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No. A positive zeroth-order coefficient can overturn the weak maximum principle for elliptic operators. On $\Omega =(0,1)$ takeThen $L\nu =0$, $\nu =0$ on $\partial \Omega$, and $\nu >0$ in $\Omega$. Thus the boundary maximum is zero while the interior maximum is one. This is a counterexample satisfying uniform ellipticity and zero drift.

The weak maximum principle for elliptic operators here statesfor $u\in C^2(\Omega )\cap C^0(\overline{\Omega })$.

To prove it, suppose the interior maximum exceeds the boundary maximum. Put $q(x)=e^{-\ell x_1}$. Since $c=0$,For sufficiently small $\epsilon >0$, $u_{\epsilon }=u+\epsilon q$ still has an interior maximum. At that point its gradient vanishes and its Hessian matrix is negative semidefinite; ellipticity gives $L{u}_{\epsilon }\le 0$. But $L{u}_{\epsilon }=Lu+\epsilon Lq>0$, a contradiction. This proves the principle.

If $u_1,u_2$ are two solutions with the same boundary values, put $w=u_1-u_2$. The mean value theorem givesTherefore $\Delta w-qw=0$ with zero boundary data. Apply the weak maximum principle for elliptic operators to $w$ and $-w$ with zeroth-order coefficient $-q\le 0$. It follows that $w=0$, proving uniqueness.

SetThen $v\ge A$ and, because $1\le e^{x_1+d}\le e^{2d}$ and $c\le 0$,Thus $(\Delta +c)(u-v)=f-(\Delta +c)v\ge 0$ and $u-v\le 0$ on the boundary. The weak maximum principle for elliptic operators gives $u\le v$. Applying the same argument to $-u$ gives $|u|\le v$, hence

The weak maximum principle for elliptic operators says that if $c\le 0$ andthenIndeed, a positive interior maximum has $Du=0$ and $\Delta u\le 0$, so the differential inequality is incompatible with a strict positive maximum. Applying this argument to a standard strictly perturbed function and then letting the perturbation tend to zero handles equality and proves the weak statement.

Assume $c\le 0$, the coefficients are bounded, and $u$ is bounded above. Choose $\kappa$ so large that the bounded positive function $q(y)=e^{\kappa y}$ satisfiesAlso set $\psi (x)=\log (1+|x{|}^2)$. Boundedness of the coefficients, $x\cdot D\psi \le 2$, and $c\le 0$ give a global upper bound $L\psi \le C$.

For $\epsilon >0$ and $0<\delta <\epsilon m/C$, the functiontends to $-\infty$ as $|x|\to \infty$ and satisfies $Lv>0$. If $v$ exceeded both zero and its values on $y=\pm 1$, it would attain a positive interior maximum. At that point $Dv=0$ and $D^{2}v\le 0$, whence $Lv\le cv\le 0$, a contradiction. Letting $\delta \downarrow 0$ and then $\epsilon \downarrow 0$ provesThis is the weak maximum principle for elliptic operators on the slab. The boundedness or a comparable growth condition is necessary because the domain is unbounded.

The weak maximum principle for elliptic operators fails without a condition at infinity. The functionis harmonic on the upper half-space, continuous on its closure, and vanishes on the boundary, but it is positive and unbounded in the interior.

For $n\ge 3$, the radial functionis harmonic on ${\mathbb{R}}^n\setminus \overline{B_1(0)}$, vanishes on the unit sphere, and is positive in the domain. It therefore violates the weak maximum principle for elliptic operators. In two dimensions the corresponding counterexample is $u(x)=\log |x|$, since the fundamental solution of the Laplace equation changes from a power to a logarithm.

TakeIts principal symbol is ${\xi }^4$, so it is elliptic, and $a_4{a}_0=-1<0$. Yet $u(x)=\sin x$ satisfies $Lu=0$ on $(0,\pi )$ and vanishes at both boundary points while remaining positive inside. Thus second-order ellipticity is essential to the usual weak maximum principle for elliptic operators.

The function $u$ solves the linear nondivergence equationThe coefficient matrix is uniformly elliptic, and part (ii) makes it $C^{0,\alpha }$ on every smaller ball. Solve the corresponding linear Dirichlet problem on $B_r⋐B_1$ with boundary value $u$. Existence gives a $C^{2,\alpha }$ solution, and the weak maximum principle for elliptic operators gives uniqueness, so that solution is $u$. Hence $u\in C_{\mathrm{loc}}^{2,\alpha }(B_1)$.

The local boundedness estimate for uniformly elliptic equations first controls $\Vert u{\Vert }_{C^0(B_r)}$ by $\Vert u{\Vert }_{L^2(B_1)}$. Applying the interior Schauder estimate and covering $B_{\theta }$ by finitely many balls gives

The curvature of $D=d+A$ isor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$. The covariant exterior derivative and the Jacobi identity give the Bianchi identity

Since the Hodge star operator satisfies ${\ast }^2=1$ on Euclidean two-forms,Using the opposite choice of sign as well givesEquality holds precisely for a self-dual or anti-self-dual Yang-Mills instanton, with the sign selected by that of $k$.

Now assume ${\partial }_0{A}_{\mu }=0$ and write $\Phi =A_0$. ThenThe spatial Bianchi identity and the self-duality equations becomewhere the upper four-dimensional sign gives the first displayed reduced sign under the orientation used here. Consequently

Gauge invariance of the inner product givesFor $w=1-|\Phi {|}^2$,Since $w\to 0$ at infinity, the weak maximum principle for elliptic operators excludes a negative interior minimum. Hence

With $d^{4}x=d{x}^0{d}^{3}x$, direct decomposition givesBecause $D_i{B}_i=0$,so the Pontryagin density reduces to the surface charge $4\pi N$ per unit $x^0$. The conventional three-dimensional energyobeys the Bogomolny bound $E_3\ge 4\pi |N|$, saturated by the Bogomolny-Prasad-Sommerfield monopole equation. The four-dimensional action density per unit $x^0$ is $2E_3$; signs relating $k$ and $N$ depend on the self-duality and orientation convention.

For the hedgehog ansatz for a monopole in the question, direct differentiation with $[T_a,T_b]=-{ϵ}_{abc}{T}_c$ givesOn a sphere of radius $r$,ThereforeThe boundary conditions $f\to 1$ and $r^2\alpha \to 1$ make the integrand equal to $r^{-2}+o(r^{-2})$. Thus

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

Forthe weak maximum principle for elliptic operators states that $Lu\ge 0$ impliesIn particular, a solution of $Lu=0$ cannot have a positive interior maximum exceeding its boundary maximum.

Because the coefficient matrix is positive definite and the closure of the smooth bounded domain is compact, strict ellipticity supplies a uniform lower bound after restricting to $\overline{\Omega }$. Rotate and translate coordinates so that $\Omega$ is bounded in the $x_1$ direction, and set $h=e^{\gamma x_1}$. For sufficiently large $\gamma$,If $u+\epsilon h$ had a positive interior maximum, its gradient would vanish and its Hessian matrix would be negative semidefinite there, giving $L(u+\epsilon h)\le 0$. This contradicts $L(u+\epsilon h)=\epsilon Lh>0$. Comparing on the boundary and sending $\epsilon \downarrow 0$ proves the assertion.

Let $v$ range over a bounded subset of $C^{1,\beta }$. All jets $(x,v,Dv)$ then lie in one compact set, so the composed coefficients have uniform $C^{0,\alpha \beta }$ bounds and one ellipticity constant. The weak maximum principle for elliptic operators bounds $T(v)$ in $C^0$ by the boundary data and the bounded forcing term. The global Schauder estimate consequently bounds $T(v)$ in $C^{2,\alpha \beta }$.

The compact embedding of Hölder spacesthen makes the image relatively compact. Hence $T$ is a compact operator on $C^{1,\beta }(\overline{\Omega })$.

If $c\le 0$, $\Delta u+cu=f$, and $u=\psi$ on the boundary of a bounded domain inside $\{|x_1|<d\}$, let $M={\sup }_{\partial \Omega }|\psi |$ and $F={\sup }_{\Omega }|f|$. The nonnegative barrier $v=M+(e^{2d}-e^{x_1+d})F$ satisfies $(\Delta +c)v\le -F$. Apply the weak maximum principle for elliptic operators to $u-v$ and $-u-v$ to get $\Vert u{\Vert }_{\infty }\le M+e^{2d}F$.