Homogeneous Quasi-linear PDE, Method of Characteristics.
$begingroup$
I want to solve the PDE:
$$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
$$
for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;
$u(t,0)=left{begin{matrix}
frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
1, t geq 2&
end{matrix}right.$
by using the method of characteristics.
The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.
For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.
For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.
In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:
$u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:
$u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.
Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.
As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).
Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).
What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
$frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$
Where the subscripts denote the value of u just to the right or jsut to the left of the shock.
I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.
pde transport-equation
$endgroup$
add a comment |
$begingroup$
I want to solve the PDE:
$$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
$$
for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;
$u(t,0)=left{begin{matrix}
frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
1, t geq 2&
end{matrix}right.$
by using the method of characteristics.
The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.
For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.
For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.
In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:
$u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:
$u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.
Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.
As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).
Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).
What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
$frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$
Where the subscripts denote the value of u just to the right or jsut to the left of the shock.
I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.
pde transport-equation
$endgroup$
add a comment |
$begingroup$
I want to solve the PDE:
$$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
$$
for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;
$u(t,0)=left{begin{matrix}
frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
1, t geq 2&
end{matrix}right.$
by using the method of characteristics.
The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.
For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.
For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.
In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:
$u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:
$u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.
Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.
As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).
Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).
What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
$frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$
Where the subscripts denote the value of u just to the right or jsut to the left of the shock.
I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.
pde transport-equation
$endgroup$
I want to solve the PDE:
$$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
$$
for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;
$u(t,0)=left{begin{matrix}
frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
1, t geq 2&
end{matrix}right.$
by using the method of characteristics.
The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.
For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.
For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.
In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:
$u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:
$u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.
Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.
As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).
Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).
What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
$frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$
Where the subscripts denote the value of u just to the right or jsut to the left of the shock.
I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.
pde transport-equation
pde transport-equation
edited Dec 20 '18 at 14:01
Harry49
6,98631238
6,98631238
asked Apr 28 '17 at 6:33
J.DoughJ.Dough
232
232
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$begingroup$
An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.
The next figure shows $u$ as a function of $x$ at various time.
$endgroup$
add a comment |
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1
active
oldest
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oldest
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votes
$begingroup$
An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.
The next figure shows $u$ as a function of $x$ at various time.
$endgroup$
add a comment |
$begingroup$
An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.
The next figure shows $u$ as a function of $x$ at various time.
$endgroup$
add a comment |
$begingroup$
An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.
The next figure shows $u$ as a function of $x$ at various time.
$endgroup$
An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.
The next figure shows $u$ as a function of $x$ at various time.
edited Apr 30 '17 at 9:21
answered Apr 29 '17 at 14:16
JJacquelinJJacquelin
44k21853
44k21853
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