Oct 25 '18 at 18:11

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