1
$\begingroup$

Prove that for every $u\in \mathcal C^2(\bar \Omega)$ where $\Omega$ is an open and bounded subset of $\mathbb R^n$ and $\partial \Omega \in C^1\;$ the following holds for every $y\in \Omega$:

$$ u(y)=\int_{\Omega} v \Delta^2 u \, dx - \int_{\partial \Omega} \left( v \frac{\partial \Delta u}{\partial \eta}-\Delta u \frac{\partial v}{\partial \eta} + \Delta v \frac{\partial u}{\partial \eta}-u\frac{\partial \Delta v}{\partial \eta} \right) dS$$

Note: $\vec{\eta}$ is the outward unit normal vector.

To begin with, I thought to substitute $u=\Delta u$ in Green's First Identity:

$$\int_{\Omega} \Delta(\Delta u) v \, dx = -\int_{\Omega} \nabla \Delta u \cdot \nabla v \, dx + \int_{\partial \Omega} v \frac{\partial \Delta u}{\partial \eta} dS $$

After that, in order to get rid of the term $\int_{\Omega} \nabla \Delta u \nabla v \, dx $ I made use of the third Green's Identity which is: $$u(y)=-\int_{\Omega} \Phi(y-x) \Delta u(y) dy + \int_{\partial \Omega} \left( \Phi(y-x) \frac{\partial u(y)}{\partial \eta}-u(y) \frac{\partial \Phi(y-x)}{\partial \eta} \right) dS(y) $$ where $\Phi(x)$ is the fundamental solution of $\Delta u=0$. I noticed that if I replace $\Phi(x)$ with $\Delta v$ in the above identity the proof is complete.

Now in favor of this substitution I thought that $\Delta v $ could be a fundamental solution of $\Delta(\Delta v)=0 $ if and only if $\Phi(y)=\Delta v(y) $.

My question is if this assumption is enough in order to do the above. It seems quite easy and I feel I'm missing something. I would really appreciate if somebody could make this more clear to me. Hints or other solutions than this are also welcome.

Thanks in advance!

$\endgroup$
  • $\begingroup$ Is $v$ a particular function, or general? $\endgroup$ – Chappers Jan 2 '17 at 14:29
  • $\begingroup$ @Chappers The exercise doesn't mention anything about $v\;$ so I suppose it's a general function.. $\endgroup$ – kaithkolesidou Jan 2 '17 at 14:46
2
$\begingroup$

This can't happen for any old $v$: for example, nothing on the right-hand side of the expression you need says anything about $y$ for a generic $v$, so $v$ must depend on $y$. You're on the right lines: since a fundamental solution just depends on the local behaviour near $y$ and satisfying the equation,your idea will work. On the other hand, since the fundamental solution is not unique unless boundary conditions are given, it's quite as simple as that. Instead, below I give a direct approach.


Integration by parts using Green's first identity in both the usual form and the vectorial form $$ \int_A u\operatorname{div}{F} = \int_{\partial A} u (n \cdot F) - \int_A \nabla u \cdot F, $$ we find $$ \int_A v \Delta^2 u = \int_{\partial A} \big( v (\partial_n \Delta u) - (\partial_n v) \Delta u + (\partial_n u) \Delta v - u (\partial_n \Delta v) \big) + \int_A u \Delta^2 v, $$ where $\partial_n = n \cdot \nabla$. This is most of the right-hand side. The problem is that we need to get to $u(y)$. Choosing $A$ to be $\Omega \setminus B(y,\varepsilon)$, the only sensible way this can be the case is if $$ -\int_{\partial B(y,\varepsilon)} \big( v (\partial_n \Delta u) - (\partial_n v) \Delta u + (\partial_n u) \Delta v - u (\partial_n \Delta v) \big) + \int_{A} u \Delta^2 v \to u(y) $$ as $\varepsilon \to 0$. Moreover, since $A$ is effectively arbitrary, we need $\int_{A} u \Delta^2 v = 0$, and since this holds for any smooth $u$, we need $\Delta^2 v = 0$ outside $B(y,\varepsilon)$. On the other hand, if we want to get $u(y)$ out of the small ball integral, it looks like the only term that should be nonzero in the limit is $ \int_{\partial B(y,\varepsilon)} u (\partial_n \Delta v) \to u(y) $.

This looks familiar: of course what we want is that $v$ is a Green's function/fundamental solution for the biharmonic equation. We don't need to worry about boundary conditions, just that the two conditions mentioned above are satisfied.


I do think that the problem is rather unhelpfully worded: Prove that there exists a $v_y$ so that [...] would be much clearer!

$\endgroup$
  • $\begingroup$ Why are three of the four terms in the second integral of the second equation zero? And is there a minus sign missing in front of it? $\endgroup$ – Open Season Feb 27 '18 at 15:59
  • $\begingroup$ I guess the minus sign is the result of reversing orientation... $\endgroup$ – Open Season Feb 27 '18 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.