Ordinary Differential Equation - Boundary Conditions Question

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

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 $\pi^2n^2=λ-1$.

• 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 $C=D=0$? – AntonySC Feb 19 '14 at 17:02
• Yes, that is exactly the result. $\sinh(0)=0$ and $\cosh(0)=1$ and $\sinh(x)>0$ for $x>0$ leaves no other option. – Dr. Lutz Lehmann Feb 19 '14 at 17:04
• 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? – AntonySC Feb 19 '14 at 17:54
• 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 $λ<1$. – Dr. Lutz Lehmann Feb 19 '14 at 17:58
• Ok, great. Starting to get to the bottom of this. Why does the first boundary condition only fix one constant and not both? – AntonySC Feb 19 '14 at 18:07