Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+y=0, \quad y(0)=0, \quad y(L)=0\)

Short answer

Answer: The boundary value problem has a nontrivial solution if and only if L is a multiple of \(\pi\).

Step-by-step solution

Find the general solution of the differential equation

We have a second-order linear homogeneous differential equation, which can be written as: \(y^{\prime \prime}+y=0\). To find its general solution, we will assume that the solutions are of the form \(y=e^{rx}\) for some constant r. Plugging this into the equation, we get: \((re^{rx})^{\prime}+e^{rx}=0\) So, we have a characteristic equation: \(r^2+1=0\) Solving for r, we get complex roots: \(r=\pm i\) So, our general solution takes the form: \(y(x) = A\cos{x} + B\sin{x}\)

Apply the boundary condition y(0) = 0

Now, we will apply the first boundary condition: \(y(0)=0\). Plugging \(x=0\) into our general solution, we get: \(0 = A\cos{0} + B\sin{0}\) Since \(\cos{0}=1\) and \(\sin{0}=0\), this simplifies to: \(0 = A\) So, we have an updated general solution: \(y(x) = B\sin{x}\)

Apply the boundary condition y(L) = 0

Now, we will apply the second boundary condition: \(y(L)=0\). Plugging \(x=L\) into our updated general solution, we get: \(0 = B\sin{L}\) In order for this equation to be true, either \(B=0\) or \(\sin{L}=0\). If \(B=0\), then our solution is trivial: \(y(x)=0\). However, it's also possible that \(\sin{L}=0\) for some nonzero value of B. In this case, L must be a multiple of \(\pi\): \(L=n\pi\) for some integer n. Therefore, our solution takes the form: \(y(x) = B\sin{\frac{n\pi x}{L}}\)

Conclusion

We have found a nontrivial solution to the given boundary value problem if \(L=n\pi\) for some integer n. Therefore, the problem has a nontrivial solution if and only if the length L is a multiple of \(\pi\). In any other case, the trivial solution \(y(x)=0\) is the only solution.

Key concepts

Second-order Linear Differential Equations

Second-order linear differential equations are fundamental in mathematical modeling because they include many natural phenomena such as mechanical vibrations and electrical circuits. The equation is of the form \( a(x)y'' + b(x)y' + c(x)y = f(x) \), where \( y'' \) is the second derivative of \( y \). In these equations, the highest derivative is squared.
A striking feature of these equations is their linearity; solutions can be added together to form new solutions.
A second-order equation can describe systems with two degrees of freedom, like springs or pendulums.
The exercise above deals with the simpler case of a homogeneous second-order equation, which we'll discuss further.

Homogeneous Differential Equations

A homogeneous differential equation has zero on the right side after simplification, meaning it can be written as \( y'' + p(x)y' + q(x)y = 0 \).
In homogeneous equations, the input is absent, and the general solution is solely determined by the system itself.
Importantly, any constant multiple of a solution is also a solution, and linear combinations of solutions yield other solutions. In the given exercise, the equation \( y'' + y = 0 \) is homogeneous. There isn't an external forcing term, so solutions emerge from the intrinsic properties of the equation itself. This concept is tied closely to the natural responses in physical systems, like free oscillations.

Complex Roots

Complex roots often arise when solving characteristic equations derived from second-order differential equations.
When solving the characteristic equation, you might encounter terms like \( r^2 + 1 = 0 \), leading to roots \( r = \pm i \).
Such complex roots indicate oscillatory solutions, typical in systems exhibiting wave or harmonic behavior. For the boundary value problem discussed, the general solution that results from complex roots is \( y(x) = A\cos{x} + B\sin{x} \). The presence of \( \cos \) and \( \sin \) functions reflects periodic behavior, essential in describing wave patterns or vibration modes.

Trivial and Nontrivial Solutions

In solving differential equations, especially with boundary conditions, you may encounter trivial and nontrivial solutions.
A trivial solution is simple, usually \( y(x) = 0 \), fulfilling all conditions without adding any meaningful insight.
Nontrivial solutions, in contrast, provide meaningful results and describe specific behaviors of a system under given conditions. The exercise illustrates both kinds: when \( B = 0 \), the solution is trivial, resulting in \( y(x) = 0 \). However, if \( \sin{L} = 0 \), and \( L \) is a multiple of \( \pi \), nontrivial solutions like \( y(x) = B\sin{\frac{n\pi x}{L}} \) arise, indicating modes of oscillation. Nontrivial solutions are crucial for understanding patterns or resonances in physical systems.