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 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: 7,664 Thanks: 2644 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: 7,664 Thanks: 2644 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 $g(0) = 1$.

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