An introduction to partial differential equations

A PDE describes a field over several independent variables

A partial differential equation relates an unknown field to its partial derivatives. For a temperature u(x,t), the position x and time t are independent inputs, while u is the dependent quantity. The heat equation u_t=κu_xx says that the local time rate is proportional to spatial curvature. In two spatial dimensions, u(x,y,t) has separate derivatives with respect to x, y and t. A subscript denotes differentiation while the other independent variables are held fixed.

An ordinary differential equation evolves finitely many state coordinates with one independent variable. A PDE can evolve an entire spatial profile, so initial data may be a function rather than a finite vector. Discretising that profile produces a large ODE system, but its grid-dependent solutions still need analysis before they represent a limit of the original field problem. An ODE local uniqueness theorem does not automatically establish existence, uniqueness or stability for an arbitrary PDE.

A classical solution has the derivatives required by the equation and satisfies it pointwise on its stated domain. For the heat equation, a common interior requirement is two continuous spatial derivatives and one continuous time derivative, written C²,¹. Boundary and initial traces require additional continuity or a specified weaker interpretation. A formula verified only at t>0 is not yet a proof that it attains a prescribed initial function in the intended sense.

Solutions can also be understood weakly, by an integral identity that transfers derivatives to test functions. This permits less smooth fields and is essential for shocks or finite-energy solutions. Classical and weak are descriptions of the solution concept, rather than automatic quality rankings. Every uniqueness claim must name its domain, boundary/initial data and admissible solution class. Throughout the geometric examples below, coordinates are dimensionless unless physical units are explicitly given.

Classify the equation using an explicit coefficient convention

The order is the highest derivative order appearing. The transport equation u_t+a u_x=0 is first order, while heat, wave and Laplace equations are second order. A linear differential operator has coefficients depending only on the independent variables and acts linearly on u. Thus u_t−κu_xx=f is linear with a prescribed source f. The equation u_t+u u_x=0 is nonlinear because a coefficient depends on the unknown field.

For a second-order scalar equation in two independent variables, write its principal part as A u_xx+2B u_xy+C u_yy. The symmetric principal matrix is [[A,B],[B,C]], and the discriminant in this convention is D=B²−AC. For a nonzero principal part, D<0 is elliptic, D>0 hyperbolic, and D=0 is the rank-one parabolic case. If the mixed term is instead written b u_xy, then b=2B and the corresponding discriminant is b²−4AC. Mixing conventions produces a factor-of-four error.

Laplace’s equation u_xx+u_yy=0 has A=C=1,B=0 and D=−1, so it is elliptic. The wave equation u_tt−c²u_xx=0, considered in variables (x,t), has A=−c²,C=1,B=0 and D=c²>0. The heat equation u_t−κu_xx=0 has principal matrix diag(−κ,0) in those variables, so its nonzero second-order part is parabolic. The first-order time term remains essential to the heat evolution even though it is outside that principal matrix.

This elementary classification is local when coefficients vary. In u_xx+xu_yy=0, the symbol is elliptic at x>0 and hyperbolic at x<0, with degeneracy at x=0. A single constant-coefficient transformation need not describe the whole domain across such a change. If A=B=C=0 at a point, the second-order principal part has vanished and the rank-one classification does not apply there. In higher dimensions, the principal matrix and the equation’s structure need their corresponding definitions.

The equation needs the right data in the right places

A PDE alone generally has many solutions. An initial condition specifies a field at a starting time, such as u(x,0)=φ(x). A Dirichlet boundary condition prescribes the field value; a Neumann condition prescribes its outward normal derivative ∂_n u. A Robin condition combines value and normal derivative. If a physical flux is −k∇u, prescribed normal derivative and prescribed outward flux differ by the sign and coefficient −k.

For heat on an interval 0<x<L, one initial temperature profile and one appropriate condition at each endpoint make a familiar evolution problem. For the wave equation, two independent time traces are needed, usually u(x,0)=φ(x) and u_t(x,0)=ψ(x). An elliptic stationary boundary problem has no distinguished starting time in these prototypes. Counting derivative orders is a useful guide, but the type, geometry and characteristic directions determine which data are legitimate.

Data at intersecting boundaries must agree if a continuous solution on the closed domain is requested. For example, heat data u(0,t)=g₀(t) and u(x,0)=φ(x) require φ(0)=g₀(0) at the corner. A solution with enough smoothness up to that corner can require further compatibility, such as κφ″(0)=g₀′(0) in the source-free case. Interior smoothing for t>0 does not erase a contradiction between two assigned corner values.

For transport with positive speed on an interval, information enters at the left endpoint and exits at the right. The inflow trace and initial profile determine the outflow. Imposing an unrelated right-endpoint value can overdetermine the problem. Conversely, prescribing data only along a characteristic can leave other characteristics undetermined. A data set must match the way the equation carries information, rather than a generic rule to fix values on every edge of a picture.

Balance laws and constitutive assumptions produce different PDEs

Let u be amount per unit length and J its rightward flux, with amount-per-time units. Conservation on a fixed interval [a,b] is d/dt ∫ₐᵇu dx=J(a,t)−J(b,t)+∫ₐᵇs dx, where s is local source per length per time. For smooth fields, differentiating locally gives u_t+J_x=s. The integral law is meaningful more broadly than its pointwise differentiated form, and the signs record outward loss at the right endpoint and inward gain at the left.

The PDE changes when a constitutive law specifies J. Constant transport velocity a gives J=au and u_t+a u_x=s, with a in length/time. A diffusion law J=−κu_x with constant κ>0 gives u_t=κu_xx+s, with κ in length²/time. Conservation alone did not select either law. Positivity, constant coefficients and the validity of the constitutive approximation are modelling assumptions, not consequences of writing a divergence.

For a thin uniform rod with temperature T, volumetric heat capacity ρc_p and conductivity k, a one-dimensional local balance is ρc_p T_t=kT_xx+Q. Dividing gives diffusivity κ=k/(ρc_p). Temperature has kelvin units, k has W/(m·K), ρc_p has J/(m³·K), and κ has m²/s. A rod’s cross-sectional area cancels only when treated consistently and constant; variable geometry or material properties can require a divergence-form operator instead.

For an ideal stretched string, transverse displacement u obeys ρu_tt=Tu_xx under the small-slope, constant-tension assumptions, where ρ is mass per length and T is tension. The speed c=√(T/ρ) has length/time units. Diffusion spreads a scalar profile dissipatively; wave evolution exchanges kinetic and elastic energy. Similar derivative symbols can therefore represent different balances, data requirements and propagation behaviour.

Follow the characteristics back to their data

For u_t+a u_x=0 on the whole real line with constant a, set x(t)=ξ+at. The chain rule gives d/dt u(x(t),t)=u_t+a u_x=0, so u is constant along that characteristic. The initial profile φ determines u(x,t)=φ(x−at). This is a classical solution for C¹ φ, verified by differentiating and checking the initial trace. Positive a moves a labelled feature rightward with speed a; it does not change the profile’s height or shape.

The characteristic construction has two parts: solve its ODEs and recover the label ξ from the observation point. Here ξ=x−at is globally available. In a variable or nonlinear field the projection from labels and time to physical coordinates can lose invertibility. Local ODE solutions alone do not ensure a single-valued smooth field after that happens. Data must also meet a noncharacteristic condition when one reconstructs a local solution from a general initial curve.

On 0<x<L with a>0 and t≥0, trace a characteristic backwards. If x−at≥0, it reaches the initial edge first and gives φ(x−at). If x−at<0, it reaches the inflow boundary at time t−x/a and gives g(t−x/a). Thus u(x,t)=φ(x−at) for x≥at and u(x,t)=g(t−x/a) for x<at, using traces where the edges meet. Continuity across x=at requires φ(0)=g(0); C¹ matching also requires g′(0)=−aφ′(0).

For example choose a=2, L=4, φ(x)=1+x and g(t)=1−2t+3t². The two matching conditions hold. The solution equals 1+x−2t in the initial-data region and 1−2(t−x/2)+3(t−x/2)² in the inflow region. The outflow trace at x=4 follows from these formulas. This piecewise construction is C¹ at the dividing characteristic even though some second derivatives differ, which is sufficient for the stated first-order classical equation.

Diffusion damps different spatial modes at different rates

For u_t=κu_xx on 0<x<L with κ>0 and zero Dirichlet endpoints, try u(x,t)=X(x)T(t). Substitution gives XT′=κX″T. Where division is valid, T′/(κT)=X″/X must be one constant. Seeking nontrivial endpoint-zero modes gives X_n(x)=sin(nπx/L), n≥1, with X_n″=−(nπ/L)²X_n. The time factor satisfies T_n′=−κ(nπ/L)²T_n and therefore decays exponentially.

Every finite sum u=Σ a_n e^(−κ(nπ/L)²t)sin(nπx/L) is directly verified by differentiating term by term. For an infinite sum, convergence and differentiability must be justified in the requested function space and time range. An expansion that converges initially in L² is not automatically a pointwise classical equality at every boundary corner. Positive-time smoothing can strengthen regularity under suitable hypotheses, but the mode calculation by itself is not that theorem.

On 0<x<π with κ=1/2, the profile 3sin x−sin3x equals 4sin³x. Its solution is u=3e^(−t/2)sin x−e^(−9t/2)sin3x. The shorter-wavelength third mode decays nine times as fast in its exponent as the first. The initial field is nonnegative and bounded by 4. The maximum principle, with zero boundary data, preserves the bounds 0≤u≤4 rather than relying on visual inspection of a difference of two sine modes.

For insulated zero-Neumann endpoints, a constant spatial mode is allowed. With initial profile 3+cos(2πx/L), the solution is 3+e^(−4κπ²t/L²)cos(2πx/L). Its integral is 3L at every time, and it approaches the mean 3. Zero Dirichlet endpoints instead permit diffusive loss through the boundary and generally do not conserve the integral. The boundary condition determines which conservation or decay statement is available.

The solution class matters even when a kernel is explicit

On the whole line, the heat kernel is K(x,t)=(4πκt)^(−1/2)exp(−x²/(4κt)) for t>0. Direct differentiation gives K_t=κK_xx, and a Gaussian substitution gives ∫ℝK dx=1. If x has length units, K has inverse-length units. Its variance is 2κt, with length-squared units, so √(2κt) is a natural spreading scale. This is a kernel with a singular initial concentration, not an ordinary finite value K(x,0).

For bounded continuous initial data φ, convolution u(x,t)=∫ℝK(x−y,t)φ(y)dy supplies a smooth solution for t>0 attaining φ locally as t decreases to 0. It is unique within the appropriate bounded classical solution class with the prescribed continuous initial trace on each finite time strip. On an unbounded domain, unrestricted smooth growth is not covered by the bounded maximum/uniqueness argument; dropping all growth control can destroy uniqueness. The equation and initial symbols alone do not name that class.

Because K is positive everywhere for t>0, nonnegative compactly supported continuous data that are not identically zero produce u(x,t)>0 everywhere at every positive time. This ideal diffusion model has instantaneous influence, unlike the finite-speed wave equation. It is a property of the continuum approximation; it should not be advertised as a microscopic relativistic transport law. Coefficient validity and the time/length scales of the model remain physical questions.

For the interval problem on 0<x<L, 0<t≤T, assume u is continuous on the closed rectangle and C²,¹ in its interior time-positive region with u_t=κu_xx. Its maximum and minimum are controlled by the parabolic boundary: the initial edge and the two spatial sides, without needing the final edge. Applying this principle to the difference of two solutions proves uniqueness and uniform continuous dependence on Dirichlet/initial data in that classical class. With a source or a reaction term, the applicable comparison statement must be adjusted.

Waves need displacement and velocity, and have finite dependence

For u_tt=c²u_xx on ℝ with c>0, initial displacement φ and velocity ψ are independent inputs. With φ∈C² and ψ∈C¹, d’Alembert’s formula is u(x,t)=[φ(x−ct)+φ(x+ct)]/2+(1/(2c))∫_{x−ct}^{x+ct}ψ(s)ds. Differentiation verifies the PDE and both initial conditions. The opposite signs in the moving arguments describe left- and right-travelling contributions; the velocity integral cannot be omitted unless the initial velocity actually vanishes.

The value at (x,t) depends only on initial data in [x−ct,x+ct]. Changing the data outside that interval cannot change this value. Equivalently, a compactly supported disturbance cannot influence points beyond its speed-c cone under this homogeneous whole-line model. This domain-of-dependence statement does not require every allowed initial profile to have compact support. Boundary reflections or forcing need their own modified construction.

Take c=2, φ(x)=x² and ψ(x)=3x in dimensionless coordinates. Averaging the two squared arguments gives x²+4t², and the integral contributes 3xt. Hence u=x²+4t²+3xt. The residual is u_tt−4u_xx=8−8=0, u(x,0)=x² and u_t(x,0)=3x. These unbounded polynomial data make a valid classical pointwise example, but its total energy on the entire real line is infinite; it is not an example for a finite-total-energy estimate.

On a finite string with fixed endpoints, reflected travelling waves or a sine-mode expansion can satisfy the boundary conditions. For example, sin(nπx/L)cos(cnπt/L) has zero displacement at x=0,L and zero initial velocity. A sine time factor supplies an independent velocity contribution. Undamped wave modes oscillate rather than exhibit the diffusive n² damping rate. Treating the two equations as interchangeable mode evolutions would erase their different physics.

Wave energy is conserved only with the right boundary balance

For a sufficiently smooth ideal string on 0<x<L, let e=(ρu_t²+Tu_x²)/2, with positive constant mass-per-length ρ and tension T. This is energy per length under the small-slope model. Using ρu_tt=Tu_xx gives e_t=T(u_tu_x)_x. Thus E(t)=∫₀ᴸe dx has derivative E′=T[u_tu_x]₀ᴸ. Equivalently the rightward energy flux is −Tu_tu_x, which puts the law in the usual e_t+J_x=0 form.

For fixed endpoints u(0,t)=u(L,t)=0 at every time, sufficient boundary smoothness implies u_t=0 there, so E′=0. Zero slope at both endpoints also kills this boundary work. Driven boundary motion can inject or remove energy, and damping or distributed forcing changes the bulk balance. On the whole line, the analogous global calculation requires finite energy and appropriate control of the boundary terms at infinity; the polynomial example deliberately did not meet that global condition.

The energy can prove uniqueness in a specified finite-energy wave problem. For two smooth solutions with equal displacement and velocity data and identical fixed-endpoint conditions, their difference w has initial energy 0 and conserved nonnegative energy. Thus w_t=w_x=0, and its zero boundary value fixes the remaining constant, giving w=0. This argument establishes at most one solution in its class; existence still needs a construction or an appropriate theorem.

A conserved squared-gradient/velocity energy does not imply that each point’s displacement is constant or that every coordinate amplitude decreases monotonically. Kinetic and elastic contributions can exchange. Nor is the spatial integral of u necessarily the physical string energy. Always retain the quantity’s definition, physical coefficients, unit and boundary flux before using the word conserved. For heat, ∫u²/2 is often a useful dissipating mathematical energy rather than the thermodynamic heat content itself.

Stationary fields are fixed by a boundary problem

The Laplacian in Cartesian space is Δu=Σ_j u_{x_jx_j}. A field is harmonic if Δu=0. For a C² harmonic function on a bounded open domain that is continuous on its closure, the maximum and minimum occur on the boundary. On a connected domain, attainment of an interior global maximum forces constancy. This is a property of harmonic fields under those hypotheses; an arbitrary source term or an unbounded domain needs its own statement.

For a Dirichlet problem, the difference of two harmonic solutions with the same boundary values is harmonic and has zero boundary values. Applying maximum and minimum principles gives difference 0. This proves uniqueness in the stated classical class, without asserting that every imaginable boundary prescription on every irregular domain has a classical solution. The existence and boundary regularity problem is a separate part of the analysis.

The polynomial u=x²−y²+2xy+3x−y is harmonic on the unit square. Its four actual boundary traces agree at their shared corners. On x=1 the trace is 4+y−y², with maximum 17/4 at y=1/2; checking all edges gives minimum −2 at (0,1). The maximum principle then gives −2≤u≤17/4 throughout the square, and the field is the unique continuous-closure harmonic solution with those full traces. Matching only four corner values would leave the boundary problem undetermined.

Poisson’s equation introduces a source; fix the sign convention as −Δu=f. On a connected bounded domain with sufficiently smooth boundary and a solution C² up to that boundary, prescribed Neumann data ∂_n u=g and the divergence theorem require ∫Ωf dx=−∫∂Ωg dS. An arbitrary additive constant remains undetermined by derivative data, so a mean or another normalization is needed for uniqueness. On the unit disk, u=x²+y²−1/2 gives f=−4, g=2 and mean 0. Its area integral −4π equals the negative boundary flux −4π. These compatibility and constant-mode issues distinguish Neumann from Dirichlet data.

Separation is a construction, with boundary and convergence conditions

For zero Dirichlet endpoints on an interval, sine functions satisfy ∫₀ᴸsin(nπx/L)sin(mπx/L)dx=0 for n≠m and L/2 for n=m. If φ has an appropriate sine expansion, the coefficients are a_n=(2/L)∫₀ᴸφ(x)sin(nπx/L)dx. The orthogonality calculation identifies coefficients; a completeness and convergence result establishes which functions the series represents and in what sense. These are distinct steps.

The same spatial eigenfunctions can carry different time equations. Heat gives a_n′=−κ(nπ/L)²a_n. Wave gives a_n″=−c²(nπ/L)²a_n and needs both a_n(0) and a_n′(0). With a source, each projected mode becomes a forced ODE. Fixed finite sums are simple exact checks, while infinite expansions need estimates justifying differentiation, traces and reconstruction. A separated product is one useful building block, not a promise that every field is a single product.

The spatial operator and its boundary conditions belong together. Changing Dirichlet endpoints to insulated Neumann endpoints adds a constant mode and replaces the sine basis by suitable cosine modes. On a connected insulated region, that constant mode encodes conserved mean. For more general coefficients, geometry or boundary laws, the eigenfunctions, weights and spectrum can differ. A familiar sine series should not be reused after silently changing the operator domain.

Separation connects PDEs to linear algebra through eigenvalues and to integration through orthogonal projection, but the infinite-dimensional setting adds convergence and domain issues. In a finite numerical basis, discarded high modes may be small for forward heat at positive time because of strong damping. The same omission can matter for sharp wave features or an inverse heat reconstruction. Approximation quality depends on the equation, time direction, data regularity and norm being measured.

A solution formula does not guarantee a stable inverse problem

Hadamard well-posedness asks for existence, uniqueness and continuous dependence on data in specified spaces. The last condition means small changes in the data norm lead to small changes in the chosen solution norm. Saying that a method returns a curve, or that an algebraic expression satisfies the PDE, answers only part of this question. The time direction, data topology and admissible solution class are part of the problem definition.

For forward heat on an interval with fixed zero Dirichlet endpoints, the maximum principle bounds the difference of two classical solutions by the initial difference in supremum norm. Thus small initial errors remain small in this measure, provided the initial/boundary regularity and compatibility needed for that class hold. This is continuous dependence and uniqueness for an existing pair of solutions, not a general existence theorem for arbitrary rough data.

Now fix T>0 and attempt to reconstruct time 0 from the temperature at time T. For n≥1, u_n(x,t)=e^(−κn²t)sin(nx) solves heat on 0<x<π with zero Dirichlet boundary. Its final data g_n=e^(−κn²T)sin(nx) tend uniformly to 0 as n→∞. Yet the reconstructed initial fields sin(nx) all have supremum norm 1. Thus the inverse final-to-initial map fails continuous dependence in these norms even though every individual member is a valid smooth solution.

This example isolates instability without claiming that every final field has a backward solution. High frequencies that forward diffusion nearly erases become strongly amplified during inversion. Regularisation can trade reconstruction resolution against noise sensitivity, but it changes how information is used and requires an explicit error model and parameter choice. Numerical precision alone cannot restore continuity of the unregularised inverse map. The same distinction between a forward model and an inverse inference appears in imaging and parameter estimation.

A smooth field can lose its gradient before its values diverge

For the one-dimensional conservation law u_t+(f(u))_x=0, smooth solutions satisfy u_t+f′(u)u_x=0. In Burgers’ equation f(u)=u²/2, the field itself determines characteristic speed. From initial label ξ with value φ(ξ), the characteristic is x=ξ+tφ(ξ) and carries u=φ(ξ). Smooth reconstruction requires a locally invertible label map, whose derivative is 1+tφ′(ξ).

Choose φ(ξ)=−sinξ. Then x=ξ−t sinξ and u=−sinξ. For 0≤t<1, the label derivative 1−t cosξ is positive everywhere, so the map is globally increasing and defines a smooth field. Its gradient is u_x=−cosξ/(1−t cosξ), which first becomes unbounded at ξ=0 as t approaches 1. The field value remains bounded between −1 and 1. Gradient blow-up differs from finite-time blow-up of the field itself.

After characteristic crossing, assigning all the characteristic values at one location does not produce a single-valued classical field. The integral conservation law suggests a weak interpretation. For a locally integrable u and flux f(u), with initial trace u₀, require ∫₀∞∫ℝ[uφ_t+f(u)φ_x]dxdt+∫ℝu₀(x)φ(x,0)dx=0 for every smooth compactly supported test function φ on space and nonnegative time. Derivatives now fall on the smooth test function. The sign of the initial term follows from integration by parts.

The weak identity can accommodate jumps and need not select a unique continuation by itself. For a piecewise smooth solution with a jump moving at speed s, conservation requires s(u_L−u_R)=f(u_L)−f(u_R). This Rankine–Hugoniot condition comes from the flux balance across a narrow moving control volume. It is not an extra arbitrary fit parameter. If the states agree, there is no jump and division by their difference is inappropriate.

Conservation gives a jump speed; admissibility selects the continuation

For Burgers’ convex quadratic flux, a shock from u_L to u_R has speed s=(u_L+u_R)/2 when the states differ. If u_L>u_R, then u_L>s>u_R: characteristics enter the moving jump from both sides. This compressive shock is the admissible entropy solution of the corresponding Riemann problem, with two constant initial half-lines. Reversing the states gives the same algebraic speed but an expansion shock that still satisfies the weak balance and is inadmissible.

For states u_L=3,u_R=−1, s=1, and the piecewise constant shock travels along x=t. The incoming characteristic speeds satisfy 3>1>−1. For the reversed data u_L=−1,u_R=3, the entropy solution at t>0 is u=−1 for x/t≤−1, u=x/t for −1<x/t<3, and u=3 for x/t≥3. The continuous fan fills the expanding gap between characteristic families. In the middle region u_t=−x/t² and uu_x=x/t², so the smooth residual is zero.

One general scalar entropy formulation requires ∂_tη(u)+∂_xq(u)≤0 in distributions for every convex entropy η and its associated flux q satisfying q′=η′f′. For Burgers, η=u²/2 gives q=u³/3. Across the compressive 3-to-(−1) shock, using [h]=h_R−h_L, the entropy measure coefficient is −s[η]+[q]=−16/3<0. The reversed jump changes that sign, violating even this particular convex entropy inequality. The full entropy solution concept retains the required family and data/solution hypotheses.

The simple incoming-speed test above is being used for the explicit convex Burgers flux and its Riemann states. General nonconvex fluxes and systems require their appropriate entropy or admissibility theory; one numerical slope inequality cannot be promoted to a universal selection theorem. A small positive viscosity u_t+(u²/2)_x=νu_xx provides a related physical regularisation, with an entropy limit under the relevant assumptions. It does not license smoothing a discontinuity and declaring the unsmoothed model classically solved.

Use the integral identity and the numerical method that were actually checked

For −Δu=f with zero Dirichlet boundary, multiplying a smooth solution by a smooth test function v that vanishes on the boundary and integrating by parts gives ∫Ω∇u·∇v dx=∫Ωfv dx. This motivates a finite-energy weak formulation with u in H¹₀(Ω). Here H¹(Ω) consists of square-integrable functions whose first weak derivatives are also square integrable, and H¹₀(Ω) is the closure of smooth compactly supported functions in the H¹ norm. On Lipschitz domains, H¹₀ is equivalently the zero-trace class. On a suitable bounded domain and with appropriate source regularity, variational existence theory makes this precise; the formal integration alone does not prove that every requested source has a solution.

Uniqueness follows from the energy identity in this class when the zero-boundary Poincaré inequality applies. If w is the difference of two weak solutions with the same source, use v=w to obtain ∫Ω|∇w|²=0, and then ||w||₂≤C||∇w||₂ gives w=0. A weak derivative, a trace and an admissible test space are mathematical definitions, not the instruction to differentiate a discontinuous field by ordinary pointwise rules. Nonzero Neumann conditions retain their different constant-mode issue.

For the heat equation on a uniform grid, forward time and centred space give u_j^{n+1}=r u_{j−1}^n+(1−2r)u_j^n+r u_{j+1}^n, where r=κΔt/Δx². If 0≤r≤1/2, this is a convex combination of three old values and supports a discrete maximum bound with properly prescribed boundaries. On an even periodic grid, the alternating mode (−1)^j is multiplied by 1−4r. At r=1 it grows with alternating sign by factor −3 per step, even though positive physical diffusion damps short wavelengths.

Consistency of a difference formula is distinct from stability and convergence. The explicit heat method is first order in time and second order in space for sufficiently smooth fields, but refining time and space along an unstable ratio can defeat convergence. Boundary treatment, forcing, resolution and the measurement norm remain part of the numerical problem. A small residual at a few grid points is not a proof of an exact field, its boundary traces or continuous dependence.

PDEs connect local conservation to global flux, eigenmodes to infinite-dimensional evolution, curvature to diffusion, and energy estimates to uniqueness and stability. Fluids, electromagnetism, elasticity and quantum fields add their own constraints, constitutive assumptions and boundary laws. The useful question is therefore precise: which field problem, solution concept and data class are being solved, with what balance, propagation and error statement? An equation label or a rendered animation alone cannot answer all of those questions.

Worked examples

Try it yourself

1. For u=u(x,y,t) with u_t=κ(u_xx+u_yy), identify the independent variables, dependent field and order.

   Show the answer and reasoning

   Independent variables x,y,t; dependent field u; second order.

   The highest derivatives are spatial second derivatives. A first time derivative does not make the entire PDE first order.

2. Which is linear with prescribed coefficients: u_t+xu_x=u, or u_t+u u_x=0?

   Show the answer and reasoning

   The first is linear; the second is nonlinear.

   The coefficient x is an independent variable. In the second equation the transport coefficient is the unknown field itself, so superposition generally fails.

3. For 2u_xx+6u_xy+2u_yy=0, compute D in the 2B convention.

   Show the answer and reasoning

   A=C=2, B=3, D=5>0, so hyperbolic.

   The mixed coefficient is 2B=6, not B=6. The principal matrix is indefinite, consistent with the positive discriminant.

4. If x is in metres and t in seconds, what are the units of κ in u_t=κu_xx?

   Show the answer and reasoning

   m²/s.

   u_t has field-units/s and u_xx has field-units/m². Multiplication by κ must restore matching units.

5. Can continuous initial data φ(0)=2 coexist with boundary value u(0,t)=3 for t≥0 in a continuous closed-rectangle heat solution?

   Show the answer and reasoning

   No. The corner assignments contradict one another.

   At (0,0) the initial edge requires 2 and the boundary edge requires 3. Interior positive-time smoothing is a different issue and does not make the two traces equal.

6. For u_t+3u_x=0 and φ(x)=2+sin(2x), give the whole-line classical solution.

   Show the answer and reasoning

   u(x,t)=2+sin(2x−6t).

   Characteristics satisfy x−3t=constant. Differentiating gives u_t=−6cos(2x−6t) and 3u_x=6cos(2x−6t), with zero sum.

7. For transport speed a<0 on 0<x<L, which endpoint supplies the inflow data?

   Show the answer and reasoning

   The right endpoint x=L.

   Information travels leftward. Tracing a future characteristic backwards reaches the right boundary; an independently prescribed left outflow can overdetermine the problem.

8. For zero-Dirichlet heat mode sin(3πx/L), what is its amplitude multiplier after time t?

   Show the answer and reasoning

   e^(−9κπ²t/L²).

   The second spatial derivative multiplies that mode by −(3π/L)², leaving the scalar amplitude equation a′=−9κπ²a/L².

9. For insulated heat evolution, must both ∫u dx and ∫u²/2 dx be constant?

   Show the answer and reasoning

   No. The mean integral is conserved, while squared-norm energy generally decreases.

   Integration gives zero net endpoint flux for the first quantity. Integration by parts gives d/dt ∫u²/2=−κ∫u_x² when boundary flux vanishes.

10. For κ=0.5 m²/s and t=4 s, what is the whole-line kernel’s standard deviation?

    Show the answer and reasoning

    2 m.

    The variance is 2κt=4 m², so its square root is 2 m. This is not the radius of a compactly supported heat front.

11. Why does u_tt=c²u_xx usually need both u(x,0) and u_t(x,0)?

    Show the answer and reasoning

    They supply independent displacement and velocity profiles.

    Each wave mode obeys a second-order time equation with two independent constants. Omitting one profile leaves the standard Cauchy problem undetermined.

12. For c=3, which initial interval can influence the whole-line wave value at x=2,t=1?

    Show the answer and reasoning

    [−1,5].

    d’Alembert’s formula uses φ at x±ct and ψ between those points, so only the interval [2−3,2+3] appears.

13. If two continuous-closure harmonic fields on the same bounded domain have identical full Dirichlet traces, what can the maximum principle prove?

    Show the answer and reasoning

    They are equal in the classical class.

    Their difference is harmonic with zero boundary values. Applying both maximum and minimum bounds makes the difference zero; existence is a separate question.

14. On the unit square, verify the source and boundary for u=sin(πx)sin(πy) in −Δu=f.

    Show the answer and reasoning

    f=2π²sin(πx)sin(πy), with zero Dirichlet values on all four sides.

    Each pure second derivative contributes −π²u. Integration by parts also gives the corresponding weak identity for zero-boundary tests.

15. For −Δu=f and ∂_n u=g, state the necessary integral compatibility and the remaining normalization issue.

    Show the answer and reasoning

    ∫Ωf=−∫∂Ωg; an additive constant remains.

    The divergence theorem applied to Δu fixes the sign. Normal derivatives and the Laplacian do not detect addition of a constant.

16. Does uniqueness of a backward heat reconstruction, when it exists in a class, imply continuous dependence on its final data?

    Show the answer and reasoning

    No. Uniqueness and continuous dependence are separate conditions.

    The sequence e^(−κn²T)sin(nx) has final sup norms tending to zero but reconstructs initial sup norm 1. An exact solution can still represent an unstable inverse problem.

17. For Burgers data φ(ξ)=−sinξ, when does the characteristic projection first lose its strictly positive derivative in forward time?

    Show the answer and reasoning

    At t=1, at labels with cosξ=1.

    The derivative is 1−t cosξ. Its first zero occurs at t=1, while u remains bounded. This is gradient blow-up and loss of smooth reconstruction, not unbounded field values.

18. For κ=2 m²/s and Δx=0.2 m, what maximum Δt meets r≤1/2 for the explicit heat scheme?

    Show the answer and reasoning

    Δt≤0.01 s, with a positive step chosen for evolution.

    κΔt/Δx²≤1/2 gives Δt≤Δx²/(2κ)=0.04/4=0.01 s. The ratio is dimensionless, and the limiting alternating mode need not decay.

What to carry forward

• A PDE solves for a field, so its domain, data and admissible solution class belong to the problem.
• Order and linearity are different properties; a coefficient depending on the unknown can make a first-order equation nonlinear.
• State the mixed-derivative convention before computing a principal discriminant.
• Initial and boundary data must match both the equation’s information direction and the requested corner regularity.
• Conservation supplies a balance, while a constitutive law selects transport, diffusion or another flux.
• Characteristics need both an ODE evolution and an invertible reconstruction to define a smooth field.
• Diffusion damps high spatial modes rapidly, and its boundary conditions decide whether the mean is conserved.
• A heat kernel needs a stated initial-trace and growth class; ideal infinite influence is a model property.
• Wave displacement and velocity data are independent and determine a finite domain of dependence.
• Wave energy has a boundary work term; constant total energy does not freeze the displacement field.
• Maximum principles qualify classical boundary uniqueness rather than prove every boundary prescription has a solution.
• Neumann Poisson data need an integral compatibility condition and a normalization for the constant mode.
• Orthogonality identifies expansion coefficients, while completeness and convergence justify reconstruction.
• Existence, uniqueness and continuous dependence are separate claims, as backward heat amplification demonstrates.
• Nonlinear characteristics can lose smoothness through a diverging gradient while field values stay bounded.
• A weak jump balance determines a speed, while entropy admissibility selects the appropriate nonlinear continuation.
• Weak energy identities and numerical stability conditions require their actual function spaces, grids, norms and boundaries.