A thin metal plate has its edge held at prescribed temperatures, hot along one stretch and cold along another, and is left alone until the temperature inside stops changing. Can the hottest point of the plate be somewhere in the interior? Physical intuition says no: a hot spot surrounded by cooler metal would lose heat to its surroundings, so it could not be in equilibrium. The intuition is right, and making it precise yields the most useful qualitative facts in the subject. They hold for every solution of \(\Delta u=0\) , and they are proved without solving anything.
The steady temperature satisfies the Laplace equation, as Section 1.2 showed. The chapter develops three properties of its solutions. A solution equals its own average over every sphere. It takes its largest and smallest values on the boundary. And it is the unique function with its boundary values, a fact with two different proofs, one by maxima and one by energy.
3.1 Harmonic functions
Definition 3.1 (harmonic, subharmonic, superharmonic). Let \(\Omega\subseteq\mathbb R^n\) be open. A function \(u\in C^2(\Omega)\) is harmonic in \(\Omega\) if \(\Delta u=0\) there, subharmonic if \(\Delta u\ge 0\) , and superharmonic if \(\Delta u\le 0\) .
In one dimension the harmonic functions are the straight lines \(u=ax+b\) , and the names make sense there: \(u''\ge 0\) means convex, and a convex function lies below its chords. In higher dimensions the supply is much richer. In the plane, \(x^2-y^2\) , \(xy\) , and \(e^{x}\sin y\) are harmonic, as one checks by differentiating; so are the real and imaginary parts of every complex-analytic function. In every dimension, linear functions are harmonic and \(\lvert x\rvert^2\) is subharmonic with \(\Delta\lvert x\rvert^2=2n\) .
The Laplacian does not change under rotations, so it is natural to look for harmonic functions that depend only on \(r=\lvert x\rvert\) . For \(u(x)=v(r)\) we have \(\partial r/\partial x_i=x_i/r\) , hence \(u_{x_i}=v'(r)x_i/r\) and
Summing over \(i\) and using \(\sum x_i^2=r^2\) ,
Setting this to zero gives \((r^{n-1}v')'=0\) , so \(v'=C/r^{n-1}\) and
These are all the radial harmonic functions on a region that excludes the origin. Unless \(B=0\) they are singular at the origin, and that singularity is the starting point of Chapter 4.
Example 3.1: Steady temperature between two concentric surfaces
The region between the circles \(r=1\) and \(r=2\) in the plane has its inner edge held at temperature \(0\) and its outer edge at temperature \(1\) . The data are radial, so try \(u=A+B\log r\) . The conditions \(A=0\) and \(A+B\log 2=1\) give
The same problem between the spheres \(r=1\) and \(r=2\) in \(\mathbb R^3\) uses \(u=A+B/r\) with \(A+B=0\) and \(A+B/2=1\) , so \(A=2\) , \(B=-2\) , and
Both solutions increase from \(0\) to \(1\) across the gap, and both lie above the straight-line profile \(r-1\) , which gives \(0.5\) at the midpoint: the temperature rises faster near the inner surface, because the same total heat flow must cross a smaller area there. That these are the only solutions, radial or not, follows from the uniqueness theorem of Section 3.4.
3.2 Green’s identities
Every integral argument about the Laplacian rests on one computation. Let \(\Omega\) be a bounded open set whose boundary is piecewise \(C^1\) , so that the divergence theorem applies, and let \(u,v\) be functions on \(\overline\Omega\) . Apply the divergence theorem to the vector field \(v\nabla u\) , whose divergence is \(\nabla v\cdot\nabla u+v\Delta u\) .
Theorem 3.2 (Green’s identities). Let \(\Omega\) be a bounded open set with piecewise \(C^1\) boundary and outward unit normal \(\nu\) . If \(u\in C^2(\overline\Omega)\) and \(v\in C^1(\overline\Omega)\) , then
If both \(u\) and \(v\) are in \(C^2(\overline\Omega)\) , then
Proof. The first identity is the divergence theorem for the \(C^1\) field \(v\nabla u\) , with \(\nabla u\cdot\nu=\partial u/\partial\nu\) on the boundary. For the second, write the first identity with the roles of \(u\) and \(v\) exchanged and subtract; the term \(\int_\Omega\nabla u\cdot\nabla v\,dx\) cancels.
The first identity is integration by parts in several variables. Taking \(v=1\) in it gives \(\int_\Omega\Delta u\,dx=\int_{\partial\Omega}\partial u/\partial\nu\,dS\) : for a harmonic function the total flux through any closed surface is zero, which is the conservation law of Section 1.2 for a steady state without sources.
3.3 The mean value property
Theorem 3.3 (mean value property). Let \(u\) be harmonic in an open set \(\Omega\subseteq\mathbb R^n\) and let \(\overline{B_r(x)}\subset\Omega\) . Then \(u(x)\) equals the average of \(u\) over the sphere \(\partial B_r(x)\) and also its average over the ball \(B_r(x)\) :
Proof. For \(0<\rho\le r\) define the spherical average, written as an integral over the unit sphere by the substitution \(y=x+\rho z\) :
Since \(u\in C^2\) on the compact ball, we may differentiate under the integral sign:
because \(z\) is the outward unit normal to \(\partial B_\rho(x)\) at the point \(x+\rho z\) . By the first Green identity with \(v=1\) the last integral equals \(\int_{B_\rho(x)}\Delta u\,dy=0\) . So \(\varphi\) is constant on \((0,r]\) . As \(\rho\to 0\) , continuity of \(u\) at \(x\) gives \(\varphi(\rho)\to u(x)\) . Hence \(\varphi(r)=u(x)\) , which is the statement for spheres.
For the ball, integrate over spheres: \(\int_{B_r(x)}u\,dy=\int_0^r\Bigl(\int_{\partial B_\rho(x)}u\,dS\Bigr)d\rho=\int_0^r\omega_n\rho^{n-1}u(x)\,d\rho=\lvert B_r\rvert\,u(x)\) .
The proof used the formula \(\varphi'(\rho)=\dfrac{1}{\lvert\partial B_\rho\rvert}\displaystyle\int_{B_\rho(x)}\Delta u\,dy\) , which holds for every \(C^2\) function, harmonic or not. It shows at once what happens otherwise: for a subharmonic function \(\varphi\) is nondecreasing, so the value at the centre is at most the average over any sphere. It also gives the converse of the theorem.
Theorem 3.4 (converse of the mean value property). If \(u\in C^2(\Omega)\) and \(u(x)\) equals the average of \(u\) over \(\partial B_r(x)\) for every ball with \(\overline{B_r(x)}\subset\Omega\) , then \(u\) is harmonic in \(\Omega\) .
Proof. Suppose \(\Delta u(x_0)>0\) for some \(x_0\) ; the opposite sign is treated by replacing \(u\) with \(-u\) . By continuity \(\Delta u>0\) on a ball \(B_r(x_0)\) with closure in \(\Omega\) . Then \(\varphi'(\rho)>0\) for \(0<\rho<r\) by the formula above, so \(\varphi(r)>\lim_{\rho\to 0}\varphi(\rho)=u(x_0)\) , contradicting the hypothesis.
So for \(C^2\) functions, being harmonic and being equal to one’s spherical averages are the same thing. The second description involves no derivatives at all, and it is the form in which the intuition about hot spots becomes a proof.
Example 3.2: Averages that need no integration
The function \(u(x,y)=e^{x}\cos y\) is harmonic in the plane. Its average over the unit circle about the origin is
an integral with no elementary antiderivative. By Theorem 3.3 its value is \(u(0,0)=1\) ; numerical quadrature confirms \(1.000000\) . Likewise the average of \(u\) over the circle of radius \(2\) about the point \((1,0)\) is \(u(1,0)=e\approx 2.71828\) .
For contrast take \(v(x,y)=x^2\) , with \(\Delta v=2\) . On the circle of radius \(\rho\) about the origin \(v=\rho^2\cos^2\theta\) , with average \(\varphi(\rho)=\rho^2/2\) . This exceeds the centre value \(v(0,0)=0\) , as it must for a subharmonic function, and \(\varphi'(\rho)=\rho\) agrees with the general formula: \(\dfrac{1}{2\pi\rho}\displaystyle\int_{B_\rho}2\,dy=\dfrac{2\pi\rho^2}{2\pi\rho}=\rho\) .
3.4 Maximum principles
Theorem 3.5 (strong maximum principle). Let \(\Omega\subseteq\mathbb R^n\) be open and connected and let \(u\) be harmonic in \(\Omega\) . If \(u\) attains its supremum over \(\Omega\) at a point of \(\Omega\) , then \(u\) is constant. The same holds for the infimum.
Proof. Let \(M=\sup_\Omega u\) be attained at \(x_0\in\Omega\) and put \(S=\{x\in\Omega:u(x)=M\}\) . Then \(S\) is nonempty, and it is closed relative to \(\Omega\) because \(u\) is continuous. It is also open. Indeed, let \(x\in S\) and choose \(r>0\) with \(\overline{B_r(x)}\subset\Omega\) . By the mean value property for the ball,
and the integrand is continuous and nonnegative. If it were positive at some point it would be positive on a small ball around that point and the integral would be positive. So \(u=M\) on all of \(B_r(x)\) , that is, \(B_r(x)\subseteq S\) . A connected open set cannot be divided into two disjoint nonempty relatively open subsets, so \(S=\Omega\) . For the infimum apply the result to \(-u\) .
The argument is the physical one. A point where \(u\) is maximal has a value equal to the average of values that are no larger; so they are all equal to it, and the set where the maximum is taken spreads until it fills the region.
For a bounded region the theorem locates the extreme values.
Theorem 3.6 (weak maximum principle). Let \(\Omega\subset\mathbb R^n\) be open and bounded and let \(u\in C^2(\Omega)\cap C(\overline\Omega)\) .
(a) If \(\Delta u\ge 0\) in \(\Omega\) , then \(\max_{\overline\Omega}u=\max_{\partial\Omega}u\) .
(b) If \(\Delta u\le 0\) in \(\Omega\) , then \(\min_{\overline\Omega}u=\min_{\partial\Omega}u\) .
In particular a harmonic function satisfies \(\min_{\partial\Omega}u\le u(x)\le\max_{\partial\Omega}u\) for all \(x\in\Omega\) .
Proof. The set \(\overline\Omega\)
is closed and bounded, hence compact (math.real-analysis.heine-borel), so the continuous function \(u\)
attains a maximum on it. Suppose first that \(\Delta u>0\)
throughout \(\Omega\)
. If the maximum were attained at an interior point \(x_0\)
, then for each \(i\)
the function \(s\mapsto u(x_0+se_i)\)
of one variable would have a local maximum at \(s=0\)
, so \(u_{x_ix_i}(x_0)\le 0\)
by the second derivative test; summing, \(\Delta u(x_0)\le 0\)
, a contradiction. So the maximum is attained on \(\partial\Omega\)
.
In the general case \(\Delta u\ge 0\) , choose \(R\) with \(\lvert x\rvert\le R\) on \(\overline\Omega\) and for \(\varepsilon>0\) put \(u_\varepsilon(x)=u(x)+\varepsilon\lvert x\rvert^2\) . Then \(\Delta u_\varepsilon=\Delta u+2n\varepsilon>0\) , so by the first case
Letting \(\varepsilon\to 0\) gives \(\max_{\overline\Omega}u\le\max_{\partial\Omega}u\) , and the reverse inequality is trivial. Part (b) follows by applying (a) to \(-u\) .
This second proof does not use the mean value property. It uses only the sign of the second derivatives at a maximum, and so it carries over to equations with variable coefficients and, in Chapter 5, to the heat equation, where no mean value property over balls is available. Notice the direction of the inequalities: a subharmonic function has its maximum on the boundary. Getting this backwards is the most common error in applying the principle.
The Dirichlet problem for the Poisson equation asks for \(u\) with
The minus sign is a convention that will earn its keep later; with it, a positive source \(f\) produces a positive solution.
Theorem 3.7 (uniqueness and continuous dependence). Let \(\Omega\) be open and bounded. The Dirichlet problem has at most one solution \(u\in C^2(\Omega)\cap C(\overline\Omega)\) . If \(u_1\) and \(u_2\) are solutions with the same \(f\) and boundary data \(g_1\) and \(g_2\) , then
Proof. The difference \(w=u_1-u_2\) is harmonic in \(\Omega\) , continuous on \(\overline\Omega\) , and equal to \(g_1-g_2\) on the boundary. By Theorem 3.6, \(w\le\max_{\partial\Omega}(g_1-g_2)\) and \(w\ge\min_{\partial\Omega}(g_1-g_2)\) everywhere, which gives the equality of the two maxima. If \(g_1=g_2\) then \(w=0\) .
Together with an existence theorem this is well-posedness in the sense of Definition 1.4, with sizes measured by the maximum. Compare Example 1.2: there, boundary data of size \(1/n\) on part of the boundary produced a solution of enormous size; here, data prescribed on the whole boundary control the solution completely.
Theorem 3.8 (a priori bound for the Poisson equation). Let \(\Omega\) be open with \(\Omega\subseteq B_R(0)\) , and let \(u\in C^2(\Omega)\cap C(\overline\Omega)\) solve \(-\Delta u=f\) in \(\Omega\) with \(u=g\) on \(\partial\Omega\) , where \(f\) is bounded. Then
Proof. Put \(K=\sup_\Omega\lvert f\rvert\) and \(w=u+\dfrac{K}{2n}\lvert x\rvert^2\) . Then \(\Delta w=-f+K\ge 0\) , so by Theorem 3.6(a), for every \(x\in\overline\Omega\) ,
The same argument applied to \(-u\) , which solves the problem with data \(-f\) and \(-g\) , bounds \(-u\) by the same quantity.
The function \(\tfrac{K}{2n}\lvert x\rvert^2\) in this proof is a barrier: an explicit function whose Laplacian dominates the source. Choosing barriers adapted to the region gives sharper information, as the second example below shows.
Example 3.3: Reading extremes off the boundary
Let \(u\) be the harmonic function in the unit disc with boundary values \(g(\theta)=\cos^2\theta\) . Without solving anything, Theorem 3.6 gives \(0\le u\le 1\) in the disc, and Theorem 3.5 sharpens this to \(0<u<1\) at interior points, since \(u\) is not constant. The mean value property gives the centre value: \(u(0,0)=\frac{1}{2\pi}\int_0^{2\pi}\cos^2\theta\,d\theta=\tfrac12\) .
In this case the solution can also be written down. Since \(\cos^2\theta=\tfrac12+\tfrac12\cos 2\theta\) and \(r^2\cos 2\theta=x^2-y^2\) is harmonic,
which equals \(\tfrac12+\tfrac12\cos2\theta\) on \(r=1\) . Its maximum over the closed disc is \(1\) , attained only at the boundary points \((\pm1,0)\) ; its minimum is \(0\) , attained only at \((0,\pm1)\) ; and \(u(0,0)=\tfrac12\) . The gradient \(\nabla u=(x,-y)\) vanishes at the origin, but the origin is a saddle point of \(u\) and not an extremum. A harmonic function may have critical points inside the region. It cannot have a strict local maximum or minimum there.
Example 3.4: Bounding a solution that is not known explicitly
Let \(u\) solve \(-\Delta u=1\) in the unit square \(\Omega=(0,1)\times(0,1)\) with \(u=0\) on the boundary. This models the temperature of a uniformly heated plate with its edge held at zero. There is no closed formula for \(u\) . How large is it at the centre?
Upper bound. The function \(w(x,y)=\tfrac12x(1-x)\) satisfies \(-\Delta w=1\) and \(w\ge 0\) on \(\partial\Omega\) . So \(w-u\) is harmonic and nonnegative on the boundary, and by Theorem 3.6, \(u\le w\) in \(\Omega\) . At the centre, \(u(\tfrac12,\tfrac12)\le\tfrac18=0.125\) .
Lower bound. The function \(z(x,y)=x(1-x)\,y(1-y)\) vanishes on \(\partial\Omega\) and has
So \(-\Delta(u-z)\ge 0\) : the function \(u-z\) is superharmonic and vanishes on the boundary, and by Theorem 3.6(b) its minimum is zero. Hence \(u\ge z\) in \(\Omega\) , and \(u(\tfrac12,\tfrac12)\ge\tfrac1{16}=0.0625\) .
So \(0.0625\le u(\tfrac12,\tfrac12)\le 0.125\) from two lines of calculus. The lower barrier also shows \(u>0\) throughout the square. The true value, computed from a Fourier series by the methods of Chapter 7, is approximately \(0.0737\) , comfortably inside the bracket.
3.5 Uniqueness by energy and Dirichlet’s principle
There is a second, entirely different route to uniqueness. It asks more smoothness of the solution up to the boundary, but it handles boundary conditions that the maximum principle does not, and it is the idea from which the weak theory of Chapters 9 and 10 grows.
Theorem 3.9 (uniqueness by the energy method). Let \(\Omega\) be a bounded, connected open set with piecewise \(C^1\) boundary, and let \(u_1,u_2\in C^2(\overline\Omega)\) both satisfy \(-\Delta u=f\) in \(\Omega\) .
(a) If \(u_1=u_2\) on \(\partial\Omega\) , then \(u_1=u_2\) in \(\Omega\) .
(b) If \(\partial u_1/\partial\nu=\partial u_2/\partial\nu\) on \(\partial\Omega\) , then \(u_1-u_2\) is constant in \(\Omega\) .
Proof. Let \(w=u_1-u_2\) , so \(\Delta w=0\) . The first Green identity with \(u=v=w\) reads
In case (a) \(w=0\) on the boundary; in case (b) \(\partial w/\partial\nu=0\) there. Either way the boundary integral vanishes, so \(\int_\Omega\lvert\nabla w\rvert^2dx=0\) . The integrand is continuous and nonnegative, so \(\nabla w=0\) everywhere, and \(w\) is constant on the connected set \(\Omega\) . In case (a) the constant is the boundary value, zero.
The quantity \(\int_\Omega\lvert\nabla w\rvert^2dx\) is called the Dirichlet energy of \(w\) . Part (b) is the best possible statement for the Neumann problem, since adding a constant to a solution gives another solution, as Exercise 1.18 showed. Harmonic functions are characterized by this energy.
Theorem 3.10 (Dirichlet’s principle). Let \(\Omega\) be as in Theorem 3.9, let \(u\in C^2(\overline\Omega)\) be harmonic in \(\Omega\) , and let \(v\in C^1(\overline\Omega)\) have the same boundary values as \(u\) . Then
with equality only if \(v=u\) .
Proof. Write \(v=u+w\) with \(w\in C^1(\overline\Omega)\) and \(w=0\) on \(\partial\Omega\) . Expanding the square,
By the first Green identity the middle integral equals \(\int_{\partial\Omega}w\,\partial u/\partial\nu\,dS-\int_\Omega w\,\Delta u\,dx=0\) . The last integral is nonnegative, and it is zero only if \(\nabla w=0\) , which with \(w=0\) on the boundary forces \(w=0\) .
Among all functions with given boundary values, the harmonic one has the least energy. Notice what the proof shows and what it does not. It shows that if a smooth harmonic function with the given boundary values exists, it is the minimizer. It does not show that a minimizer exists. Nineteenth-century mathematicians, Riemann among them, took existence for granted, and Weierstrass pointed out that an infimum need not be attained. Repairing the argument requires a space of functions in which minimizing sequences converge, and that is the story of Chapter 9.
Example 3.5: The cost of not being harmonic
On the unit disc take the boundary values \(g=\cos\theta\) . The harmonic function with these values is \(u=x\) , with \(\lvert\nabla u\rvert^2=1\) and Dirichlet energy \(\int_\Omega\lvert\nabla u\rvert^2dx=\pi\) .
Every function \(v_c=x+c\,x(1-x^2-y^2)\) has the same boundary values, because the added term vanishes on \(r=1\) . With \(w=x(1-x^2-y^2)\) we have \(\nabla w=(1-3x^2-y^2,\,-2xy)\) , and integrating in polar coordinates gives \(\int_\Omega\lvert\nabla w\rvert^2dx=\tfrac{2\pi}{3}\) , while the cross term \(\int_\Omega\nabla u\cdot\nabla w\,dx=\int_\Omega(1-3x^2-y^2)\,dx=\pi-\tfrac{3\pi}{4}-\tfrac{\pi}{4}=0\) , as the theorem predicts. Therefore
smallest exactly when \(c=0\) . Any departure from the harmonic function, in either direction, is paid for in energy, and the price is quadratic in the size of the departure.
3.6 What can go wrong on unbounded regions
The mean value property has a striking consequence for functions harmonic in all of space.
Theorem 3.11 (Liouville). A bounded harmonic function on \(\mathbb R^n\) is constant.
Proof. Let \(\lvert u\rvert\le M\) and fix two points \(x,y\) at distance \(d=\lvert x-y\rvert\) . For \(R>d\) the mean value property over balls gives
The contributions of the common part \(B_R(x)\cap B_R(y)\) cancel. What remains is an integral over the part of each ball outside the other, and both of those parts lie in the shell \(\{z:R-d\le\lvert z-x\rvert\le R+d\}\) : a point of \(B_R(y)\) is within \(R+d\) of \(x\) , and a point within \(R-d\) of \(x\) is within \(R\) of \(y\) . Hence
The right side tends to zero as \(R\to\infty\) , so \(u(x)=u(y)\) .
Liouville’s theorem is a uniqueness statement for the whole space, and the hypothesis of boundedness cannot be dropped: every linear function is harmonic. The same caution applies to the maximum principle.
Common mistake. Applying the weak maximum principle on an unbounded region. The function \(u(x,y)=y\) is harmonic in the upper half-plane \(y>0\) , continuous on its closure, and zero on the boundary line \(y=0\) . It is not zero. Theorem 3.6 assumes that \(\Omega\) is bounded, and the proof used this twice: to obtain a maximum on the compact set \(\overline\Omega\) and to bound \(\lvert x\rvert^2\) . On an unbounded region the behaviour at infinity is part of the boundary data. Uniqueness is restored by a growth condition; for instance, a bounded harmonic function in the half-plane that vanishes on the boundary line is zero. Before invoking the maximum principle, check that the region is bounded and that the function is continuous up to the boundary.
The Laplace equation has now been shown to have at most one solution for given boundary values, and that solution has been located between the extremes of its data, all without a formula. The next chapter supplies the formula. The radial solutions \(\log r\) and \(r^{2-n}\) of Section 3.1 were set aside because of their singularity at the origin. What does the Laplacian of such a function look like if the singular point is not excluded?
Exercises
-
Decide which of these functions are harmonic in the region indicated, showing the computation: (a) \(x^3-3xy^2\) in \(\mathbb R^2\) ; (b) \(e^{-x}\sin y\) in \(\mathbb R^2\) ; (c) \(\log(x^2+y^2)\) in \(\mathbb R^2\) without the origin; (d) \((x^2+y^2)^{-1/2}\) in \(\mathbb R^2\) without the origin; (e) \((x^2+y^2+z^2)^{-1/2}\) in \(\mathbb R^3\) without the origin.
Answer
(a) Harmonic: \(u_{xx}=6x\) , \(u_{yy}=-6x\) . (b) Harmonic: \(u_{xx}=u\) , \(u_{yy}=-u\) . (c) Harmonic: it is \(2\log r\) , a radial solution for \(n=2\) . (d) Not harmonic in the plane: with \(v=r^{-1}\) and \(n=2\) , \(v''+v'/r=2r^{-3}-r^{-3}=r^{-3}\ne 0\) . (e) Harmonic: \(r^{2-n}\) with \(n=3\) .
-
Using \(x=r\cos\theta\) and \(y=r\sin\theta\) , show that the Laplacian in polar coordinates is \(u_{rr}+\tfrac1ru_r+\tfrac{1}{r^2}u_{\theta\theta}\) , and deduce that \(r^k\cos k\theta\) and \(r^k\sin k\theta\) are harmonic for every positive integer \(k\) .
Hint
Write \(u_r=u_x\cos\theta+u_y\sin\theta\) and \(u_\theta=-u_xr\sin\theta+u_yr\cos\theta\) , differentiate each once more, and combine \(u_{rr}+\tfrac1{r^2}u_{\theta\theta}\) ; the mixed terms cancel and \(-\tfrac1ru_r\) is left over. For \(u=r^k\cos k\theta\) the three terms are \(k(k-1)r^{k-2}\cos k\theta\) , \(kr^{k-2}\cos k\theta\) , and \(-k^2r^{k-2}\cos k\theta\) , which sum to zero.
-
Find the steady temperature between the spheres \(r=1\) and \(r=3\) in \(\mathbb R^3\) when the inner sphere is held at temperature \(1\) and the outer at \(0\) , evaluate it at \(r=2\) , and compute the total heat flux \(-\int\partial u/\partial\nu\,dS\) through each of the two spheres, with \(\nu\) pointing away from the origin, to confirm that they agree.
Answer
\(u=A+B/r\) with \(A+B=1\) and \(A+B/3=0\) , so \(u(r)=\dfrac{3}{2r}-\dfrac12\) and \(u(2)=\tfrac14\) . Since \(\partial u/\partial r=-\tfrac{3}{2r^2}\) , the flux through the sphere of radius \(r\) is \(\tfrac{3}{2r^2}\cdot4\pi r^2=6\pi\) , the same for \(r=1\) and \(r=3\) : no heat is created or lost in between.
-
Without evaluating an integral, find the average of \(u(x,y)=x^3-3xy^2+5\) over the circle of radius \(2\) centred at \((1,1)\) and over the disc it bounds, and justify the answer.
Answer
\(x^3-3xy^2\) is harmonic and so is the constant \(5\) , so \(u\) is harmonic in the whole plane. By Theorem 3.3 both averages equal the value at the centre, \(u(1,1)=1-3+5=3\) .
-
Let \(u\in C^2(\Omega)\) be subharmonic and \(\overline{B_r(x)}\subset\Omega\) . Prove that \(u(x)\) is at most the average of \(u\) over \(\partial B_r(x)\) and at most its average over \(B_r(x)\) .
Solution
Use the spherical average \(\varphi(\rho)\) from the proof of Theorem 3.3, defined for \(0<\rho\le r\) . The computation there, which did not use harmonicity until its last step, gives
\[ \varphi'(\rho)=\frac{1}{\lvert\partial B_\rho\rvert}\int_{B_\rho(x)}\Delta u\,dy\ \ge 0 \]because \(\Delta u\ge 0\) . So \(\varphi\) is nondecreasing on \((0,r]\) , and since \(\varphi(\rho)\to u(x)\) as \(\rho\to 0\) by continuity, \(u(x)\le\varphi(\rho)\) for every \(\rho\in(0,r]\) . With \(\rho=r\) this is the statement for the sphere. For the ball, integrate the inequality \(u(x)\le\varphi(\rho)\) against \(\omega_n\rho^{n-1}d\rho\) from \(0\) to \(r\) :
\[ \lvert B_r\rvert\,u(x)\le\int_0^r\omega_n\rho^{n-1}\varphi(\rho)\,d\rho=\int_0^r\int_{\partial B_\rho(x)}u\,dS\,d\rho=\int_{B_r(x)}u\,dy . \]Dividing by \(\lvert B_r\rvert\) gives the claim. A subharmonic function lies below its averages, just as a convex function of one variable lies below its chords.
-
Compute the average of the harmonic polynomial \(u=x^4-6x^2y^2+y^4\) over the square \([-1,1]\times[-1,1]\) and compare it with \(u(0,0)\) . What does the result say about the role of balls in Theorem 3.3?
Answer
\(\int_{-1}^1\int_{-1}^1x^4\,dx\,dy=\tfrac45\) , the same for \(y^4\) , and \(\int\int x^2y^2=\tfrac49\) , so the integral of \(u\) is \(\tfrac85-\tfrac83=-\tfrac{16}{15}\) and the average over the square of area \(4\) is \(-\tfrac4{15}\) , not \(u(0,0)=0\) . The mean value property is a statement about balls and spheres; it relies on the rotational symmetry of the Laplacian and fails for other shapes.
-
Find the maximum and minimum of \(u=xy\) on the closed unit disc and all points where they are attained. Then find the critical point of \(u\) in the open disc and say what kind of critical point it is.
Answer
\(u\) is harmonic, so its extremes over the closed disc are on the circle, where \(u=\tfrac12\sin2\theta\) . The maximum \(\tfrac12\) is attained at \(\pm(1/\sqrt2,1/\sqrt2)\) and the minimum \(-\tfrac12\) at \(\pm(1/\sqrt2,-1/\sqrt2)\) . The only critical point is the origin, where \(\nabla u=(y,x)=0\) ; it is a saddle.
-
Let \(u\) be harmonic in the unit disc and continuous on its closure, with boundary values \(g(\theta)=\lvert\sin\theta\rvert\) . Prove that \(0<u<1\) at every interior point and find \(u(0,0)\) exactly.
Solution
The boundary values satisfy \(0\le g\le 1\) , with \(g=0\) at \(\theta=0,\pi\) and \(g=1\) at \(\theta=\pm\pi/2\) . By the weak maximum principle, Theorem 3.6, \(0\le u\le 1\) on the closed disc. Suppose \(u(x_0)=1\) at an interior point \(x_0\) . Then \(u\) attains its supremum over the open disc, which is connected, at a point of it, and by the strong maximum principle, Theorem 3.5, \(u\) is constant in the disc; by continuity it would be constant on the closed disc, contradicting the fact that \(g\) takes both values \(0\) and \(1\) . The same argument with the infimum excludes \(u(x_0)=0\) . Hence \(0<u<1\) inside.
For the centre value use the mean value property on the circles of radius \(\rho<1\) and let \(\rho\to1\) ; since \(u\) is uniformly continuous on the closed disc, the averages converge to the average of \(g\) :
\[ u(0,0)=\frac{1}{2\pi}\int_0^{2\pi}\lvert\sin\theta\rvert\,d\theta=\frac{1}{2\pi}\cdot4=\frac{2}{\pi}\approx0.637 . \] -
Prove directly from Theorem 3.6 that if \(u,v\in C^2(\Omega)\cap C(\overline\Omega)\) on a bounded open set satisfy \(-\Delta u\le-\Delta v\) in \(\Omega\) and \(u\le v\) on \(\partial\Omega\) , then \(u\le v\) in \(\Omega\) . This is the comparison principle.
Hint
Put \(w=u-v\) . Then \(\Delta w=\Delta u-\Delta v\ge 0\) , so \(w\) is subharmonic, and \(w\le 0\) on \(\partial\Omega\) . Apply Theorem 3.6(a).
-
A student argues: “Let \(-\Delta u=f\) in \(\Omega\) with \(f>0\) and \(u=0\) on \(\partial\Omega\) . By the maximum principle the maximum of \(u\) is on the boundary, so \(u\le 0\) in \(\Omega\) .” Find the error, state the correct conclusion, and confirm it on the example \(u=(1-\lvert x\rvert^2)/(2n)\) in the unit ball.
Solution
The error is in the direction of the principle. The hypothesis \(-\Delta u=f>0\) means \(\Delta u<0\) : the function is superharmonic. Theorem 3.6(a), which puts the maximum on the boundary, requires \(\Delta u\ge 0\) and does not apply. What does apply is part (b): a superharmonic function attains its minimum over \(\overline\Omega\) on the boundary. Since \(u=0\) there, the correct conclusion is \(u\ge 0\) in \(\Omega\) , the opposite of the student’s claim. This is also what the physics says: a body heated throughout, with its surface held at zero, is warmer inside.
For the example, \(u=(1-\lvert x\rvert^2)/(2n)\) vanishes on the unit sphere, and \(\Delta\lvert x\rvert^2=2n\) gives \(-\Delta u=1>0\) . Inside the ball \(\lvert x\rvert<1\) , so \(u>0\) , with maximum \(1/(2n)\) at the centre. The maximum is in the interior, which a subharmonic function could never have.
-
Let \(-\Delta u=f\) in the unit disc of the plane with \(u=0\) on the boundary and \(0\le f\le 1\) . Use the comparison principle of Exercise 9 with a radial barrier to prove \(0\le u(x)\le\tfrac14(1-\lvert x\rvert^2)\) .
Solution
The radial barrier is \(v(x)=\tfrac14(1-\lvert x\rvert^2)\) . In the plane \(\Delta\lvert x\rvert^2=4\) , so \(-\Delta v=1\) , and \(v=0\) on the unit circle.
Upper bound. In the disc \(-\Delta u=f\le 1=-\Delta v\) , and on the circle \(u=0=v\) . The comparison principle of Exercise 3.9, with the pair \((u,v)\) , gives \(u\le v\) , that is, \(u(x)\le\tfrac14(1-\lvert x\rvert^2)\) . Spelled out: \(w=u-v\) has \(\Delta w=-f+1\ge 0\) , so \(w\) is subharmonic, vanishes on the boundary, and by Theorem 3.6(a) is at most zero.
Lower bound. The zero function satisfies \(-\Delta 0=0\le f=-\Delta u\) and equals \(u\) on the circle. The comparison principle with the pair \((0,u)\) gives \(0\le u\) .
Both bounds are attained: \(f=0\) gives \(u=0\) and \(f=1\) gives \(u=v\) . In particular \(u(0)\le\tfrac14\) , and the bound uses nothing about \(f\) beyond its range.
-
Let \(\Omega\) be as in Theorem 3.9 and \(\alpha>0\) a constant. Prove by the energy method that the Robin problem \(-\Delta u=f\) in \(\Omega\) , \(\partial u/\partial\nu+\alpha u=g\) on \(\partial\Omega\) has at most one solution in \(C^2(\overline\Omega)\) . Where does the proof break if \(\alpha<0\) ?
Solution
Let \(u_1,u_2\in C^2(\overline\Omega)\) be two solutions and \(w=u_1-u_2\) . Then \(\Delta w=0\) in \(\Omega\) and \(\partial w/\partial\nu+\alpha w=0\) on \(\partial\Omega\) , so \(\partial w/\partial\nu=-\alpha w\) there. Green’s first identity with \(u=v=w\) gives
\[ \int_\Omega\lvert\nabla w\rvert^2\,dx=\int_{\partial\Omega}w\,\frac{\partial w}{\partial\nu}\,dS=-\alpha\int_{\partial\Omega}w^2\,dS . \]The left side is nonnegative and, because \(\alpha>0\) , the right side is nonpositive. So both are zero: \(\nabla w=0\) in \(\Omega\) , hence \(w\) is constant on the connected set \(\Omega\) , and \(\int_{\partial\Omega}w^2dS=0\) , hence that constant is zero. Thus \(u_1=u_2\) .
If \(\alpha<0\) the two sides have the same sign and the identity yields no contradiction; the argument gives nothing. In fact uniqueness can then fail. On the interval \((0,1)\) the outward normal derivative is \(-w'(0)\) at the left end and \(w'(1)\) at the right end. Take \(\alpha=-2\) and \(w=x-\tfrac12\) . Then \(w''=0\) , at \(x=0\) we get \(-w'(0)+\alpha w(0)=-1-2\cdot(-\tfrac12)=0\) , and at \(x=1\) we get \(w'(1)+\alpha w(1)=1-2\cdot\tfrac12=0\) : a nonzero solution of the homogeneous problem.
-
On the interval \((0,1)\) the one-dimensional harmonic function with \(u(0)=0\) , \(u(1)=1\) is \(u=x\) . Compute \(\int_0^1(v')^2dx\) for \(v=x+c\sin(\pi x)\) and verify Dirichlet’s principle for this family.
Answer
\(v'=1+c\pi\cos(\pi x)\) , so \(\int_0^1(v')^2dx=1+2c\pi\int_0^1\cos(\pi x)\,dx+c^2\pi^2\int_0^1\cos^2(\pi x)\,dx=1+\tfrac12\pi^2c^2\) . The cross term vanishes, as in Theorem 3.10, and the energy is least, equal to \(1\) , exactly when \(c=0\) .
-
Let \(u\) be harmonic on \(\mathbb R^n\) with \(\lvert u(x)\rvert\le C(1+\lvert x\rvert)^{1/2}\) for all \(x\) . Adapt the proof of Liouville’s theorem to show that \(u\) is constant.
Solution
Fix \(x,y\) with \(d=\lvert x-y\rvert\) and let \(R>d\) . As in the proof of Theorem 3.11,
\[ \lvert u(x)-u(y)\rvert\le\frac{1}{\lvert B_R\rvert}\int_{S_R}\lvert u(z)\rvert\,dz,\qquad S_R=\{z:R-d\le\lvert z-x\rvert\le R+d\}. \]The only change is that \(u\) is no longer bounded by a fixed constant. On the shell \(S_R\) we have \(\lvert z\rvert\le\lvert x\rvert+R+d\) , so \(\lvert u(z)\rvert\le C(1+\lvert x\rvert+R+d)^{1/2}\) . Therefore
\[ \lvert u(x)-u(y)\rvert\le C\,(1+\lvert x\rvert+R+d)^{1/2}\left[\Bigl(1+\frac dR\Bigr)^{n}-\Bigl(1-\frac dR\Bigr)^{n}\right]. \]For \(R\ge 2d\) the bracket is at most \(c_n\,d/R\) with a constant \(c_n\) depending only on \(n\) , by the mean value theorem applied to \(s\mapsto s^n\) on \([\tfrac12,\tfrac32]\) . So the right side is at most a constant times \(R^{1/2}\cdot R^{-1}=R^{-1/2}\) for large \(R\) , which tends to zero. Hence \(u(x)=u(y)\) . The same proof works for any growth bound \(C(1+\lvert x\rvert)^{\beta}\) with \(\beta<1\) ; for \(\beta=1\) it fails, as it must, since linear functions are harmonic.
-
Show that \(u=e^{x}\sin y\) is harmonic in the strip \(0<y<\pi\) , vanishes on both boundary lines, and is not zero. Which hypothesis of Theorem 3.7 fails, and how does \(u\) behave at infinity?
Answer
\(u_{xx}=u\) and \(u_{yy}=-u\) , and \(\sin0=\sin\pi=0\) . The strip is unbounded, so Theorem 3.7 does not apply: both \(u\) and the zero function are harmonic with zero boundary values. The function grows like \(e^{x}\) as \(x\to+\infty\) ; it is the missing condition at infinity that lets it exist.
-
On the rectangle \(0<x<\pi\) , \(0<y<1\) consider the harmonic function with boundary values \(\tfrac1n\sin(nx)\) on the side \(y=0\) and zero on the other three sides. Find it explicitly, verify that its maximum is \(1/n\) as Theorem 3.7 predicts, and contrast this with Example 1.2, where data of the same size \(1/n\) were imposed as initial conditions on the line \(y=0\) alone.
Solution
Try \(u=\tfrac1n\sin(nx)\,Y(y)\) . Harmonicity requires \(Y''=n^2Y\) , the side \(y=1\) requires \(Y(1)=0\) , and the side \(y=0\) requires \(Y(0)=1\) . The combination of \(e^{\pm ny}\) that vanishes at \(y=1\) is \(\sinh(n(1-y))\) , so
\[ u(x,y)=\frac1n\,\sin(nx)\,\frac{\sinh\bigl(n(1-y)\bigr)}{\sinh n}. \]It vanishes at \(x=0\) and \(x=\pi\) because \(\sin(nx)\) does. It is harmonic, continuous on the closed rectangle, and by Theorem 3.7 it is the only solution. Since \(0\le\sinh(n(1-y))\le\sinh n\) for \(0\le y\le1\) , we get \(\lvert u\rvert\le1/n\) with equality on the side \(y=0\) : the maximum is \(1/n\) , on the boundary, exactly as predicted.
In Example 1.2 data of size \(1/n\) on the line \(y=0\) , with nothing prescribed at \(y=1\) , produced the factor \(\sinh(ny)\) , which grows from \(y=0\) . Here the condition at \(y=1\) selects the factor \(\sinh(n(1-y))/\sinh n\) , which decays from \(y=0\) . Surrounding the region with boundary conditions removes the exponentially growing mode. That is the difference between an ill-posed and a well-posed problem for the same equation.
-
Let \(u\in C^2(\overline\Omega)\) be harmonic in a bounded region with piecewise \(C^1\) boundary. Use Green’s first identity to prove that \(\int_\Omega\lvert\nabla u\rvert^2dx=\int_{\partial\Omega}u\,\partial u/\partial\nu\,dS\) , and check the identity for \(u=x^2-y^2\) on the unit disc.
Answer
Green’s first identity with \(v=u\) gives \(\int_\Omega u\,\Delta u\,dx=-\int_\Omega\lvert\nabla u\rvert^2dx+\int_{\partial\Omega}u\,\partial u/\partial\nu\,dS\) , and the left side is zero. For \(u=x^2-y^2\) : \(\lvert\nabla u\rvert^2=4r^2\) , with integral \(4\cdot2\pi\cdot\tfrac14=2\pi\) over the disc; on the circle \(u=\cos2\theta\) and \(\partial u/\partial\nu=u_r=2\cos2\theta\) , with integral \(\int_0^{2\pi}2\cos^22\theta\,d\theta=2\pi\) .