# Is there a metric on Euclidean space that turns the Helmholtz equation into the Laplace equation?

Is there a Riemannian metric $$\tilde g$$ on $$\mathbb R^d$$ such that $$\tag{1} \Delta_{\tilde g}=e^f(\Delta +1),$$ for some $$f\in C^\infty(\mathbb R^d)$$? Here $$\Delta=\partial_{x_1}^2+\ldots+\partial_{x_d}^2.$$ (Answer: no, as (1) fails on constants. See Edit below).

If there is such a $$\tilde g$$, it cannot be conformal to the standard Euclidean metric $$g=\delta_{ij}$$. Indeed, if $$\tilde g = e^{2\phi}g$$, then $$\Delta_{\tilde g} = e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x_j}\frac{\partial}{\partial x_i}\right),$$ and either $$d=2$$, or the second summand in the round brackets is constant only in the trivial case $$\nabla \phi=0$$. In both cases (1) cannot be satisfied.

EDIT. The equation (1) cannot hold verbatim, as it clearly fails on constant functions (thanks Terry Tao for this remark). Instead, let us consider $$\tag{1b} L_{\tilde g} = e^f(\Delta +1),$$ where $$L_{\tilde g}=\frac{d-1}{4(d-2)} \Delta_{\tilde g} - \mathrm{Scal}_{\tilde g}$$ is the conformal Laplacian. The additive term is the scalar curvature of $$\tilde g$$.

In this case, the fact that $$\tilde g$$ cannot be conformal to the Euclidean metric is even more apparent, as $$L_{\tilde g}$$ is conformally invariant.

• Applying (1) to the constant function 1 gives $0=e^f$, which is absurd. Commented Oct 28, 2020 at 18:49
• @TerryTao: Right, that was easy. I actually oversimplified my question. What I am really interested in is not the standard Laplacian $\Delta_{\tilde g}$, but rather the conformal Laplacian $L_{\tilde g}= \Delta_{\tilde g} + C \mathrm{Scal}_{\tilde g}$. I will edit. Commented Oct 28, 2020 at 19:06
• Even with the lower order term, a comparison of top order terms in (1) shows that $\tilde g^{ij} = e^f \delta^{ij}$ (possibly up to a normalising constant depending on your definition of conformal Laplacian), so your analysis of the conformal case applies. Commented Oct 28, 2020 at 19:15

After Terry Tao's comments, I came to the conclusion that the only possible choice of a metric $$\tilde g$$ and of an operator $$T_{\tilde g}= \Delta_{\tilde g} + \text{lower order terms}$$ that will give $$\tag{1} T_{\tilde g}=e^f(\Delta +1)$$ is the following, in Cartesian coordinates: $$\tag{2} \tilde g_{ij}= e^{2\phi}\delta_{ij},\qquad T_{\tilde g}=\Delta_{\tilde g}-e^{-2\phi}\delta^{ij}\partial_i \phi \partial_j + e^{-2\phi}.$$ That is, $$\tilde g$$ must be conformal to the standard Euclidean metric, with an arbitrary conformal factor $$e^{2\phi}$$. I have no idea whether the operator $$T_{\tilde g}$$ has a geometric meaning, or if this small computation can ever be useful. In any case I am posting it here.
We can write (1) as $$\tilde{g}^{ij}\partial_i\partial_j + a^k\partial_k +c= e^f \delta^{ij}\partial_i \partial_j + e^f,$$ for some scalar fields $$a_k$$ and $$c$$. This clearly implies $$\tilde g^{ij}=e^f \delta^{ij}$$, which is the first equation in (2), and also $$a_k=0, c=e^{-2\phi}$$. Since $$\Delta_{\tilde g}= e^{-2\phi} \left(\Delta + (d-2)g^{ij}\frac{\partial \phi}{\partial x_j}\frac{\partial}{\partial x_i}\right),$$ the only possibility is to define $$T_{\tilde g}$$ like in the second equation in (2).