ODE, PDE & Applied Mathematics

Partial Differential Equations

27 previous-year questions →every option and the correct answer, free Practice this topic →one question at a time, no clock, check as you go Partial Differential Equations — Worked Problems →the worked-problem companion to this topic

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

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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 free

In plain words

A first-order PDE is an ordinary equation hiding along certain curves. Find those curves, carry the initial data along them, and watch for the moment two of them collide.

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

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

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)
$u u_{x} + u_y = 0, u(x,0) = x$	characteristics $x = x_{0} +$ uy, $u = x_{0} \Rightarrow u = x/(1 + y)$
$u u_{x} + u_y = 1, u(x,0) =$ kx	u = kx $_{0} + y, x = x_{0} +$ kx $_{0}y + y^{2}/2 \Rightarrow$ needs 1 + ky ≠ 0
$\partial_{x}u + 2\partial_yu = 0$	u constant on y − 2x = const

Existence trap: the solution exists for all y > 0 exactly when the Jacobian $\partial{}x/\partial{}x_{0} = 1 +$ ky never vanishes there — i.e. k ≥ 0. Negative k makes characteristics cross and the solution breaks down (shock).

Data on a characteristic

If the initial curve is itself a characteristic, there is either no solution or infinitely many. Always check that the data curve is non-characteristic: $a\cdot{}n_{1} + b\cdot{}n_{2} \ne 0$ along it.

Charpit (fully nonlinear F(x, y, u, p, q) = 0)

dx/F_p = dy/F_q = du/(pF_p + qF_q) = −dp/(F_x + pF_u) = −dq/(F_y + qF_u).

Key takeaways

Characteristics carry the data; crossing characteristics = loss of existence.
Check the non-characteristic condition on the initial curve before anything else.
Burgers-type $u u_{x} + u_y = 0$ with increasing data is fine; decreasing data forms a shock.

Lagrange's method on a first-order PDEinteractive

This visual is part of Trap Analysis

The auxiliary system, two first integrals, and the general solution — worked end to end.

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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.

Practice

Part B2025 Juneshow ▾

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

(y + u)\partial{}u/\partial{}x + y \partial{}u/\partial{}y = x - y, x \in \mathbb{R}, y > 0, u(x, 1) = 1 + x, x \in \mathbb{R}

. Then which of the following statements is true?

A.u(1, 1) = 2✓ Correct
B.u(2, 2) = 4
C.u(3, 3) = 3/2
D.u(4, 4) = 2/3

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Consider the Cauchy problem (CP

) x \partial{}u/\partial{}x + \partial{}u/\partial{}y = 1, u(0, y) = e^y, y \in \mathbb{R}

. Then which of the following statements are true?

A.There is NO neighbourhood of the origin on which (CP) has a solution✓ Correct
B.(CP) has a unique solution defined on some neighbourhood of the origin
C.(CP) has a unique solution defined on some neighbourhood of the point (0, 1) in the xy-plane
D.(CP) has an infinite number of solutions, each of which is defined on some neighbourhood of the origin

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾

Consider the initial-boundary value problem (IBVP

) u_{t} + u_{x} = 0, 0 < x < \infty, t > 0, u(x,0) = e

ˣ,

0 < x < \infty, u(0,t) = 1 - \sin t, t > 0

. Then which of the following statements is true?

A.There exists a unique solution u(x,t) of IBVP such that u(1,t) = e − sin t, for all t < 1.
B.There does NOT exist a solution u(x,t) of IBVP such that u(1,t) = e − sin t, for all t > 1.✓ Correct
C.There exists a solution u(x,t) of IBVP such that u(x,t) = e^(x−t), for all t > x.
D.There exists a solution u(x,t) of IBVP such that u(x,t) = e − sin(t−x), for all t < x.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

Let u = u(x, t) be the solution of the following initial value problem:

u_{t} + 2024u_{x} = 0

for

x \in \mathbb{R}, t > 0

, and

u(x, 0) = u_{0}(x)

for

x \in \mathbb{R}

, where

u_{0} : \mathbb{R} \to \mathbb{R}

is an arbitrary

C^{1}

function. Consider the following statements:

S_{1}

: If

A_{t} :=

{

x \in \mathbb{R} : u(x, t) < 1

} and |

A_{t}

| denotes the Lebesgue measure of

A_{t}

for every t ≥ 0, then |

A_{t}

| = |

A_{0}

\forall{}t > 0. S_{2}

: If

u_{0}

is Lebesgue integrable, then for every t > 0, the function x ↦ u(x, t) is Lebesgue integrable. Then

A.both $S_{1}$ and $S_{2}$ are true✓ Correct
B. $S_{1}$ is true but $S_{2}$ is false
C. $S_{2}$ is true but $S_{1}$ is false
D.both $S_{1}$ and $S_{2}$ are false

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

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

x \partial{}u/\partial{}x + y \partial{}u/\partial{}y = u, (x, y) \ne (0, 0)

, with

u(x, 1) = \sqrt{1 + x^{2}}, x \in \mathbb{R}

. Then which of the following statements is true?

A.u(1, 0) = 0
B. $u(x_{1}, y_{1}) = u(x_{2}, y_{2})$ whenever $x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2}$ ✓ Correct
C. $u(1, y) = \sqrt{2}$ for all $y \in \mathbb{R}$
D. $u(x_{1}, y_{1}) = u(x_{2}, y_{2})$ whenever $x_{1} + y_{1} = x_{2} + y_{2}$

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

The problem

\partial{}u/\partial{}x + \partial{}u/\partial{}y = u^{2}, u(x, 0) = x^{2}, \forall{}x \in \mathbb{R}

has a solution on an open set containing the line {

(x, y) \in \mathbb{R}^{2}

: ax + by = 0} if

A.a = 1 and b = 0
B.a = 1 and b = −1✓ Correct
C.a = 2 and b = 1
D.a = 1 and b = 2

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Consider the initial boundary value problem (IBVP

) u_{t} + u_{x} = 2u

for x > 0, t > 0; u(0, t) = 1 + sin t for t > 0; u(x, 0) = eˣ cos x for x > 0. If u is the solution of the IBVP, then the value of

u(2\pi, \pi)/u(\pi, 2\pi)

A. $e^\pi$
B. $e^(-\pi)$
C. $-e^\pi$ ✓ Correct
D. $-e^(-\pi)$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Juneshow ▾

Let

u : \mathbb{R}^{2} \to \mathbb{R}

solve

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

\mathbb{R}^{2}

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

v : \mathbb{R}^{2} \to \mathbb{R}

solve

\partial_{x}v + 2\partial_yv = 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.✓ Correct
C.v changes sign in the interior of S.✓ Correct
D.v vanishes along a line in S.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Consider the Cauchy problem

u\cdot{}u_{x} + u_y = 1

for

(x, y) \in \mathbb{R} \times (0, \infty)

, with u(x, 0) = kx,

x \in \mathbb{R}

, for a real parameter k. For which of the following values of k does the problem have a solution defined on

\mathbb{R} \times (0, \infty)

A.k = 0✓ Correct
B.k = −2
C.k = 4✓ Correct
D.k = 1✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾

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

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

for

x \in \mathbb{R}, y > 0

, with u(x, 0) = x for

x \in \mathbb{R}

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

A.2
B.3
C.1/2✓ Correct
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 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.✓ Correct
B. $u(x, y) = x^{2}$ is a solution.✓ Correct
C.u(x, y) = xy is a solution.✓ Correct
D.The general solution is $u = x^{2}\cdot{}F(y/x)$ for an arbitrary differentiable F.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Solve p + q = z by Lagrange's method, where

p = \partial{}z/\partial{}x, q = \partial{}z/\partial{}y

. Which of the following are true?

A.u = x − y is a first integral.✓ Correct
B.v = z·e⁻ˣ is a first integral.✓ Correct
C.The general solution can be written z = eˣ $\cdot\varphi(x - y)$ for arbitrary $\varphi$ .✓ Correct
D.z = e^(x+y) is a particular solution of p + q = z.

Solution

The auxiliary equations dx/1 = dy/1 = dz/z give $x - y = c_{1}$ from the first pair, and z·e⁻ˣ $= c_{2}$ from dz/z = dx — (1)–(2). Setting $v = \varphi(u)$ and solving gives z = eˣ $\varphi(x-y)$ ; direct substitution confirms p + q = eˣ $\varphi + e$ ˣ $\varphi' - e$ ˣ $\varphi' = e$ ˣ $\varphi = z$ — (3). (4) is false: for z = e^(x+y), p = q = e^(x+y), so p + q = 2z ≠ z — it does not satisfy the equation at all.

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Part Cshow ▾

Consider the first-order linear PDE P(x,y,z)p + Q(x,y,z)q = R(x,y,z), where

p = \partial{}z/\partial{}x, q = \partial{}z/\partial{}y

. Which of the following are true (Lagrange's method)?

A.The general solution is obtained from the auxiliary equations dx/P = dy/Q = dz/R.✓ Correct
B.If $u = c_{1}$ and $v = c_{2}$ are two independent first integrals of the auxiliary equations, the general solution is F(u, v) = 0 for arbitrary F.✓ Correct
C.Away from points where P, Q vanish together, characteristic curves projected onto the xy-plane never cross each other.✓ Correct
D.Charpit's method applies only to linear first-order PDEs, not to nonlinear ones.

Solution

(1)–(2) are Lagrange's method for this linear form. (3) follows from uniqueness of solutions to the characteristic ODE dy/dx = Q/P. (4) is false, and backwards: Charpit's method is the general method for the nonlinear equation F(x,y,z,p,q) = 0; Lagrange's method here is the special case for the linear form Pp + Qq = R.

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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✓ Correct
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

The worked solution and the per-option analysis are in Trap Analysis.

2. Classification and canonical forms

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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 free

In plain words

One number picks one of three families, and the family decides the behaviour. Elliptic smooths everywhere, parabolic smooths forwards in time, and hyperbolic carries sharp features along unblunted.

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 $u_{xx} + u_yy = 0$	none real
= 0	parabolic	heat $u_t = u_{xx}$	one family
> 0	hyperbolic	wave $u_tt = u_{xx}$	two families

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

Worked signs

PDE	$B^{2} - 4$ AC	Type
$u_{xx} + 2u_{x}$ ᵧ + u_yy = 0	4 − 4 = 0	parabolic everywhere
$x^{2}u_{xx} - 2$ xy $u_{x}$ ᵧ $- 3y^{2}u_yy = 0$	$4x^{2}y^{2} + 12x^{2}y^{2} = 16x^{2}y^{2} > 0 ($ xy ≠ 0)	hyperbolic
$u_{xx} + x u_yy = 0$	−4x	elliptic x > 0, hyperbolic x < 0

Canonical forms

Solve A(dy $)^{2} - B$ dx dy + C(dx $)^{2} = 0, i.e$ . dy/dx $= (B \pm \surd(B^{2} - 4$ AC))/(2A), to get the characteristic families $\xi(x,y) = c_{1}, \eta(x,y) = c_{2}$ . Then:

hyperbolic: $u_\xi\eta = \dots ($ or u_ss − u_tt = …)
parabolic: $u_\eta\eta = \dots ($ take $\xi =$ the single family, $\eta$ anything independent)
elliptic: complex characteristics; take real and imaginary parts to get $u_\xi\xi + u_\eta\eta = \dots$

Well-posedness

Problem	Well-posed?
Laplace with Dirichlet data on a closed boundary	✓
Laplace with Cauchy data (Hadamard)	✗ — not continuous in the data
heat forward in t	✓
heat backward in t	✗
wave with Cauchy data	✓

Key takeaways

Compute $B^{2} - 4$ AC and remember it can change sign with position.
Characteristic ODE gives the change of variables to canonical form.
Elliptic wants boundary data, hyperbolic wants Cauchy data; mixing them is ill-posed.

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

Step 1 / 4The discriminant

For $Au_{xx} + Bu_{xy} + Cu_{yy} + \cdots = 0$ , compute $B^2-4AC$ :

$<0$ elliptic (Laplace)
$=0$ parabolic (heat)
$>0$ hyperbolic (wave)

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.

Practice

Part C2024 Decembershow ▾

Consider the partial differential equation

4(\partial^{2}u/\partial{}x^{2}) + 6(\partial^{2}u/\partial{}x\partial{}y) + 3(\partial^{2}u/\partial{}y^{2}) = 0

. Which of the following partial differential equations can be obtained by a change of independent variables given by

(\xi, \eta)

ᵗ = A(x, y)ᵗ for some 2 × 2 invertible matrix A with real entries?

A. $\partial^{2}w/\partial\xi^{2} + \partial^{2}w/\partial\xi\partial\eta + \partial^{2}w/\partial\eta^{2} = 0$ ✓ Correct
B. $\partial^{2}w/\partial\xi^{2} + \partial^{2}w/\partial\eta^{2} = 0$ ✓ Correct
C. $\partial^{2}w/\partial\xi^{2} + 4(\partial^{2}w/\partial\xi\partial\eta) + \partial^{2}w/\partial\eta^{2} = 0$
D. $2(\partial^{2}w/\partial\xi^{2}) + 2(\partial^{2}w/\partial\xi\partial\eta) + \partial^{2}w/\partial\eta^{2} = 0$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

The partial differential equation

x^{2} u_{xx} - 2

u_{x}

ᵧ

- 3y^{2} u_yy + 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}✓ Correct
D.parabolic in {(x, y) : xy ≠ 0}

Solution

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

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Part Cshow ▾

The PDE u_xx + 2u_xy + u_yy = 0 is

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

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider the second-order linear PDE Au_xx + Bu_xy + Cu_yy + (lower order) = 0. Which of the following are true?

A.The PDE is elliptic at a point where $B^{2} - 4$ AC < 0.✓ Correct
B.The PDE is hyperbolic at a point where $B^{2} - 4$ AC > 0.✓ Correct
C.The PDE is parabolic at a point where $B^{2} - 4$ AC = 0.✓ Correct
D.Laplace's equation u_xx + u_yy = 0 is hyperbolic.

Solution

(1)–(3) are the standard classification by discriminant. (4) is false: for Laplace's equation A=1, B=0, C=1, so $B^{2} - 4$ AC = −4 < 0 — elliptic, the textbook example of the type, not hyperbolic.

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

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Consider the Tricomi-type equation y·u_xx + u_yy = 0 on

\mathbb{R}^{2}

. Which of the following are true?

A.The equation is elliptic for y > 0.✓ Correct
B.The equation is hyperbolic for y < 0.✓ Correct
C.The equation is parabolic along the line y = 0.✓ Correct
D.The equation has the same type at every point of $\mathbb{R}^{2}$ .

Solution

Here A = y, B = 0, C = 1, so $B^{2} - 4$ AC = −4y. For y > 0 this is negative (elliptic), for y < 0 positive (hyperbolic), and zero exactly on y = 0 (parabolic) — (1)–(3). (4) is false: this equation is the standard example of mixed type, changing classification across the plane, which is exactly what makes it interesting (it models transonic flow).

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Part Bshow ▾

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

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

The worked solution and the per-option analysis are in Trap Analysis.

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.✓ Correct
B.The equation is hyperbolic everywhere.
C.The characteristic curves satisfy dy/dx = 2.✓ Correct
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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3. Laplace, heat and wave equations: separation of variables

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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

In plain words

Guess space times time and the equation splits in two. The space half is the same for heat and for waves, so only the time factor differs — and that is all the difference.

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

d'Alembert on $\mathbb{R}: u(x,t) =$ ½ $[\varphi(x +$ ct $) + \varphi(x -$ ct $)] + (1/2c)\int_{x-ct}^{x+ct}$ $\psi(s)$ 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: $u = \sum (a_{n} \cos(n\pi$ ct $/L) + b_{n} \sin(n\pi$ ct $/L)) \sin(n\pi{}x/L)$ .

Heat equation $u_t = k u_{xx}$

$u = \sum c_{n}$ $e^{-k n^{2}\pi^{2}t/L^{2}}$ $\sin(n\pi{}x/L)$ on [0, L]; on $\mathbb{R}$ , 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 $\Delta{}u = 0$

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 $\mathbb{R}^{n}$ 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.

d'Alembert: one bump becoming two wavesinteractive

This visual is part of Trap Analysis

Watch the initial profile split into halves travelling at ±c, with the light cone marking where the string has not moved yet.

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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).

Practice

Part C2025 Juneshow ▾

Suppose u = u(x, y) is the solution of the boundary value problem

\partial^{2}u/\partial{}x^{2} + \partial^{2}u/\partial{}y^{2} + \partial{}u/\partial{}x = 0

in {

(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} < 1

u(x, y) = 1 + 2x^{2}y^{2}

on {

(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} = 1

}. Then which of the following statements are true?

A.The minimum value of u is 1✓ Correct
B.The maximum value of u is 3
C.The minimum value of u is 2
D.The maximum value of u is 3/2✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Consider the boundary value problem (BVP

) u_{xx}+u_yy=0

\Omega=

{

(x,y)\in\mathbb{R}^{2}:x^{2}+y^{2}<1

}, u(x,y)=e^(x+y) on

\partial\Omega=

{

(x,y)\in\mathbb{R}^{2}:x^{2}+y^{2}=1

}. Then which of the following statements are true?

A.There exists a unique solution to BVP.✓ Correct
B.The BVP does NOT have a solution.
C.There exists a solution u to BVP such that u(x,y)=(1+e)/2 for some $(x,y)\in\Omega\cup\partial\Omega$ .✓ Correct
D.There exists a solution u to BVP such that $u(x,y)=(1+e^{3})/2$ for some $(x,y)\in\Omega\cup\partial\Omega$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Let u(x,t) be the solution of the initial-boundary value problem

u_{tt} - u_{xx} + u_{t} = 0, 0<x<\pi, t>0, u(0,t)=0, u(\pi,t)=0, t>0, u(x,0)=\sin x, u_{t}(x,0)=0, 0<x<\pi

. For t≥0, define

E(t) = \int_{0}^\pi (u_{t}^{2}+u_{x}^{2})

dx. Then which of the following statements are true?

A. $E(t) \ge \pi$ for all t>0
B. $E(t) \le \pi$ for all t>0✓ Correct
C. $E(t) \ge \pi/2$ for all t>0
D. $E(t) \le \pi/2$ for all t>0✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾

Let u(x,t) be the solution of the partial differential equation

u_{t} = 16u_{xx} + 2, 0 < x < 7, t > 0

, satisfying the conditions

u_{x}(0,t) = u(7,t) = 0

for t > 0, and u(x,0) = 0 for 0 < x < 7. Then which of the following statements is true?

A.For each x ∈ (0,7), u(x,t) → 0 as $t \to \infty$ .
B.For each $x \in (0,7), u(x,t) \to x^{2}(7-x)^{2}/16$ as $t \to \infty$ .
C.For each $x \in (0,7), u(x,t) \to (x^{2}/16)(7-x)$ as $t \to \infty$ .
D.For each $x \in (0,7), u(x,t) \to (1/16)(49-x^{2})$ as $t \to \infty$ .✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾

Let u(x,t) be the solution of the initial value problem

u_{tt} - u_{xx} = e^(-t), x \in \mathbb{R}, t > 0, u(x,0) = \cos x, u_{t}(x,0) = 0, x \in \mathbb{R}

. Then which of the following statements is true?

A.For each $x_{0} \in \mathbb{R}, e^t\cdot{}u(x_{0},t) \to 0$ as $t \to \infty$ .
B.For each $x_{0} \in \mathbb{R}$ , | $u(x_{0},t)$ | → 0 as $t \to \infty$ .
C.For each $x_{0} \in \mathbb{R}$ , there exists $a t_{0} > 0$ such that $u(x_{0},t_{0}) \ge 4$ .✓ Correct
D.For each $x_{0} \in \mathbb{R}$ , there exists $a t_{0} > 0$ such that $u(x_{0},t_{0}) \le -4$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

Let u = u(x, t) be the solution of

\partial^{2}u/\partial{}t^{2} - \partial^{2}u/\partial{}x^{2} = 0, x \in \mathbb{R}, t > 0, u(x, 0) = 1 + x^{2}, x \in \mathbb{R}, \partial{}u/\partial{}t(x, 0) = x + 1, x \in \mathbb{R}

. Then the value of u(1, 1) is

A.2
B.3
C.4
D.5✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

The problem

\partial^{2}u/\partial{}x^{2} + \partial^{2}u/\partial{}y^{2} = 0

in {

(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} > 1

}, u(x, y) = 1 on {

(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} = 1

} has

A.no solution
B.exactly one solution
C.exactly two solutions
D.infinitely many solutions✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let u = u(x, t) be the solution of the initial-boundary value problem

\partial{}u/\partial{}t = \partial^{2}u/\partial{}x^{2}

(0, 1) \times (0, \infty)

, with u(x, 0) = 4x(1 − x) for x ∈ [0, 1] and u(0, t) = u(1, t) = 0 for t ≥ 0. Then which of the following statements are true?

A. $\lim(t\to\infty) u(x, t) = 0$ for all x ∈ (0, 1)✓ Correct
B.u(x, t) = u(1 − x, t) for all x ∈ (0, 1), t > 0✓ Correct
C. $\int_{0}^{1}(u(x, t))^{2}$ dx is a non-increasing function of t✓ Correct
D. $\int_{0}^{1}(u(x, t))^{2}$ dx is a non-decreasing function of t

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Let B(0,2) = {

(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} < 4

}, and

\partial{}B

denote the boundary of B(0,2). Assume

(\alpha, \beta) \ne (0,0), k \in \mathbb{R}

, and u is any solution to

-\Delta{}u = 0

B(0,2), \alpha{}u(x, y) + \beta(\partial{}u/\partial\nu)(x, y) = 1 + (x^{2} + y^{2})k

\partial{}B

, where

\nu(x, y)

is the unit outward normal to B(0,2) at

(x, y) \in \partial{}B

. Consider the following statements:

S_{1}

: If

\beta = 0

, then there exists

a (x_{0}, y_{0}) \in B(0,2)

such that |

u(x_{0}, y_{0})

| = |1 + 4k|/|

\alpha

S_{2}

: If

\alpha = 0

, then k = −1/4. Then

A. $S_{1}$ is true but $S_{2}$ is false
B. $S_{2}$ is true but $S_{1}$ is false
C.both $S_{1}$ and $S_{2}$ are true✓ Correct
D.both $S_{1}$ and $S_{2}$ are false

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

If u = u(x, t) is the solution of the initial value problem

u_{t} = u_{xx}

for

x \in \mathbb{R}, t > 0

, with u(x, 0) = sin(4x) + x + 1 for

x \in \mathbb{R}

, satisfying |u(x, t)|

< 3e^(x^{2})

for all

x \in \mathbb{R}

and t > 0, then

A. $u(\pi/8, 1) + u(-\pi/8, 1) = 2$ ✓ Correct
B. $u(\pi/8, 1) = u(-\pi/8, 1)$
C. $u(\pi/8, 1) + 2u(-\pi/8, 1) = 2$
D. $u(\pi/8, 1) = -u(-\pi/8, 1)$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let u = u(x, t) be the solution to the initial value problem

\partial^{2}u/\partial{}t^{2} - \partial^{2}u/\partial{}x^{2} = 0, x \in \mathbb{R}, t > 0, u(x, 0) = f(x), x \in \mathbb{R}, (\partial{}u/\partial{}t)(x, 0) = 1/\sqrt{1 + x^{2}}, x \in \mathbb{R}

, where f is a twice continuously differentiable function defined on

\mathbb{R}

satisfying f(x) → 0 as |x|

\to \infty

. Then which of the following statements are true?

A.For every $x \in \mathbb{R}, \lim(t\to\infty) u(x, t) = 0$
B.For every $x \in \mathbb{R}, \lim(t\to\infty) u(x, t) = \infty$ ✓ Correct
C.For every $t \in (0, \infty), \lim(x\to\infty) u(x, t) = 0$ ✓ Correct
D.For every $t \in (0, \infty), \lim(x\to-\infty) u(x, t) = \infty$

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let u = u(x, t) be a solution of the wave equation

\partial^{2}u/\partial{}t^{2} - \partial^{2}u/\partial{}x^{2} = 0, x \in \mathbb{R}, t > 0

satisfying the condition u(0, t) = 0, ∀t ≥ 0. Then which of the following statements is true?

A.u(x, t) = 0, whenever x = t
B.u(x, t) = 0, whenever x = −t
C.u(−x, t) = u(x, t), whenever x > 0, t > 0
D.u(−x, t) = −u(x, t), whenever 0 < x ≤ t✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let u = u(x, y) be the solution of the boundary value problem

\partial^{2}u/\partial{}x^{2} + \partial^{2}u/\partial{}y^{2} = 0

on (0, 1) × (0, 1), with

u(x, 0) = e^(\pi{}x), u(x, 1) = -e^(\pi{}x)

for

x \in [0, 1], u(0, y) = \cos(\pi{}y) + \sin(\pi{}y)

and

u(1, y) = e^\pi(\cos(\pi{}y) + \sin(\pi{}y))

for y ∈ [0, 1]. Then there exists a point

(x_{0}, y_{0}) \in (0, 1) \times (0, 1)

such that

A. $u(x_{0}, y_{0}) = \sqrt{2} e^\pi$
B. $u(x_{0}, y_{0}) = e^\pi$ ✓ Correct
C. $u(x_{0}, y_{0}) = -1$ ✓ Correct
D. $u(x_{0}, y_{0}) = -e^\pi$

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Let B be the open unit disc in

\mathbb{R}^{2}, \partial{}B

its boundary and

{\bar{B}} = B \cup \partial{}B

. For

\lambda \in (0, \infty)

, let

S_\lambda

be the set of twice continuously differentiable functions in B, continuous on B̄, satisfying

(u_{x})^{2} + \lambda(u_y)^{2} = 1

in B and u = 0 on

\partial{}B

. Which of the following statements are true?

A. $S_{1} = \emptyset$ ✓ Correct
B. $S_{2} = \emptyset$ ✓ Correct
C. $S_{1}$ has exactly one element and $S_{2}$ has exactly two elements.
D. $S_{1}$ and $S_{2}$ are both infinite.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

Consider the Cauchy problem for the wave equation

u_tt - 4u_{xx} = 0

\mathbb{R} \times (0, \infty)

, with u(x, 0) =

e^{-1/x^{2}}

for x ≠ 0 (and 0 at x = 0), and u_t(x, 0) = x

e^{-x^{2}}

. Which one of the following is true?

A. $lim_{t\to\infty}$ u(5, t) = 1✓ Correct
B. $lim_{t\to\infty}$ u(5, t) = 2
C. $lim_{t\to\infty}$ u(5, t) = 1/2
D. $lim_{t\to\infty}$ u(5, t) = 0

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let u solve

u_tt = c^{2}u_xx

\mathbb{R}

with u(x, 0) = f(x), u_t(x, 0) = g(x), where f, g are smooth and vanish outside [−1, 1]. Which of the following are true?

A.u(x, t) = ½[f(x − ct) + f(x + ct $)] + (1/2c)\int_{x-ct}^{x+ct}$ g(s) ds.✓ Correct
B. $u(x_{0}, t_{0})$ depends only on the initial data in $[x_{0} -$ ct $_{0}, x_{0} +$ ct $_{0}]$ .✓ Correct
C.u(x, t) = 0 whenever |x| > 1 + ct.✓ Correct
D.Initial disturbances influence all of space instantaneously.

Solution

(1) is d'Alembert's formula and (2) its immediate consequence — the domain of dependence.

(3) is finite propagation speed: outside the cone reached from [−1, 1] at speed c, both terms vanish.

(4) belongs to the heat equation, whose kernel is positive everywhere for t > 0 — the standard contrast with the wave equation.

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Part Bshow ▾

Let u solve the heat equation u_t = u_xx on 0 < x < L, 0 < t ≤ T, with continuous data. By the maximum principle, the maximum of u over the closed rectangle is attained

A.on the parabolic boundary: the bottom edge t = 0 or the sides x = 0, x = L✓ Correct
B.only at interior points
C.only on the top edge t = T
D.at the centre of the rectangle

Solution

The parabolic maximum principle: heat flows downhill, so new interior maxima cannot appear after time 0 — extremes are inherited from the initial data or pushed in through the lateral boundary.

The top edge t = T is excluded from the 'boundary' precisely because it is determined by the past.

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Partial Differential Equations

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

First-order PDE: characteristics, Lagrange and Charpit free

In plain words

Quasilinear: aux+buy=ca u_{x} + b u_y = caux​+buy​=c

The standard worked patterns

Data on a characteristic

Charpit (fully nonlinear F(x, y, u, p, q) = 0)

Key takeaways

Lagrange's method on a first-order PDEinteractive

2. Classification and canonical forms

Classification of second-order PDE free

In plain words

The discriminant

Worked signs

Canonical forms

Well-posedness

Key takeaways

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

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

The three classical equations free

In plain words

Wave equation utt=c2uxxu_tt = c^{2}u_{xx}ut​t=c2uxx​

Heat equation ut=kuxxu_t = k u_{xx}ut​=kuxx​

Laplace Δu=0\Delta{}u = 0Δu=0

Contrast table

Key takeaways

d'Alembert: one bump becoming two wavesinteractive

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

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

Heat equation $u_t = k u_{xx}$

Laplace $\Delta{}u = 0$