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Consider the differential equation:

$$u''(x) = \lambda \ u(x), \ \ \ 0 <x < \pi$$

I want to find solutions $u_1, u_2$ such that the following data are satisfied:

$$u_1(0) = 0, \ \ u_1'(0) = 1; \ \ \ u_2(\pi) = 0, \ u_2'(\pi) = 1 $$

We may allow $u_1, u_2, \lambda$ to be complex valued. This seems to me like a 2nd order ODE with constant coefficients with general solution:

$$y = c\exp(x\sqrt{\lambda} ) + d\exp(-x\sqrt{\lambda} )$$

This cannot be the case otherwise the solution would only be the trivial one. Any suggestions, solutions, hints will be deeply appreciated.

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That's an awful lot of "initial" conditions for a second order equation! – André Nicolas Mar 7 '13 at 6:25
It's under the chapter of boundary-value problems, if this helps. Any suggestions at all? – user44069 Mar 7 '13 at 6:27
@AndréNicolas I think he's looking for two solutions, one with initial conditions for $u_1$ and another for initial conditions specified as $u_2$. That said, the initial conditions $f(0) = c_1$, $f'(0) = c_2$ and $f''(x) = \lambda f(x)$ for all $x$ has only one solution by a general existence-uniqueness theorem. – Andrew Salmon Mar 7 '13 at 6:30
Still though, there are way to many conditions. – user44069 Mar 7 '13 at 6:33
Not really. The existence-uniqueness theorem guarantees the existence of at least one solution satisfying the differential equation and two initial conditions and the fact that that solution is unique. – Andrew Salmon Mar 7 '13 at 6:35

If $\lambda<0$ then we denote $-\lambda=w^2$ then the general solution is $$y=c\cos(wx)+d\sin(wx).$$ With $u_1(0)=0$ we have $c=0$ and $u'_1(0)=1$ gives $d=\frac{1}{w}$.

For $u_2$ we must solve the system: $$\left\{\begin{array}{ll} c\cos(w\pi)+d\sin(w\pi)&=0\\ -cw\sin(w\pi)+dw\cos(w\pi)&=1\end{array}\right.$$

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That's the way to do it. – Kaster Mar 7 '13 at 6:41
What happens though when $\lambda > 0$? – user44069 Mar 7 '13 at 6:48
If $\lambda>0$ then the solution is expressed with exponantial function so if it vanishes at a point it vanishes everywhere. – user63181 Mar 7 '13 at 6:54
Nice! That was my thought too! Thanks for all the help! All the best! – user44069 Mar 7 '13 at 6:57
You're welcome. – user63181 Mar 7 '13 at 6:58

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