Partial Differential Equations

1. First-order PDE: Lagrange, Charpit, characteristics

Exam focus: Write the characteristic ODEs, carry the initial data along them, and check whether characteristics cross — that crossing is what makes the solution fail to exist globally.

Lec-22 Introduction to First Order PDE

NPTEL · Numerical Methods of ODE and PDE

Lec-37 Method of characteristics for Hyperbolic PDEs - I

NPTEL · Numerical Methods of ODE and PDE

Characteristics, and what happens when they cross.

First-order PDE: characteristics, Lagrange and Charpit

Quasilinear: $a u_{x} + b u_{y} = c$

Lagrange's method: solve dx/a = dy/b = du/c. Two independent integrals $φ, ψ$ give the general solution $F (φ, ψ) = 0$ .

Characteristic form: dx/dt = a, dy/dt = b, du/dt = c. The solution is constant (or evolves by c) along characteristic curves.

The standard worked patterns

PDE, data	Solution
$u_{x} + u_{y} = 0, u (x, 0) = f (x)$	u = f(x − y)

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Part B2023 Juneshow ▾

Let u(x, y) be the solution of the Cauchy problem

u \cdot u_{x} + u_{y} = 0

for

x \in R, y > 0

, with u(x, 0) = x for

x \in R

. Which of the following is the value of u(2, 3)?

A.2
B.3
C.1/2✓
D.1/3

Solution

Characteristics: dx/dy = u with u constant along them, so $x = x_{0} + u \cdot y$ and $u = x_{0}$ . Hence u = x/(1 + y), and u(2, 3) = 1/2.

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Part C2023 Juneshow ▾

Let

u : R^{2} \to R

solve

\partial_{x} u + 2 \partial_{y} u = 0

R^{2}

with u(x, y) = sin x on the line y = 3x + 1, and let

v : R^{2} \to R

solve

\partial_{x} v + 2 \partial_{y} v = 0

with v(x, 0) = sin x. Let S = [0,1] × [0,1]. Which of the following statements are true?

A.u changes sign in the interior of S.
B.u(x, y) = v(x, y) along a line in S.✓
C.v changes sign in the interior of S.✓
D.v vanishes along a line in S.✓

Solution

Both are constant on lines y − 2x = c. v = sin(x − y/2), which vanishes on x = y/2 and changes sign across it inside S. For u, the data line y = 3x + 1 meets the characteristic through (x, y) at $x_{0} = y - 2 x - 1$ , so u = sin(y − 2x − 1); on S the argument lies in $[- 3, 0] \subset (- π, 0]$ , so u ≤ 0 and does not change sign. u = v where y − 2x − 1 = x − y/2, i.e. on the line y = 2x + 2/3, which crosses S.

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Part C2023 Decembershow ▾

Previous-year question — unlock to view

Part Bshow ▾

The general solution of the PDE u_x + u_y = 0 is

A.u(x, y) = F(x − y) for an arbitrary differentiable F✓
B.u(x, y) = F(x + y) for an arbitrary differentiable F
C.u(x, y) = F(xy) for an arbitrary differentiable F
D.u(x, y) = c, constants only

Solution

The characteristics of dx/1 = dy/1 are the lines x − y = const, and u is constant along them, so u = F(x − y). Check: u_x + u_y = F′ − F′ = 0. F(x + y) instead solves u_x − u_y = 0.

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Part Cshow ▾

Consider the PDE x·u_x + y·u_y = 2u on x > 0. Which of the following are true?

A.The projected characteristic curves are the rays y = cx.✓
B. $u (x, y) = x^{2}$ is a solution.✓
C.u(x, y) = xy is a solution.✓
D.The general solution is $u = x^{2} \cdot F (y / x)$ for an arbitrary differentiable F.✓

Solution

Lagrange's auxiliary system dx/x = dy/y = du/2u gives first integrals y/x and $u / x^{2}$ , so $u = x^{2} F (y / x)$ . Checks: $u = x^{2}$ is $F \equiv 1 (x \cdot 2 x = 2 x^{2}); u =$ xy is F(t) = t (xy + yx = 2xy). By Euler's relation these are exactly the functions positively homogeneous of degree 2 — all four statements hold.

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2. Classification and canonical forms

Exam focus: B² − 4AC decides the type pointwise; the sign can change with the point (Tricomi). Reduce to canonical form by solving the characteristic ODE dy/dx = (B ± √(B²−4AC))/2A.

Lec-23 Introduction to Second Order PDE

NPTEL · Numerical Methods of ODE and PDE

The B² − 4AC classification; remember it is pointwise.

Classification of second-order PDE

The discriminant

For $A u_{xx} + B u_{x}$ ᵧ + C u_yy + (lower order) = 0:

$B^{2} - 4$ AC	Type	Model	Characteristics
< 0	elliptic	Laplace $u_{xx} + u_{y} y = 0$	none real
= 0	parabolic	heat $u_{t} = u_{xx}$	one family
> 0	hyperbolic	wave $u_{t} t = u_{xx}$	two families

**The type is pointwise. $* * y u_{xx} + u_{y} y = 0 ($ Tricomi) is elliptic for y > 0, parabolic on y = 0, hyperbol

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Part B2023 Decembershow ▾

The partial differential equation

x^{2} u_{xx} - 2

u_{x}

ᵧ

- 3 y^{2} u_{y} y + u_{x} - u_{y} = 0

A.elliptic in {(x, y) : y > 0}
B.parabolic in {(x, y) : x > 0, y > 0}
C.hyperbolic in {(x, y) : xy ≠ 0}✓
D.parabolic in {(x, y) : xy ≠ 0}

Solution

$B^{2} - 4$ AC $= 4 x^{2} y^{2} + 12 x^{2} y^{2} = 16 x^{2} y^{2} > 0$ wherever xy ≠ 0: hyperbolic.

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Part Bshow ▾

The second-order PDE u_xx + 3u_xy + u_yy = 0 is

A.elliptic
B.parabolic
C.hyperbolic✓
D.of mixed type

Solution

With A = 1, B = 3, C = 1 the discriminant is $B^{2} - 4$ AC = 9 − 4 = 5 > 0 at every point: hyperbolic everywhere, hence not of mixed type.

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Part Cshow ▾

The PDE u_xx + 2u_xy + u_yy = 0 is

A.elliptic
B.parabolic✓
C.hyperbolic
D.of the same type everywhere in $R^{2}$ ✓

Solution

$B^{2} - 4$ AC = 4 − 4 = 0 everywhere ⇒ parabolic throughout.

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Part Cshow ▾

Consider the second-order PDE u_xx + 4u_xy + 4u_yy = 0. Which of the following are true?

A.The equation is parabolic everywhere.✓
B.The equation is hyperbolic everywhere.
C.The characteristic curves satisfy dy/dx = 2.✓
D.The equation has two distinct families of real characteristics.

Solution

With A = 1, B = 4, C = 4 the discriminant is $B^{2} - 4$ AC = 16 − 16 = 0, so the equation is parabolic everywhere — (1) holds, (2) fails. The characteristics satisfy A(dy/dx $)^{2} - B ($ dy/dx) + C = 0, i.e. (dy/dx $)^{2} - 4 ($ dy/dx) + 4 = 0, giving the repeated root dy/dx = 2 and confirming (3). Because the root is repeated there is only one family of real characteristics, so (4) fails — that is precisely what distinguishes the parabolic case from the hyperbolic one.

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Part Cshow ▾

Consider the PDE y·u_xx + u_yy = 0 (Tricomi-type). Which of the following are true?

A.The equation is elliptic in the half-plane y > 0.✓
B.The equation is hyperbolic in the half-plane y < 0.✓
C.The equation is parabolic exactly on the line y = 0.✓
D.The type of the equation is the same at every point of the plane.

Solution

With A = y, B = 0, C = 1 the discriminant is $B^{2} - 4$ AC = −4y: negative (elliptic) for y > 0, positive (hyperbolic) for y < 0, zero on y = 0. The failure of (4) is the whole point of Tricomi-type equations — the type changes across a line, which is what makes transonic-flow models hard.

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3. Laplace, heat and wave equations: separation of variables

Exam focus: Know d'Alembert cold, the separation-of-variables series for the three classical equations, and the qualitative differences: smoothing (heat), finite speed (wave), mean value (Laplace).

Lec-24 Finite Difference Approximations to Parabolic PDEs

NPTEL · Numerical Methods of ODE and PDE

The heat equation and its smoothing behaviour.

Lec-31 Finite Difference Approximations to Elliptic PDEs - I

NPTEL · Numerical Methods of ODE and PDE

Laplace's equation and the maximum principle.

Lec-35 Finite Difference Approximations to Hyperbolic PDEs - I

NPTEL · Numerical Methods of ODE and PDE

The wave equation: finite propagation speed.

The three classical equations free

Wave equation $u_{t} t = c^{2} u_{xx}$

d'Alembert on $R : u (x, t) =$ ½ $[φ (x +$ ct $) + φ (x -$ ct $)] + (1/2 c) \int_{x - c t}^{x + c t}$ $ψ (s)$ ds.

Finite propagation speed c; the domain of dependence is [x − ct, x + ct].
No smoothing: singularities in the data travel forever.
Energy is conserved; the equation is reversible in time.

On [0, L] with Dirichlet data: $u = \sum (a_{n} cos (nπ$ ct $/ L) + b_{n} sin (nπ$ ct $/ L)) sin (nπ x / L)$ .

Heat equation $u_{t} = k u_{xx}$

$u = \sum c_{n}$ $e^{- k n^{2} π^{2} t / L^{2}}$ $sin (nπ x / L)$ on [0, L]; on $R$ , convolution with the Gaussian kernel.

Infinite propagation speed, instant smoothing to $C^{\infty}$ for t > 0.
Irreversible: the backward problem is ill-posed.
Maximum principle: max over the parabolic boundary; no interior max.
Solutions decay to the steady state as $t \to \infty$ .

Laplace $Δ u = 0$

**Mean value property $* * : u (x_{0})$ is the average over any ball/sphere centred at $x_{0}$ .
Maximum principle: no interior max or min on a domain unless constant.
Harmonic $\Rightarrow C^{\infty} ($ indeed real-analytic).
Liouville: a bounded harmonic function on $R^{n}$ is constant.
Uniqueness: the Dirichlet problem on a bounded domain has at most one solution; the Neumann problem is unique up to an additive constant and needs $\oint \partial u / \partial n = 0$ for solvability.

Contrast table

	Wave	Heat	Laplace
Speed	finite	infinite	—
Smoothing	none	instant	—
Time reversible	✓	✗	—
Maximum principle	✗	✓	✓

Key takeaways

d'Alembert for the wave equation on the line; separation of variables on an interval.
Heat smooths and forgets; the wave keeps everything and travels at speed c.
Maximum principle holds for heat and Laplace, never for the wave.

Partial Differential Equations

1. First-order PDE: Lagrange, Charpit, characteristics

Lagrange's method on a first-order PDEinteractive

First-order PDE: characteristics, Lagrange and Charpit

Quasilinear: $a u_{x} + b u_{y} = c$

The standard worked patterns

2. Classification and canonical forms

Classifying a second-order PDE, and the type that changesinteractivefree

Classification of second-order PDE

The discriminant

3. Laplace, heat and wave equations: separation of variables

d'Alembert: one bump becoming two wavesinteractive

The three classical equations free

Wave equation $u_{t} t = c^{2} u_{xx}$

Heat equation $u_{t} = k u_{xx}$

Laplace $Δ u = 0$

Contrast table

Key takeaways

Partial Differential Equations

1. First-order PDE: Lagrange, Charpit, characteristics

Lagrange's method on a first-order PDEinteractive

First-order PDE: characteristics, Lagrange and Charpit

Quasilinear: aux​+buy​=c

The standard worked patterns

2. Classification and canonical forms

Classifying a second-order PDE, and the type that changesinteractivefree

Classification of second-order PDE

The discriminant

3. Laplace, heat and wave equations: separation of variables

d'Alembert: one bump becoming two wavesinteractive

The three classical equations free

Wave equation ut​t=c2uxx​

Heat equation ut​=kuxx​

Laplace Δu=0

Contrast table

Key takeaways

Quasilinear: $a u_{x} + b u_{y} = c$

Wave equation $u_{t} t = c^{2} u_{xx}$

Heat equation $u_{t} = k u_{xx}$

Laplace $Δ u = 0$