Question

determine the normalized eigenfunctions of the given problem. $$ y^{\prime \prime}-2 y^{\prime}+(1+\lambda) y=0, \quad y(0)=0, \quad y(1)=0 $$

Short answer

Answer: The normalized eigenfunctions for the given differential equation and boundary conditions are: $$ y_1(x) = \frac{1}{\sqrt{\frac{2}{3} e^{1.5} - 2 + \frac{2}{3} e^{0.5}}} \left( e^{(1 + \sqrt{0.25})x} - e^{(1 - \sqrt{0.25})x} \right) $$ $$ y_2(x) = \frac{-1}{\sqrt{\frac{2}{3} e^{1.5} - 2 + \frac{2}{3} e^{0.5}}} \left( e^{(1 + \sqrt{0.25})x} - e^{(1 - \sqrt{0.25})x} \right) $$

Step-by-step solution

Solve the given differential equation

The given differential equation is, $$ y^{\prime \prime}-2 y^{\prime}+(1+\lambda) y=0 $$ To solve this equation, let's assume a solution of the form \(y(x) = e^{rx}\), where r is a constant to be determined. Plugging the solution into the differential equation, we get: $$ (r^2-2r+1+\lambda)e^{rx}=0 $$ Since \(e^{rx}\) is never zero, we can divide both sides by it, which yields: $$ r^2-2r+1+\lambda=0 -\lambda, -(1+\lambda) $$ Now, let's find all the possible eigenfunctions \(y(x) = Ae^{rx}\) using these values of r:

Determine the corresponding eigenfunctions

For the value of \(r = -\lambda\), the eigenfunction is: $$ y(x) = Ae^{-\lambda x} $$ Applying the boundary condition \(y(0) = 0\), we have: $$ Ae^{-\lambda \cdot 0} = A = 0 $$ This results in the trivial solution y(x) = 0, which does not represent a valid eigenfunction. Now, for the value of \(r = -(1+\lambda)\), the eigenfunction is: $$ y(x) = Be^{-(1+\lambda) x} $$ Applying the boundary condition \(y(0) = 0\), we have: $$ Be^{-(1+\lambda) \cdot 0} = B = 0 $$ This also results in the trivial solution y(x) = 0, which does not represent a valid eigenfunction. In either case, we have failed to find any non-trivial eigenfunctions, so now let's consider if we missed something in the analysis.

Reexamine the characteristic equation

Looking back at the characteristic equation, $$ r^2-2r+1+\lambda=0 $$ Notice that we can rewrite the quadratic expression as \((r-1)^2 = \lambda\). Then, the roots are given by \(r_{1,2} = 1 \pm \sqrt{\lambda}\). Therefore, the general solution is given by $$ y(x) = A e^{(1 + \sqrt{\lambda_x})} + B e^{(1 - \sqrt{\lambda_x})} $$ Applying the first boundary condition \(y(0) = 0\), we have: $$ 0 = A e^{(1 + \sqrt{\lambda})\cdot 0} + B e^{(1 - \sqrt{\lambda})\cdot 0} $$ which simplifies to $$ 0 = A + B $$ That means we have \(B = -A\).

Apply the second boundary condition

Now we apply the second boundary condition \(y(1) = 0\), with \(B = -A\): $$ 0 = A e^{(1 + \sqrt{\lambda})} - A e^{(1 - \sqrt{\lambda})} $$ We include a condition that \(A\neq 0\) (otherwise, it leads to a trivial solution). Thus, we can divide both sides by A and obtain $$ e^{\sqrt{\lambda}} = e^{1-\sqrt{\lambda}} $$ Taking the natural logarithm of both sides, $$ \sqrt{\lambda} = 1 - \sqrt{\lambda} $$ Therefore, we have \(\sqrt{\lambda} = 0.5\). So, \(\lambda =0.25\).

Find the normalized eigenfunction

Now, with \(\lambda=0.25\) and \(B=-A\), the eigenfunction is $$ y(x) = A \left( e^{(1 + \sqrt{0.25})x} - e^{(1 - \sqrt{0.25})x} \right) $$ We must normalize the eigenfunction, so we need to find A such that: $$ 1 = \int_{0}^{1} y^2(x) dx $$ Plugging in the eigenfunction and simplifying the expression, we get: $$ 1 = A^2 \int_{0}^{1} \left( e^{1.5x} - 2 + e^{0.5x} \right) dx $$ Solving the integral, we obtain: $$ 1 = A^2 \left(\frac{2}{3} e^{1.5} - 2 + \frac{2}{3} e^{0.5}\right) $$ Now, we can solve for A: $$ A = \pm \frac{1}{\sqrt{\frac{2}{3} e^{1.5} - 2 + \frac{2}{3} e^{0.5}}} $$

Write down the normalized eigenfunctions

Finally, the normalized eigenfunctions are given by the two possible values of A: $$ y_1(x) = \frac{1}{\sqrt{\frac{2}{3} e^{1.5} - 2 + \frac{2}{3} e^{0.5}}} \left( e^{(1 + \sqrt{0.25})x} - e^{(1 - \sqrt{0.25})x} \right) $$ $$ y_2(x) = \frac{-1}{\sqrt{\frac{2}{3} e^{1.5} - 2 + \frac{2}{3} e^{0.5}}} \left( e^{(1 + \sqrt{0.25})x} - e^{(1 - \sqrt{0.25})x} \right) $$

Key concepts

Characteristic Equation

The characteristic equation is a fundamental tool in solving differential equations, especially when dealing with linear homogeneous differential equations with constant coefficients. It is obtained by assuming a solution in the form of an exponential function, typically expressed as erx, where r represents the roots of the characteristic equation. The roots are obtained by substituting the assumed solution into the differential equation and isolating the polynomial that dictates the behavior of the exponent r.

For the exercise in question, the characteristic equation is derived from the differential equation y'' - 2y' + (1 + λ)y = 0. By substituting the assumed solution erx and canceling out common terms, we find the roots through r2 - 2r + 1 + λ = 0. Incorrect roots can lead to trivial or incorrect solutions, as discovered when reexamining the equation, where the correct form turns out to be (r - 1)2 = λ, leading to two distinct eigenfunctions based on the sign of the square root of λ.

Boundary Value Problems

Boundary value problems (BVPs) involve differential equations along with a set of constraints called boundary conditions. These conditions provide the necessary information to determine a unique solution over a defined interval—in this case, the interval is from x = 0 to x = 1.

In the given exercise, the boundary conditions are y(0) = 0 and y(1) = 0, which lead to a system of equations that helps in discovering the coefficients of the eigenfunctions. The boundary conditions are applied to the general solution of the differential equation, placing limits on the possible values of constants A and B. This process furthers the refinement of potential solutions, filtering out those that do not fit the criteria set by the stated boundaries.

Normalization of Eigenfunctions

Normalization is the process of adjusting the magnitude of an eigenfunction so that its total probability (in quantum mechanics) or, more generally in mathematics, its total 'weight' over a defined interval equals one. It ensures that the solution to the differential equation has a meaningful and standard scale for comparison and interpretation.

In practice, to normalize an eigenfunction, we find a constant A such that the integral of the square of the eigenfunction over the interval of interest is equal to one: 1 = t_{0}^{1} y^2(x) dx. The use of normalization in the exercise was a crucial step in deriving the non-trivial solutions, ultimately leading to expressions for y_1(x) and y_2(x) that satisfy both the original differential equation and the boundary conditions.

Second Order Linear Differential Equations

Second order linear differential equations are equations of the form a(x)y'' + b(x)y' + c(x)y = 0, where y'' is the second derivative of y with respect to x. Solving these kinds of equations often involves finding eigenvalues and eigenfunctions, as is the case in the original exercise.

In the context of this problem, y'' - 2y' + (1 + λ)y = 0 is a second-order linear homogeneous differential equation with constant coefficients. The solutions to such equations can be broad, encompassing exponential, sinusoidal, or polynomial functions, depending on the nature of the characteristic equation's roots. The technique and approach for solving these equations depend on the form of the coefficients and the specifics of the boundary or initial conditions. In this example, the solutions required a careful analysis of the characteristic equation, along with precise application of the given boundary conditions, to successfully ascertain meaningful eigenfunctions.