# Find the eigenvalue and eigenfunction successions for S-L BVP

The problem in question is

$$(1+x)y''(x)+\frac12y'(x) + \lambda y(x)=0$$

subject to $$y(0)=y(3)=0$$.

Thanks to a suggested change of variable ($$y(x)=u(\sqrt{1+x})$$), I've managed to find the general solution to the equation: $$y(x)=c_1\cos(2\sqrt{\lambda}\sqrt{1+x})+c_ 2\sin(2\sqrt{\lambda}\sqrt{1+x})$$ for $$c_1, c_2 \in \mathbb{R}$$.

However, the boundary values don't help much to find the problem's solution (is it even possible?), so I don't know how to proceed in finding the eigenvalues and eigenfunctions. Does anyone have any suggestion?

Edit: this sums up to finding $$\lambda$$ such that the system $$Sc = 0$$ below admits solutions other than $$c=0$$:

$$Sc=0\Leftrightarrow\begin{bmatrix}\cos(2\sqrt{\lambda}) & \sin(2\sqrt{\lambda}) \\ \cos(4\sqrt{\lambda}) & \sin(4\sqrt{\lambda})\end{bmatrix}\begin{bmatrix}c_1 \\ c_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$

Taking the determinant of $$S$$, we get $$\cos(2\sqrt{\lambda})\sin(4\sqrt{\lambda}) - \cos(4\sqrt{\lambda})\sin(2\sqrt{\lambda})=\sin(2\sqrt{\lambda})$$. If $$Sc=0$$ for $$c\neq0$$, then $$\det(S)=0$$, so we're pretty much done:

$$\sin(2\sqrt{\lambda})=0\Leftrightarrow\lambda=\frac{k^2\pi^2}4,\, k\in\mathbb{Z}.$$

After this, we can find that $$c_1=0$$ and $$c_2$$ can be anything, so the solution to this problem would be $$y(x) = c_2\sin(k\pi\sqrt{1+x})$$.

• What do you mean? $y(0)=0$ lets us get rid of the $\cos(k\pi\sqrt{1+x})$ term, because then $c_1 = 0$. $y(3)=0$ gives us $0=c_2\sin(2k\pi)$, which means $c_2$ can be anything. What am I not understanding correctly? – AstlyDichrar Jun 8 at 13:16
• You are right, for the eigenvalues my phase difference is a multiple of $\pi$ and thus can be removed from the solution formula. – LutzL Jun 8 at 13:20

You get by introducing a specific phase constant into the solution formula that the arguments of the sine and cosine are zero at $$x=0$$ and thus the solution satisfying the left boundary condition as $$y(x)=c\sin(2\sqrtλ(\sqrt{1+x}-1)).$$ Then for the right boundary condition you need $$0=y(3)=c\sin(2\sqrtλ)\implies 2\sqrtλ=k\pi,~~k\in\Bbb N_{>0}.$$
Why the phase term: The function class $$y(x)=c_1\cos(ϕ(x))+c_2\sin(ϕ(x))$$ is equal by basic trigonometric identities to the class $$y(x)=d_1\cos(ϕ(x)+A)+d_2\sin(ϕ(x)+A)$$. Now set $$A=-ϕ(0)$$ to get that $$y(0)=0$$ implies $$d_1=0$$, so that $$y(x)=d_2\sin(ϕ(x)-ϕ(0))$$.