Equation $\Delta u=f$

Question

If we consider $f \in C^{\infty}(\mathbb{R}^n)$, does always exist $u \in C^{\infty}(\mathbb{R}^n)$ such that $$ \Delta u = f,$$ where $\Delta$ is the laplacian ?

If the dimension is $n=1$, the answer is yes, but is it still true for $n \geq 2$ ? The function $f$ could, for instance, growth exponentially fast.

Thanks !

Answer 1

Though this is not a full solution, but I will post some partial results which may help lead to one.

Theorem: If $u \in D'(\mathbb R^n)$ solves the equation $\Delta u = f$ in the distributional sense, so we have,

$$ \langle \Delta u, \phi \rangle = \langle f,\phi \rangle, $$

for all $\phi \in D(\mathbb R^n),$ then $u \in C^{\infty}(\mathbb R^n).$

To prove this, we define the space $H^s_{\mathrm{loc}}(\mathbb R^n)$ to be the set of distributions $u \in D'(\mathbb R^n)$ such that for all $\phi \in D(\mathbb R^n)$ we have,

$$ \lVert u\phi\rVert_{H^s(\mathbb R^n)}^2 = \int_{\mathbb R^n} |\widehat{u\phi}|(1+|\xi|^2)^s\,\mathrm{d}\xi < \infty.$$

Here $\hat{}$ denotes the Fourier transform. We will use the following theorem.

Theorem: If $\Delta u = f$ in the distributional sense and $f \in H^s_{\mathrm{loc}}(\mathbb R^n),$ then $u \in H^{s+2}_{\mathrm{loc}}(\mathbb R^n).$

Iterating this and using Sobolev embedding implies the previous theorem. I will not spell out the details here, but one can consult for example Folland's 'Introduction to Partial Differential Equations' text, where this is proved in chapter 6.D (p215) in the general case for constant coefficient hypoelliptic operators (in particular theorem 6.36, (H4) => (H5)). Since this theorem is attributed to Hörmander it's probably somewhere in one of his volumes also.

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We have the fundmental solution $E$ of the Laplacian is given by,

$$ E(x):=\begin{cases}\frac1{2\pi}\log\left(\frac1{|x|}\right), & \text{ for }n=2,\\\frac{1}{(n-2)\omega_n}\frac1{|x|^{n-2}}, & \text{ for }n>2.\end{cases} $$

One can show these are locally integrable and hence lie in $D'(\mathbb R^n).$ Now if $u = E \ast f$ defines an element of $D'(\mathbb R^n)$ and we get it satisfies the equation in a distributional sense, then by above we get a smooth solution. Since each $E$ is locally integrable and is bounded away from the origin, this certainly holds if $f \in \mathcal{S}'(\mathbb R^n).$

Some more detailed analysis will probably show the solution exists for a more general class of functions $f,$ however this will likely not lead to a complete solution.

Answer 2

The answer is yes for any $n.$

I'll use the fact that a harmonic function on a ball can be approximated uniformly by harmonic polynomials. This follows from the fact that the harmonic polynomials are dense on the unit sphere, which is mentioned on Wikipedia here: https://en.wikipedia.org/wiki/Spherical_harmonics#Higher_dimensions.

By taing a partition of unity, the function $f$ can be decomposed as a sum $f=\sum_{k\geq 0} f_k$ of smooth functions with $f_k$ supported in $k-1<|r|<k+1$ say. You already know there are smooth $u_k$ such that $\Delta u_k=f_k.$ We can guarantee $|u_k|\leq 2^{-k}$ within the ball of radius $k-1$ (where $u_k$ is harmonic) by adding a harmonic polynomial. The sum $u=\sum u_k$ then converges uniformly on compact sets to a smooth function $u$ satisfying $\Delta u=f.$