3) Consider the non-linear initial value problem | { Ofu – Ozu = (d,u)² – (d;u)², x,t e R, - (2) u(x, 0) = 0, du(x, 0) = g(x), g : R → R given. (a) Prove that if w = w(x, t) satisfies Ofw – 0w = 0, x,t e R, - %3D w(x, 0) = 1, dw(x, 0) = g(x), then w(x, t) = e"(z,t). %3D (b) Show that ra+t u(x, t) = log (1+5/. 9(»)ds). (c) Prove that if (3) 19(*) ds < 2,
3) Consider the non-linear initial value problem | { Ofu – Ozu = (d,u)² – (d;u)², x,t e R, - (2) u(x, 0) = 0, du(x, 0) = g(x), g : R → R given. (a) Prove that if w = w(x, t) satisfies Ofw – 0w = 0, x,t e R, - %3D w(x, 0) = 1, dw(x, 0) = g(x), then w(x, t) = e"(z,t). %3D (b) Show that ra+t u(x, t) = log (1+5/. 9(»)ds). (c) Prove that if (3) 19(*) ds < 2,
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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Subject: Partial differential equations
Kindly solve it as soon as possible , kindly give it a try at least Thank you!!
Transcribed Image Text:3) Consider the non-linear initial value problem
Ofu – 0zu = (du)² – (d¿u)²,
x, t e R,
(2)
u(x, 0) = 0,
du(x, 0) = g(x),
g :R → R given.
(a) Prove that if w =
w(x, t) satisfies
{
Ofw – aw = 0,
w(x,0) = 1,
dw(x, 0) = g(x),
x,t e R,
then
w(x, t) = e"(z,t).
(b) Show that
pI+t
u(7, t) = log (1 + 5 9(e)ds).
(c) Prove that if
(3)
| I9(s) ds < 2,
then the solution u(x, t) of (2) is defined for all t e R. Notice that if
(3) fails, then the solution u(x, t) of (2) may only exist in a finite time
interval [0,T), T > 0.
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