Quasi-linear pde $u_t + x u u_x = 0$, find shock time
$begingroup$
Using the characteristics method, show that the Cauchy Problem for the
quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
future and one in the past. Determine $t^*_pm $.
The characteristic curves are the solutions to the ODE system
$$ dot t = 1 qquad t(0) = 0$$
$$ dot x = ux qquad x(0) = x_0$$
$$ dot u = 0 qquad u(0) = phi (x_0)$$
This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
$$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?
pde characteristics
edited Dec 6 '18 at 13:52
Harry49
6,17331132
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
add a comment |
$begingroup$
Using the characteristics method, show that the Cauchy Problem for the
quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
future and one in the past. Determine $t^*_pm $.
The characteristic curves are the solutions to the ODE system
$$ dot t = 1 qquad t(0) = 0$$
$$ dot x = ux qquad x(0) = x_0$$
$$ dot u = 0 qquad u(0) = phi (x_0)$$
This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
$$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?
pde characteristics
edited Dec 6 '18 at 13:52
Harry49
6,17331132
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
add a comment |
$begingroup$
Using the characteristics method, show that the Cauchy Problem for the
quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
future and one in the past. Determine $t^*_pm $.
The characteristic curves are the solutions to the ODE system
$$ dot t = 1 qquad t(0) = 0$$
$$ dot x = ux qquad x(0) = x_0$$
$$ dot u = 0 qquad u(0) = phi (x_0)$$
This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
$$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?
pde characteristics
edited Dec 6 '18 at 13:52
Harry49
6,17331132
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
Using the characteristics method, show that the Cauchy Problem for the
quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
future and one in the past. Determine $t^*_pm $.
The characteristic curves are the solutions to the ODE system
$$ dot t = 1 qquad t(0) = 0$$
$$ dot x = ux qquad x(0) = x_0$$
$$ dot u = 0 qquad u(0) = phi (x_0)$$
This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
$$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?
pde characteristics
pde characteristics
edited Dec 6 '18 at 13:52
Harry49
6,17331132
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
edited Dec 6 '18 at 13:52
Harry49
6,17331132
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
edited Dec 6 '18 at 13:52
Harry49
6,17331132
edited Dec 6 '18 at 13:52
Harry49
6,17331132
edited Dec 6 '18 at 13:52
Harry49
6,17331132
6,17331132
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
asked Sep 14 '18 at 15:19
Riccardo OrlandoRiccardo Orlando
1,654517
1,654517
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As outlined by OP's self-answer,
$$
x(t) = x_0exp(tphi(x_0)) , .
$$
Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
$$
t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
$$
Thus, for positive times, the classical solution breaks at the breaking time
$$t^*_+ = inf_{x_0>0} t_S = 2, .$$
Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
$endgroup$
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
add a comment |
$begingroup$
There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
$$ x = x_0 exp(tphi) $$
From here on, the method I was trying to use, and that Harry49 explained in his answer.
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As outlined by OP's self-answer,
$$
x(t) = x_0exp(tphi(x_0)) , .
$$
Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
$$
t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
$$
Thus, for positive times, the classical solution breaks at the breaking time
$$t^*_+ = inf_{x_0>0} t_S = 2, .$$
Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
$endgroup$
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
add a comment |
$begingroup$
As outlined by OP's self-answer,
$$
x(t) = x_0exp(tphi(x_0)) , .
$$
Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
$$
t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
$$
Thus, for positive times, the classical solution breaks at the breaking time
$$t^*_+ = inf_{x_0>0} t_S = 2, .$$
Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
$endgroup$
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
add a comment |
$begingroup$
As outlined by OP's self-answer,
$$
x(t) = x_0exp(tphi(x_0)) , .
$$
Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
$$
t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
$$
Thus, for positive times, the classical solution breaks at the breaking time
$$t^*_+ = inf_{x_0>0} t_S = 2, .$$
Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
$endgroup$
As outlined by OP's self-answer,
$$
x(t) = x_0exp(tphi(x_0)) , .
$$
Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
$$
t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
$$
Thus, for positive times, the classical solution breaks at the breaking time
$$t^*_+ = inf_{x_0>0} t_S = 2, .$$
Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
edited Sep 18 '18 at 18:33
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
answered Sep 15 '18 at 10:47
Harry49Harry49
6,17331132
6,17331132
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
add a comment |
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
$begingroup$
Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
$endgroup$
– Riccardo Orlando
Sep 15 '18 at 11:10
add a comment |
$begingroup$
There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
$$ x = x_0 exp(tphi) $$
From here on, the method I was trying to use, and that Harry49 explained in his answer.
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
add a comment |
$begingroup$
There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
$$ x = x_0 exp(tphi) $$
From here on, the method I was trying to use, and that Harry49 explained in his answer.
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
add a comment |
$begingroup$
There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
$$ x = x_0 exp(tphi) $$
From here on, the method I was trying to use, and that Harry49 explained in his answer.
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
$endgroup$
There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
$$ x = x_0 exp(tphi) $$
From here on, the method I was trying to use, and that Harry49 explained in his answer.
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
answered Sep 18 '18 at 14:05
Riccardo OrlandoRiccardo Orlando
1,654517
1,654517
add a comment |
add a comment |
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