A field equation asks for a function whose derivatives obey a rule. It does not usually determine that function until we prescribe data. This chapter defines those statements and proves the elementary uniqueness and propagation results needed for electromagnetism.
Read chapter 2 for partial derivatives, gradients, the Laplacian, and vector identities, and chapter 3 for integrals and the divergence theorem. No differential-equation knowledge is assumed. The standing background is elementary real analysis, the Euclidean metric and basic topology, and complex arithmetic. We use real scalar functions here; apply statements component by component to vector fields. Here a domain means a nonempty open connected subset of Euclidean space; the maximum principle below also applies to nonempty open sets with several components.
4.1. Equations, solutions, and data
Definition 4.1. Partial differential equations
A partial differential equation (PDE) is a relation between an unknown function of several variables and its partial derivatives. Its order is the highest derivative order appearing. A classical solution on an open set is a function with the continuous derivatives required by the equation, satisfying it at every point. The notation \(C^k\) means that all partial derivatives through order \(k\) exist and are continuous.
A differential operator \(L\) is linear if \(L(au+bv)=aLu+bLv\) for scalar constants \(a,b\). A linear equation \(Lu=f\) is homogeneous when \(f=0\), and inhomogeneous otherwise. An equation is nonlinear if its dependence on the unknown and its derivatives cannot be written this way.
Alphabeta Math: linear and nonlinear PDEs
Examples. \(u_x+u_y=0\) is first order and linear; \(u=x-y\) solves it. \(u_{xx}+u_{yy}=0\) is second order; \(u=x^2-y^2\) solves it. \(u_x^2+u_y^2=1\) is nonlinear: \(u=x\) solves it, but \(2u\) does not. A variable coefficient such as \(x u_x\) still gives a linear operator, because \(x\) is prescribed rather than unknown.
Definition 4.2. Laplace, Poisson, and wave equations
For \(x=(x_1,\ldots,x_d)\), let \(\Delta u=\sum_{j=1}^d\partial_{x_j}^2u\). The Laplace equation is \(\Delta u=0\); its \(C^2\) solutions are called harmonic. The Poisson equation is \(-\Delta u=f\), with prescribed source \(f\). The sign is a convention. See Alphabeta Math’s Laplacian definition. For an additional time variable \(t\) and a constant \(c>0\), the wave equation is \(u_{tt}-c^2\Delta u=0\).
Alphabeta Math: wave equation and speed
Examples. Constants and affine functions \(a+b\cdot x\) are harmonic. In three dimensions \(-\Delta(-|x|^2/6)=1\), since each \(x_j^2\) has second derivative two. The sine and cosine definitions and derivative laws are proved in the complex-arithmetic chapter. The function \(u(x,t)=\cos(kx)\cos(ckt)\) solves the one-dimensional wave equation for every real \(k\), as two differentiations verify. These are equations for unknown fields, not formulas selecting a unique field.
Definition 4.3. Initial and boundary conditions
An initial condition specifies information at a starting time. For the second-order wave equation the usual pair is \(u(x,0)=f(x)\), \(u_t(x,0)=g(x)\): initial displacement and velocity. A boundary condition specifies information on the spatial boundary. Dirichlet data prescribe \(u\); Neumann data prescribe the outward normal derivative \(\partial_nu=\nabla u\cdot n\). A solution with boundary values is understood to have the corresponding continuous boundary extension.
Compatibility means that all imposed conditions agree wherever they overlap and with consequences of the equation. If \(u(0,t)=0\) and \(u(x,0)=f(x)\), continuity requires \(f(0)=0\). If derivatives also extend continuously to the corner, \(u_t(0,t)=0\) forces \(g(0)=0\). Higher smoothness can impose further conditions.
Alphabeta Math: Cauchy data
Why data matter. On \((0,1)\), every \(u=a+bx\) solves \(u''=0\). The conditions \(u(0)=2\), \(u(1)=5\) select \(u=2+3x\). Derivative conditions \(u'(0)=u'(1)=3\), corresponding to outward Neumann values \(-3\) at zero and \(3\) at one, leave the constant \(a\) free. The wave solutions \(u=0\) and \(u=t\) have the same displacement at zero but different velocities. Neither the PDE alone nor displacement alone selects one.
Definition 4.4. Well-posedness
A problem is well posed in specified spaces and on a specified time interval if (i) each admissible datum has a solution, (ii) that solution is unique, and (iii) solutions depend continuously on the data: convergence in the declared data distance implies convergence in the declared solution distance. Local well-posedness for an evolution problem asserts this on a positive time interval, possibly depending on the data. “Continuous dependence” has no precise meaning until those distances are stated.
Worked example for Definition 4.4
Counterexample: Small harmonic Cauchy data need not control the interior
Example. The interval Dirichlet problem just solved is well posed for endpoint data with distance \(|a-\widetilde a|+|b-\widetilde b|\) and the uniform solution norm: linear interpolation gives \(\sup|u-\widetilde u|\le\max(|a-\widetilde a|,|b-\widetilde b|)\). If both Dirichlet and contradictory Neumann values are prescribed, existence can fail; Neumann data alone above give a failure of uniqueness. We will prove a uniform continuous-dependence estimate for the wave initial-value problem below.
4.2. A maximum principle and static uniqueness
Theorem 4.5. Weak maximum principle
Let \(\Omega\subset\mathbb R^d\) be nonempty, bounded, and open, and let \(u\in C^2(\Omega)\cap C(\overline\Omega)\). If \(\Delta u\ge0\), then \(\sup_{\overline\Omega}u=\sup_{\partial\Omega}u\). In particular a harmonic function attains neither a larger maximum nor a smaller minimum than its boundary values. No smoothness of the boundary is needed for this statement.
Proof.
The closure is compact, so a continuous function attains its maximum. Put \(u_\varepsilon=u+\varepsilon|x|^2\) for \(\varepsilon>0\). Its Laplacian is \(\Delta u+2d\varepsilon>0\). At an interior maximum every coordinate second derivative is nonpositive. To justify that one-variable fact, restrict the function to a coordinate line through the maximum and write its value as \(f(h)\), with the maximum at \(h=0\). The derivative definition gives \(f'(0)=0\). Twice applying the fundamental theorem gives \(f(h)=f(0)+hf'(0)+h^2\int_0^1(1-s)f''(sh)\,ds\). Thus the symmetric quotient \([f(h)+f(-h)-2f(0)]/h^2\) tends to \(f''(0)\); its numerator is nonpositive at a maximum, so \(f''(0)\le0\). Applying this to every coordinate shows that the Laplacian at an interior maximum would be nonpositive. Thus its maximum lies on the boundary. Choose \(R\) with \(|x|\le R\) on the closure. For every \(x\in\Omega\),
\[u(x)\le u_\varepsilon(x)\le\sup_{\partial\Omega}u+\varepsilon R^2.\]Let \(\varepsilon\downarrow0\). Boundary points already satisfy the bound; the reverse inequality follows from inclusion. For a harmonic function apply the argument to both \(u\) and \(-u\).
□Corollary 4.6. Dirichlet uniqueness and stability
Two classical solutions of \(-\Delta u=f\) on a bounded domain, continuous up to the boundary and with the same boundary data, are equal. For the same source but different boundary values \(h,\widetilde h\), their difference obeys \(\sup_{\overline\Omega}|u-\widetilde u|\le\sup_{\partial\Omega}|h-\widetilde h|\). This asserts uniqueness and a stability estimate whenever solutions exist; it does not assert general existence.
Alphabeta Math: Dirichlet uniqueness
Worked example for Corollary 4.6
Counterexample: Small harmonic Cauchy data need not control the interior
Proof.
Subtract the equations. The difference is harmonic; its maximum and the maximum of its negative are bounded by the corresponding boundary maxima, by Theorem 4.5.
□Examples. On the ball \(|x|<R\) in \(\mathbb R^d\), \(u=(R^2-|x|^2)/(2d)\) is the unique classical solution of \(-\Delta u=1\) with zero boundary data. Direct differentiation proves existence, and the corollary proves uniqueness. On an interval, the solution of \(-u''=f\) with \(u(0)=a\), \(u(L)=b\), for continuous \(f\), is
\[u(x)=a+Cx-\int_0^x(x-s)f(s)\,ds,\qquad C=\frac{b-a+\int_0^L(L-s)f(s)\,ds}{L}.\]The fundamental theorem of calculus gives \(u''=-f\); substitution checks both endpoints. With zero boundary values, changing \(f\) by \(\delta f\) changes \(u\) by at most \(L^2\sup|\delta f|\), using the two displayed integrals and \(0\le x\le L\). Thus this particular Poisson problem is well posed in uniform norms.
A boundary-free counterexample. The unbounded domain \(\mathbb R^d\) has no boundary and admits the nonconstant harmonic function \(x_1\). Bounded-domain uniqueness cannot simply be asserted there. A condition at infinity can restore uniqueness: if a harmonic \(v\) on all of \(\mathbb R^d\) tends uniformly to zero as \(|x|\to\infty\), apply the maximum principle on balls of radius \(R\), then let \(R\to\infty\), to obtain \(v=0\).
4.3. Solving the one-dimensional wave equation
Theorem 4.7. d’Alembert’s initial-value formula
For \(c>0\), \(f\in C^2(\mathbb R)\), and \(g\in C^1(\mathbb R)\), the unique \(C^2\) solution on \(\mathbb R\times\mathbb R\) of \(u_{tt}=c^2u_{xx}\), \(u(x,0)=f(x)\), \(u_t(x,0)=g(x)\) is
\[u(x,t)=\frac{f(x-ct)+f(x+ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}g(s)\,ds.\]For negative \(t\), the integral has its usual oriented meaning.
Alphabeta Math: d’Alembert formula · general solution · initial-data verification
Proof.
Set \(\xi=x-ct\), \(\eta=x+ct\), and write \(v(\xi,\eta)=u((\xi+\eta)/2,(\eta-\xi)/(2c))\). The chain rule gives \(u_{tt}-c^2u_{xx}=-4c^2v_{\xi\eta}\). Hence the equation is \(v_{\xi\eta}=0\). For fixed \(\xi\), the derivative \(v_\xi\) is constant in \(\eta\), by the fundamental theorem of calculus. Integrating in \(\xi\) gives \(v=F(\xi)+H(\eta)\) for \(C^2\) functions \(F,H\). Initial data therefore imply
\[F(x)+H(x)=f(x),\qquad-cF'(x)+cH'(x)=g(x).\]Differentiate the first relation and solve the two scalar linear equations: \(F'=(f'-g/c)/2\), \(H'=(f'+g/c)/2\). Integration and the first relation give the displayed formula; the arbitrary opposite constants cancel. Conversely, take an antiderivative \(Q\) of \(g\). Since \(g\in C^1\), \(Q\in C^2\); the integral is \(Q(x+ct)-Q(x-ct)\). Thus the formula is \(C^2\) and is a sum of functions of \(x-ct\) and \(x+ct\), so it solves the PDE. At \(t=0\), its value is \(f\), and differentiating in time gives \(g\). The derivation applies to every \(C^2\) solution, proving uniqueness.
□Examples. If \(g=0\), a displacement splits into equal copies travelling left and right. If \(f=0\), \(g=1\), the integral gives \(u=t\). For \(f(x)=\cos(kx)\), \(g=0\), the addition formula for cosine gives \(u=\cos(kx)\cos(ckt)\). A travelling profile \(F(x-ct)\) has initial velocity \(-cF'\), explaining why displacement alone does not specify direction.
Corollary 4.8. Continuous dependence and finite propagation
For \(0\le t\le T\), the value \(u(x,t)\) depends only on the initial data in \([x-ct,x+ct]\). If two initial data pairs agree there, their solutions agree at \((x,t)\). For bounded differences of data,
\[\sup_{x,\,0\le t\le T}|u-\widetilde u|\le\sup_x|f-\widetilde f|+T\sup_x|g-\widetilde g|.\]For the data space of bounded \(C^2\) displacements and bounded \(C^1\) velocities, with distance \(\sup|f-\widetilde f|+\sup|g-\widetilde g|\), this specifies continuous dependence in the uniform norm of the field on every finite time interval. Together with Theorem 4.7 it proves well-posedness in these specified distances. It does not claim control of derivative norms by these weaker data norms.
Alphabeta Math: one-dimensional domain of dependence
Proof.
Inspect the endpoints and integral in the formula. The interval has length \(2ct\), so the integral of the velocity difference contributes at most \(t\sup|g-\widetilde g|\); the two displacement terms together contribute at most \(\sup|f-\widetilde f|\).
□Example. If both initial functions vanish outside \([-R,R]\), the solution vanishes when \(|x|>R+ct\), for \(t\ge0\). Information cannot reach these points at speed exceeding \(c\). A field may oscillate rapidly inside the interval; finite propagation concerns where an initial change can have an effect.
4.4. Energy and propagation in three dimensions
The one-dimensional formula does not solve the three-dimensional equation. The following argument nevertheless proves its domain of dependence without an existence formula. A ball is \(B_R(x_0)=\{x:|x-x_0|<R\}\).
Theorem 4.9. The shrinking-ball energy estimate
Let \(u\in C^2\) solve \(u_{tt}=c^2\Delta u\) on a neighborhood of the closed cone \(0\le t\le T<R/c\), \(|x-x_0|\le R-ct\). If \(u(x,0)=u_t(x,0)=0\) on \(B_R(x_0)\), then \(u=0\) throughout the open cone. Consequently two such solutions with equal initial data on that ball agree in the cone. In particular classical solutions with the same initial data on all of space are unique on any common interval of existence on all of space.
Alphabeta Math: energy and flux · local conservation · cone energy identity · finite propagation · energy uniqueness
Proof: Theorem 4.9. The shrinking-ball energy estimate
Proof.
Define the nonnegative energy density \(e=(u_t^2+c^2|\nabla u|^2)/2\) and the vector \(q=c^2u_t\nabla u\) (the outward energy-current vector is \(-q\)). Product differentiation and the PDE give
\[\partial_t e=u_tu_{tt}+c^2\nabla u\cdot\nabla u_t=\nabla\cdot q.\]For \(r(t)=R-ct\), let \(E(t)=\int_{B_{r(t)}(x_0)}e(x,t)\,dV\). Radial integration, the fundamental theorem of calculus for the moving upper radius, and differentiation of an integrand with continuous \(e\) and \(\partial_t e\) on a common compact neighborhood give
\[E'(t)=\int_{B_{r(t)}}\partial_t e\,dV-c\int_{\partial B_{r(t)}}e\,dS=\int_{\partial B_{r(t)}}(q\cdot n-ce)\,dS.\]To justify integral differentiation, the fundamental theorem gives \([e(x,t+h)-e(x,t)]/h=\int_0^1\partial_t e(x,t+sh)\,ds\). Uniform continuity makes these quotients converge uniformly on the fixed compact spatial region to \(\partial_t e(x,t)\). The integral bound then permits passage to the limit. Radial integration supplies the moving-boundary term. The last equality is the divergence theorem. Its sphere case is proved in chapter 3, and the radial integration element is derived in its coordinate section. Pointwise on the sphere, writing \(a=u_t\), \(b=\partial_nu\),
\[q\cdot n-ce=c^2ab-\frac c2(a^2+c^2|\nabla u|^2)\le-\frac c2(a-cb)^2\le0,\]because \(|b|\le|\nabla u|\). Thus \(E\) is nonincreasing. Initial displacement zero implies initial spatial gradient zero inside the ball, and initial velocity zero gives \(E(0)=0\). Nonnegativity yields \(E(t)=0\). A nonnegative continuous integrand with zero integral must vanish: if positive at one interior point it is bounded below on a small box of positive volume. Therefore \(u_t=0\) and \(\nabla u=0\) on each shrinking ball. A fixed point in the ball at time \(t\) lies in every earlier ball; integrating \(u_t(x,s)=0\) from zero to \(t\) gives \(u(x,t)=0\). Apply this to the difference of two solutions. For any specified later point choose a sufficiently large initial ball; this proves uniqueness on all of space for later times. For earlier times apply the same argument to \(u(x,-t)\).
□Example. A change confined initially to \(B_a(0)\) cannot change a classical solution outside \(B_{a+ct}(0)\): for \(|x|>a+ct\), choose \(R\) with \(ct<R<|x|-a\) and apply the theorem in the ball centered at \(x\). The theorem proves uniqueness and finite propagation for existing solutions. General three-dimensional existence requires additional construction; the explicit plane waves in chapter 6 will be verified directly.
For a sufficiently smooth solution with support confined to a bounded region throughout the time interval, integration of \(\partial_t e=\nabla\cdot q\) over a larger fixed ball gives conserved total energy: the flux vanishes on its boundary. This follows from the same computation with a fixed radius; compare Alphabeta Math’s total-energy theorem. Without boundary decay or an integrability hypothesis, a total-energy integral can be infinite, so we do not assert conservation of a finite number in that case.
The next chapter applies Poisson’s equation to electrostatic potentials. A point charge will require a broader meaning of derivative, because its field is singular at the charge.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Equations, solutions, and boundary data
Applies to: Definition 4.1, Definition 4.2, Definition 4.3.
On the unit ball in \(\mathbb R^3\), \(u(x)=1-|x|^2\) is smooth and satisfies the linear second-order Poisson equation \(-\Delta u=6\), with Dirichlet boundary value zero. Its outward normal derivative on the sphere is \(-2\), illustrating a Neumann datum; the interior source and boundary values are different pieces of information. The function \(v=x_1\) is harmonic, \(\Delta v=0\). For the one-dimensional wave equation with \(c=1\), \(w(x,t)=\sin x\cos t\) satisfies \(w_{tt}=w_{xx}\), initial displacement \(f=\sin x\), and velocity \(g=0\). On the half-line it also has boundary condition \(w(0,t)=0\), compatible with \(f(0)=g(0)=0\). All required derivatives exist, so these are classical, rather than weak, solutions of the defined equations.
Worked example 2. Well-posedness in specified uniform distances
Applies to: Definition 4.4, Theorem 4.7, Corollary 4.8.
For \(c>0\), choose bounded smooth data \(f(x)=\cos x,g(x)=0\). D’Alembert gives \(u(x,t)=\tfrac12(\cos(x-ct)+\cos(x+ct))=\cos x\cos ct\), which has the declared initial data and satisfies \(u_{tt}=c^2u_{xx}\). The uniqueness theorem applies on all of \(\mathbb R\times\mathbb R\). Perturb the data by constants \(\widetilde f=f+\varepsilon,\widetilde g=\delta\). The new solution is \(\widetilde u=u+\varepsilon+\delta t\), so on \(0\le t\le T\), \(\sup|\widetilde u-u|\le|\varepsilon|+T|\delta|\), exactly the stated continuous-dependence bound in the uniform field distance and uniform data distances. This is a concrete instance of the specified well-posed problem.
For finite propagation take a smooth initial bump supported in \((3,4)\) and zero initial velocity. At \((x,t)=(0,1/c)\), the dependence interval is \([-1,1]\), where both data vanish, so the value is zero; direct evaluation of the two shifted bumps gives the same answer. Support and the time interval, not just a coefficient’s units, establish this conclusion.
Further proof details
Uniqueness for this whole-line classical wave problem can be derived in characteristic coordinates. For a difference \(w\) of two \(C^2\) solutions with the same data, put \(\xi=x-ct,\eta=x+ct\). The equation gives \(w_{\xi\eta}=0\), so \(\partial_\xi w\) is independent of \(\eta\), and integrating gives \(w=F(\xi)+G(\eta)\) for \(C^2\) functions. Initial displacement gives \(F(x)+G(x)=0\), and initial velocity gives \(-cF'(x)+cG'(x)=0\). Differentiating the first equality and combining forces \(F'=G'=0\). Their opposite constants cancel, yielding \(w=0\). This proves the uniqueness invoked in the cosine example with exactly its all-space and \(C^2\) hypotheses.
Worked example 3. Maximum principle and boundary stability
Applies to: Theorem 4.5, Corollary 4.6.
On the unit ball, \(v(x)=|x|^2\) has \(\Delta v=6\ge0\); its supremum 1 occurs on the boundary, as the weak maximum principle states. The harmonic function \(h=x_1\) has maximum 1 and minimum \(-1\), both on the sphere. For Poisson source 6, the functions \(u=1-|x|^2\) and \(\widetilde u=1-|x|^2+\varepsilon x_1\) have boundary values 0 and \(\varepsilon x_1\). Their uniform difference on the closed ball is \(|\varepsilon|\), exactly the uniform boundary difference. The functions are \(C^2\) inside and continuous on the compact closure, with the same source, checking all hypotheses of the uniqueness and stability statement. Setting the boundary difference to zero forces the difference to vanish.
Worked example 4. A smooth wave front outside a shrinking cone
Applies to: Theorem 4.9.
Define \(\eta(s)=0\) for \(s\le R\), and \(\eta(s)=e^{-1/(s-R)^2}\) for \(s>R\), with \(R>0\). It is smooth: every right derivative near \(R\) is a polynomial in \(1/(s-R)\) times the decaying exponential, hence tends to zero. In three dimensions \(u(x,t)=\eta(x_1-ct)\) satisfies \(u_{tt}=c^2\Delta u\) and has \(u=u_t=0\) on the initial ball \(|x|\le R\). For \(0\le t<R/c\) and \(|x|<R-ct\), \(x_1-ct<R\), so \(u=0\), directly verifying the theorem. The wave is nonzero elsewhere. The cone energy \(\tfrac12\int_{B_{R-ct}}(u_t^2+c^2|\nabla u|^2)\) is identically zero here because both derivatives vanish there, explicitly checking the energy argument in a nonzero global solution.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Small harmonic Cauchy data need not control the interior
Related statements: Definition 4.4, Corollary 4.6.
Claim refuted. The stability estimate for a Dirichlet problem also holds when only a function and its normal derivative on one boundary side are prescribed.
Proof.
On \(\Omega=(-1,1)\times(0,1)\), consider the smooth harmonic functions \(u_n(x,y)=e^{-\sqrt n}\cos(nx)\sinh(ny)\). Differentiation gives \((u_n)_{xx}=-n^2u_n\), \((u_n)_{yy}=n^2u_n\), so \(\Delta u_n=0\). On the bottom side \(y=0\), \(u_n=0\) and \((u_n)_y=ne^{-\sqrt n}\cos(nx)\), whose uniform norm tends to zero. Even every fixed-order tangential derivative tends uniformly to zero, since its norm is at most \(n^{k+1}e^{-\sqrt n}\); the exponential-series bound shows this beats each fixed power of \(n\).
Nevertheless \(u_n(0,1/2)=e^{-\sqrt n}\sinh(n/2)\to\infty\): for large \(n\), it is at least \(e^{n/2-\sqrt n}/4\). Thus these small Cauchy data on one side cannot give a uniform bound even at a fixed interior point. The omitted information is not the Poisson source, which is zero for every \(n\), but the boundary conditions on the rest of \(\partial\Omega\). The full-boundary Dirichlet estimate in Corollary 4.6 remains valid; the uncontrolled other sides have large values here.
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