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I was told to use the maximum principle $$\max_{\bar{\Omega}}u=\max_{\partial\Omega}u$$ for subharmonic functions to prove that the problem $\Delta u=u^3$ in $\Omega=B(0, 1)$, $u=0$ in $\partial \Omega$, has unique solution ($u=0$). So assume that $v$ is another solution. Then define $w=u-v$. Then $w=0$ in $\partial \Omega$ and $$\Delta w=\Delta u-\Delta v=(u^2+uv+v^2)w,$$ where $u^2+uv+v^2\geq 0$. I should somehow conclude that $\Delta w\geq 0$ to be able to apply the maximum principle, and I believe I must also use the minimum principle. So I should end with $w\leq 0$ and $w\geq 0$ in $\Omega$ and hence $w=0\Leftrightarrow u=v$. Any help is appreciated.

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Suppose $u$ has a minimum at some point $p \in \Omega$. Then

$u(p) \ge 0; \tag 1$

otherwise, we have

$u(p) < 0, \tag 2$

whence,

$\nabla^2 u(p) = \Delta u(p) = u^3(p) < 0; \tag 3$

(3) asserts that the trace of the Jacobean matrix $J_u$ of $u$, which is

$J_u = \left [ \dfrac{\partial^2 u}{\partial x_i \partial x_j} \right ], \; 1 \le i, j \le n, \tag 4$

obeys

$\text{Tr} J_u(p) < 0, \tag 5$

since

$\text{Tr} J_u(p) = \displaystyle \sum_1^n \dfrac{\partial^2 u}{\partial x_i^2} = \Delta u(p) < 0; \tag 6$

here the $x_i$ are the usual cartesian coordinates on $\Bbb R^n$. By virtue of (6), some eigenvalue $\lambda$ of $J_u(p)$ must satisfy

$\lambda < 0; \tag 7$

this in turn implies $u$ is decreasing in the direction of an eigenvector $\vec v \ne 0$ corresponding to $\lambda$,

$J_u(p) \vec v = \lambda \vec v, \tag 8$

and hence we cannot have

$u(p) < 0 \tag 9$

at a minimum $p \in \Omega$ of $u$. So

$u(p) \ge 0, \tag {10}$

which in turn implies

$u(x) \ge 0, \; \forall x \in \Omega. \tag{11}$

Now by virtue of (11),

$\Delta u(x) = u^3(x) \ge 0 \tag{12}$

for $x \in \Omega$, so $u$ is subharmonic. The fact that $u$ is subharmonic in turn implies that

$\max_{\bar{\Omega}}u=\max_{\partial\Omega}u, \tag{13}$

as pointed out by our OP Infinitebig in the text of the question. Thus, since $u(x) = 0$ for $x \in \partial \Omega$, we have

$u(x) \le 0 \tag{14}$

for $x \in \Omega$. (11) and (14) together show that

$u(x) = 0, \; \forall x \in \Omega, \tag{15}$

and thus the uniqueness of the solution $u(x) = 0$ is established.

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    $\begingroup$ Thank you for detailed solution. Only part I don't remember is that if you have negative/positive eigenvalue for Jacobian, then function must be decreasing/increasing in that direction. Maybe this is trivial, but it's been a while since I had calculus of extremas. $\endgroup$ – Infinitebig Sep 28 '17 at 16:14
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    $\begingroup$ @Infinitebig: yes, that part could use more exposition, but if one diagonalizes $J_u(p)$, which is possible by an orthogonal change of coordinates since $J_u(p)$ is symmetric, it is easier to see. Thanks for the acceptance! Cheers! $\endgroup$ – Robert Lewis Sep 28 '17 at 16:21
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    $\begingroup$ @Infibitebig: I'm pretty sure that the assertion follows from Taylor's theorem. $\endgroup$ – Robert Lewis Sep 28 '17 at 23:34
  • $\begingroup$ Okay, thanks, I'll take a look at that. $\endgroup$ – Infinitebig Sep 29 '17 at 7:22

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