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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 January 21st, 2015, 12:29 AM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 Hi, can someone please help me to solve this problem? Consider an initial value problem for transport equation with non-linear source ut − ux = u2 x ∈ R, t > 0, u(x, t = 0) = g(x). Where g(x) is non-negative continuous function with continuous derivative that attains its maximum at a single point x = 0 and g(0) = 1. a) solve initial value problem. b) Show that solution becomes unbounded in finite time and thus solution for the problem exists only on a finite time interval. Find maximal interval of existence. c) Find coordinate of a point (x∗, t∗) at which solution of this problem becomes unbounded. I tried with the first question and I find u(x,t)= g(x+t)/(1-tg(x+t)) . Is it the right answer? If it is, please help me to solve the following questions. Thanks. Last edited by skipjack; January 21st, 2015 at 01:14 AM. January 21st, 2015, 04:54 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra I agree with you. \begin{align*} &\text{Write} & {\mathrm d t \over \mathrm d s} &= 1 \qquad t(0) = 0 \\ &\text{and} & {\mathrm d x \over \mathrm d s} &= -1 \qquad x(0) = \epsilon \\ &\text{then we have} & {\mathrm d U \over \mathrm d s} &= u^2(x,t) \qquad U(0) = g(\epsilon) \\[12pt] & \text{so} & \left\{\begin{array}{l}t = s \\ x = \epsilon - s \end{array}\right\} &\implies \begin{cases} s=t \\ \epsilon = x + t \end{cases} \\ && -{1 \over U} &= s - {1 \over g(\epsilon)} = {sg(\epsilon)-1 \over g(\epsilon)} \\ && U(s) &= {g(\epsilon) \over 1-sg(\epsilon)} \\ && u(x,t) = U\big(x(s,\epsilon),t(s, \epsilon)\big) &= {g(x+t) \over 1-tg(x+t)} \end{align*} To confirm the solution: \begin{align*} u_t &= {g'(x+t)\big(1-tg(x+t)\big) + g(x+t)\big(g(x+t) + tg'(x+t)\big) \over \big(1-tg(x-t)\big)^2} \\ &= {g^2(x,t) + g'(x,t) \over \big(1-tg(x-t)\big)^2}\\ u_x &= {g'(x+t)\big(1-tg(x+t)\big) + tg(x+t)g'(x+t) \over \big(1-tg(x-t)\big)^2} \\ &= {g'(x,t) \over \big(1-tg(x-t)\big)^2}\\ u_t - u_x &= {g^2(x-t) \over \big(1-sg(x-t)\big)^2} = u^2(x,t) \quad \text{as required}\end{align*} Last edited by v8archie; January 21st, 2015 at 05:12 AM. January 21st, 2015, 05:10 AM #3 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 differential equation hi v8archie thanks for your help.can you please help me with the following questions? thanks January 21st, 2015, 06:27 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra I'm struggling with b). I can't help thinking that there are functions for which $1-tg(x+t)$ is never zero (for some $x$). Clearly, $(x,t) = (-1,1)$ is one place that the solution explodes, but I don't see how we prove that it explodes for all $x$. It seems to me that, for positive $x$, $g(x+t)$ could be decreasing faster than $t$ increases which would mean that $tg(x+t) \to 0$. One such example is $g(x+t) = \mathrm e^{-(x+t)}$ which can only explode for negative $x$. Note that for negative $x$ the solution always explodes at $t = -x \gt 0$ because $g(0) = 1$. Tags differential, differentiel, equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Sonprelis Calculus 6 August 6th, 2014 10:07 AM PhizKid Differential Equations 0 February 24th, 2013 10:30 AM tsl182forever8 Differential Equations 2 March 14th, 2012 10:15 PM billmmmm Differential Equations 1 June 29th, 2010 09:10 AM baby_Nway Differential Equations 4 November 21st, 2007 03:24 PM

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