# Using Green's identity to show that a harmonic function with zero boundary values is identically zero

I am confused how to do this question. I need to use Green's first identity and if $\nabla(f)=0$ then $f$ is constant on $\Omega$ since $\Omega$ is path connected.

I have subbed in the information into green's identity but I don't get anything useful.

$$\iiint_\Omega \nabla f\cdot \nabla g \,dV =\iint_{\partial\Omega} f\nabla g\cdot n \,dA - \iiint_\Omega f\cdot \Delta g \,dV$$

We get:

$$\iiint_\Omega \nabla f\cdot \nabla g \,dV =-\iiint_\Omega f\cdot \Delta g \,dV$$

$$\iiint_\Omega \nabla. (f \nabla g) \,dV =0$$ $$\iint_{\partial\Omega} f\nabla g\cdot n \,dA=0$$

• Can you show us how far you got when you subbed in? – tomi Nov 19 '15 at 15:22
• Yes edited it into first post – Santon Nov 19 '15 at 15:38
• Still need help with this – Santon Nov 19 '15 at 22:48

Edit:More direct method, use integration by parts on the Dirichlet energy, $$\iiint |\nabla f |^2 dV = \underbrace{\iint f \nabla f \cdot n dS}_{f=0 \text{ on } \partial \Omega} - \underbrace{\iiint f \Delta f dV}_{\Delta f = 0 \text{ on } \Omega} =0$$ Thus this means that $\nabla f =0$, so you can use your other result.
• I don't understand the last part. How do I show $\nabla f$ is 0? And how does the last bit imply f=0? – Santon Nov 19 '15 at 16:51
• Why do you choose g? It says in the hint use the result of another question - which states suppose $\nabla f =0$. Then f is constant if omega is path connected – Santon Nov 19 '15 at 19:44
• There is more than one way to show this result. I choose $g$ since that's how you seemed to approach the question... the best choice would be $g \in C^ \infty_0$, i.e. something like a bump function. – Jeb Nov 19 '15 at 20:28