# Dirichlet problem with odd function.

Let $$\Omega \subset \mathbb{R}^2$$ be open, bounded and symmetric with respect to the origin. Let$$f:\partial \Omega \to \mathbb{R}$$ be odd and continuous function. Show that if $$u$$ is solution of:

$$\begin{cases} \Delta u(x)&=0,&\text{ if }x \in \Omega \\ \phantom{-}u(x)&=f(x),&\text{ if }x \in \partial\Omega \end{cases}$$

then $$u$$ is odd.

Hint: study of the Laplacian's invariance in relation to $$T(x,y)=(-x,-y)$$

I tried to show that $$\Delta u= \Delta (u\circ T)=0$$, but how to finish?

If you show that $$\Delta (u\circ T)=0$$ in $$\Omega$$ you're done by uniqueness.

Indeed, first note that since $$\Omega$$ is bounded and symmetric with respect to the origin then $$(x,y)\in\partial\Omega$$ iff $$(-x,-y)\in\partial\Omega$$. Then, using that $$f$$ is odd and $$u$$ a solution we have for all $$(x,y)\in\partial\Omega$$ $$u(x,y)=f(x,y)=-f(-x,-y)=-u(-x,-y)=-(u\circ T)(x,y).$$ Now, as you noted $$\Delta(u\circ T)=0$$, which follows by symmetry of $$\Omega$$ and the identities for all $$(x,y)\in\Omega$$ $$\partial^2_x(u\circ T)(x,y)=\partial^2_xu(-x,-y),\\\partial^2_y(u\circ T)(x,y)=\partial^2_yu(-x,-y).$$ Finally, this implies that both $$u$$ and $$-(u\circ T)$$ are solutions to your Dirichelt problem. The fact the $$f$$ is continuous implies that you only have one solution, hence $$u(x,y)=-(u\circ T)(x,y)=-u(-x,-y)$$ for all $$(x,y)\in\mathbb{R}^2$$.

Note: While I was writing this, another answer was posted. The solution is essentially the same: maxima principle $$\Rightarrow$$ uniqueness.

• I lost a bit in the derivatives. $\frac{\partial u}{\partial x}(-x,-y)=-\frac{\partial u}{\partial x}(x,y)$, Right? – Lucas May 15 at 22:04
• The chain rule reads $\partial_x(u\circ T)(x,y)=\partial_xu(T(x,y))\partial_xT(x,y)=-\partial_xu(-x,-y)$. – pipenauss May 16 at 12:26

If you already showed that $$\Delta u(-x, - y) =0$$, take $$v=u(x, y) +u(-x, - y)$$.

By the fact that f is odd, $$v$$ is solution for the problem: $$$$\begin{cases} \Delta v=0 \\ v=0, \ \ \ in \ \partial \Omega \end{cases}$$$$

So, by the maxima principle $$v=0$$ in $$\Omega$$ then $$u(x, y)=-u(-x, - y)$$ in $$\Omega$$.