Question

Find the eigenvalues and eigenfunctions of the given boundary value problem. Assume that all eigenvalues are real. \(y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(0)=0, \quad y^{\prime}(L)=0\)

Short answer

The eigenvalues for the given boundary value problem are λn = (n²π²)/L² for n = 1, 2, 3, ... The corresponding eigenfunctions are yn(x) = cos(nπx/L).

Step-by-step solution

Based on the given differential equation, we can write the characteristic equation as \(r^2+\lambda = 0\). So, we have \(r = \pm \sqrt{-\lambda}\). We will consider three cases: \(\lambda > 0\), \(\lambda < 0\) and \(\lambda = 0\). #Step 2: Solve the differential equation for \(\lambda > 0\)#

From the roots \(r = \pm \sqrt{-\lambda}\), we get \(r = \pm i\sqrt{\lambda}\). The general solution for this case can be written as \(y(x) = C_1\cos(\sqrt{\lambda}x) + C_2\sin(\sqrt{\lambda}x)\). #Step 3: Apply boundary conditions for \(\lambda > 0\)#

Applying the boundary condition \(y^{\prime}(0) = 0\), we have \(0 = -C_1\sqrt{\lambda}\sin(\sqrt{\lambda}0) + C_2\sqrt{\lambda}\cos(\sqrt{\lambda}0) = C_2\sqrt{\lambda}\). Since \(\lambda > 0\), we get \(C_2 = 0\). So, the solution in this case is \(y(x) = C_1\cos(\sqrt{\lambda}x)\). Now, applying the boundary condition \(y^{\prime}(L) = 0\), we get \(0 = -C_1\sqrt{\lambda}\sin(\sqrt{\lambda}L)\). From this, we get \(\sin(\sqrt{\lambda}L) = 0\). So, we have \(\sqrt{\lambda}L = n\pi\), where \(n = 1, 2, 3, \cdots\). Thus, we have \(\lambda_n = \frac{n^2\pi^2}{L^2}\) for \(n = 1, 2, 3, \cdots\) and the corresponding eigenfunctions are \(y_n(x) = \cos(\frac{n\pi x}{L})\). #Step 4: Solve the differential equation for \(\lambda < 0\)#

Let \(\lambda = -\mu^2\), with \(\mu > 0\). The roots for this case are \(r = \pm \sqrt{-(-\mu^2)} = \pm \mu\). The general solution is \(y(x) = C_1e^{\mu x} + C_2e^{-\mu x}\). #Step 5: Apply boundary conditions for \(\lambda < 0\)#

Applying the boundary condition \(y^{\prime}(0) = 0\), we get \(0 = C_1\mu e^{\mu 0} - C_2\mu e^{-\mu 0} = C_1\mu - C_2\mu\). As this does not uniquely determine the constants, there are no eigenvalues and eigenfunctions in this case. #Step 6: Solve the differential equation for \(\lambda = 0\)#

In this case, the roots are \(r = 0\). The general solution is \(y(x) = C_1 + C_2x\). #Step 7: Apply boundary conditions for \(\lambda = 0\)#

Applying the boundary condition \(y^{\prime}(0) = 0\), we get \(0 = C_2\). So, the solution is \(y(x) = C_1\). However, applying the second boundary condition \(y^{\prime}(L) = 0\) does not give any further information or constraints on the constants, so there are no eigenvalues and eigenfunctions in this case. #Summary of the solution#

Thus, for the given boundary value problem, the eigenvalues and eigenfunctions are found only for \(\lambda > 0\). The eigenvalues are \(\lambda_n = \frac{n^2\pi^2}{L^2}\) for \(n = 1, 2, 3, \cdots\) and the corresponding eigenfunctions are \(y_n(x) = \cos(\frac{n\pi x}{L})\).

Key concepts

Boundary Value Problems

Boundary value problems are a type of differential equation problem where you need to find a function that satisfies a differential equation along with certain conditions at the boundaries of the interval. These boundary conditions essentially help us nail down the exact function we are looking for. Typically, you might see problems that involve physical processes like heat conduction or wave propagation. In these problems, knowing how the system behaves at the boundaries, like at the start and end of a physical domain, is crucial. For instance, in this exercise, we are asked to find an appropriate function that complies with the equation \(y^{\prime \prime}+\lambda y=0\) and meets the specific boundary conditions \(y^{\prime}(0)=0\) and \(y^{\prime}(L)=0\). Understanding how boundary conditions work is a fundamental step in determining the correct solution.

Characteristic Equation

The characteristic equation is derived from the differential equation part of a boundary value problem. It is an algebraic equation, which appears from assuming solutions in terms of exponential functions—usually a starting point for finding the general solution. For the differential equation \(y^{\prime \prime}+\lambda y=0\), the characteristic equation is \(r^2 + \lambda = 0\). From this, you solve for \(r\), which gives you the potential forms for the solutions. In this problem, the solution of the characteristic equation involves considering different cases for \(\lambda\):

• \(\lambda > 0\)
• \(\lambda < 0\)
• \(\lambda = 0\)

Each of these cases results in different forms of solutions and necessitates checking how these solutions adhere to the given boundary conditions.

General Solution

Once the characteristic equation is solved, the next step is to formulate the general solution for each case of \(\lambda\). These solutions describe a family of functions that solve the differential equation. Here’s how they materialize:

• For \(\lambda > 0\), the solutions are sinusoidal: \(y(x) = C_1\cos(\sqrt{\lambda}x) + C_2\sin(\sqrt{\lambda}x)\).
• For \(\lambda < 0\), the solutions are exponential: \(y(x) = C_1e^{\mu x} + C_2e^{-\mu x}\), assuming \(\lambda = -\mu^2\).
• For \(\lambda = 0\), the solution is linear: \(y(x) = C_1 + C_2x\).

These general solutions embody combinations of basic functions that solve the differential equations for any values of \(C_1\) and \(C_2\). The actual solutions to the boundary value problems will depend on how these constants play out when boundary conditions are applied.

Boundary Conditions

Boundary conditions are constraints that allow us to find specific solutions from the general family of solutions. They define how the solution behaves at the edges of the domain. In the step-by-step solution of this exercise, they are given as \(y^{\prime}(0) = 0\) and \(y^{\prime}(L) = 0\), meaning the derivative of the solution must vanish at both endpoints. Applying these to the general solution refines it to a specific form, by eliminating constants where needed. For example, in the case of \(\lambda > 0\), applying these conditions steered us to the conclusion that \(y_n(x) = \cos(\frac{n\pi x}{L})\), a much more specific form than the general solution. The solution for other \(\lambda\) cases showed that no valid solutions exist for them under the given boundary conditions. This highlights the critical role boundary conditions play in resolving boundary value problems to unique solutions.