The equation uₓₓ + 4uₓ_{y} + 3u_{yy} + uₓ = 0 is

of different types in different regions

Which of the following equations are hyperbolic at every point of ℝ²?

uₓₓ + 2uₓ_{y} + u_{yy} = 0

Under ξ = x + y, η = x − y, the equation uₓₓ − u_{yy} = 0 becomes

u_{ξη} = 0, with general solution F(x + y) + G(x − y)

For u_{t} + cuₓ = 0 with c a non-zero constant, which statements are true?

every solution is constant along each line x − ct = const

the solution with u(x, 0) = f(x) is u = f(x − ct)

every solution is constant along each line x − ct = const

the solution with u(x, 0) = f(x) is u = f(x − ct)

the line t = 0 is a characteristic

prescribing u on the line x = ct determines a unique solution

Which statements about harmonic functions are true?

a non-constant harmonic function on a bounded domain, continuous on its closure, attains its maximum only on the boundary

the Dirichlet problem for Laplace’s equation on a bounded domain has at most one solution continuous on the closure

a harmonic function on all of ℝ² that is bounded above is constant

a non-constant harmonic function on a bounded domain, continuous on its closure, attains its maximum only on the boundary

the Dirichlet problem for Laplace’s equation on a bounded domain has at most one solution continuous on the closure

a harmonic function on all of ℝ² that is bounded above is constant

the value of a harmonic function at the centre of a disc equals its maximum on the boundary circle

For the Cauchy problem u_{t} = uₓₓ on ℝ with u(x, 0) = f(x) ≥ 0, f continuous, bounded, and positive only on [0, 1], the solution satisfies

u(x, t) > 0 for every x ∈ ℝ and every t > 0

u(x, t) = 0 outside [0, 1] for all t

u is not differentiable for t > 0

Partial Differential Equations: Characteristics, Classification, and the Laplace, Heat and Wave Equations

Partial Differential Equations is Section 9 of the GATE Mathematics (MA) paper. Its first-order half is one idea — the method of characteristics, which turns a linear or quasilinear equation into ordinary differential equations along curves, and which also explains when a Cauchy problem is solvable and when a quasilinear solution breaks down. Its second-order half is three equations in two variables and the classification that separates them: the discriminant decides whether an equation is hyperbolic (the wave equation), parabolic (the heat equation) or elliptic (Laplace’s equation), and each type has its own canonical form, its own correct boundary data and its own qualitative behaviour. The chapter covers separation of variables for Laplace’s equation in Cartesian and polar coordinates with the maximum principle, the heat and wave equations on an interval, the Cauchy problem for the wave equation with d’Alembert’s formula, domains of dependence and influence and the non-homogeneous equation, the Cauchy problem for the heat equation, and the Laplace and Fourier transform methods.

1. First-order linear and quasilinear equations: the method of characteristics

For the quasilinear equation a(x, y, u)uₓ + b(x, y, u)u_y = c(x, y, u), the graph of a solution is a union of characteristic curves, the solutions of dx/a = dy/b = du/c (Lagrange’s auxiliary equations). If φ(x, y, u) = c₁ and ψ(x, y, u) = c₂ are two independent integrals, the general solution is F(φ, ψ) = 0. For the Cauchy problem — u prescribed along a curve Γ — a unique local solution exists when Γ is non-characteristic: its tangent is nowhere parallel to the projected characteristic direction (a, b).

Worked characteristics
Problem	Characteristics	Solution
u_t + cuₓ = 0, u(x, 0) = f(x)	x − ct = const, u constant on each	u = f(x − ct), a travelling wave
uₓ + 2u_y = 0, u(x, 0) = sin x	y − 2x = const	u = sin(x − y/2)
xuₓ + yu_y = 2u, u(x, 1) = x	y/x = c₁, u/x² = c₂	u = x² f(y/x) with f(s) = s, so u = xy
u_t + uuₓ = 0, u(x, 0) = f(x) (inviscid Burgers)	straight lines x = x₀ + f(x₀)t carrying u = f(x₀)	implicitly u = f(x − ut); for f(x) = x, u = x/(1 + t)

⚠️ Quasilinear solutions can break down in finite time

For u_t + uuₓ = 0 the characteristic from x₀ has speed f(x₀). If f is decreasing somewhere, faster characteristics start behind slower ones and they cross; the classical solution ceases to exist at t∗ = −1/min f′. For f(x) = −2x, u = −2x/(1 − 2t), which blows up (all characteristics meet) at t = 1/2. A linear equation with smooth coefficients never does this: its characteristics do not depend on u.

2. Second-order equations: classification and canonical forms

For a uₓₓ + b uₓ_y + c u_yy + (lower-order terms) = 0 with a, b, c functions of (x, y), the type at a point is fixed by the discriminant Δ = b² − 4ac: hyperbolic if Δ > 0, parabolic if Δ = 0, elliptic if Δ < 0. The type is unchanged by any smooth invertible change of variables. The characteristic curves satisfy a(dy)² − b dx dy + c(dx)² = 0, i.e. dy/dx = (b ± √Δ)/(2a).

Types and canonical forms
Type	Characteristics	Canonical form	Model
hyperbolic, Δ > 0	two real families ξ, η	u_ξη + … = 0 (or u_αα − u_ββ + … = 0)	u_tt = c²uₓₓ
parabolic, Δ = 0	one real family	u_ηη + … = 0	u_t = kuₓₓ
elliptic, Δ < 0	none real	u_αα + u_ββ + … = 0	uₓₓ + u_yy = 0

uₓₓ + 4uₓ_y + 3u_yy = 0: Δ = 16 − 12 = 4 > 0, hyperbolic. uₓₓ + 2uₓ_y + u_yy = 0: Δ = 0, parabolic. uₓ_y = 0: Δ = 1, hyperbolic, already canonical, with general solution F(x) + G(y).
The Tricomi equation y uₓₓ + u_yy = 0 has Δ = −4y: elliptic for y > 0, parabolic on y = 0, hyperbolic for y < 0. The type can change across a curve.
For uₓₓ − u_yy = 0, ξ = x + y, η = x − y give −4u_ξη = 0, so u = F(x + y) + G(x − y).

3. Laplace’s equation: separation of variables and the maximum principle

Rectangle 0 < x < a, 0 < y < b, with u = 0 on three sides and u(x, b) = f(x): separating u = X(x)Y(y) gives X″ + λX = 0 with X(0) = X(a) = 0, so Xₙ = sin(nπx/a), and Yₙ = sinh(nπy/a). Then u = Σ bₙ sin(nπx/a) sinh(nπy/a)/sinh(nπb/a), with bₙ the Fourier sine coefficients of f. Disc r < R in polar coordinates: u_rr + u_r/r + u_θθ/r² = 0 gives Θ periodic, so Θ = cos nθ, sin nθ and R(r) = rⁿ (r⁻ⁿ and log r are discarded for boundedness at 0): u = a₀/2 + Σ (r/R)ⁿ(aₙ cos nθ + bₙ sin nθ), with aₙ, bₙ the Fourier coefficients of the boundary data.

Example. On the unit disc with u(1, θ) = cos²θ = ½ + ½ cos 2θ, u = ½ + ½r² cos 2θ = ½ + ½(x² − y²). The centre value ½ is the average of the boundary data (the mean value property), and u(1/2, 0) = ½ + ⅛ = 0.625.

🎯 The maximum principle and uniqueness

If Δu = 0 in a bounded domain Ω and u is continuous on the closure, then max and min of u are attained on ∂Ω; if they are attained inside, u is constant (strong form, from the mean value property). Consequence: two solutions of the Dirichlet problem with the same boundary data differ by a harmonic function vanishing on ∂Ω, whose max and min are 0, so the solution is unique. The heat equation has the same principle on the parabolic boundary (the base and sides of the space–time rectangle).

4. Heat and wave equations in one space variable

Heat on an interval. u_t = kuₓₓ on 0 < x < L, u(0, t) = u(L, t) = 0, u(x, 0) = f(x): u = Σ bₙ e^{−k(nπ/L)²t} sin(nπx/L). Higher modes decay faster; with L = π, k = 1 and f = sin x + 3 sin 2x, u = e^−t sin x + 3e^−4t sin 2x. With insulated ends uₓ = 0 the modes are cos(nπx/L) and u tends to the mean of f.
Wave on an interval. u_tt = c²uₓₓ, fixed ends: u = Σ [aₙ cos(nπct/L) + bₙ sin(nπct/L)] sin(nπx/L), with aₙ from u(x, 0) and bₙ(nπc/L) from u_t(x, 0). The modes oscillate without decay.

Cauchy problem for the wave equation on ℝ — d’Alembert. u_tt = c²uₓₓ, u(x, 0) = f(x), u_t(x, 0) = g(x) has the unique solution u(x, t) = ½[f(x − ct) + f(x + ct)] + (1/2c)∫_x−ct^x+ct g(s) ds. The value at (x, t) depends only on the data on the domain of dependence [x − ct, x + ct]; data at x₀ affects only the domain of influence |x − x₀| ≤ ct: finite speed of propagation. For the non-homogeneous equation u_tt − c²uₓₓ = F with zero data, Duhamel’s principle adds (1/2c)∫₀ᵗ∫_{x−c(t−τ)}^x+c(t−τ) F(s, τ) ds dτ; for F ≡ 2 and c = 1 this is t².

Cauchy problem for the heat equation on ℝ. u_t = kuₓₓ, u(x, 0) = f(x) bounded and continuous: u(x, t) = (4πkt)^−1/2∫ e^{−(x−y)²/(4kt)} f(y) dy, the convolution with the heat kernel. It is C^∞ for t > 0 even if f is only continuous, and a positive bump in f makes u > 0 everywhere at once: infinite speed of propagation, the opposite of the wave equation. Bounded solutions are unique.

5. Laplace and Fourier transform methods

Fourier transform in x (on ℝ): with û(ω, t) = ∫ u e^−iωx dx, the heat equation becomes û_t = −kω²û, so û = f̂(ω)e^−kω²t, and the inverse transform is the heat-kernel formula. For f = e^−x² and k = 1, u = (1 + 4t)^−1/2 e^{−x²/(1 + 4t)}; at x = 0, t = 3/4 this is 1/2.
Laplace transform in t (for t > 0, often on a half-line in x): L{u_t} = sU − u(x, 0) turns the PDE into an ODE in x with parameter s. For u_t = uₓₓ on x > 0 with u(x, 0) = 0, u(0, t) = 1 and u bounded, sU = U″ gives U = e^−x√s/s, whose inverse is erfc(x/(2√t)).

Key takeaways

Characteristics dx/a = dy/b = du/c solve first-order quasilinear equations; data must be given on a non-characteristic curve, and quasilinear solutions can break at t∗ = −1/min f′.
For a uₓₓ + b uₓ_y + c u_yy, Δ = b² − 4ac > 0 hyperbolic, = 0 parabolic, < 0 elliptic, pointwise; Tricomi changes type across y = 0.
Laplace on a disc: u = a₀/2 + Σ rⁿ(aₙ cos nθ + bₙ sin nθ); harmonic functions obey the mean value property and the maximum principle, which gives uniqueness.
d’Alembert: u = ½[f(x − ct) + f(x + ct)] + (1/2c)∫g over [x − ct, x + ct], the domain of dependence; waves travel at speed c, heat at infinite speed.
Separation of variables gives e^{−k(nπ/L)²t} sin(nπx/L) for heat and trigonometric time factors for waves; Fourier in x and Laplace in t turn PDEs into ODEs.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The equation uₓₓ + 4uₓ_y + 3u_yy + uₓ = 0 is
1. hyperbolic everywhere
2. parabolic everywhere
3. elliptic everywhere
4. of different types in different regions
Show answer
Answer: A — hyperbolic everywhere
With a = 1, b = 4, c = 3 the discriminant is b² − 4ac = 16 − 12 = 4 > 0 at every point, so the equation is hyperbolic everywhere; the first-order term plays no part in the type. Using b² − ac with the coefficient of uₓ_y as b would give 13, still positive, but the two conventions must not be mixed.
Which of the following equations are hyperbolic at every point of ℝ²?
1. uₓₓ − u_yy = 0
2. uₓₓ + 2uₓ_y + u_yy = 0
3. uₓₓ + x²u_yy = 0
4. uₓ_y = 0
Show answer
Answer: A — uₓₓ − u_{yy} = 0; D — uₓ_{y} = 0
Δ = b² − 4ac. (1) 0 − 4(1)(−1) = 4 > 0. (2) 4 − 4 = 0: parabolic. (3) −4x² < 0 for x ≠ 0 (elliptic) and 0 on x = 0 (parabolic). (4) 1 − 0 = 1 > 0; it is the canonical form of the wave equation.
The Tricomi equation y uₓₓ + u_yy = 0 is hyperbolic exactly in the region
1. y < 0
2. y > 0
3. y = 0
4. x < 0
Show answer
Answer: A — y < 0
a = y, b = 0, c = 1, so Δ = −4y: positive iff y < 0. It is elliptic for y > 0 and parabolic on the line y = 0. The type depends only on the principal coefficients, here only on y.
Under ξ = x + y, η = x − y, the equation uₓₓ − u_yy = 0 becomes
1. u_ξη = 0, with general solution F(x + y) + G(x − y)
2. u_ξξ + u_ηη = 0
3. u_ξξ = 0
4. u_ξ = u_η
Show answer
Answer: A — u_{ξη} = 0, with general solution F(x + y) + G(x − y)
uₓ = u_ξ + u_η, u_y = u_ξ − u_η, so uₓₓ = u_ξξ + 2u_ξη + u_ηη and u_yy = u_ξξ − 2u_ξη + u_ηη. The difference is 4u_ξη, so u_ξη = 0 and u = F(ξ) + G(η). The new coordinates are the characteristics x ± y = const.
Let u solve uₓ + 2u_y = 0 with u(x, 0) = sin x. The value of u(π, π) is ____.
Numerical answer — type the value.
Show answer
Answer: 1
The characteristics are dy/dx = 2, i.e. y − 2x = const, and u is constant on each. The one through (π, π) meets y = 0 at x = π − π/2 = π/2, so u = sin(x − y/2) and u(π, π) = sin(π/2) = 1.
Let u solve xuₓ + yu_y = 2u for x > 0 with u(x, 1) = x. The value of u(2, 3) is ____.
Numerical answer — type the value.
Show answer
Answer: 6
dx/x = dy/y = du/(2u) gives y/x = c₁ and u/x² = c₂, so u = x²f(y/x). On y = 1: x²f(1/x) = x, so f(s) = s and u = x²(y/x) = xy. Check: x·y + y·x = 2xy = 2u. u(2, 3) = 6.
For u_t + uuₓ = 0 with u(x, 0) = −2x, the classical solution exists for 0 ≤ t < T. The largest such T is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
u is constant along x = x₀ + f(x₀)t = x₀(1 − 2t), so x₀ = x/(1 − 2t) and u = −2x/(1 − 2t). All characteristics meet at x = 0 when t = 1/2, where u blows up. The general rule t∗ = −1/min f′ = −1/(−2) = 1/2 agrees.
For u_t + cuₓ = 0 with c a non-zero constant, which statements are true?
1. every solution is constant along each line x − ct = const
2. the solution with u(x, 0) = f(x) is u = f(x − ct)
3. the line t = 0 is a characteristic
4. prescribing u on the line x = ct determines a unique solution
Show answer
Answer: A — every solution is constant along each line x − ct = const; B — the solution with u(x, 0) = f(x) is u = f(x − ct)
Along x = x₀ + ct, du/dt = u_t + cuₓ = 0, giving (1) and (2). (3) is false: characteristics have direction (c, 1) in the (x, t)-plane, not parallel to t = 0 since c is finite. (4) is false: x = ct is itself a characteristic, so data there must be constant to be consistent and then leave u undetermined off it.
u is harmonic in the unit disc, continuous up to the boundary, with u(1, θ) = cos²θ in polar coordinates. The value of u at the point (x, y) = (1/2, 0) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.625
cos²θ = ½ + ½ cos 2θ, and each Fourier mode cos nθ extends as rⁿ cos nθ, so u = ½ + ½r² cos 2θ = ½ + ½(x² − y²). At r = 1/2, θ = 0: ½ + ½ · ¼ = 0.625. The centre value ½ is the boundary average, as the mean value property requires.
Which statements about harmonic functions are true?
1. a non-constant harmonic function on a bounded domain, continuous on its closure, attains its maximum only on the boundary
2. the Dirichlet problem for Laplace’s equation on a bounded domain has at most one solution continuous on the closure
3. a harmonic function on all of ℝ² that is bounded above is constant
4. the value of a harmonic function at the centre of a disc equals its maximum on the boundary circle
Show answer
Answer: A — a non-constant harmonic function on a bounded domain, continuous on its closure, attains its maximum only on the boundary; B — the Dirichlet problem for Laplace’s equation on a bounded domain has at most one solution continuous on the closure; C — a harmonic function on all of ℝ² that is bounded above is constant
(1) is the strong maximum principle. (2) The difference of two solutions is harmonic with zero boundary values, so its max and min are 0. (3) If u ≤ M then M − u is a non-negative harmonic function on ℝ², hence constant (Liouville for harmonic functions; in ℝ² write u = Re f and apply Liouville to e^f). (4) is false: the centre value is the boundary average, e.g. u = x on the unit disc has centre value 0 and boundary maximum 1.
u_t = uₓₓ on 0 < x < π with u(0, t) = u(π, t) = 0 and u(x, 0) = sin x + 3 sin 2x. The value of u(π/2, 1), correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.37
Each mode sin nx decays as e^−n²t, so u = e^−t sin x + 3e^−4t sin 2x. At x = π/2, sin 2x = sin π = 0, so u(π/2, 1) = e⁻¹ = 0.368, i.e. 0.37. The second mode, though larger initially, vanishes at the midpoint for all t.
For the Cauchy problem u_t = uₓₓ on ℝ with u(x, 0) = f(x) ≥ 0, f continuous, bounded, and positive only on [0, 1], the solution satisfies
1. u(x, t) > 0 for every x ∈ ℝ and every t > 0
2. u(x, t) = 0 for |x| > 1 + t
3. u(x, t) = 0 outside [0, 1] for all t
4. u is not differentiable for t > 0
Show answer
Answer: A — u(x, t) > 0 for every x ∈ ℝ and every t > 0
u = (4πt)^−1/2∫₀¹ e^{−(x−y)²/(4t)} f(y) dy, and the Gaussian kernel is positive everywhere, so u > 0 at every point once t > 0: infinite speed of propagation. A support bound like |x| ≤ 1 + t is the wave equation’s behaviour. The heat kernel makes u C^∞ for t > 0.
u solves u_tt = 4uₓₓ on ℝ with u(x, 0) = x² and u_t(x, 0) = 0. The value of u(1, 1) is ____.
Numerical answer — type the value.
Show answer
Answer: 5
c = 2, so d’Alembert gives u = ½[(x − 2t)² + (x + 2t)²] = x² + 4t². Check: u_tt = 8 = 4 · 2 = 4uₓₓ. u(1, 1) = 1 + 4 = 5. Taking c = 4 instead of √4 gives 17.
u solves u_tt = uₓₓ on ℝ with u(x, 0) = 0 and u_t(x, 0) = x². The value of u(1, 3) is ____.
Numerical answer — type the value.
Show answer
Answer: 12
u = ½∫_x−t^x+t s² ds = [(x + t)³ − (x − t)³]/6 = x²t + t³/3. At (1, 3): 3 + 9 = 12. Directly: ½∫₋₂⁴ s² ds = ½(64 + 8)/3 = 12.
For u_tt = 9uₓₓ on ℝ, the value u(2, 1) depends only on the initial data on an interval of the x-axis. The length of that interval is ____.
Numerical answer — type the value.
Show answer
Answer: 6
c = 3, and the domain of dependence of (x, t) is [x − ct, x + ct] = [2 − 3, 2 + 3] = [−1, 5], of length 2ct = 6. Using c = 9 gives 18.
u solves u_tt − uₓₓ = 2 on ℝ with u(x, 0) = 0 and u_t(x, 0) = 0. The value of u(5, 3) is ____.
Numerical answer — type the value.
Show answer
Answer: 9
Duhamel: u = ½∫₀ᵗ∫_x−(t−τ)^x+(t−τ) 2 ds dτ = ½∫₀ᵗ 4(t − τ) dτ = t². Check: u = t² has u_tt = 2, uₓₓ = 0, zero data. So u(5, 3) = 9, independent of x, as the x-independent forcing suggests.
u solves u_t = uₓₓ on ℝ with u(x, 0) = e^−x². The value of u(0, 3/4) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
Fourier in x: û(ω, t) = f̂(ω)e^−ω²t with f̂ = √π e^−ω²/4, so û = √π e^{−ω²(1 + 4t)/4}, the transform of (1 + 4t)^−1/2 e^{−x²/(1 + 4t)}. At x = 0, t = 3/4: (1 + 3)^−1/2 = 0.5. The Gaussian spreads while its integral √π stays fixed.