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Let $u(\mathbf{x})=u(x_1, x_2)$ be a harmonic function in $\mathbb{R}^2$. That is $$\Large \Delta_\mathbf{x} u(\mathbf{x})=0\tag1$$ I must prove that, fixed $\mathbf{y}\in\mathbb{R}^2$, the function $$u(\mathbf{x}-\mathbf{y})$$ is also harmonic in $\mathbb{R}^2.$ That is we must prove that $$\Large\Delta_\mathbf{x}[u(\mathbf{x}-\mathbf{y})]=0\tag2$$

We consider the trasformation $h\colon \mathbb{R}^2\to \mathbb{R}^2$ defined as $$h(\mathbf{x})=\mathbf{\tilde{x}}=\mathbf{x}-\mathbf{y}.$$ therefore $$u(\mathbf{x}-\mathbf{y})=u\circ h(\mathbf{x})$$

Then we have that $$\Large \frac{\partial u(\mathbf{x-y})}{\partial x_i}=\frac{\partial u}{\partial\tilde{x_i}}(\mathbf{x}-\mathbf{y}),\quad i=1,2.$$ And therefore

$$\Large \Delta_\mathbf{x} [u(\mathbf{x}-\mathbf{y})]=[\Delta_{\tilde{\mathbf{x}}}u](\mathbf{x}-\mathbf{y}) $$

Can I conclude that $$\Large [\Delta_{\tilde{\mathbf{x}}}u](\mathbf{x}-\mathbf{y})=[\Delta_{\mathbf{\tilde{x}}}u](\mathbf{\tilde{x}})=0?$$ Since $(1)$ holds, how can i deduce this?

It is true or false that $$\Large [\Delta_{\mathbf{\tilde{x}}}u](\mathbf{\tilde{x}})=0?$$ If yes, why?

who are the old and new variables?

NatMath
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  • The $x$ is supposed to be a bound variable, so it doesn't matter whether you decide to say the function $v$ such that for all $x\in\Bbb{R}^n$, $v(x):=u(x-y)$, or whether you decide to say "the function $v$ such that for all $\tilde{x}\in\Bbb{R}^n$, $v(\tilde{x}):=u(\tilde{x}-y)$", or whether you say "the function $v$ such that for all $@\in\Bbb{R}^n$, $v(@):=u(@-y)$". So, it doesn't strictly make sense to say "$u=u(x)$", because the LHS is a function, the RHS is a function value. But yes, to prove that translates of harmonic functions are harmonic, one simply uses the chain rule (twice!). – peek-a-boo Nov 28 '22 at 07:27
  • @peek-a-boo Yes, but I wanted a relationship between the old variables and the new ones. That is, at a notational level, when I define $v$ if I don't change the name of the variable how can I apply the chain rule without running into errors? – NatMath Nov 28 '22 at 08:47
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    See Notation for partial derivative of functions of functions. It is a long answer (and thread of comments), but you should read it in detail. You simply define a new map $h:\Bbb{R}^2\to\Bbb{R}^2$, $h(x_1,x_2)=(x_1-y_1,x_2-y_2)$, i.e $h(x)=x-y$. Then, $v=u\circ h$, and it is easy to write the chain rule as $\partial_iv=(\partial_1u)\circ h\cdot \partial_ih_1+ (\partial_2u)\circ h\cdot \partial_ih_2$. – peek-a-boo Nov 28 '22 at 10:46
  • @peek-a-boo I've edited the question trying to write in more detail. Thanks for your interest. – NatMath Nov 28 '22 at 11:42
  • Remember, $\mathbf x-\mathbf y = \tilde{\mathbf x}$, so you have $\Delta_{\tilde{\mathbf{x}}}u( \tilde{\mathbf x}) = 0$ But that is just $\Delta_\mathbf{x} u(\mathbf x)=0$ with a different name given to the variable. – Paul Sinclair Nov 28 '22 at 12:20
  • @PaulSinclair So, we can say that the Laplacian is invariant under changes of variables? – NatMath Nov 28 '22 at 12:52
  • That is what you just calculated! Or were you just proclaiming above where you claimed the partial differentials were the same, but hadn't actually verified it? (The Laplacian makes use of that calculation twice, since it involves 2nd derivatives). Your confusion occurred because when you changed variables, you perversely only changed one use of the variable, while leaving the other occurrence unchanged. When you change variables, change them everywhere. – Paul Sinclair Nov 28 '22 at 14:31
  • @PaulSinclair I'm sorry, but I don't know how to do this then. I can't write it well. Let me just show that $u(x-y)$ is harmonic. – NatMath Nov 28 '22 at 14:39
  • I don't know, I don't understand. Could someone explain to me where I'm wrong? – NatMath Nov 28 '22 at 14:51

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