Consider the eigenvalue problem $$((1+x^4)y')'+\lambda y=0 \ , \ x\in(0,1) \\ y(0)=0 \ , \ y(1)+2y'(1)=0$$ Then which of the following
are true?$(a)$ All the eigenvalues are negative.
$(b)$ All the eigenvalues are positive.
$(c)$ There exist some positive eigenvalues and some negative
eigenvalues.$(d)$ There are no eigenvalues.
All Sturm-Liouville problems that I have seen till now, requires to put $\lambda=-k^2,0,k^2$ where $k\neq 0$ and then solve the equation and using the boundary conditions for a non-trivial solution to find the eigenvalues of the problem. But in this case I am unable to solve the equation at hand for the cases $\lambda=-k^2$ and $k^2$. I want to know if there is a theorem in Sturm-Liouville theory by using which I can calculate that how many positive or negative eigenvalues do a SL-BVP possesses without explicity finding the solution, as there may be many typical equations at hand which are not possible to solve by hand, like this one. Any help is appreciated.
Best Answer
Multiplying both sides by $y$ and then integrating from 0 to 1, one has $$ \int_0^1((1+x^4)y')'ydx=-\lambda \int_0^1y^2dx. \tag{1}$$ Since \begin{eqnarray} \int_0^1((1+x^4)y')'ydx&=&\int_0^1yd((1+x^4)y')\\ &=&y(1+x^4)y'\bigg|_0^1-\int_0^1(1+x^4)(y')^2dx\\ &=&2y(1)y'(1)-\int_0^1(1+x^4)(y')^2dx\\ &=&-y^2(1)-\int_0^1(1+x^4)(y')^2dx \end{eqnarray} then (1) becomes $$ y^2(1)+\int_0^1(1+x^4)(y')^2dx=\lambda \int_0^1y^2dx. \tag{2}$$ Since the LHS is non-negative, so one has $\lambda\ge0$. If $\lambda=0$, then (2) becomes $$ y^2(1)+\int_0^1(1+x^4)(y')^2dx=0 $$ which implies $y(1)=0$ and $$ \int_0^1(1+x^4)(y')^2dx=0. $$ This gives $y'=0$ and $y'(1)=0$. So $y\equiv0$ in $[0,1]$ and one has trivial solution, i.e., $\lambda=0$ is not an eigenvalue. Hence $\lambda>0$ and you have choose (b).