# 1D Wave equation with free end: limitation of Neumann's boundary condition?

Consider a slinky whose one end is fixed and the other is free. The textbook approach for analyzing its longitudinal vibrations is to use the wave equation along with Neumann's boundary condition, i.e. assume that at the free end the slinky is neither stretched nor compressed. However, this is easily violated by initially stretching the whole slinky and then letting it vibrate freely. How do we analyze the motion in this case?

There is nothing special about the slinky. The same issue arises with a vibrating string where one end is fixed and the other is free to slide without friction on a vertical pole. If we initially stretch the string to form a right triangle with the horizontal and the pole and then let it go, the same issue arises.

Note that the standard justification for Neumann's boundary condition for the 1D wave equation with free end is that without this condition, the free end will have infinite acceleration, which is claimed to be "unphysical" and thus the condition is implied. However, the previous example is completely physical and also violates the condition.

This issue was presented in a comment here but there was no answer.

• The “free-end” description, is an idealised mathematical construct (since the ideal string does not exist anyway). The argument is that the actual end of the string, has zero mass. In the “calculus sense”, you can go arbitrarily close to the end of the string and only when you reach the end of the string the Neumann condition must hold. Of course, you understand that even the end of the string is a mathematical construct since you’ll never be able to reach it in the strict (mathematical) sense (calculus allows its use through its tools). At this end point the Neumann condition holds. Jan 15 at 23:28

For the comment and the triangle initial condition, this is not inconsistent with Neumann boundary conditions (NBC’s). The BC’s only are applied at later times and not the initial condition. Mathematically, say you model the field by the 1D wave equation: $$\partial_t^2u=\partial_x^2u\\ u(x=0)=\partial_x u(x=L)=0$$ Given “any” initial condition: $$u(t=0)=f(x)\\ \partial_t u(t=0)=g(x)$$ then there is a unique solution to the Cauchy problem. To get the formula, one trick is to use the method of images by extending $$f,g$$ to the entire line. This is done by imposing antisymmetry at $$x=0$$ and symmetry at $$x=L$$ and using the usual propagation on the entire line.

For your example, the initial condition would be a triangular wave whose period is $$4L$$. Assuming $$\partial_tu(t=0)$$, at later times you have two half copies of the wave propagating at equal and opposite velocity. By superposition, they sum up to a solution satisfying NBC’s. You basically have a piecewise affine function with a single kink propagating at celerity $$c$$ bouncing back and forth.

Btw, the same objection could apply for Dirichlet boundary conditions (DBC’s) as well. You could very well start with the initial condition $$u(t=0)=1$$ and $$\partial_tu(t=0)=0$$ which has a perfectly defined evolution.

In general, the justification of the NBC’s can be argued by taking the contiuum limit of a discrete model. In the discrete case, you rather have: $$\ddot u_n=\frac{u_{n+1}+u_{n-1}-2u_n}{a^2}$$ by imagining many springs in series and $$a$$ the lattice spacing. The coninuum is obtained by sending $$a\to0$$ and $$u_n=u(x=na)$$. However, at the free end, you only have one spring: $$\ddot u_N=\frac{u_{N-1}-u_N}{a^2}$$ Since: $$\frac{u_{N-1}-u_N}{a}\to-\partial_xu(x=L)$$ having $$\partial_xu\neq0$$ gives a singular acceleration, i.e. it’s a Dirac delta. Physically, this just means that $$\partial_tu$$ is discontinuous and the jump is given by $$-\partial_xu$$.

You can check that this is the case for your initial condition. Even if you can start it with $$\partial_u(t=0)=0$$, it will develop a finite value at arbitrarily short time $$\partial_tu(0^+)\neq0$$ (you can calculate this value pretty easily).

In summary, the NBC is not always satisfied in the naive way. When it is not satisfied, this translated into a jump in time derivative.

Hope this helps.