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.