On the existence of global classical non-zero solutions of a linear elliptic equation

Question

Does the equation $$-\Delta u +u=0$$ have any non-zero classical, i.e., $C^2$, solutions on $\mathbb{R}^d$? How about if $\mathbb{R}^d$ is replaced with half-space? How about solutions of polynomial growth at infinity?

Answer 1

As I said in the comments, there are many non-trivial solutions. For example, $u(x)=e^{x_1}$ is one.

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To answer your question about polynomial growth:

Theorem. Suppose that $\Delta u=u$ in $\mathbb R^n$ and there exists $C_0>0$, $k \in \mathbb Z^+$ such that $\vert u(x) \vert \leqslant C(\vert x \vert +1)^k$ for all $x\in \mathbb R^n$. Then $u \equiv 0$.

First a lemma:

Lemma. Suppose that $p$ is a polynomial such that $\Delta p = p $ in $\mathbb R^n$. Then $p=0$.

Proof. Denote by $m$ the degree of $p$. If $m\geqslant 2$ then $\Delta p $ has degree $m-2$, but $p=\Delta p$ so $p$ has degree $m-2$ which is a contradiction. If $m=0,1$ then $\Delta p=0$ so $p=0$. $\square$

Proof of Theorem. Recall that the Cacciopolli inequality tells us that if $v\Delta v \geqslant 0$ in $B_{2r}$ then $$ \int_{B_r} \vert \nabla v \vert^2\,dx \leqslant \frac 4 {r^2} \int_{B_{2r}\setminus B_r} v^2 \, dx. $$ In particular, this means that $$\int_{B_r} \vert \partial_i v \vert^2\,dx \leqslant \frac 4 {r^2} \int_{B_{2r}\setminus B_r} v^2 \, dx \leqslant \frac 4 {r^2} \int_{B_{2r}} v^2 \, dx \tag{$\ast$} $$ Differentiating $\Delta u=u$ (using that $u$ is analytic by standard regularity theory) we have $\Delta u_\alpha =u_\alpha$ where $u_\alpha:= D^\alpha u $ and $\alpha $ is a multiindex. Hence, $u_\alpha\Delta u_\alpha =u_\alpha^2\geqslant0$, so iterating ($\ast$) we conclude that for $\vert \alpha \vert \geqslant 1$ (recall for a multiindex $\vert \alpha \vert = \alpha_1+\dots+\alpha_n$), $$ \int_{B_r} \vert D^\alpha u \vert^2\,dx \leqslant \frac C {r^{2\vert \alpha \vert}} \int_{B_{2^{\vert \alpha \vert}r}\setminus B_{2^{\vert\alpha\vert-1} r}} v^2 \, dx. \tag{$\ast\ast$}$$ Then $$\int_{B_{2^{\vert \alpha \vert}r}\setminus B_{2^{\vert\alpha\vert-1} r}} v^2 \, dx \leqslant C\int_{B_{2^{\vert \alpha \vert}r}\setminus B_{2^{\vert\alpha\vert-1} r}} (\vert x \vert+1)^{2k} \, dx \leqslant Cr^{n+2k}.$$ Then by $(\ast\ast$), $$ \int_{B_r} \vert D^\alpha u \vert^2\,dx \leqslant C r^{n+2k-2\vert \alpha \vert}. $$ Thus, for $\vert \alpha \vert$ sufficiently large, sending $r \to \infty$ gives $D^\alpha u\equiv 0$ in $\mathbb R^n$. This implies that $u$ is a polynomial which further implies $u=0$ by the lemma. $\square$

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To mention briefly about halfspaces, every solution in $\mathbb R^n$ will also be a solution in a halfspace. However, the above theorem does not hold in a halfspace, say $\{x_1>0\}$, if $\mathbb R^n$ is simply replaced by $\{x_1>0\}$. Indeed, $u(x)=e^{-x_1}$ is bounded in $\{x_1>0\}$ and solves the PDE.