prove $\left\{\begin{array}{l}X^{\prime \prime}+\lambda X=0,00$ without solving

Question

Can you help me some idea to prove it or some similar question about it. I know prove $\lambda>0$ is easy when solving it, it's just like normal differential equation.

Answer 1

The idea is that the differential equation either has sines and cosines for solutions (if $\lambda > 0$) or exponentials (if $\lambda < 0$) or linear ($\lambda = 0$).

The second boundary condition looks a bit strange. If it had been $X(1) = 0$, that would have been enough to conclude it must be trig functions right there.

The $\lambda = 0$ case can be ruled out because $X(1) + X'(1) = 0$ means that there must be a positive slope to a negative value, or vice versa, which doesn't work starting from the origin except with a zero solution.

Answer 2

We compute the Rayleigh quotient. Multiply your ODE by $X$ (assuming $X \ne 0$ is non-trivial) and integrate over the domain $x \in [0, 1]$

\begin{align} - \lambda X^{2} &= X X'' \\ \implies - \lambda \int_{0}^{1} X^{2} dx &= \int_{0}^{1} X X'' dx \\ &= XX' \bigg \lvert_{0}^{1} - \int_{0}^{1} X'^{2} dx \\ &= - X(1)^{2} - \int_{0}^{1} X'^{2} dx \end{align}

where we have used integration by parts and the boundary conditions to get the last line. Hence

$$\lambda = \frac{X(1)^{2} + \int_{0}^{1} X'^{2} dx}{\int_{0}^{1} X^{2} dx} > 0$$

as the integral of a non-negative function is non-negative and the numerator is always positive.

Answer 3

This is a problem of eigenvalues. We attach the characteristic polynomial: $r^2+\lambda=0\Rightarrow r^2=-\lambda.$ We talk about the parameter $\lambda.$

1. If $\lambda>0\Rightarrow r_{1}=i\sqrt{\lambda}, r_{2}=-i\sqrt{\lambda}\in \mathbb{C}.$ In this case the solution will be: $X(x)=c_1cos(\sqrt{\lambda} x)+c_2sin(\sqrt{\lambda} x)$. We set the conditions to the limit and we have: $X(0)=c_1=0$ and $X(1)=c_1cos(\sqrt{\lambda})+c_2sin(\sqrt{\lambda})$ and $X'(x)=-c_1sin(\sqrt{\lambda}x)\sqrt{\lambda}+c_2cos(\sqrt{\lambda}x)\sqrt{\lambda}\Rightarrow X'(1)=-c_1sin(\sqrt{\lambda})\sqrt{\lambda}+c_2cos(\sqrt{\lambda})\sqrt{\lambda}.$

$X(1)+X'(1)=c_1cos(\sqrt{\lambda})+c_2sin(\sqrt{\lambda})+c_1sin(\sqrt{\lambda})\sqrt{\lambda}+c_2cos(\sqrt{\lambda})\sqrt{\lambda}=0$, $c_1=0 $ so $X(1)+X'(1)=c_2sin(\sqrt{\lambda}) +c_2cos(\sqrt{\lambda})\sqrt{\lambda}=0$ and solve this equation and find the eigenvalues.

2. If $\lambda=0\Rightarrow r_1=r_2=0$. In this case the solution will be:$X(x)=c_1+c_2x.$

$X(0)=c_1=0$ and $X(1)=c_2$, $X'(x)=c_2\Rightarrow X'(1)=c_2.\Rightarrow X(1)+X'(1)=c_2+c_2=0\Rightarrow c_1=c_2=0$$\Rightarrow X(x)=0$ so $\lambda=0$ it's not a eigenvalue.

3. If $\lambda<0\Rightarrow r_1=\sqrt{-\lambda}, r_2=-\sqrt{-\lambda}.$ In this case the solution will be: $X(x)=c_1e^{r_1}+c_2e^{r_2}.$ And we put the initial conditions in this case as well and we see if $\lambda<0$ is a eigenvalue.