Taylor Church

2022-09-06

$\mathrm{\Delta}(x\cdot \mathrm{\nabla}u)=\mathrm{\Delta}u+x\cdot (\mathrm{\nabla}(\mathrm{\Delta}u))$

Does this hold for $x\in {\mathbb{R}}^{n}$ and $u:{\mathbb{R}}^{n}\to \mathbb{R}$ ? If it does, why does it?

I calculated the above formula by writing out all the variables from ${x}_{1}$ to ${x}_{2}$ and guessing that it must be correct. But I am not familiar with 'direct' calculations using $\mathrm{\nabla}$ and $\mathrm{\Delta}$, so I need some help.

Does this hold for $x\in {\mathbb{R}}^{n}$ and $u:{\mathbb{R}}^{n}\to \mathbb{R}$ ? If it does, why does it?

I calculated the above formula by writing out all the variables from ${x}_{1}$ to ${x}_{2}$ and guessing that it must be correct. But I am not familiar with 'direct' calculations using $\mathrm{\nabla}$ and $\mathrm{\Delta}$, so I need some help.

barquegese2

Beginner2022-09-07Added 11 answers

Since there is no standard writing of the derivative of a vector field I don't know of a way to do this in a coordinate free manner. I'm coming to the following result which is different from your formula.

The j in the calculation is fixed, and ${\delta}_{ij}$ is the Kronecker delta. I write ${x}_{i}$ for the ${i}^{\mathrm{t}\mathrm{h}}$ component of x and ${u}_{.i}$ for the partial derivative of the scalar function u with respect to ${x}_{i}$. Then

$\begin{array}{rl}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{j}}\left(\sum _{i}{x}_{i}{u}_{.i}\right)& =\sum _{i}{\textstyle (}{\delta}_{ij}{u}_{.i}+{x}_{i}{u}_{.ij}{\textstyle )}\\ & ={u}_{.j}+\sum _{i}{x}_{i}{u}_{.ij}\text{}.\end{array}$

Doing this a second time we obtain

$\begin{array}{rl}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{x}_{j}^{2}}\left(\sum _{i}{x}_{i}{u}_{.i}\right)& ={u}_{.jj}+\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{j}}\sum _{i}{x}_{i}{u}_{.ij}\\ & ={u}_{.jj}+\sum _{i}{\textstyle (}{\delta}_{ij}{u}_{.ij}+{x}_{i}{u}_{.ijj}{\textstyle )}\\ & =2{u}_{.jj}+\sum _{i}{x}_{i}{u}_{.ijj}\text{}.\end{array}$

Summing over j we find that in fact

$\begin{array}{}\text{(1)}& \mathrm{\Delta}(x\cdot \mathrm{\nabla}u)=2\mathrm{\Delta}u+x\cdot \mathrm{\nabla}(\mathrm{\Delta}u)\text{}.\end{array}$

Note that in the one-dimensional case we have

$(}x{u}^{\prime}(x){{\textstyle )}}^{\u2033}={\textstyle (}{u}^{\prime}(x)+x{u}^{\u2033}(x){{\textstyle )}}^{\prime}=2{u}^{\u2033}(x)+x{\textstyle (}{u}^{\prime}(x){{\textstyle )}}^{\u2033}\text{},$

as obtained in (1)

The j in the calculation is fixed, and ${\delta}_{ij}$ is the Kronecker delta. I write ${x}_{i}$ for the ${i}^{\mathrm{t}\mathrm{h}}$ component of x and ${u}_{.i}$ for the partial derivative of the scalar function u with respect to ${x}_{i}$. Then

$\begin{array}{rl}\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{j}}\left(\sum _{i}{x}_{i}{u}_{.i}\right)& =\sum _{i}{\textstyle (}{\delta}_{ij}{u}_{.i}+{x}_{i}{u}_{.ij}{\textstyle )}\\ & ={u}_{.j}+\sum _{i}{x}_{i}{u}_{.ij}\text{}.\end{array}$

Doing this a second time we obtain

$\begin{array}{rl}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{x}_{j}^{2}}\left(\sum _{i}{x}_{i}{u}_{.i}\right)& ={u}_{.jj}+\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{j}}\sum _{i}{x}_{i}{u}_{.ij}\\ & ={u}_{.jj}+\sum _{i}{\textstyle (}{\delta}_{ij}{u}_{.ij}+{x}_{i}{u}_{.ijj}{\textstyle )}\\ & =2{u}_{.jj}+\sum _{i}{x}_{i}{u}_{.ijj}\text{}.\end{array}$

Summing over j we find that in fact

$\begin{array}{}\text{(1)}& \mathrm{\Delta}(x\cdot \mathrm{\nabla}u)=2\mathrm{\Delta}u+x\cdot \mathrm{\nabla}(\mathrm{\Delta}u)\text{}.\end{array}$

Note that in the one-dimensional case we have

$(}x{u}^{\prime}(x){{\textstyle )}}^{\u2033}={\textstyle (}{u}^{\prime}(x)+x{u}^{\u2033}(x){{\textstyle )}}^{\prime}=2{u}^{\u2033}(x)+x{\textstyle (}{u}^{\prime}(x){{\textstyle )}}^{\u2033}\text{},$

as obtained in (1)

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