I am a second-year undergraduate taking an introductory course on partial differential equations. I am trying to apply the method of separation of variables to find a solution to the wave equation
$$
u_{tt} – c^2 u_{xx} = 0, \qquad 0 < x < l, \quad t > 0,
$$
with initial conditions $u(x, 0) = 0$, $u_t(x, 0) = 1$, and Neumann boundary conditions $u_x(0, t) = 0 = u_x(l , t)$.
I start by assuming the solution is of the form $u(x, t) = X(x) T(t)$, and substitute it into the wave equation, and separate the variables to get two eigenvalue problems:
$$
X''(x) + \lambda X(x) = 0, \qquad T''(t) + c^2 \lambda T(t) = 0.
$$
The given boundary conditions imply that $X'(0) = 0 = X'(l)$, so when $\lambda < 0$ there is no non-trivial solution, so I assume $\lambda \geq 0$.
When $\lambda > 0$, we have $X(x) = A \cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$, and the boundary conditions imply $B = 0$ and $\lambda = n^2\pi^2 / l$. So we get the eigenfunction $X_n(x) = A_n \cos(n \pi x / l)$ for the eigenvalue $\lambda_n = n^2 \pi^2 / l$. For this eigenvalue, we solve the other EVP to get $T_n(t) = A_n' \cos(n \pi c t / l) + B_n'(n \pi c t / l)$.
When $\lambda = 0$, we have $X(x) = A + Bx$ and the boundary conditions imply $B = 0$, so we get the eigenfunction $X_0(x) = A_0$ for the eigenvalue $\lambda_0 = 0$. For this eigenvalue, we solve the other EVP to get $T_0(t) = A_0' + B_0't$.
Hence, $u_n(x, t) = X_n(x) T_n(t)$ is a solution to the wave equation with the given boundary conditions for every $n \geq 0$, and so is $u(x, t) = \sum_{n = 0}^\infty u_n(x, t)$ (provided the initial data is "nice" enough, though I'm not fully sure what those conditions are).
Combining the constants together, we get
$$
u(x, t) = a_0 + b_0 t + \sum_{n = 1}^\infty \cos(n \pi x / l) \bigl( a_n \cos(n \pi c t / l) + b_n \sin(n \pi c t / l) \bigr).
$$
Now, applying the initial condition $u(x, 0) = 0$, we get $0 = a_0 + \sum_{n = 1}^\infty a_n \cos(n \pi x / l)$, which imply $a_n = 0$ for all $n \geq 0$. Next, applying the initial condition $u_t(x, 0) = 1$, we get $1 = b_0 + \sum_{n = 1}^\infty (b_n n \pi c / l) \cos(n \pi x / l)$. This is a Fourier cosine series, so the formulae for the coefficients are:
$$
b_0 = \frac{1}{l} \int_0^l 1\, dx = 1, \qquad \frac{b_n n \pi c}{l} = \frac{2}{l} \int_0^l 1 \cdot \cos(n \pi x / l)\, dx = 0 \implies b_n = 0\ \ \forall\ n \geq 1.
$$
Thus, we get the solution $u(x, t) = t$.
Can someone please verify for me whether I have solved this problem correctly? I have only followed the outline given in my lecture notes for using the method of separation of variables, but the final answer puzzles me.
Best Answer
You applied the method correctly, and it produced a solution to the problem, as you noticed by just substituting $u=t$ and seeing that it obviously satisfies all the requirements. (a) Why your TA thought it might be complicated--I think you would have to ask him or her--possibly your TA made a mistake about that; we all do. (b) How to understand the solution--there are at least two ways. One way is to think of some application of the problem and see whether the solution fits your expectation of the physics. For example, the pde can describe vibration of a string but the boundary conditions say the the ends of the string at $x=0,\ell$ are not pinned down, This is a little bit nonphysical but since the string starts with uniform velocity $u_t(x,0) = 1$ upward, and no curvature since $u(x,0) = 0$ implies that $u_{xx}(x,0) = 0$. So without curvature there is no restoring force initially. Maybe that makes it plausible that there is nothing to restrain the string and so it keeps moving upward at speed 1. That argument is ok, but not great in my view.
Another way to understand the solution, more mathematically, is to ask whether there could be any other solution besides the one you found. This is asking whether the problem has a unique solution. It does, but this is not usually addressed in an introductory course. A proof of uniqueness can be based on conservation of energy: let $$E(t) = \int_0^\ell (\tfrac{1}{2}u_t^2+\tfrac{1}{2}c^2u_x^2)dx.$$ Here is a way-too-short outline of the argument: It is not too hard to prove that $E'(t) = 0$. Taking $u$ to represent the difference of two suposed solutions, one finds that the energy of $u$ is initially and therefore always $0$, and deduce that $u$ is identically zero, thus the two solutions are the same. I.E. the solution you found is the only one there is.
A last comment is that once you know this, you can sometimes avoid a long computation. With hindsight you might think that $u_t(x,0) = 1$ suggests $u=t$ at least initially, and then realize by subtituting that you have already found the answer! Of course you don't think this way when you are first learning the separation method.