Newton found that a point mass attracts with a force proportional to the inverse square of the distance, and Coulomb found the same law for a point charge. In the language of potentials both say that a unit point source at the origin of \(\mathbb R^3\) produces the potential \(1/(4\pi\lvert x\rvert)\) . An extended body is a collection of point sources, and since the governing equation is linear, its potential should be the sum of theirs:
Is this function really a solution of \(-\Delta u=f\) ? The question is delicate, because the function \(1/\lvert x\rvert\) being summed is harmonic wherever it is defined, so a careless differentiation under the integral sign gives \(\Delta u=0\) . The source term \(f\) has to come out of the singularity.
This chapter makes the superposition idea precise. It identifies the right singular solution in every dimension, proves that convolution with it solves the Poisson equation in the whole space, and then modifies it to respect a boundary, which produces Green’s function and the explicit Poisson formulas for a half-space and a ball.
4.1 The potential of a point source
Section 3.1 found the radial harmonic functions: \(A+B\log r\) in the plane and \(A+Br^{2-n}\) in \(\mathbb R^n\) for \(n\ge 3\) . We fix the constants so that the total flux out of the origin is one.
Definition 4.1 (fundamental solution). The fundamental solution of \(-\Delta\) in \(\mathbb R^n\) , \(n\ge 2\) , is the function defined for \(x\ne 0\) by
In particular \(\Phi(x)=\dfrac{1}{4\pi\lvert x\rvert}\) in \(\mathbb R^3\) .
Lemma 4.2 (three properties of the fundamental solution). (a) \(\Phi\) is harmonic in \(\mathbb R^n\setminus\{0\}\) . (b) Its gradient is \(\nabla\Phi(x)=-\dfrac{1}{\omega_n}\dfrac{x}{\lvert x\rvert^{n}}\) ; consequently on every sphere \(\partial B_r(0)\) , with \(\nu\) the normal pointing away from the origin, \(\partial\Phi/\partial\nu=-1/(\omega_nr^{n-1})\) and
(c) \(\Phi\) is integrable near the origin: for \(0<\varepsilon<1\) , \(\int_{B_\varepsilon(0)}\lvert\Phi\rvert\,dx\) equals \(\tfrac12\varepsilon^2\lvert\log\varepsilon\rvert+\tfrac14\varepsilon^2\) when \(n=2\) and \(\dfrac{\varepsilon^2}{2(n-2)}\) when \(n\ge 3\) .
Proof. (a) was shown in Section 3.1. (b) Writing \(\Phi(x)=v(r)\) , we have \(v'(r)=-1/(\omega_nr^{n-1})\) in both cases, since \(\omega_2=2\pi\) , and \(\nabla\Phi=v'(r)\,x/r\) . On the sphere \(\nu=x/r\) , so \(\partial\Phi/\partial\nu=v'(r)\) , and the sphere has area \(\omega_nr^{n-1}\) . (c) In polar coordinates \(\int_{B_\varepsilon}\lvert\Phi\rvert\,dx=\int_0^\varepsilon\lvert v(r)\rvert\,\omega_nr^{n-1}\,dr\) . For \(n\ge 3\) this is \(\int_0^\varepsilon\frac{r}{n-2}\,dr\) ; for \(n=2\) it is \(\int_0^\varepsilon r\lvert\log r\rvert\,dr\) , and both integrals are elementary.
Property (b) is the normalization. Think of \(-\nabla\Phi\) as a flux, in the sense of Section 1.2. The same amount, one unit per unit time, crosses every sphere about the origin, however small. Nothing is created away from the origin, because \(\Phi\) is harmonic there. So all of it is created at the origin: \(\Phi\) is the steady response to a unit point source. Property (c) says that the singularity, although infinite, is mild enough to integrate.
4.2 Solving the Poisson equation in the whole space
Theorem 4.3 (the Newtonian potential solves the Poisson equation). Let \(n\ge 2\) and let \(f\in C^2(\mathbb R^n)\) vanish outside a bounded set. Then
belongs to \(C^2(\mathbb R^n)\) and satisfies \(-\Delta u=f\) in \(\mathbb R^n\) .
Proof. Substituting \(y\mapsto x-y\) gives \(u(x)=\int\Phi(y)f(x-y)\,dy\) , which puts the dependence on \(x\) into the smooth factor. Since \(\Phi\) is integrable on bounded sets and the second derivatives of \(f\) are bounded, uniformly continuous, and vanish outside a bounded set, we may differentiate twice under the integral sign, and
is a continuous function of \(x\) . Fix \(x\) and \(\varepsilon>0\) and split the integral into the part \(I_\varepsilon\) over \(B_\varepsilon(0)\) and the part \(J_\varepsilon\) over its complement.
The part near the singularity. By Lemma 4.2(c), \(\lvert I_\varepsilon\rvert\le\max\lvert\Delta f\rvert\int_{B_\varepsilon}\lvert\Phi\rvert\,dy\to 0\) as \(\varepsilon\to 0\) .
The part away from it. Note \(\Delta_xf(x-y)=\Delta_yf(x-y)\) . Choose \(R\) so large that \(y\mapsto f(x-y)\) vanishes near \(\partial B_R(0)\) and outside it, and apply Green’s second identity on the region \(U=B_R(0)\setminus\overline{B_\varepsilon(0)}\) to the functions \(\Phi\) and \(y\mapsto f(x-y)\) . Since \(\Delta\Phi=0\) in \(U\) and nothing contributes on \(\partial B_R\) ,
where \(\nu\) is the unit normal pointing out of \(U\) , hence toward the origin. The first term is bounded by \(\max\lvert\nabla f\rvert\cdot\lvert v(\varepsilon)\rvert\cdot\omega_n\varepsilon^{n-1}\) , which is a constant times \(\varepsilon\lvert\log\varepsilon\rvert\) or \(\varepsilon\) and tends to zero. For the second, the normal points inward, so by Lemma 4.2(b) \(\partial\Phi/\partial\nu=+1/(\omega_n\varepsilon^{n-1})\) on \(\partial B_\varepsilon(0)\) , and
This is minus the average of \(f\) over the sphere of radius \(\varepsilon\) about \(x\) , which tends to \(-f(x)\) because \(f\) is continuous. Letting \(\varepsilon\to 0\) in \(\Delta u(x)=I_\varepsilon+J_\varepsilon\) gives \(\Delta u(x)=-f(x)\) .
The proof shows where the source comes from. Away from the singularity the integrand contributes nothing, because \(\Phi\) is harmonic. The entire value \(-f(x)\) is produced on the shrinking sphere, by the unit flux of Lemma 4.2(b). Chapter 8 will compress the whole computation into the statement that \(-\Delta\Phi\) is the Dirac delta.
Two remarks on hypotheses. The solution is not unique: adding any harmonic function gives another. In \(\mathbb R^3\) the Newtonian potential tends to zero at infinity, and by Liouville’s theorem it is the only solution that does (Exercise 15). And the smoothness of \(f\) can be reduced, but not to mere continuity: there are continuous \(f\) for which no \(C^2\) solution exists. The weak theory of Chapter 9 avoids this difficulty by changing what “solution” means.
Example 4.1: The potential of a uniformly charged ball
Let \(f=1\) in the ball \(B_R(0)\subset\mathbb R^3\) and \(f=0\) outside. This \(f\) is not continuous, so Theorem 4.3 does not apply as stated, but the integral \(u(x)=\frac{1}{4\pi}\int_{B_R}\lvert x-y\rvert^{-1}dy\) makes sense and can be computed exactly with the mean value property.
Integrate over spheres \(\lvert y\rvert=s\) . For fixed \(x\) with \(\lvert x\rvert=r\) , we claim that the average of \(\lvert x-y\rvert^{-1}\) over the sphere \(\lvert y\rvert=s\) equals \(1/\max(r,s)\) . If \(s<r\) , the function \(y\mapsto\lvert x-y\rvert^{-1}\) is harmonic in a ball of radius larger than \(s\) about the origin, so by Theorem 3.3 its average over the sphere equals its value at the centre, \(1/r\) . If \(s>r\) , the average is, by symmetry, the same as the average of \(\lvert x-y\rvert^{-1}\) over all positions of \(x\) on the sphere of radius \(r\) with \(y\) fixed, and \(x\mapsto\lvert x-y\rvert^{-1}\) is harmonic in a ball of radius larger than \(r\) ; its centre value is \(1/s\) . Therefore
Outside the ball, \(r\ge R\) , this is \(\dfrac{1}{r}\displaystyle\int_0^Rs^2ds=\dfrac{R^3}{3r}\) . Inside, \(r<R\) , it is \(\dfrac1r\displaystyle\int_0^rs^2ds+\int_r^Rs\,ds=\dfrac{r^2}{3}+\dfrac{R^2-r^2}{2}\) . So
Outside, the potential is that of a point source of strength \(\tfrac43\pi R^3\) , the volume of the ball, placed at the centre: \(\tfrac43\pi R^3\cdot\tfrac{1}{4\pi r}\) . This is Newton’s theorem that a uniform ball attracts as if its mass were concentrated at its centre. Inside, \(u''+\tfrac2ru'=-\tfrac13-\tfrac23=-1\) , so \(-\Delta u=1\) , and outside \(u\) is harmonic. The two pieces fit together with a continuous first derivative, \(-R/3\) on both sides at \(r=R\) , but the second derivative jumps there, as it must since \(f\) does. For \(R=1\) the potential at \(r=\tfrac12\) is \(\tfrac12-\tfrac1{24}=\tfrac{11}{24}\approx 0.4583\) and at \(r=2\) it is \(\tfrac16\) ; direct numerical integration of the defining integral reproduces both.
4.3 A representation formula for bounded regions
On a bounded region the solution of the Poisson equation is pinned down by boundary values, and the whole-space potential ignores them. The first step toward including them is an identity that expresses any smooth function through its Laplacian and its boundary behaviour.
Theorem 4.4 (Green’s representation formula). Let \(\Omega\) be a bounded open set with piecewise \(C^1\) boundary and let \(u\in C^2(\overline\Omega)\) . Then for every \(x\in\Omega\) ,
Proof. Fix \(x\) and let \(\varepsilon\) be so small that \(\overline{B_\varepsilon(x)}\subset\Omega\) . Apply Green’s second identity on \(\Omega_\varepsilon=\Omega\setminus\overline{B_\varepsilon(x)}\) to \(u\) and \(v(y)=\Phi(y-x)\) , which is harmonic there:
On the small sphere the normal points toward \(x\) . Exactly as in the proof of Theorem 4.3, the first term there tends to zero and the second tends to \(-u(x)\) . The left side tends to \(\int_\Omega v\,\Delta u\,dy\) because \(v\) is integrable. Rearranging gives the formula.
If \(u\) is harmonic the first integral disappears and \(u\) inside is expressed by \(u\) and \(\partial u/\partial\nu\) on the boundary. But this cannot yet be used to solve a boundary value problem. By Theorem 3.7 the boundary values of \(u\) alone determine a harmonic function, so its normal derivative is not ours to prescribe, and we do not know it. The term containing \(\partial u/\partial\nu\) must be eliminated.
4.4 Green’s function
The fundamental solution is determined only up to the addition of a harmonic function. We use that freedom to make it vanish on the boundary.
Definition 4.5 (Green’s function for the Dirichlet problem). Let \(\Omega\) be as in Theorem 4.4. Suppose that for each \(x\in\Omega\) there is a corrector \(h^{x}\in C^2(\overline\Omega)\) with
The Green’s function of \(\Omega\) is \(G(x,y)=\Phi(y-x)-h^{x}(y)\) for \(x,y\in\Omega\) with \(x\ne y\) .
So \(y\mapsto G(x,y)\) has the same singularity as the fundamental solution at \(y=x\) , is harmonic elsewhere, and vanishes on \(\partial\Omega\) . Physically it is the potential of a unit point source at \(x\) inside a grounded conducting shell.
Theorem 4.6 (representation by Green’s function). Let \(u\in C^2(\overline\Omega)\) solve \(-\Delta u=f\) in \(\Omega\) with \(u=g\) on \(\partial\Omega\) . Then for every \(x\in\Omega\) ,
Proof. Green’s second identity for \(u\) and \(h^{x}\) , both in \(C^2(\overline\Omega)\) , gives
because \(\Delta h^{x}=0\) . On the boundary \(h^{x}=\Phi(\cdot-x)\) , so this identity expresses \(\int_{\partial\Omega}\Phi(y-x)\,\partial u/\partial\nu\,dS\) as \(\int_\Omega h^{x}\Delta u\,dy+\int_{\partial\Omega}u\,\partial h^{x}/\partial\nu\,dS\) . Substituting into Theorem 4.4 and collecting terms,
Now use \(-\Delta u=f\) and \(u=g\) .
The function \(K(x,y)=-\partial G/\partial\nu_y(x,y)\) , defined for \(x\in\Omega\) and \(y\in\partial\Omega\) , is the Poisson kernel of \(\Omega\) . The theorem says: weigh the source by \(G\) , weigh the boundary data by \(K\) , and add.
Two cautions keep the theorem honest. First, it is a formula for a solution assumed to exist and to be smooth up to the boundary; it is not by itself an existence theorem. Second, finding the corrector \(h^{x}\) is itself a Dirichlet problem, one for each \(x\) . The gain is that it has to be solved once for the region, after which every choice of \(f\) and \(g\) is handled by integration. For regions with enough symmetry the corrector can be written down, and then one can verify directly that the formula does produce a solution.
Theorem 4.7 (symmetry and positivity). For \(x\ne y\) in \(\Omega\) , \(G(x,y)=G(y,x)\) and \(G(x,y)>0\) when \(\Omega\) is connected.
Proof. Symmetry. Fix \(x\ne y\) and put \(v(z)=G(x,z)\) , \(w(z)=G(y,z)\) . Both are harmonic away from their singular points and vanish on \(\partial\Omega\) . Apply Green’s second identity on \(\Omega\) with small balls about \(x\) and \(y\) removed. The integral over \(\partial\Omega\) vanishes, and so does the volume integral. On the sphere about \(x\) , the function \(w\) is smooth and \(v\) has the singularity of \(\Phi(z-x)\) , so the limiting computation of Theorem 4.4 gives the contribution \(-w(x)\) from the term \(-w\,\partial v/\partial\nu\) and zero from the other; on the sphere about \(y\) the roles are exchanged and the contribution of \(v\,\partial w/\partial\nu\) is \(+v(y)\) . Hence \(v(y)=w(x)\) , which is \(G(x,y)=G(y,x)\) .
Positivity. Since \(h^{x}\) is bounded and \(\Phi(z-x)\to+\infty\) as \(z\to x\) , there is \(\varepsilon>0\) with \(v>0\) on \(\overline{B_\varepsilon(x)}\setminus\{x\}\) . On the region \(\Omega\setminus\overline{B_\varepsilon(x)}\) the function \(v\) is harmonic, is zero on \(\partial\Omega\) , and is positive on \(\partial B_\varepsilon(x)\) . By the weak maximum principle \(v\ge 0\) there. For \(n\ge 2\) removing a small closed ball leaves a connected region, and \(v\) is not identically zero on it, so by the strong maximum principle \(v\) cannot attain the minimum value zero at an interior point: \(v>0\) .
Symmetry is a reciprocity law: the potential at \(y\) due to a source at \(x\) equals the potential at \(x\) due to a source at \(y\) . Positivity says a positive source produces a positive potential, the pointwise form of the remark after Theorem 3.6.
Example 4.2: Green’s function of an interval
In one dimension the same construction works with \(\Phi(x)=-\tfrac12\lvert x\rvert\) , whose derivative jumps by \(-1\) at the origin. Take \(\Omega=(0,1)\) . The corrector \(h^{x}\) must satisfy \((h^{x})''=0\) , so it is linear in \(y\) , with \(h^{x}(0)=-\tfrac12x\) and \(h^{x}(1)=-\tfrac12(1-x)\) . Thus \(h^{x}(y)=-\tfrac12x+\bigl(x-\tfrac12\bigr)y\) and
For \(y\ge x\) this simplifies to \(x-xy\) , and for \(y\le x\) to \(y-xy\) :
It is symmetric, positive inside, zero when \(y=0\) or \(y=1\) , and its \(y\) -derivative jumps from \(1-x\) to \(-x\) at \(y=x\) , a jump of \(-1\) .
Now solve \(-u''=f\) on \((0,1)\) with \(u(0)=u(1)=0\) by \(u(x)=\int_0^1G(x,y)f(y)\,dy\) . For \(f=1\) ,
For \(f(y)=y\) , the same splitting gives \(u(x)=(1-x)\dfrac{x^3}{3}+x\Bigl(\dfrac16-\dfrac{x^2}{2}+\dfrac{x^3}{3}\Bigr)=\dfrac{x(1-x^2)}{6}\) . Both are confirmed by differentiating twice. The boundary term of Theorem 4.6 is also visible here: the outward derivative of \(G\) at \(y=1\) is \(-x\) and at \(y=0\) is \(-(1-x)\) , so boundary values \(g(0)\) and \(g(1)\) contribute \((1-x)\,g(0)+x\,g(1)\) , the straight line through them, which is the harmonic function with those boundary values.
4.5 The method of images
For a region with a mirror symmetry the corrector is the potential of a fictitious source placed outside the region, at the mirror image of the real one.
Half-space. Let \(\Omega=\{x\in\mathbb R^n:x_n>0\}\) and for \(x\in\Omega\) let \(\tilde x=(x_1,\dots,x_{n-1},-x_n)\) be its reflection. The function \(y\mapsto\Phi(y-\tilde x)\) is harmonic in \(\Omega\) , since its singularity is outside, and for \(y\) on the boundary plane \(\lvert y-\tilde x\rvert=\lvert y-x\rvert\) . So it is the corrector, and
The outward normal on the boundary is \(-e_n\) , so the Poisson kernel is \(K=\partial G/\partial y_n\) at \(y_n=0\) . From Lemma 4.2(b), \(\partial_{y_n}\Phi(y-x)=-\dfrac{y_n-x_n}{\omega_n\lvert y-x\rvert^n}\) and \(\partial_{y_n}\Phi(y-\tilde x)=-\dfrac{y_n+x_n}{\omega_n\lvert y-\tilde x\rvert^n}\) ; at \(y_n=0\) their difference is
The half-space is unbounded, so Theorem 4.6 does not apply as stated. Instead one checks directly that the resulting formula gives a solution. In the plane the result is as follows.
Theorem 4.8 (Poisson’s formula for the half-plane). Let \(g\) be bounded and continuous on \(\mathbb R\) . For \(x_2>0\) define
Then \(u\) is harmonic in the upper half-plane, \(\lvert u\rvert\le\sup\lvert g\rvert\) , and \(u(x)\to g(s_0)\) as \(x\to(s_0,0)\) for every \(s_0\in\mathbb R\) .
Proof. Write \(P(x_1,x_2)=\dfrac{x_2}{\pi(x_1^2+x_2^2)}\) , so that \(u(x)=\int P(x_1-s,x_2)\,g(s)\,ds\) . Three properties of \(P\) do all the work.
It is harmonic in the upper half-plane. Indeed \(P=\tfrac1\pi\,\partial_{x_2}\log\lvert x\rvert\) , and a derivative of a harmonic function is harmonic. Since \(P\) and its derivatives up to order two are bounded by an integrable function of \(s\) , locally uniformly in \(x\) , differentiation under the integral gives \(\Delta u=0\) .
It is positive with integral one. \(\displaystyle\int_{-\infty}^{\infty}P(x_1-s,x_2)\,ds=\frac1\pi\Bigl[\arctan\frac{s-x_1}{x_2}\Bigr]_{-\infty}^{\infty}=1\) . Hence \(\lvert u(x)\rvert\le\sup\lvert g\rvert\) .
It concentrates. For \(\delta>0\) , \(\displaystyle\int_{\lvert s-x_1\rvert\ge\delta}P(x_1-s,x_2)\,ds=1-\frac2\pi\arctan\frac{\delta}{x_2}\to 0\) as \(x_2\to 0\) .
Now fix \(s_0\) and \(\eta>0\) , and choose \(\delta>0\) with \(\lvert g(s)-g(s_0)\rvert<\eta\) for \(\lvert s-s_0\rvert<2\delta\) . Because the kernel has integral one, \(u(x)-g(s_0)=\int P(x_1-s,x_2)\bigl(g(s)-g(s_0)\bigr)ds\) . If \(\lvert x_1-s_0\rvert<\delta\) , every \(s\) with \(\lvert s-x_1\rvert<\delta\) has \(\lvert s-s_0\rvert<2\delta\) , so
The last term is less than \(\eta\) once \(x_2\) is small enough. Hence \(u(x)\to g(s_0)\) .
The pattern of this proof, a positive kernel of total mass one that concentrates at a point, is called an approximate identity. It will return in Chapter 5 with a different kernel.
Example 4.3: Harmonic measure of a segment
In the half-plane let \(g=1\) on the segment \(-1<s<1\) and \(g=0\) elsewhere. The formula gives
The quantity in parentheses is the angle under which the segment \([-1,1]\) is seen from the point \(x\) : each arctangent is the angle between the vertical through \(x\) and the line from \(x\) to one endpoint. So \(u(x)\) is that angle divided by \(\pi\) . At \(x=(0,1)\) the segment subtends a right angle and \(u=\tfrac12\) . At \(x=(2,1)\) , \(u=\tfrac1\pi(\arctan(-1)+\arctan3)\approx 0.1476\) . As \(x\) approaches a point of the open segment the angle tends to \(\pi\) and \(u\to1\) ; as it approaches a boundary point outside the closed segment the angle tends to \(0\) . The data here are discontinuous at \(s=\pm1\) , where the theorem makes no claim; at those two points the limit depends on the direction of approach, taking every value between \(0\) and \(1\) .
Ball. For the ball \(B_R(0)\) the mirror is inversion in the sphere: the image of \(x\ne 0\) is \(\tilde x=R^2x/\lvert x\rvert^2\) , which lies outside the ball on the same ray. A source at \(\tilde x\) of a suitably rescaled strength is the corrector:
The second term is harmonic in \(y\) inside the ball. To see that \(G\) vanishes for \(\lvert y\rvert=R\) , compute
Differentiating \(G\) along the normal \(y/R\) gives the Poisson kernel of the ball (Exercise 11).
Theorem 4.9 (Poisson’s formula for the ball). Let \(g\) be continuous on \(\partial B_R(0)\) . The function
is harmonic in \(B_R(0)\) and extends continuously to the closed ball with boundary values \(g\) . In the plane, in polar coordinates,
We do not write out the proof. It has the same three steps as that of Theorem 4.8: the kernel is harmonic in \(x\) , it is positive with integral one over the sphere, and it concentrates near \(y\) as \(x\to y\) . Together with Theorem 3.7 this settles the Dirichlet problem for the Laplace equation in a ball completely: for every continuous \(g\) there is exactly one solution, it depends continuously on \(g\) in the maximum norm, and it is given by an explicit integral. Setting \(x=0\) in the formula recovers the mean value property.
Example 4.4: A disc heated on its upper half
Let \(u\) be harmonic in the unit disc with boundary values \(g=1\) on the upper semicircle \(0<\phi<\pi\) and \(g=0\) on the lower one. Evaluating the Poisson integral in closed form gives
One checks directly that this function is harmonic, that it tends to \(\tfrac12+\tfrac1\pi\cdot\tfrac\pi2=1\) as \(x^2+y^2\to1\) with \(y>0\) , and to \(0\) with \(y<0\) . On the horizontal diameter \(u=\tfrac12\) , as symmetry demands. At the point \((0,\tfrac12)\) ,
and numerical evaluation of the Poisson integral at that point gives the same value. The point lies halfway between the centre, where \(u=\tfrac12\) , and the top of the hot arc, where \(u=1\) , and its temperature is above the halfway value \(0.75\) : the kernel weighs the nearby hot arc more heavily than the distant cold one, but both count.
Common mistake. Using the whole-space potential on a bounded region. To solve \(-\Delta u=1\) in the disc of radius \(2\) with \(u=0\) on the circle, a student extends \(f\) by zero and writes \(u=\Phi\ast f\) . That function does satisfy \(-\Delta u=1\) inside the disc; for the plane it equals \(-2\log2+\tfrac14(4-r^2)\) there. But on the circle \(r=2\) its value is \(-2\log 2\approx-1.386\) , not zero. The fundamental solution knows nothing about the boundary. The correct solution subtracts the harmonic function with the boundary values of the potential, here the constant \(-2\log2\) , leaving \(u=\tfrac14(4-r^2)\) . That subtraction, done once for a point source, is the corrector in Green’s function. A related slip is a sign error: \(\Phi\) is the fundamental solution of \(-\Delta\) , so \(\Delta(\Phi\ast f)=-f\) .
Elliptic problems now have explicit solutions in the whole space, the half-space, and the ball, and a structural formula in general. The next two chapters apply the same strategy to equations with time: find the response to a point disturbance and superpose. For the heat equation the point response is not singular at all, and the question becomes how something as rough as a point source could turn smooth instantly.
Exercises
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Compute \(-\int_{\partial B_r(0)}\partial\Phi/\partial\nu\,dS\) directly from the formulas for \(\Phi\) in dimensions \(2\) and \(3\) , and explain, using the divergence theorem on the region between two spheres, why the answer cannot depend on \(r\) .
Answer
For \(n=2\) : \(\partial\Phi/\partial\nu=-1/(2\pi r)\) on a circle of length \(2\pi r\) , giving \(1\) . For \(n=3\) : \(\partial\Phi/\partial\nu=-1/(4\pi r^2)\) on a sphere of area \(4\pi r^2\) , giving \(1\) . Between two spheres \(\Phi\) is harmonic, so the divergence theorem applied to \(\nabla\Phi\) on the shell gives \(\int_{\partial B_{r_2}}\partial\Phi/\partial\nu\,dS-\int_{\partial B_{r_1}}\partial\Phi/\partial\nu\,dS=\int\Delta\Phi\,dx=0\) .
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Let \(f\in C^2(\mathbb R)\) vanish outside a bounded interval and let \(u(x)=-\tfrac12\int_{-\infty}^{\infty}\lvert x-y\rvert f(y)\,dy\) . Show by splitting the integral at \(y=x\) and differentiating that \(-u''=f\) .
Hint
Write \(u(x)=-\tfrac12\int_{-\infty}^x(x-y)f(y)\,dy-\tfrac12\int_x^{\infty}(y-x)f(y)\,dy\) . Differentiating once, the boundary terms vanish because the integrands are zero at \(y=x\) , leaving \(u'(x)=-\tfrac12\int_{-\infty}^xf\,dy+\tfrac12\int_x^{\infty}f\,dy\) . Differentiate again.
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For \(f\) equal to \(1\) on \([-1,1]\) and \(0\) elsewhere, compute \(u(x)=-\tfrac12\int_{-1}^1\lvert x-y\rvert\,dy\) for all real \(x\) , and verify that \(u\) is \(C^1\) , that \(-u''=1\) for \(\lvert x\rvert<1\) , and that \(u''=0\) for \(\lvert x\rvert>1\) .
Solution
For \(\lvert x\rvert\le 1\) split the integral at \(y=x\) :
\[ u(x)=-\frac12\left(\int_{-1}^{x}(x-y)\,dy+\int_x^1(y-x)\,dy\right)=-\frac12\left(\frac{(x+1)^2}{2}+\frac{(1-x)^2}{2}\right)=-\frac12\,(1+x^2). \]For \(x\ge 1\) every \(y\) in \([-1,1]\) is to the left of \(x\) , so \(u(x)=-\tfrac12\int_{-1}^1(x-y)\,dy=-x\) ; for \(x\le-1\) similarly \(u(x)=x\) . Altogether \(u(x)=-\lvert x\rvert\) for \(\lvert x\rvert\ge 1\) .
Inside, \(u''=-1\) , so \(-u''=1=f\) . Outside, \(u\) is linear and \(u''=0=f\) . At \(x=1\) the inner formula gives \(u=-1\) and \(u'=-x=-1\) , and the outer gives \(u=-1\) and \(u'=-1\) ; by symmetry the same holds at \(x=-1\) . So \(u\) is \(C^1\) on the line, while \(u''\) jumps from \(-1\) to \(0\) at \(x=\pm1\) , exactly where \(f\) jumps. This is the one-dimensional picture of Example 4.1: outside the source the potential is that of a point source of total strength \(2\) at the origin, \(-\tfrac12\cdot2\,\lvert x\rvert\) .
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For the potential \(u\) of Example 4.1 compute the field \(-\nabla u\) inside and outside the ball, and verify that its flux through the sphere of radius \(r\) equals the total source enclosed, for \(r<R\) and for \(r>R\) .
Answer
Inside, \(u'=-r/3\) , so the field \(-\nabla u\) is \(\tfrac r3\) in the outward radial direction, with flux \(\tfrac r3\cdot4\pi r^2=\tfrac43\pi r^3\) through the sphere of radius \(r\) : the volume, hence the total source, enclosed. Outside, \(u'=-R^3/(3r^2)\) and the flux is \(\tfrac{R^3}{3r^2}\cdot4\pi r^2=\tfrac43\pi R^3\) , the whole source.
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Find the potential \(u=\Phi\ast f\) in the plane when \(f=1\) on the disc of radius \(R\) and \(f=0\) outside, by the method of Example 4.1: first show that the average of \(\log\lvert x-y\rvert\) over the circle \(\lvert y\rvert=s\) is \(\log\max(\lvert x\rvert,s)\) . Verify that \(-\Delta u=1\) inside the disc.
Solution
The averaging claim. Fix \(x\) with \(\lvert x\rvert=r\) . If \(s<r\) , the function \(y\mapsto\log\lvert x-y\rvert\) is harmonic in the disc \(\lvert y\rvert<r\) , so by the mean value property its average over \(\lvert y\rvert=s\) is its value at \(y=0\) , namely \(\log r\) . If \(s>r\) , the average over \(\lvert y\rvert=s\) with \(x\) fixed depends only on the distribution of the angle between \(x\) and \(y\) , so it equals the average over \(\lvert x\rvert=r\) with \(y\) fixed; and \(x\mapsto\log\lvert x-y\rvert\) is harmonic in \(\lvert x\rvert<s\) with centre value \(\log s\) . In both cases the average is \(\log\max(r,s)\) .
The potential. With \(\Phi=-\tfrac{1}{2\pi}\log\lvert\cdot\rvert\) and circles of length \(2\pi s\) ,
\[ u(x)=-\int_0^Rs\,\log\max(r,s)\,ds . \]For \(r\ge R\) this is \(-\tfrac12R^2\log r\) . For \(r<R\) , using \(\int s\log s\,ds=\tfrac12s^2\log s-\tfrac14s^2\) ,
\[ u(x)=-\frac{r^2}{2}\log r-\Bigl[\tfrac12s^2\log s-\tfrac14s^2\Bigr]_r^R=-\frac{R^2}{2}\log R+\frac{R^2-r^2}{4}. \]Check. Inside, \(u'=-r/2\) and \(u''=-\tfrac12\) , so \(-\Delta u=-(u''+u'/r)=1\) . Outside \(u\) is a multiple of \(\log r\) , hence harmonic, and at \(r=R\) both formulas give \(-\tfrac12R^2\log R\) with derivative \(-R/2\) . Unlike the three-dimensional potential, this one does not tend to zero at infinity; it behaves like the total source \(\pi R^2\) times \(\Phi\) .
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Using the Green’s function of Example 4.2, prove that the solution of \(-u''=f\) on \((0,1)\) with \(u(0)=u(1)=0\) satisfies \(u\ge 0\) when \(f\ge 0\) and \(\max\lvert u\rvert\le\tfrac18\max\lvert f\rvert\) for continuous \(f\) , and show that the constant \(\tfrac18\) cannot be improved.
Solution
By Example 4.2, \(u(x)=\int_0^1G(x,y)f(y)\,dy\) with \(G(x,y)\ge 0\) on the square. This representation holds for the unique solution because \(u''=-f\) can be verified by differentiating the split integral \((1-x)\int_0^xyf(y)\,dy+x\int_x^1(1-y)f(y)\,dy\) twice, and the boundary values are visibly zero.
If \(f\ge 0\) the integrand is nonnegative, so \(u\ge 0\) . For the bound,
\[ \lvert u(x)\rvert\le\max\lvert f\rvert\int_0^1G(x,y)\,dy=\max\lvert f\rvert\cdot\frac{x(1-x)}{2}\le\frac18\max\lvert f\rvert , \]where the middle equality is the case \(f=1\) computed in Example 4.2 and the last step uses \(x(1-x)\le\tfrac14\) . The constant is sharp: for \(f=1\) the solution is \(u=\tfrac12x(1-x)\) , and \(u(\tfrac12)=\tfrac18=\tfrac18\max\lvert f\rvert\) . This is the one-dimensional case of Theorem 3.8 with the best possible constant for this interval.
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Find the Green’s function of the interval \((0,L)\) for \(-u''\) with Dirichlet conditions, verify its symmetry, and show that \(\partial G/\partial y\) jumps by \(-1\) across \(y=x\) .
Answer
\(G(x,y)=\dfrac{x(L-y)}{L}\) for \(x\le y\) and \(G(x,y)=\dfrac{y(L-x)}{L}\) for \(y\le x\) . Exchanging \(x\) and \(y\) swaps the two formulas, so \(G\) is symmetric. For \(y<x\) , \(\partial G/\partial y=(L-x)/L\) ; for \(y>x\) , \(\partial G/\partial y=-x/L\) ; the jump is \(-x/L-(L-x)/L=-1\) .
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For the half-plane, write \(G(x,y)\) explicitly with logarithms, verify that it vanishes when \(y_2=0\) , and derive the kernel of Theorem 4.8 by differentiating in \(y_2\) .
Hint
\(G(x,y)=-\tfrac{1}{4\pi}\log\bigl((y_1-x_1)^2+(y_2-x_2)^2\bigr)+\tfrac{1}{4\pi}\log\bigl((y_1-x_1)^2+(y_2+x_2)^2\bigr)\) . The two logarithms coincide at \(y_2=0\) . The outward normal is \(-e_2\) , so \(K=\partial G/\partial y_2\) at \(y_2=0\) ; each term contributes \(\dfrac{x_2}{2\pi\bigl((y_1-x_1)^2+x_2^2\bigr)}\) .
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Solve the Dirichlet problem in the upper half-plane with boundary values \(g(s)=1/(1+s^2)\) . Verify that your answer is harmonic and has the right boundary values. A useful observation is that \(g\) is itself \(\pi\) times the kernel \(P\) evaluated at height one.
Solution
With \(P(x_1,x_2)=\dfrac{x_2}{\pi(x_1^2+x_2^2)}\) we have \(g(s)=\dfrac{1}{1+s^2}=\pi P(s,1)\) : the boundary datum is the trace on the line \(x_2=1\) of the function \(\pi P\) , which is harmonic and bounded in the half-plane \(x_2>0\) . This suggests shifting: the function
\[ u(x_1,x_2)=\pi P(x_1,1+x_2)=\frac{1+x_2}{x_1^2+(1+x_2)^2} \]is harmonic for \(x_2>-1\) , in particular in the upper half-plane and on its boundary, where it equals \(1/(1+x_1^2)=g(x_1)\) . It is bounded, since \(0<u\le 1/(1+x_2)\le 1\) .
It remains to see that this is the function given by Poisson’s formula. Both are bounded and harmonic in the half-plane with the same continuous boundary values, so their difference is a bounded harmonic function vanishing on the boundary line, which is zero by the uniqueness statement quoted at the end of Section 3.6. Direct verification of harmonicity: \(u\) is a translate of \(\pi P\) , and \(P=\tfrac1\pi\partial_{x_2}\log\lvert x\rvert\) is harmonic. The solution at height \(x_2\) is the datum spread out: its maximum is \(1/(1+x_2)\) and its width grows like \(1+x_2\) .
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Let \(u\) be harmonic and nonnegative in the disc \(B_R(0)\subset\mathbb R^2\) and continuous on its closure. Use Poisson’s formula to prove Harnack’s inequality \(\dfrac{R-r}{R+r}\,u(0)\le u(x)\le\dfrac{R+r}{R-r}\,u(0)\) for \(\lvert x\rvert=r<R\) .
Hint
Since \(-1\le\cos(\theta-\phi)\le 1\) , the denominator of the kernel lies between \((R-r)^2\) and \((R+r)^2\) , so the kernel lies between \(\dfrac{R-r}{R+r}\) and \(\dfrac{R+r}{R-r}\) . Multiply by \(g\ge 0\) , integrate, and use \(\tfrac{1}{2\pi}\int_0^{2\pi}g\,d\phi=u(0)\) .
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In the plane, differentiate the Green’s function of the disc \(B_R(0)\) in the direction of the outward normal at a boundary point \(y\) and show that \(-\partial G/\partial\nu_y=\dfrac{R^2-\lvert x\rvert^2}{2\pi R\,\lvert x-y\rvert^2}\) .
Hint
\(G=-\tfrac{1}{2\pi}\log\lvert y-x\rvert+\tfrac{1}{2\pi}\log\bigl(\tfrac{\lvert x\rvert}{R}\lvert y-\tilde x\rvert\bigr)\) , so \(\nabla_yG=-\tfrac{1}{2\pi}\Bigl(\dfrac{y-x}{\lvert y-x\rvert^2}-\dfrac{y-\tilde x}{\lvert y-\tilde x\rvert^2}\Bigr)\) . On \(\lvert y\rvert=R\) use \(\lvert y-\tilde x\rvert^2=R^2\lvert y-x\rvert^2/\lvert x\rvert^2\) to combine the two fractions; the terms in \(x\) cancel and what is left is a multiple of \(y\) . Then take the inner product with \(\nu=y/R\) .
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A student solves \(-\Delta u=1\) in the unit ball of \(\mathbb R^3\) with \(u=0\) on the sphere by taking \(u=\Phi\ast f\) with \(f\) extended by zero, and reports \(u=\tfrac12-\tfrac16r^2\) . Show that this satisfies the equation and violates the boundary condition, identify what is missing, and give the correct solution.
Solution
By Example 4.1 with \(R=1\) the whole-space potential of the extended source is indeed \(\tfrac12-\tfrac16r^2\) inside the ball. It satisfies the equation: \(u''+\tfrac2ru'=-\tfrac13-\tfrac23=-1\) , so \(-\Delta u=1\) . But on the sphere \(r=1\) its value is \(\tfrac12-\tfrac16=\tfrac13\) , not \(0\) .
What is missing is the corrector. The convolution with \(\Phi\) solves the equation in all of space and takes whatever boundary values it happens to take. By Theorem 3.7 the solution of the Dirichlet problem is unique, and it is obtained by subtracting from the potential the harmonic function in the ball that has the same boundary values. Here those boundary values are the constant \(\tfrac13\) , and the harmonic function with constant boundary values is that constant. So the correct solution is
\[ u=\frac12-\frac{r^2}{6}-\frac13=\frac{1-r^2}{6}, \]which still satisfies \(-\Delta u=1\) and vanishes at \(r=1\) . In terms of Theorem 4.6, the student integrated \(f\) against \(\Phi(y-x)\) where \(G(x,y)=\Phi(y-x)-h^{x}(y)\) was required.
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Use three image sources to write the Green’s function of the first quadrant \(\{y_1>0,\ y_2>0\}\) of the plane, and verify that it vanishes on both bounding half-lines.
Answer
With \(x'=(x_1,-x_2)\) , \(x''=(-x_1,x_2)\) , and \(-x=(-x_1,-x_2)\) : \(G(x,y)=\Phi(y-x)-\Phi(y-x')-\Phi(y-x'')+\Phi(y+x)\) . On \(y_2=0\) , \(\lvert y-x\rvert=\lvert y-x'\rvert\) and \(\lvert y-x''\rvert=\lvert y+x\rvert\) , so the terms cancel in pairs; on \(y_1=0\) , \(\lvert y-x\rvert=\lvert y-x''\rvert\) and \(\lvert y-x'\rvert=\lvert y+x\rvert\) . All three image points lie outside the open quadrant.
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Prove that if \(f\ge 0\) is continuous on \(\overline\Omega\) and \(u\in C^2(\overline\Omega)\) solves \(-\Delta u=f\) with \(u=0\) on \(\partial\Omega\) , then \(u\ge 0\) , in two ways: from Theorem 4.6 and Theorem 4.7, and directly from Theorem 3.6.
Solution
From Green’s function. By Theorem 4.6 with \(g=0\) , \(u(x)=\int_\Omega G(x,y)f(y)\,dy\) . By Theorem 4.7, \(G(x,y)\ge 0\) for \(x\ne y\) ; the positivity argument there gives \(G\ge 0\) on each component of \(\Omega\) even without connectedness. Since \(f\ge 0\) , the integrand is nonnegative and so \(u(x)\ge 0\) .
From the maximum principle. We have \(\Delta u=-f\le 0\) in \(\Omega\) , so \(u\) is superharmonic, and \(u\in C^2(\overline\Omega)\subset C^2(\Omega)\cap C(\overline\Omega)\) . By Theorem 3.6(b) the minimum of \(u\) over \(\overline\Omega\) is attained on \(\partial\Omega\) , where \(u=0\) . Hence \(u\ge 0\) throughout.
The second proof is shorter and needs no Green’s function, only boundedness of \(\Omega\) . The first gives more when it applies: it shows how much each part of the source contributes at each point, and that the contribution of a source near \(y\) to the value at \(x\) is the same as that of a source near \(x\) to the value at \(y\) .
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Let \(f\in C^2(\mathbb R^3)\) vanish outside a ball. Prove that \(u=\Phi\ast f\) tends to zero as \(\lvert x\rvert\to\infty\) , and use Liouville’s theorem to prove that it is the only solution of \(-\Delta u=f\) in \(\mathbb R^3\) with that property.
Solution
Let \(f\) vanish outside \(B_\rho(0)\) and put \(m=\int\lvert f\rvert\,dy\) . For \(\lvert x\rvert>\rho\) and \(y\) in the support of \(f\) we have \(\lvert x-y\rvert\ge\lvert x\rvert-\rho\) , so
\[ \lvert u(x)\rvert\le\frac{1}{4\pi}\int\frac{\lvert f(y)\rvert}{\lvert x-y\rvert}\,dy\le\frac{m}{4\pi\,(\lvert x\rvert-\rho)}\longrightarrow 0\quad(\lvert x\rvert\to\infty). \]By Theorem 4.3, \(u\in C^2\) and \(-\Delta u=f\) .
Now let \(\tilde u\) be another \(C^2\) solution that tends to zero at infinity. Then \(w=u-\tilde u\) is harmonic on \(\mathbb R^3\) and tends to zero at infinity. A continuous function with a limit at infinity is bounded: it is small outside a large ball and bounded on that ball by compactness. By Liouville’s theorem \(w\) is constant, and the constant is its limit at infinity, zero. Hence \(\tilde u=u\) . In the plane the argument fails at the first step, since the logarithmic potential does not decay; there uniqueness needs a different normalization.
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Prove the claim used in Example 4.1 in the case \(s>r\) : for fixed \(y\) with \(\lvert y\rvert=s\) , the average of \(\lvert x-y\rvert^{-1}\) over the sphere \(\lvert x\rvert=r\) equals \(1/s\) . Then explain why this is the same number as the average over \(\lvert y\rvert=s\) with \(x\) fixed.
Hint
For fixed \(y\) with \(\lvert y\rvert=s\) the function \(x\mapsto\lvert x-y\rvert^{-1}\) is harmonic in the ball \(\lvert x\rvert<s\) , which contains the closed ball of radius \(r\) ; apply Theorem 3.3 at the centre. For the second part, \(\lvert x-y\rvert\) depends only on \(r\) , \(s\) , and the angle between \(x\) and \(y\) , and a rotation carrying one configuration to the other preserves surface measure, so both averages are the same integral over that angle.
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In the upper half-plane let \(g=1\) for \(s>0\) and \(g=0\) for \(s<0\) . Show that the Poisson formula gives \(u=1-\theta/\pi\) , where \(\theta\in(0,\pi)\) is the polar angle of \(x\) , check that this is harmonic, and describe the limits of \(u\) at the origin along different rays.
Answer
\(u=\tfrac1\pi\int_0^\infty\dfrac{x_2\,ds}{(s-x_1)^2+x_2^2}=\tfrac12+\tfrac1\pi\arctan\dfrac{x_1}{x_2}\) . With \(x_1=r\cos\theta\) , \(x_2=r\sin\theta\) this is \(\tfrac12+\tfrac1\pi\bigl(\tfrac\pi2-\theta\bigr)=1-\theta/\pi\) . It is harmonic because \(\theta\) is, having no \(r\) -dependence and vanishing second \(\theta\) -derivative in the polar Laplacian. Along the ray of angle \(\theta\) the limit at the origin is \(1-\theta/\pi\) : every value between \(0\) and \(1\) occurs.