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How to show that the Laplacian is the only 2nd order operator that is translation and rotation invariant such that $L0=0$?

I have shown that it is rotation and translation invariant but I could not show the uniqueness.

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  • $\begingroup$ As a first step, translation invariance implies that the operator is homogeneous of order two. $\endgroup$ – Michael Joyce Apr 16 '13 at 12:20
  • $\begingroup$ Shouldn't the hypothesis be that $L(1) = 0$? I assumed that operator meant linear operator. $\endgroup$ – Michael Joyce Apr 16 '13 at 12:54
  • $\begingroup$ He does mean linear. Otherwise there are entire families, such as the p-Laplacian, which fit the description. $\endgroup$ – Ray Yang Apr 16 '13 at 14:29
  • $\begingroup$ Please do not deface your posts. $\endgroup$ – Martin Apr 20 '13 at 11:26
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As I was suspecting, this question is not clear.

The question can be written more precisely as: Let $L[u](x)=\sum_{ij} a_{ij}(x)D_{ij}u(x)+\sum_i b_i(x)D_iu(x)+c(x)u(x)$ be a translation and rotation invariant operator, show that $L[u]=\lambda\Delta u+cu$ whre $\lambda$ and $c$ are constants.

I have just shown that $b=0$ and $c=k=constant$ but I couldn't prove yet that $a_{ij}=\lambda \delta_{ij}$. First of all, being translation and rotation invariant means that $$ L[u\circ T](x)=Lu\circ T(x)\;(*) $$ for all translation or rotation $T$. Taking $u\equiv k=constant$ and using the hypothesis (*) we get $c(x)\cdot k=c(T(x))\cdot k$, for any traslation $T$ so $c$ is constant. Now taking $u(x)=x_k$, $k=1,...,n$ and $T$ to be the translation $T_v(x)=x+v$ we obtain $u\circ T(x)=u(x+v)=x_k+v_k$ whence $D_k(u\circ T)=1$ and, as before, $b(x)=b(T(x))$, so $b$ is constant. Now to conclude that $b_k=0$ by applying the hypothesis to the rotation $T(x)=-x$ and $u(x)=x_k$ again.

It remains to show that $a_{ij}=\lambda \delta_{ij}$.

The idea is is to take $u$ as somtehing with degree two, let say, $u(x)=x_k\cdot x_l$ or sometring like this and the rotation as being $T(x)=(x_1,...,-x_k,...,x_n)$, that is, the reflection about the $k$-th axis.

I still did not figure out how to conclude, could someone please give me some help?

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I guess you figured out the answer by now, but if someone still wants to know you might proceed by showing something like that if R is a rotation matrix acting on x, then for all functions u, $L[u\circ R]=\sum_{ij} a_{ij}(x)D_{ij}u(Rx)=\sum_{ij}a_{ij}\sum_{kl}R_{ik}R_{jl}D_{kl}u(x)=\sum_{kl}\left(\sum_{ij}R_{ik}R_{jl}a_{ij}\right)D_{kl}u(x)$; substituting some nice test functions with specified quadratic behavior at some point $x_0$ then ought to give $a_{kl}=\left(\sum_{ij}R_{ik}R_{jl}a_{ij}\right)$ so that the result follows rotational invariance of $a_{ij}$.

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