Partial Differential Equations MCQ Quiz - Objective Question with Answer for Partial Differential Equations - Download Free PDF

Explanation:

The given PDE is

$\sqrt{p}+\sqrt{q}=1$

i.e., $\sqrt{p}+\sqrt{q}-1=0$ which id of the form f(p, q) = 0

so, the solution is

z = ax + by + c where

$\sqrt{a}+\sqrt{b}=1$

i.e., $\sqrt{b}=1-\sqrt{a}$

i.e., b = $(1-\sqrt{a}{)}^2$

So, the solution is

z = ax + $(1-\sqrt{a}{)}^2$y + c

(2), (3) are correct.

Concept:

A PDE of the form 

Au_xx + Bu_xy + Cu_yy + f(x, y, u, u_x, u_y) = 0 is

(i) hyperbolic if B² - 4AC > 0

(ii) parabolic if B2 - 4AC = 0

(iii) elliptic if B2 - 4AC < 0

Explanation:

Given PDE

$(1-\sqrt{xy})u_{xx}-2u_{xy}+(1+\sqrt{xy})u_{yy}=0$

B² - 4AC = (-2)² - 4$(1-\sqrt{xy})(1+\sqrt{xy})$ = 4 - (4 - 4xy) = 4xy

So, B2 - 4AC > 0 in first and third quadrant as x > 0, y > 0 in 1st quadrant and x < 0, y < 0 in 3rd quadrant.

So, PDE is hyperbolic in first and third quadrant.

(1) and (3) are true.

and B2 - 4AC < 0 in second and fourth quadrant as x < 0, y > 0 is 2nd quadrant and x > 0, y < 0 in 4th quadrant.

So, PDE is elliptic in second and fourth quadrant.

(2) and (4) are true.

Concept:

The Lagrange's auxiliary equation of the PDE of the for Pp + qq = R is

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

Explanation:

Given PDE is

z(px - qy) = y2 - x2

⇒ xzp - yzq = y2 - x2

 Integrating we get

Also, $\frac{d(x+y)}{z(x-y)}=\frac{dz}{y^2-x^2}$

⇒ $\frac{d(x+y)}{z}=\frac{dz}{-(x+y)}$

⇒ (x + y) d(x + y) + zdz = 0 

Integrating both sides we get

$(x+y{)}^2+z^2=c_2$....(iii)

Taking (i) and (iii) we get teh solution

x2 + y2 + z2 = $f((x+y{)}^2+z^2)$

(3) is true.

Taking (ii) and (ii) we get the solution

$(x+y{)}^2+z^2=f(xy)$

(4) is true.

Concept:

Quasi-linear PDE of the form Pp + Qq = R satisfy Lagrange's equation

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

Explanation:

Lagrange’s auxiliary equation 

Concept:

The non-linear PDE of the form

f(x, y, z, p, q) = 0 satisfy Charpit equation

$\frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{p{f}_p+q{f}_q}$$=\frac{dp}{-(f_x+p{f}_z)}=\frac{dq}{-(f_y+q{f}_z)}$

Explanation:

Here f(x, y, z, p, q) = (p2 + q2)y - qz 

Using Charpit formula

$\frac{dx}{2py}=\frac{dy}{2qy-z}$$=\frac{dz}{2p^{2}y+2q^{2}y-qz}$$=\frac{dp}{0+pq}=\frac{dq}{-p^2-q^2+q^2}$

⇒ $\frac{dx}{2py}=\frac{dy}{2qy-z}$$=\frac{dz}{2p^{2}y+2q^{2}y-qz}$$=\frac{dp}{pq}=\frac{dq}{-p^2}$

Taking

$\frac{dp}{pq}=\frac{dq}{-p^2}$

⇒ pdp + qdq = 0

Integrating

p² + q² = a²....(i)

Putting in the given equation

a²y = qz

⇒ q = $\frac{a^{2}y}{z}$

Putting in (i)

p = $\sqrt{a^2-q^2}$ = $\frac{a}{z}\sqrt{z^2-a^2{y}^2}$

Putting p and q in

dz = pdx + qdy

⇒ dz = $\frac{a}{z}\sqrt{z^2-a^2{y}^2}$dx + $\frac{a^{2}y}{z}$dy

⇒ $\frac{zdz-a^{2}ydy}{\sqrt{z^2-a^2{y}^2}}$ = a dx

Integrating

$\sqrt{z^2-a^2{y}^2}$ = ax + b

Squaring both sides

z² = a²y² + (ax+b)²

Both (1) and (2) are correct.

Hence option (3) is true.

Explanation:

Given 

utt − uxx = 0, 0 < x < 2, t > 0

u(0, t) = 0 = u(2, t), ∀ t > 0, 

u(x, 0) = sin(πx) + 2sin(2πx), 0 ≤ x ≤ 2,

ut(x, 0) = 0, 0 ≤ x ≤ 2.

which is a wave equation of finite length. So solution is

u(x, t) = $\sum D_n\sin (\frac{n\pi x}{l})\cos (\frac{n\pi ct}{l})$ where

$D_n=\frac{2}{l}{\int }_0^{l}f(x)\sin (\frac{n\pi x}{l})dx$

Here c = 1, l = 2, f(x) = sin(πx) + 2sin(2πx)

So, $D_n=\frac{2}{2}{\int }_0^2(\sin (\pi x)+2\sin (2\pi x))\sin (\frac{n\pi x}{2})dx$

and u(x, t) = $\sum D_n\sin (\frac{n\pi x}{2})\cos (\frac{n\pi ct}{2})$

u(x, 0) = $\sum D_n\sin (\frac{n\pi x}{2})$ = sin(πx) + 2sin(2πx)

Comparing we get

D₂ = 1, D₄ = 2, D_n = 0 for other natural number n

Hence we get

u(x, t) = sin(πx) cos(πt) + 2sin(2πx)cos(2πt)

Then u(1, 1) = 0

u(1/2, 1) = -1

u(1/2, 2) = 1

u(1/2, 1/2) = 0

Option (3) is correct, other are false.

Explanation:

Given z² = kxy, k ∈ ℝ ← system surface

⇒ $\mathrm{k}=\frac{{\mathrm{z}}^2}{\mathrm{x}\mathrm{y}}=\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})$      ........(i)

Now, we will solve it using lagrange A.E - 

(Recall: Pp + Qq = R)

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=\frac{{\mathrm{z}}^2}{\mathrm{y}}(-\frac{1}{{\mathrm{x}}^2})=\frac{-{\mathrm{z}}^2}{{\mathrm{x}}^2\mathrm{y}}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{{\mathrm{z}}^2}{\mathrm{x}}(-\frac{1}{{\mathrm{y}}^2})=\frac{-{\mathrm{z}}^2}{\mathrm{x}{\mathrm{y}}^2}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{z}}=\frac{2\mathrm{z}}{\mathrm{x}\mathrm{y}}$

So, $\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\mathrm{p}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\mathrm{q}=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}$ (p = zx, q = zy)

⇒ $\left(\frac{-{\mathrm{z}}^2}{{\mathrm{x}}^2\mathrm{y}}\right)\mathrm{p}+\left(\frac{-{\mathrm{z}}^2}{\mathrm{x}{\mathrm{y}}^2}\right)\mathrm{q}=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}=\frac{2\mathrm{z}}{\mathrm{x}\mathrm{y}}$

⇒ $-\frac{\mathrm{z}}{\mathrm{x}\mathrm{y}}[\frac{\mathrm{z}}{\mathrm{x}}\mathrm{p}+\frac{\mathrm{z}}{\mathrm{y}}\mathrm{q}]=\frac{\mathrm{z}}{\mathrm{x}\mathrm{y}}[2]$

⇒ $\frac{\mathrm{z}}{\mathrm{x}}\mathrm{p}+\frac{\mathrm{z}}{\mathrm{y}}\mathrm{q}=-2$

⇒ (zy)p + (xz)q = -2xy (on taking LCM)     (ii)

(zy)p → P

(xz)q → Q

-2xy → R

So, by lagrange, A.E -

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{P}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{Q}}=\frac{\mathrm{d}\mathrm{z}}{\mathrm{R}}$   

$\Rightarrow \frac{\mathrm{d}\mathrm{z}}{\mathrm{z}\mathrm{y}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{z}\mathrm{x}}=\frac{\mathrm{d}\mathrm{z}}{(-2\mathrm{x}\mathrm{y})}$

Now,

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{z}\mathrm{y}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{z}\mathrm{x}}$

⇒ x dx = y dy

on integrating

$\frac{{\mathrm{x}}^2}{2}=\frac{{\mathrm{y}}^2}{2}+\mathrm{c}$

⇒ x² - y² = c₁

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{z}\mathrm{y}}=\frac{-\mathrm{d}\mathrm{z}}{2\mathrm{x}\mathrm{y}}$

⇒ 2x dx = -z dz

on integrating

${\mathrm{x}}^2=\frac{-\mathrm{z}62}{2}+\mathrm{c}$

⇒ 2x² + z² = c₂

So, general solⁿ will be ϕ (c₁, c₂) = 0

⇒ ϕ (x² - y², 2x² + z²) = 0 ⇒ option (3) correct.

other possible form of solution and when c₁ & c₂ calculated.

ϕ(c1, c2), c₁ = ϕ(c2), c₂ = ϕ(c2), $\frac{{\mathrm{c}}_1}{2}=\varphi ({\mathrm{c}}_2)$...

Concept:

Let Pp + Qq = R be a PDE where P, Q, R are functions of x, y, z then by Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

Explanation:

Given 

uux + uy = 0, x ∈ ℝ, y > 0,

u(x, 0) = x, x ∈ ℝ.

Using Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

⇒ $\frac{dx}{u}$ = $\frac{dy}{1}$ = $\frac{du}{0}$

Solving this we get 

u = c₁...(i)

and putting u = c1 we get from first two terms

$\frac{dx}{c_1}$ = $\frac{dy}{1}$  

dx = c₁dy
⇒ x - c₁y = c₂

⇒ x - uy = c₂...(ii)

 From (i) and (ii) we get the general solution as

u = ϕ(x - uy)

Using u(x, 0) = x we get

x = ϕ(x)  so ϕ(x - uy) = x - uy

Hence solution is

u = x - uy ⇒  u(1 + y) = x ⇒ u = $\frac{x}{1+y}$

Therefore u(2, 3) = $\frac{2}{4}=\frac{1}{2}$

Option (3) is correct

Explanation:

Given 

$u_t=u_{xx},x\in \mathbb{R},t>0$ with the initial condition

$u(x,0)=\sin (4x)+x+1,x\in \mathbb{R}$ 

which is a heat equation for infinite domain.

So solution is

u(x, t) = $\frac{1}{\sqrt{4\pi ct}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(x-y{)}^2}{4c^{2}t}}f(y)dy$

Here f(x) = sin 4x + x + 1 and c = 1 then

u(x, t) = $\frac{1}{\sqrt{4\pi t}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(x-y{)}^2}{4t}}(\sin 4y+y+1)dy$....(i)

(1): $\mathrm{u}(\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(\frac{\pi }{8}-y{)}^2}{4}}(\sin 4y+y+1)dy$

Let $\frac{\pi }{8}-y=u\Rightarrow dy=-du$ so

$\mathrm{u}(\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\sin 4(\frac{\pi }{8}-u)+(\frac{\pi }{8}-u)+1)du$

$\mathrm{u}(\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}\cos 4udu+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\frac{\pi }{8}-u)+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du]$......(ii)

and $\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(-\frac{\pi }{8}-y{)}^2}{4}}(\sin 4y+y+1)dy$

Let $\frac{\pi }{8}+y=u\Rightarrow dy=du$ so

$\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\sin 4(-\frac{\pi }{8}+u)+(-\frac{\pi }{8}+u)+1)du$

$\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}[-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}\cos 4udu-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\frac{\pi }{8}-u)+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du]$.....(iii)

adding (ii) and (iii) we get

$\mathrm{u}(\frac{\pi }{8},1)+\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{2}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du$

                                 = $\frac{1}{\sqrt{\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du$

                               = $\frac{1}{\sqrt{\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-p^2}2dp$ (let u = 2p then du = 2dp)

                              = $\frac{1}{\sqrt{\pi }}.2\sqrt{\pi }(\because {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-p^2}dp=\sqrt{\pi })$

                             = 2

(1) is true.

From (ii) and (iii) we can see that (2), (3), (4) are false.

Given -

Consider the Cauchy problem

$\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{u};$

𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

Concept -

If the PDE is the form of pP + qQ = R then 

(i) $\frac{p(t)}{x^{\prime }}=\frac{q(t)}{y^{\prime }}=\frac{R}{u^{\prime }(t)}$ ⇒ Infinite solutions 

(ii) $\frac{p(t)}{x^{\prime }}\ne \frac{q(t)}{y^{\prime }}\ne \frac{R}{u^{\prime }(t)}$ ⇒ Unique solution

(iii) $\frac{p(t)}{x^{\prime }}=\frac{q(t)}{y^{\prime }}\ne \frac{R}{u^{\prime }(t)}$ ⇒ No solution

where p = x, q = y and R = u  

Explanation -

we have the PDE is $\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{u};$ and 𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

So x = t, y = t and u = f(t) 

Now for Statement (P) -

$\frac{t}{1}=\frac{t}{1}\ne \frac{2t+1}{2}$

Hence there is no solution. So Statement P is false.

Now for Statement (Q) -

$\frac{t}{1}=\frac{t}{1}\ne \frac{2t-1}{2}$

Hence there is no solution. So Statement Q is false.

Hence option (4) is true.

Explanation:

Given: $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$ i.e. xp + yq = 0

on comping with Pp + Qq = R, we have-

P = x, Q = y & R = 0

so, By Lagrange auxillary equation

 $\frac{dx}{p}=\frac{dy}{Q}=\frac{dz}{R}$

$\Rightarrow \frac{dx}{x}=\frac{dy}{y}=\frac{dz}{0}$

Now dz = 0

⇒ z = c₁

using first and 2^nd term

 $\frac{dx}{x}=\frac{dy}{y}$

Integrating, 

log |z| = log |y| + log c₂

$\Rightarrow \log \left(\frac{|x|}{|y|}\right)=\log c_2$

$\Rightarrow c_2=\frac{|x|}{|y|}$

Hence, general sol is -

c₁ ϕ(c₂) or c₂ = ϕ(c₁) or ϕ(c₁ c₂) = o

⇒ z = ϕ $\left(\frac{|x|}{|y|}\right)$ 

⇒ option (1) correct

Explanation:

Given 

utt − uxx = 0, 0 < x < 2, t > 0

u(0, t) = 0 = u(2, t), ∀ t > 0, 

u(x, 0) = sin(πx) + 2sin(2πx), 0 ≤ x ≤ 2,

ut(x, 0) = 0, 0 ≤ x ≤ 2.

which is a wave equation of finite length. So solution is

u(x, t) = $\sum D_n\sin (\frac{n\pi x}{l})\cos (\frac{n\pi ct}{l})$ where

$D_n=\frac{2}{l}{\int }_0^{l}f(x)\sin (\frac{n\pi x}{l})dx$

Here c = 1, l = 2, f(x) = sin(πx) + 2sin(2πx)

So, $D_n=\frac{2}{2}{\int }_0^2(\sin (\pi x)+2\sin (2\pi x))\sin (\frac{n\pi x}{2})dx$

and u(x, t) = $\sum D_n\sin (\frac{n\pi x}{2})\cos (\frac{n\pi ct}{2})$

u(x, 0) = $\sum D_n\sin (\frac{n\pi x}{2})$ = sin(πx) + 2sin(2πx)

Comparing we get

D₂ = 1, D₄ = 2, D_n = 0 for other natural number n

Hence we get

u(x, t) = sin(πx) cos(πt) + 2sin(2πx)cos(2πt)

Then u(1, 1) = 0

u(1/2, 1) = -1

u(1/2, 2) = 1

u(1/2, 1/2) = 0

Option (3) is correct, other are false.

Explanation:

From the trial and error method, we can say that u = 16(x2 + y2) - 16 will be the solution of the given partial differential equation as

u_x = 32x ⇒ $\frac{{\partial }^{2}u}{\partial x^2}$ = 32 and uy = 32y ⇒ $\frac{{\partial }^{2}u}{\partial y^2}$ = 32 so

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=64$

Also, it satisfies the given boundary condition as on the boundary of a unit disk u = 16 × 1 - 16 = 0, so u vanishes.

Hence u $(\frac{1}{4},\frac{1}{\sqrt{2}})=16(\frac{1}{16}+\frac{1}{2})$ - 16 = 16.$\frac{9}{16}$ - 16 = 9 -16 = - 7

Option (3) is correct

Concept:

If u(x, t) is a solution of a homogeneous PDE then u(x- a, t-b) is also a solution of the PDE for a, b ∈ $\mathbb{R}$ and u(ax, bt) is also a solution if a = b is real number

Explanation:

(∗) $\frac{{\partial }^{2}u}{\partial t^2}-\frac{{\partial }^{2}u}{\partial x^2}=0$ x, t) ∈ ℝ2

u(x, t) is a solution of (∗) 

then u(x - θ, t) is also a solution of (i) for any fixed θ ∈ ℝ.

u(3x, 3t) is also a solution of (∗)

u(3x, 9t) is not a solution of (∗) as 3 ≠ 9

(3) is FALSE