a) Show that the equation and boundary conditions $\frac{d}{dx}(x^2\frac{dy}{dx})+\lambda xy=0, y(1)=0, y'(2)=0$ form a regular Sturm-Liouville system. Show further that the system can be written as a constrained variational problem with functional $S[y]=\int_{1}^{2}x^2y'^2dx, y(1)=0$, and constraint $C[y]=\int_{1}^{2}xy^2dx=1$.
b) Assume that the eigenvalues $\lambda_{k}$ and eigenfunctions $y_{k}, k=1, 2, …,$ exist. Working from equation ($\frac{d}{dx}(x^2\frac{dy}{dx})+\lambda xy=0, y(1)=0, y'(2)=0)$, derive the relationship $\lambda_{k}=\int_{1}^{2}x^2y_{k}'^2dx, k=1, 2, …$.
c) Using the trial function $z=Asin(\pi(x-1)/2)$, show that the smallest eigenvalue, $\lambda_{1}$, satisfies the inequality $\lambda_{1}\leq\frac{(7\pi^2-18)\pi^2}{6(4+3\pi^2)}$. Justify your answer briefly.
Here's my work:
a) Consider the equation and boundary conditions $\frac{d}{dx}(x^2\frac{dy}{dx})+\lambda xy=0, y(1)=0, y'(2)=0$.
By definition, a regular Sturm-Liouville system is defined to be the linear, homogeneous, second-order differential equation $\frac{d}{dx}(p(x)\frac{dy}{dx})+(q(x)+\lambda w(x))y=0$ defined on a finite interval of the real axis $a\leq x\leq b$, together with the homogeneous boundary conditions $A, y(a)+A_{2}y'(a)=0$ and $B, y(b)+B_{2}y'(b)=0$, with $A_{1}, A_{2}, B_{1}$ and $B_2$ real constants, and the two cases $A_{1}=A_{2}=0$ and $B_{1}=B_{2}=0$ are excluded. Let $p(x)=x^2, q(x)=0$ and $w(x)=x$. Note that the functions $p(x)=x^2, q(x)=0$ and $w(x)=x$ are real and continuous for $a\leq x\leq b$ where $a=1$ and $b=2$.
Also, $p(x)=x^2$ and $w(x)=x$ are strictly positive for $a\leq x\leq b, p'(x)=2x$ exists and is continuous for $a\leq x\leq b$.
Thus, the equation and boundary conditions $\frac{d}{dx}(x^2\frac{dy}{dx})+\lambda xy=0, y(1)=0, y'(2)=0$ form a regular Sturm-Liouville system.
Now we will consider the functional $S[y]=\int_{1}^{2}x^2y'^2dx, y(1)=0$, and constraint $C[y]=\int_{1}^{2}xy^2dx=1$, where $\lambda$ is the Lagrange multiplier.
Then the auxiliary functional is $S[y]-\lambda C[y]=\int_{1}^{2}(x^2y'^2-\lambda xy^2)dx, y(1)=0, y'(2)=0$.
Note that the Euler-Lagrange equation for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$ is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$.
Observe that $\frac{\partial F}{\partial y'}=2x^2y', \frac{\partial F}{\partial y}=-2\lambda xy$, and $\frac{d}{dx}(\frac{\partial F}{\partial y'})=2x^2y''+4xy'$.
Hence, the Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 2x^2y''+4xy'+2\lambda xy=0\implies x^2y''+2xy'+\lambda xy=0$.
Therefore, the equation and boundary conditions $\frac{d}{dx}(x^2\frac{dy}{dx})+\lambda xy=0, y(1)=0, y'(2)=0$, form a regular Sturm-Liouville system and the system can be written as a constrained variational problem with functional $S[y]=\int_{1}^{2}x^2y'^2dx, y(1)=0$, and constraint $C[y]=\int_{1}^{2}xy^2dx=1$.
c) Consider the trial function $z=Asin(\pi(x-1)/2)$, where $A$ is a parameter.
By using the Rayleigh-Ritz method in the theory of Sturm-Liouville systems, note that the smallest eigenvalue $\lambda_{1}$ is given by $\lambda_{1}=inf S[y]$, where the infimum is taken over all admissible functions $y$ with $C[y]=1$.
Observe that $1=\int_{1}^{2}xz^2dx=\int_{1}^{2}x[Asin(\pi(x-1)/2)]^2$
From here I'm stuck. How should I proceed and find the integral above? Also, how should I do part b) of this problem? Is part a) correct?
Best Answer
For part (b), multiplying $(x^2y')' + \lambda xy = 0$ by $y$ and integrating by parts, we find \begin{align*} \lambda\int_1^2 xy^2\, dx & = -\int_1^2 (x^2y')'y\, dx = -x^2y'y\bigg|_1^2 + \int_1^2 x^2(y')^2\, dx = \int_1^2 x^2(y')^2\, dx, \end{align*} where the boundary term vanishes due to the boundary conditions. For part (c), note that the trigonometric identity for $\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$ permits us to rewrite $z = -A\cos(\pi x/2)$. You can compute the integral $\int_1^2 x\cos^2(\pi x/2)\, dx$ using the identity $\cos^2\theta = \dfrac{1 + \cos(2\theta)}{2}$ and integration by parts.
Added: The integral becomes $\int_1^2 \frac{x}{2}\left(1 + \cos(\pi x)\right) dx = \int_1^2 \frac{x}{2}\, dx + \int_1^2 \frac{x}{2}\cos(\pi x)\, dx$. The first integral is now straightforward. For the second integral, integrating by parts with $u = x/2$ and $dv = \cos(\pi x)\, dx$, we find \begin{align*} \int_1^2 \frac{x}{2}\cos(\pi x)\, dx = \frac{x}{2}\cdot \frac{\sin(\pi x)}{\pi}\bigg|_1^2 - \int_1^2 \frac{\sin(\pi x)}{2\pi}\, dx. \end{align*}