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I have:

$$v^2 \frac {\partial^2 f} {\partial x^2} + 2 v \frac {\partial^2 f} {\partial x \partial t} + \frac {\partial^2 f}{\partial t^2} = 0$$

Why does that imply that $\dfrac {\partial^2 f} {\partial x^2} = \dfrac {\partial^2 f} {\partial x \partial t} = \dfrac {\partial^2 f}{\partial t^2} = 0$?

Here, $f$ is a function of $x$ and $t$, and $v = \dfrac {\mathrm d x} {\mathrm d t}$, if it matters.

The explanation I came up with was that we can consider it a polynomial in $v$, by considering paths where the coefficients are constant but letting v vary. There are infinitely many such points (as one can check) so we must have the zero polynomial. Is this correct? Is there another way of proving it in any case?

hardmath
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user85798
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1 Answers1

1

It is not sufficient to have this be true for only one curve $x(t)$, as the following example shows, even if typical smoothness on $f$ is assumed (so that mixed partials commute).

We can write the condition on $v$ with second partial derivatives of $f$ as coefficients in the following Hessian matrix form:

$$ \begin{pmatrix} v & 1 \end{pmatrix} \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial t} \\ \frac{\partial^2 f}{\partial x \partial t} & \frac{\partial^2 f}{\partial t^2} \end{pmatrix} \begin{pmatrix} v \\ 1 \end{pmatrix} = 0 $$

Consider the "curve" $x(t) = t$ and the "field" $f(x,t) = x^2 - t^2$. The second partials of $f$ are not identically zero, but the "bilinear" condition above is satisfied at each point on the curve.

Since this first example yields only a single value for $v$, it might reasonably be objected that it does not address the OP's intended argument about polynomial identity. However here is a second example in which $v$ attains infinitely many values.

Let $f(x,t) = x^4 - t^4$, so that the Hessian is the diagonal matrix:

$$ H = \begin{pmatrix} 12x^2 & 0 \\ 0 & -12t^2 \end{pmatrix} $$

Now consider the (semi)circular trajectory $x^2 + t^2 = 1$ for $t \in (-1,1)$, where:

$$ v = \frac{dx}{dt} = -\frac{t}{x} $$

Then:

$$ \begin{pmatrix} v & 1 \end{pmatrix} H \begin{pmatrix} v \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{t}{x} & 1 \end{pmatrix} \begin{pmatrix} -12xt \\ -12t^2 \end{pmatrix} = 0 $$

Note that the slope $v$ along the trajectory takes all values $(-\infty,+\infty)$, but the second derivatives of $f$ do not simultaneously vanish even at a single point of the trajectory.

On the other hand if it is supposed that the bilinear form vanishes for every trajectory, then the vanishing of the Hessian is easily established by considering the trajectories of constant but arbitrary slope $v$ through point $(x,t)$. The polynomial identity argument outlined in the Question is then justified, since $v$ can take infinitely many values for the same point $(x,t)$.

Here is an alternative argument for the same conclusion. Assuming (per the OP's comment) that the Hessian is symmetric, it must have real eigenvalues and a corresponding basis of orthogonal eigenvectors. It suffices to argue that at every point of the domain $(x,t)$, $H$ has only zero eigenvalues, and thus $H=0$ there.

Let $\lambda$ be the real eigenvalue of $H$ of greatest absolute value. According to the theory of Rayleigh quotients, $|\lambda|$ can be approximated as closely as we wish by:

$$ \left| \frac{\begin{pmatrix} v & 1 \end{pmatrix} H \begin{pmatrix} v \\ 1 \end{pmatrix}}{ \begin{pmatrix} v & 1 \end{pmatrix} \begin{pmatrix} v \\ 1 \end{pmatrix} } \right| $$

But these quotients are identically zero, so $\lambda = 0$.

hardmath
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