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

\({dx\over P}={dy\over Q}={dz\over R}\)

Explanation:

Given PDE is

z(px - qy) = y2 - x2

⇒ xzp - yzq = y2 - x2

 Integrating we get

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

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

⇒ (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

\({dx\over P}={dy\over Q}={dz\over R}\)

Explanation:

Lagrange’s auxiliary equation 

Concept:

The non-linear PDE of the form

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

\({dx\over f_p}={dy\over f_q}={dz\over pf_p+qf_q}\)\(={dp\over -(f_x+pf_z)}={dq\over -(f_y+qf_z)}\)

Explanation:

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

Using Charpit formula

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

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

Taking

\({dp\over pq}={dq\over -p^2}\)

⇒ pdp + qdq = 0

Integrating

p² + q² = a²....(i)

Putting in the given equation

a²y = qz

⇒ q = \(a^2y\over z\)

Putting in (i)

p = \(\sqrt{a^2-q^2}\) = \(\frac az\sqrt{z^2-a^2y^2}\)

Putting p and q in

dz = pdx + qdy

⇒ dz = \(\frac az\sqrt{z^2-a^2y^2}\)dx + \(a^2y\over z\)dy

⇒ \(zdz-a^2ydy\over\sqrt{z^2 -a^2 y^2}\) = a dx

Integrating

\(\sqrt{z^2-a^2y^2}\) = 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) = \(\sum D_n \sin({n\pi x\over l})\cos({n\pi ct\over l})\) where

\(D_n=\frac2l\int_0^lf(x)\sin({n\pi x\over l})dx\)

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

So, \(D_n=\frac22\int_0^2(\sin(\pi x)+2\sin (2\pi x))\sin({n\pi x\over 2})dx\)

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

u(x, 0) = \(\sum D_n \sin({n\pi x\over 2})\) = 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

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

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

(Recall: Pp + Qq = R)

\(\rm \frac{\partial f}{\partial x}=\frac{z^2}{y}\left(-\frac{1}{x^2}\right)=\frac{-z^2}{x^2y}\)

\(\rm \frac{\partial f}{\partial y}=\frac{z^2}{x}\left(-\frac{1}{y^2}\right)=\frac{-z^2}{xy^2}\)

\(\rm \frac{\partial f}{\partial z}=\frac{2z}{xy}\)

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

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

⇒ \(\rm -\frac{z}{xy}\left[\frac{z}{x}p+\frac{z}{y}q\right]=\frac{z}{xy}[2]\)

⇒ \(\rm \frac{z}{x}p+\frac{z}{y}q=-2\)

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

(zy)p → P

(xz)q → Q

-2xy → R

So, by lagrange, A.E -

\(\rm \frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\)   

\(\rm ⇒ \frac{dz}{zy}=\frac{dy}{zx}=\frac{dz}{(-2xy)}\)

Now,

\(\rm\frac{dx}{zy}=\frac{dy}{zx}\)

⇒ x dx = y dy

on integrating

\(\rm \frac{x^2}{2}=\frac{y^2}{2}+c\)

⇒ x² - y² = c₁

\(\rm \frac{dx}{zy}=\frac{-dz}{2xy}\)

⇒ 2x dx = -z dz

on integrating

\(\rm x^2=\frac{-z62}{2}+c\)

⇒ 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), \(\rm \frac{c_1}{2}=\phi(c_2)\)...

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) = \(\frac{2}{4}=\frac12\)

Option (3) is correct

Explanation:

Given 

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

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

which is a heat equation for infinite domain.

So solution is

u(x, t) = \({1\over \sqrt{4\pi ct}}\int_{-\infty}^{\infty}e^{-{(x-y)^2\over 4c^2t}}f(y)dy\)

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

u(x, t) = \({1\over \sqrt{4\pi t}}\int_{-\infty}^{\infty}e^{-{(x-y)^2\over 4t}}(\sin 4y+y+1)dy\)....(i)

(1): \(\rm u\left(\frac{\pi}{8}, 1\right)\) = \({1\over \sqrt{4\pi }}\int_{-\infty}^{\infty}e^{-{(\frac{\pi}{8}-y)^2\over 4}}(\sin 4y+y+1)dy\)

Let \({\pi\over 8}-y=u\Rightarrow dy=-du\) so

\(\rm u\left(\frac{\pi}{8}, 1\right)\) = \({1\over \sqrt{4\pi }}\int_{-\infty}^{\infty}e^{-{u^2\over 4}}(\sin 4(\frac{\pi}{8}-u)+(\frac{\pi}{8}-u)+1)du\)

\(\rm u\left(\frac{\pi}{8}, 1\right)\) = \({1\over \sqrt{4\pi }}\left[\int_{-\infty}^{\infty}e^{-{u^2\over 4}}\cos 4udu+\int_{-\infty}^{\infty}e^{-{u^2\over 4}}(\frac{\pi}{8}-u)+\int_{-\infty}^{\infty}e^{-{u^2\over 4}}du\right]\)......(ii)

and \(\rm u\left(-\frac{\pi}{8}, 1\right)\) = \({1\over \sqrt{4\pi }}\int_{-\infty}^{\infty}e^{-{(-\frac{\pi}{8}-y)^2\over 4}}(\sin 4y+y+1)dy\)

Let \({\pi\over 8}+y=u\Rightarrow dy=du\) so

\(\rm u\left(-\frac{\pi}{8}, 1\right)\) = \({1\over \sqrt{4\pi }}\int_{-\infty}^{\infty}e^{-{u^2\over 4}}(\sin 4(-\frac{\pi}{8}+u)+(-\frac{\pi}{8}+u)+1)du\)

\(\rm u\left(-\frac{\pi}{8}, 1\right)\) = \({1\over \sqrt{4\pi }}\left[-\int_{-\infty}^{\infty}e^{-{u^2\over 4}}\cos 4udu-\int_{-\infty}^{\infty}e^{-{u^2\over 4}}(\frac{\pi}{8}-u)+\int_{-\infty}^{\infty}e^{-{u^2\over 4}}du\right]\).....(iii)

adding (ii) and (iii) we get

\(\rm u\left(\frac{\pi}{8}, 1\right)+u\left(-\frac{\pi}{8},1\right)\) = \({2\over \sqrt{4\pi }}\int_{-\infty}^{\infty}e^{-{u^2\over 4}}du\)

                                 = \({1\over \sqrt{\pi }}\int_{-\infty}^{\infty}e^{-{u^2\over 4}}du\)

                               = \({1\over \sqrt{\pi }}\int_{-\infty}^{\infty}e^{-p^2}2dp\) (let u = 2p then du = 2dp)

                              = \({1\over \sqrt{\pi }}.2\sqrt \pi\,\,(\because\int_{-\infty}^{\infty}e^{-p^2}dp=\sqrt\pi)\)

                             = 2

(1) is true.

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

Given -

Consider the Cauchy problem

\(\rm x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=u;\)

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) \(\frac{p(t)}{x'}= \frac{q(t)}{y'} = \frac{R}{u'(t)}\) ⇒ Infinite solutions 

(ii) \(\frac{p(t)}{x'} \neq \frac{q(t)}{y'} \neq \frac{R}{u'(t)}\) ⇒ Unique solution

(iii) \(\frac{p(t)}{x'} = \frac{q(t)}{y'} \neq \frac{R}{u'(t)}\) ⇒ No solution

where p = x, q = y and R = u  

Explanation -

we have the PDE is \(\rm x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=u;\) 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) -

\(\frac{t}{1}= \frac{t}{1} \neq \frac{2t+1}{2}\)

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

Now for Statement (Q) -

\(\frac{t}{1}= \frac{t}{1} \neq \frac{2t-1}{2}\)

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

 \(\frac{d x}{p}=\frac{d y}{Q}=\frac{d z}{R}\)

\(\Rightarrow \frac{d x}{x}=\frac{d y}{y}=\frac{d z}{0}\)

Now dz = 0

⇒ z = c₁

using first and 2^nd term

 \(\frac{d x}{x}=\frac{d y}{y}\)

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) = \(\sum D_n \sin({n\pi x\over l})\cos({n\pi ct\over l})\) where

\(D_n=\frac2l\int_0^lf(x)\sin({n\pi x\over l})dx\)

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

So, \(D_n=\frac22\int_0^2(\sin(\pi x)+2\sin (2\pi x))\sin({n\pi x\over 2})dx\)

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

u(x, 0) = \(\sum D_n \sin({n\pi x\over 2})\) = 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 ⇒ \(\frac{\partial ^2u}{\partial x^2}\) = 32 and uy = 32y ⇒ \(\frac{\partial ^2u}{\partial y^2}\) = 32 so

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

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 \(\left(\frac{1}{4},\frac{1}{\sqrt2} \right) =16(\frac1{16}+\frac12)\) - 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