Laplace Equation, Heat and Wave Equations MCQ Quiz in తెలుగు - Objective Question with Answer for Laplace Equation, Heat and Wave Equations - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Solution of the heat equation 

u_t = ku_xx 0 < x < a, t > 0

with boundary condition

u(0, t) = u(a, t) = 0, t > 0 

and initial condition

u(0, x) = f(x) is

u(x, t) = \(\sum_{n=1}^{\infty}D_n \sin({n\pi x\over a})e^{-k{n^2\pi^2t\over a^2}}\)

where

D_n = \(\frac2a\int_0^af(x)\sin({n\pi x\over a})dx\)

Explanation:

If u(x, t) = Ae-t sin x is the solution of initial boundary value problem

\(\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}\), 0 < x < π, t > 0 

u(0, t) = u(π, t) = 0, t > 0 

\(u(x, 0)= \begin{cases}60, & 0

Solution of the given heat equation is

u(x, t) = \(\sum_{n=1}^{\infty}D_n \sin({nx})e^{-{n^2t}}\) (here k = 1, a = π)

Given  u(x, t) = Ae-t sin x

Comparing both we get  

D₁ = A for n = 1

So, A = D₁ = \(\frac2{\pi}\int_0^{\pi}f(x)\sin({ x})dx\)

                 = \(\frac2{\pi}\int_0^{\pi\over 2}60\sin({ x})dx\) + \(\frac2{\pi}\int_{\pi\over 2}^{\pi}40\sin({ x})dx\)

                = \(\frac2{\pi}.60.1+\frac{2}{\pi}.40.1\) = \(\frac{200}{\pi}\)

Hence πA = 200

Concept:

A second-order PDE of the form

Auxx + Buxy + Cuyy + f(x, y, z, u_x, u_y) = 0 is 

(i) Hyperbolic if B2 - 4AC > 0

(ii) Elliptic if B2 - 4AC < 0

(ii) Parabolic if B2 - 4AC = 0

Explanation:

auxx + buxy + auyy = 0 in ℝ2 for a, b ∈ ℝ.

B² - 4AC = b² - 4a.a = b² - 4a²

Now, PDE is hyperbolic if

b2 - 4a2 > 0 ⇒ (b - 2a)(b + 2a) > 0 ⇒ b > 2a and b > -2a

(1), (4) are false

PDE is elliptic if

b2 - 4a2 < 0 ⇒ (b - 2a)(b + 2a) < 0 

⇒ b < 2a and b > -2a or b > 2a and b < -2a

⇒  |b| < 2|a|

(3) is true

PDE is parabolic if

b2 - 4a2 = 0 ⇒ (b - 2a)(b + 2a) = 0 

⇒ b = 2a or b = -2a

(2) is false