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The following problem has brought up some misunderstandings for me -

Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:

$$ {u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0 $$

Solution:

Try $ u\left ( x \right )=Ae^{rx} $, which gives roots $ r=-1\pm \sqrt{1-\lambda } $

Solution is altered with $$ \lambda <1\; ,\; \; \lambda =1\; ,\; \; \lambda >1 $$

For the first case $ \lambda <1 $ the general solution is $$ u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x} $$ $$ u\left ( x \right )=C\cosh \left ( -1+\sqrt{1-\lambda } \right )x+D\sinh \left ( -1-\sqrt{1-\lambda } \right )x $$

Applying boundaries: (this is where my question lies - how to correctly apply BCs)

$$ u\left ( 0 \right )=0 \; \; \Rightarrow \; \; C+D=0 $$ (some cases i've seen the conclusion that only $ C=0 $).

Do i assume that as $ \cosh $ is never zero that $ C=0 $ and therefore it must be that $ D=0 $. Or do i only take the result $ C=0 $ from the first BC and then apply the second BC to see what happens to $ D $? The latter (assuming $ C=0 $) gives $$ D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0 $$

So either $ D=0 $ or $ \sinh \left ( -1-\sqrt{1-\lambda } \right )=i\pi n $

I'm confused by the what the rules are for BCs. Can anyone point out how to proceed? Thanks

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Since $\sinh(0)=0$, the first condition is obviously $0=u(0)=C\cosh(0)=C$. The sum applies to the exponential representation, i.e., $0=u(0)=Ae^0+Be^0=A+B$.

At the point $1$ you get to

$$0=u(1)=C\cosh(−1−\sqrt{1−λ})+D\sinh(−1−\sqrt{1−λ})=D\sinh(−1−\sqrt{1−λ})$$

where neither factor reduces to zero, but $C=0$ can be inserted. So one concludes $D=0$.

Which is also what one would expect, that eigenfunctions will be in terms of the trigonometric sin and cos and not the hyperbolic functions.


Using $v(x)=e^{x}u(x)$ one gets $$v'(x)=e^x(u'(x)+u(x))$$ and $$v''(x)=e^x(u''(x)+2u'(x)+u(x))=(1-λ)v(x)$$ with the boundary conditions $v(0)=0=v(1)$ carrying over. This now is the classical problem of the vibrating string with the known eigenfunctions $v_n(x)=\sin(\pi nx)$ where $\pi^2n^2=λ-1$.

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  • $\begingroup$ I don't see how for $u\left ( 0 \right )=0$ that $C\cosh(0)=C$. How does one interpret that as $\cosh $ doesn't exist at $0$. Surely, doesn't this give the result that $C=D=0$? $\endgroup$ – AntonySC Feb 19 '14 at 17:02
  • $\begingroup$ Yes, that is exactly the result. $\sinh(0)=0$ and $\cosh(0)=1$ and $\sinh(x)>0$ for $x>0$ leaves no other option. $\endgroup$ – Dr. Lutz Lehmann Feb 19 '14 at 17:04
  • $\begingroup$ Ok. So back to the correct use of the boundary conditions... Having applied the first BC of $u(0)=0$ and found that $C=D=0$ does that now make the application of the second BC $u(1)=0$ redundant? Or do i go on to get a result from this? $\endgroup$ – AntonySC Feb 19 '14 at 17:54
  • $\begingroup$ The first boundary condition only fixes one constant, $C=0$. The second is needed for $D=0$. And of course this all is only valid for the requested case $λ<1$. $\endgroup$ – Dr. Lutz Lehmann Feb 19 '14 at 17:58
  • $\begingroup$ Ok, great. Starting to get to the bottom of this. Why does the first boundary condition only fix one constant and not both? $\endgroup$ – AntonySC Feb 19 '14 at 18:07

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