I am trying to derive the 2d wave equation on a membrane. Instead of considering a small $\Delta x\times \Delta y$ rectangle around $x_0$, I consider a small circle with radius $r$ around $x_0$. But I don't know why the total force obtained is only half. What is wrong with my derivation?

The derivation is as follows,

  1. Denote the parameter of the circle as $\omega$. Then on a small segment $d\omega$, the force on it will be $Sd\omega$, where $S$ is the surface tension.
  2. Denoting the wave as $u$, the vertical component of the force $Sd\omega$ is approximately $\frac{u(x)-u(x_0)}{r}Sd\omega$ where $x$ is the location of the segment $d\omega$. (using $\sin\theta\approx\theta$).
  3. Then the total force is, $$ \oint \frac{u(x)-u(x_0)}{r}Sd\omega $$
  4. Using the conclusion here (based on Taylor's expansion of $u$), the total force is approximately, $$ \frac{r}{4}\omega S\nabla^2u $$
  5. Considering the mass of the circle is $\frac{1}{2}\omega r\rho$, by Newton's second law, we have, $$ \frac{r}{4}\omega S\nabla^2u = \frac{1}{2}\omega r\rho \ddot u $$ So the result is, $$ \color{red}{\frac{1}{2}}\nabla^2u = \frac{\rho}{S}\ddot u $$

My question is: why is there a $\frac{1}{2}$ in the equation? What's is wrong with my thoughts?



1 Answer 1


The problem is the approximation that the vertical force on segment $d\omega$ is $\frac{u(x)-u(x_0)}{r}Sd\omega$. This is too crude an approximation, based on a Taylor expansion keeping only the linear terms, and the problem goes away once we include higher order terms.

The exact expression for the vertical force on $d\omega$ is $$dF = Sd\omega\left[\vec\nabla u(x)\cdot \hat r\right]=\frac{Sd\omega}{r}\left[\vec\nabla u(x)\cdot (\vec x - \vec x_0)\right].$$ Hopefully this makes sense, but let me know if it doesn't. I'll let $\vec y=\vec x-\vec x_0$ and use Einstein's index notation (repeated indices are summed over) to make things a bit less cumbersome. In this notation,

$$dF = \frac{Sd\omega}{r}\partial_i u(y)y^i.$$

Now expanding $u(y)$ about $y=0$ to second order, $$u(y) = u(0)+\partial_iu(0)y^i+\frac{1}{2}\partial_i\partial_ju(0)y^iy^j \tag1\label1$$ $$\partial_ku(y)=\partial_ku(0)+\partial_k\partial_iu(0)y^i \tag2\label2$$ where in $\eqref2$ we used the equality of mixed partials ($\partial_i\partial_ju=\partial_j\partial_iu).$ $$\partial_i u(y)y^i=\partial_iu(0)y^i+\partial_i\partial_ju(0)y^iy^j$$ Substituting $\partial_i u(0) y^i$ from $\eqref1$, $$\partial_i u(y)y^i=u(y)-\frac12\partial_i\partial_ju(0)y^iy^j+\partial_i\partial_ju(0)y^iy^j=u(y)-u(0)+\frac12\partial_i\partial_ju(0)y^iy^j$$ $$dF=\frac{Sd\omega}{r}\left[u(y)-u(0)+\frac12\partial_i\partial_ju(0)y^iy^j\right].$$ The first two terms are exactly what you used in your integral, and the third term is what was missing. We can now take the Laplacian of the integrand to apply the same Laplacian theorem. We have $$\nabla^2\left[u(y)-u(0)+\frac12\partial_i\partial_ju(0)y^iy^j\right]=\nabla^2u(y) + \frac12\partial_i\partial_ju(0)\partial_k\partial_k(y^iy^j)$$ $$=\nabla^2u(y) + \partial_k\partial_ju(0)\partial_ky^j$$ $$=\nabla^2u(y) + \partial_k\partial_ku(0)$$ $$=\nabla^2u(y)+\nabla^2u(0).$$ When evaluated at $y = 0$ (i.e. $x = x_0$), this gives $\color{red}{2}\nabla^2u(0)$, and that's the missing factor of 2. This was a bit roundabout but hopefully it shows what was missing in your derivation.

  • 1
    $\begingroup$ Cool, thanks for the answer! I think you are right, the issue is in step 2. $\frac{u(x)-u(x_0)}{r}$ should approximate the gradient in the middle between $x$ and $x_0$ whereas what we need is the gradient at $x$. If we assume $u$ is approximated by a quadratic function locally around $u(x_0)$ (i.e. use up-to second order Taylor expansion), the gradient should be a linear function and the gradient at $x$ should be 2 times the gradient at $\frac{x+x_0}{2}$. So I underestimated the gradient by a factor of 2. $\endgroup$
    – user112758
    May 28 at 9:31

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