Proving Laplace Equation in $\Omega_{(a,b)}$

[Julio1]

Let $\Omega$ an open domain in $\mathbb{R}^2.$ Suppose that $u$ is an application $C^2$ that satisfy the Laplace equation $\Delta u=0$ in $\Omega.$ Let $\Omega_{(a,b)}=\{(x+a,y+b): (x,y)\in \Omega\}$ and define $v(X,Y)=u(X-a,Y-b)$ for all $(X,Y)\in \Omega_{(a,b)}.$ Show that $v$ is an application $C^2$ on $\Omega_{(a,b)}$ and satisfy the equation $\dfrac{\partial^2 v}{\partial X^2}+\dfrac{\partial^2 v}{\partial Y^2}=0$ in $\Omega_{(a,b)}.$

Hello MHB :). My question is, can use Chain Rule for this case? and how or that form can be used?

 

[Julio1]

Can someone help me out?

 

[Euge]

Hi Julio,

Let $x = X-a$ and $y=Y-b$, so that $v(X,Y)=u(x,y)$. Since $v$ is the composition of $u$ with the translation map $(X,Y) \mapsto (X-a,Y-b)$ and $u$ is $C^2$ in $\Omega$, then $v$ is $C^2$ in $\Omega_{(a,b)}$. By the chain rule, argue that $$\frac{\partial^2 v} {\partial X^2} + \frac{\partial^2 v}{\partial Y^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.$$

 

[Julio1]

Hi Julio,

Let $x = X-a$ and $y=Y-b$, so that $v(X,Y)=u(x,y)$. Since $v$ is the composition of $u$ with the translation map $(X,Y) \mapsto (X-a,Y-b)$ and $u$ is $C^2$ in $\Omega$, then $v$ is $C^2$ in $\Omega_{(a,b)}$. By the chain rule, argue that $$\frac{\partial^2 v} {\partial X^2} + \frac{\partial^2 v}{\partial Y^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.$$

Thanks Euge :).

But for show that $v\in C^2(\Omega_{(a,b)})$ I don't can show that $v$ has continuous derivate? It is necessary for this case?

 

[Julio1]

Can help me? :(

 

[Euge]

Julio, please do not bump threads. Since $v$ is the composition of $C^2$ maps (i.e., the map $u$ and the translation map $(X,Y) \mapsto (X-a,Y-b)$), it is $C^2$.