Question

(a) Write the homogeneous Sturm-Liouville eigenvalue equation for which \(y(a)=y(b)=0\) as $$ \mathscr{L}(y ; \lambda) \equiv\left(p y^{\prime}\right)^{\prime}+q y+\lambda \rho y=0 $$ where \(p(x), q(x)\) and \(\rho(x)\) are continuously differentiable functions. Show that if \(z(x)\) and \(F(x)\) satisfy \(\mathcal{L}(z ; \lambda)=F(x)\) with \(z(a)=z(b)=0\) then $$ \int_{a}^{b} y(x) F(x) d x=0 $$ (b) Demonstrate the validity of result (a) by direct calculation for the case in which \(p(x)=\rho(x)=1, q(x)=0, a=-1, b=1\) and \(z(x)=1-x^{2}\).

Short answer

If both \(y(x)\) and \(z(x)\) satisfy the given Sturm-Liouville problem, their product integrations with \(F(x)\) result to zero because boundary conditions and symmetry eliminate contributions.

Step-by-step solution

Write the homogeneous Sturm-Liouville equation

The homogeneous Sturm-Liouville equation is given by oindent\[ \mathscr{L}(y ; \lambda) \equiv(p y^{\prime})^{\prime}+q y+\lambda \rho y=0 \]where \(p(x)\), \(q(x)\), and \(\rho(x)\) are continuously differentiable functions, and \(y(a) = y(b) = 0\).

Write the inhomogeneous equation

Given \(z(x)\) and \(F(x)\), the inhomogeneous Sturm-Liouville equation is oindent\[ \mathcal{L}(z ; \lambda) = F(x) \]with the boundary conditions \( z(a) = z(b) = 0 \).

Multiply the homogeneous equation by \(z(x)\)

Multiply the homogeneous equation \( \mathcal{L}(y ; \lambda) = 0 \) by \(z(x)\): oindent\[ z(x) (p y^{\prime})^{\prime} + z(x) q y + z(x) \lambda \rho y = 0. \]

Multiply the inhomogeneous equation by \(y(x)\)

Multiply the inhomogeneous equation \( \mathcal{L}(z ; \lambda) = F(x) \) by \(y(x)\): oindent\[ y(x) (p z^{\prime})^{\prime} + y(x) q z + y(x) \lambda \rho z = y(x) F(x). \]

Integrate both equations over \([a,b]\)

Integrate both equations over \([a,b]\): oindent\[ \int_{a}^{b} y(x) (p z^{\prime})^{\prime} + y(x) q z + y(x) \lambda \rho z \, dx = \int_{a}^{b} y(x) F(x) \, dx. \]oindent\[ \int_{a}^{b} z(x) (p y^{\prime})^{\prime} + z(x) q y + z(x) \lambda \rho y \, dx = 0. \]

Use boundary conditions

Since \( y(a) = y(b) = 0 \) and \( z(a) = z(b) = 0 \), all terms involving boundary contributions (from integration by parts) will vanish. Thus: oindent\[ \int_{a}^{b} y(x) (p z^{\prime})^{\prime} dx = \int_{a}^{b} z(x) (p y^{\prime})^{\prime} dx, \] which after canceling and rearranging leads to oindent\[ \int_{a}^{b} y(x) F(x) \, dx = 0. \]

Verify for specific functions

Consider \( p(x) = \rho(x) = 1, \) \( q(x) = 0 \), \( a = -1 \) and \( b = 1 \), and \( z(x) = 1 - x^2 \). The example works as follows: \( (py')' = y''\) and applying it to inhomogeneous form: oindent\[ y'' \cdot (1 - x^2) = F(x). \] Now integrate considering \( F(x) = -2, \) yielding: oindent\[ \int_{-1}^{1} y(x) F(x) dx= \int_{-1}^{1} y(x) (-2) dx = 0, \] due to symmetry and zero boundary conditions for continuous and differentiable \(y\).

Key concepts

Eigenvalue Problems

Eigenvalue problems are fundamental in many areas of mathematics and physics because they involve finding characteristic values (eigenvalues) that simplify complex systems. In the context of Sturm-Liouville theory, we solve the equation: \[ \mathscr{L}(y ; \lambda) \equiv(p y^{\t////prime})^{\t////prime} + q y + \lambda \rho y = 0 \] for certain functions and boundary conditions. Here, \( y \) is called the eigenfunction corresponding to the eigenvalue \( \lambda \. \) \In simpler terms, an eigenvalue problem helps us find special values (eigenvalues) that permit simple scaling of the function (eigenfunction) rather than altering its direction.\