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In Evans' book, PDE, it is asked to show that if $O$ is a $n\times n$ orthogonal matrix, and we define$$ v(x):=u(Ox) $$ then $\Delta v=0$. I tried to compute $\Delta v=D_xv\cdot D_xv$ and expand it using the fact that $O^T=O^{-1}$, but I get stuck after a while. Help!!

  • Use linearity of differentiation. – amd Oct 19 '17 at 00:31
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    ok, i will try. –  Oct 19 '17 at 00:37
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    $\Delta v \neq D_x v \cdot D_x v $. This is equivalent to saying $v'' = v' \cdot v'$ – user770687 Oct 25 '20 at 14:01
  • Alternative solution: Try to show that: Let $u$ be $C^2$ and $A$ be an $n \times n$ matrix. If $v$ is a function defined by $v(x) = u(Ax)$, then $\Delta v = tr (A^T Hu A)$, where $Hu$ is the Hessian matrix of $u$. – E.E. Oct 28 '21 at 05:38

3 Answers3

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A function $u \in C^2(\Omega)$ is harmonic iff it satisfies the mean value property. Then note that for any $B_r(x) \subset \Omega$ there holds $Q(B_r(x))=B_r(Qx)$. Then:
$$ \int_{B_r(x)}u(Qy)dy=\int_{B_r(x)}u(Qy)|\det(Q)|dy=\int_{Q(B_r(x))}u(y)dy=\int_{B_r(Qx)}u(y)dy=u(Qx) $$

F. Conrad
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Another way to see this is to resort to indices:

Let $O$ be the orthogonal matrix. Then

\begin{align} \Delta v & = \sum_{i=1}^n \frac{\partial ^2}{\partial x_i^2} u(Ax) \end{align}

Define $y_k = \sum_j A_{kj}x_j$. Note that $$ \frac{\partial v}{\partial x_i} = \sum_{j=1}^n \frac{\partial u}{\partial y_j}\frac{\partial y_j}{\partial x_i} = \sum_{j=1}^n \frac{\partial u}{\partial y_j} a_{ji}$$

Then we have \begin{align} \Delta v & = \sum_{i}\frac{\partial}{\partial x_i} \left(\sum_{j=1} \frac{\partial u}{\partial y_j}a_{ji}\right)\\ & = \sum_{ij}a_{ji}\frac{\partial }{\partial x_i}\frac{\partial u}{\partial y_j} \\ & = \sum_{ij}a_{ji}\sum_{k}\frac{\partial u}{\partial y_j \partial y_k}\frac{\partial y_k}{\partial x_i} \\ & = \sum_{ij}a_{ji}\sum_{k}\frac{\partial u}{\partial y_j \partial y_k}a_{ki}\\ & = \sum_{jk}u_{y_j,y_k}\sum_i a_{ji}a^T_{ik}\\ & = \sum_{j} u_{jj} = 0 \end{align}

Since $\sum_i a_{ji}a^T_{ik} = 1$ if and only if $j=k$ by orthogonality.

trautner
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Using your notation, let $O=(a_{ij})$ and $y\equiv Ox$. Then $\Delta v(x)=\Delta u(Ox)=\Delta u(y)$, with$$ y_j=\sum_{i=1}^n a_{ji}x_i. $$ We then have$$ \frac{\partial v}{\partial x_i}=\sum_{j=1}\frac{\partial u}{\partial y_j} \frac{\partial y_j}{\partial x_i}=\sum_{j=1}\frac{\partial u}{\partial y_j} a_{ji}. $$ Thus\begin{align*} \begin{pmatrix} \frac{\partial v}{\partial x_1}\\ \vdots\\ \frac{\partial v}{\partial x_n}\end{pmatrix}&=\begin{pmatrix} a_{11}&\cdots&a_{n1}\\ \vdots &\ddots &\vdots\\ a_{1n}&\cdots&a_{nn} \end{pmatrix}\begin{pmatrix} \frac{\partial u}{\partial y_1}\\ \vdots\\ \frac{\partial u}{\partial y_n} \end{pmatrix}=O^T\begin{pmatrix} \frac{\partial u}{\partial y_1}\\ \vdots\\ \frac{\partial u}{\partial y_n} \end{pmatrix} \end{align*} which can be simply rewritten as $D_x\cdot v=O^TD_y\cdot u $. Then, it follows that\begin{align*} \Delta v&=D_xv\cdot D_xv \\ &= (O^TD_yu)\cdot(O^TD_y u)\\ &= (O^TD_yu)^TO^TD_yu \\ &= (D_yu)^T(O^T)^TO^TD_yu\\ &= (D_yu)^TOO^TD_yu \\ &= (D_yu)^TD_yu \\ &= (D_yu)\cdot (D_yu) \\ &= \Delta u(y)\\ &= 0, \end{align*} since $O$ is orthogonal and thus $O^T=O^{-1}$.

sam wolfe
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