Let's say that we have some IVP for the wave equation in $1$ dimension. I.e that $u_{tt} - u_{xx} = 0$ on the entire real line, at time $t>0$ and $u(x,0) = f(x)$, and $u_t(x,0) = g(x)$. Let us consider another solution $v$ which satisfies the same initial conditions. Does D'Alembert's formula: $$ u(x,t) = f(x+t) - f(x-t) + \int_{x- t}^{x+t} g(y) \mathrm{d}y $$ not imply that $v = u$ everywhere? There are some proofs of uniqueness that use energy methods, but why are these necessary? Also, why can this not be generalized to the spherical means solution? What am I missing?

  • $\begingroup$ @LeeDavidChungLin yes, sorry that should be a $g$ in the integrand. $\endgroup$ May 7, 2018 at 4:53
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    $\begingroup$ Energy methods give you access to better tools. Just because you have A solution, it does not mean you have THE solution. You could have solution sets that are piecewise based on some parameter, for example. $\endgroup$ May 7, 2018 at 5:54

1 Answer 1


D'Alembert's solution is entirely determined by the initial values and the solution is symmetric about the starting point.

Taking $c = 1$;

$$u(x,t) = \frac{1}{2}(f(x+t)+f(x-t))+\frac{1}{2}\int_{x-t}^{x+t}g(\lambda)d\lambda$$

If there is another solution $v(x,t)$ that satisfies the wave equation and the conditions, then $u(x,t)=v(x,t)$ $\forall$ $(x,t)$ as D'Alembert's solution is the general solution depending on $f(x)$ and $g(x)$.

Proving uniqueness removes any thought that there are other solutions remaining. If it's unique and we have a solution, then that solution is the only solution.

In the case of a generalized spherical D'Alembert's solution, the domain changes.

Instead of $-\infty < x < \infty$, it becomes $0 < r < \infty$

  • $\begingroup$ This makes sense, thank you. What about the generalizations to $R^n$?Those solutions (the ones with spherical means) seem to be uniquely determined by initial conditions as well. $\endgroup$ May 7, 2018 at 5:50
  • $\begingroup$ That's a good question! I'm not too certain how waves behave in $\mathbb R^6$ for example, but the more dimensions we introduce, the more complicated the Laplacian becomes. The solution will change, but will still depend on the initial conditions. $\endgroup$
    – mallan
    May 7, 2018 at 6:05
  • $\begingroup$ Turns out that Evans has an energy methods proof for the general case. Thanks for talking this out with me. $\endgroup$ May 7, 2018 at 6:07
  • $\begingroup$ Of course! It was fun, I got to dig out my old books $\endgroup$
    – mallan
    May 7, 2018 at 6:08

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