3.) Use the Maximum/Minimum Principles to deduce that the solution u of the problemD.E. u = kuz 0 0.

given differential equation ut=kuxx ...i where 0≤x≤π, t≥0 boundary condition u0,t=0 uπ,t=0…

Answered: 3.) Use the Maximum/Minimum Principles…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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3.) Use the Maximum/Minimum Principles to deduce that the solution u of the problem
D.E. u = kuz 0 <r < T, t N 0
B.C. u(0, t) = 0
u (π, t) 0
I.C. u(x, 0) = sin(x) + sin(2x)
%3D
satisfies 0 < u(, t) <V3 for all 0 < ¤ < T, t > 0.

Expert Solution

Step 1

given

differential equation

$u_{t} = k u_{x x} . . . (i)$

where $0 \leq x \leq π, t \geq 0$

boundary condition

$u (0, t) = 0 u (π, t) = 0$

initial condition

$u (x, 0) = \sin x + \frac{1}{2} \sin 2 x$

satisfies

$0 \leq u (x, t) \leq \frac{3}{4} \sqrt{3}$ for all $0 \leq x \leq π, t \geq 0$

Step 2

solution

let the solution be of the form

$u (x, t) = X (x) T (t)$

differentiate it with respect to $x and t$

$u_{x} = X^{'} (x) T (t)$

$u_{t} = X (x) T^{'} (t)$

again differentiate $u_{x} = X^{'} (x) T (t)$ with respect to $x$

$u_{x} = X^{''} (x) T (t)$

substitute the values in equation $(i)$

we get

$k X^{''} (x) T (t) = X (x) T^{'} (t) \Rightarrow X^{''} (x) T (t) = \frac{1}{k} X (x) T^{'} (t)$

Step 3

divide both sides by $X (x) T (t)$

$\frac{X^{''} (x)}{X (x)} = \frac{1}{k} \frac{T^{'} (t)}{T (t)}$

let

$\frac{X^{''} (x)}{X (x)} = \frac{1}{k} \frac{T^{'} (t)}{T (t)} = - p^{2}$

p is any constant

we have

$\frac{X^{''} (x)}{X (x)} = - p^{2} \Rightarrow X^{''} (x) + p^{2} X (x) = 0 . . . (i i)$

and

$\frac{1}{k} \frac{T^{'} (t)}{T (t)} = - p^{2} \Rightarrow \frac{T^{'} (t)}{T (t)} = - p^{2} k . . . (i i i)$

Step 4

solving equation (ii)

$X^{''} (x) + p^{2} X (x) = 0$

auxillary equation

$m^{2} + p^{2} = 0 m^{2} = - p^{2} \Rightarrow m = \pm i p$

we have

$X (x) = c_{1} \cos p x + c_{2} \sin p x$

integrate equation (iii)

we get

$\int \frac{T^{'} (t)}{T (t)} = \int - p^{2} k d t \ln T (t) = - p^{2} k t T (t) = e^{- p^{2} k t}$

thus the solution becomes

$u (x, t) = (c_{1} \cos p x + c_{2} \sin p x) e^{- p^{2} k t}$

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3.) Use the Maximum/Minimum Principles to deduce that the solution u of the problem D.E. ut = kuzr 00 B.C. u(0, t) = 0 u(7, t) = 0 I.C. u(x, 0) = sin(x) + ; sin(2x) %3D satisfies 0 < u(x, t) 0.