Final answer:
The one-dimensional wave equation \(\partial_{t}^{2} u = \partial_{x}^{2} u\) with boundary conditions \(u|_{x=0} = 0\) and \(\partial_{x} u |_{x=2} = 0\) is solved by separation of variables into ODEs, applying boundary conditions to determine the spatial solution, solving for the temporal part, and superimposing solutions for each mode.
Step-by-step explanation:
The partial differential equation given is \(\partial_{t}^{2} u = \partial_{x}^{2} u\) with boundary conditions \(u|_{x=0} = 0\) and \(\partial_{x} u |_{x=2} = 0\). This is a one-dimensional wave equation, and it requires a method of separation of variables to solve. By assuming a solution of the form \(u(x, t) = X(x)T(t)\), we can separate the time and spatial parts into two ordinary differential equations (ODEs).
Applying the boundary conditions will help us determine the form of the solution for X(x). The condition \(u|_{x=0} = 0\) implies that X(0) = 0. Additionally, \(\partial_{x} u |_{x=2} = 0\) suggests that the derivative of X with respect to x, \(\partial_{x}X\), is zero at x = 2, indicating a node of the derivative at that point. These conditions lead us to a trigonometric function for X(x), typically a sine function that equals zero at x = 0 and whose derivative is zero at x = 2. Using the ansatz leads to a characteristic equation whose solutions give the allowable wavenumbers k_n.
Once the spatial part is solved, we can proceed to find the temporal solution T(t), which will likely involve harmonic functions, sine, and cosine, since the second derivative concerning time is equal to the second derivative concerning space. Lastly, we superimpose the solutions for each mode n to obtain the final series solution that satisfies the initial conditions given in the problem.