Transport equation $p_t - (xp)_x = 0$ for density of substance
$begingroup$
So I'm trying to do the following:
i) Solve
$$p_t - (xp)_x = 0 quadtext{for}quad (t,x) in (0, infty ) times mathbb{R}$$
$$p(0,x) = {p_0}(x) quadtext{for}quad x in mathbb{R}, {p_0} in C_0^1 ( mathbb{R})$$
ii) Calculate $$lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx quadtext{for}quad phi in C( mathbb{R})$$
iii) Assuming that $p_0 geq 0$ and $p$ is the density of a substance at the time $t$. Interpet i ) and ii) from a physical perspective! Is the mathematical result of ii) the same as one would expect physically?
Ok, so I think I've managed to solve i) with the method of characteristics and got $p = e^t p_0(xe^t)$.
For ii), I was thinking you could substitute $u=x e^t$ to get
begin{aligned}
lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx &= lim_{t to infty} int_{mathbb{R}} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} lim_{t to infty} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} phi (0) p_0(u)
end{aligned}
but I am not sure if it is finished or even correct.
I also need help with iii), I know that $p$ is the density and that the integral over the density can give you the volume, but I am not sure how to physically interpret the formulas in i) and ii).
pde hyperbolic-equations transport-equation
edited Dec 6 '18 at 13:45
Harry49
6,17331132
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
$endgroup$
add a comment |
$begingroup$
So I'm trying to do the following:
i) Solve
$$p_t - (xp)_x = 0 quadtext{for}quad (t,x) in (0, infty ) times mathbb{R}$$
$$p(0,x) = {p_0}(x) quadtext{for}quad x in mathbb{R}, {p_0} in C_0^1 ( mathbb{R})$$
ii) Calculate $$lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx quadtext{for}quad phi in C( mathbb{R})$$
iii) Assuming that $p_0 geq 0$ and $p$ is the density of a substance at the time $t$. Interpet i ) and ii) from a physical perspective! Is the mathematical result of ii) the same as one would expect physically?
Ok, so I think I've managed to solve i) with the method of characteristics and got $p = e^t p_0(xe^t)$.
For ii), I was thinking you could substitute $u=x e^t$ to get
begin{aligned}
lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx &= lim_{t to infty} int_{mathbb{R}} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} lim_{t to infty} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} phi (0) p_0(u)
end{aligned}
but I am not sure if it is finished or even correct.
I also need help with iii), I know that $p$ is the density and that the integral over the density can give you the volume, but I am not sure how to physically interpret the formulas in i) and ii).
pde hyperbolic-equations transport-equation
edited Dec 6 '18 at 13:45
Harry49
6,17331132
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
$endgroup$
add a comment |
$begingroup$
So I'm trying to do the following:
i) Solve
$$p_t - (xp)_x = 0 quadtext{for}quad (t,x) in (0, infty ) times mathbb{R}$$
$$p(0,x) = {p_0}(x) quadtext{for}quad x in mathbb{R}, {p_0} in C_0^1 ( mathbb{R})$$
ii) Calculate $$lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx quadtext{for}quad phi in C( mathbb{R})$$
iii) Assuming that $p_0 geq 0$ and $p$ is the density of a substance at the time $t$. Interpet i ) and ii) from a physical perspective! Is the mathematical result of ii) the same as one would expect physically?
Ok, so I think I've managed to solve i) with the method of characteristics and got $p = e^t p_0(xe^t)$.
For ii), I was thinking you could substitute $u=x e^t$ to get
begin{aligned}
lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx &= lim_{t to infty} int_{mathbb{R}} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} lim_{t to infty} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} phi (0) p_0(u)
end{aligned}
but I am not sure if it is finished or even correct.
I also need help with iii), I know that $p$ is the density and that the integral over the density can give you the volume, but I am not sure how to physically interpret the formulas in i) and ii).
pde hyperbolic-equations transport-equation
edited Dec 6 '18 at 13:45
Harry49
6,17331132
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
$endgroup$
So I'm trying to do the following:
i) Solve
$$p_t - (xp)_x = 0 quadtext{for}quad (t,x) in (0, infty ) times mathbb{R}$$
$$p(0,x) = {p_0}(x) quadtext{for}quad x in mathbb{R}, {p_0} in C_0^1 ( mathbb{R})$$
ii) Calculate $$lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx quadtext{for}quad phi in C( mathbb{R})$$
iii) Assuming that $p_0 geq 0$ and $p$ is the density of a substance at the time $t$. Interpet i ) and ii) from a physical perspective! Is the mathematical result of ii) the same as one would expect physically?
Ok, so I think I've managed to solve i) with the method of characteristics and got $p = e^t p_0(xe^t)$.
For ii), I was thinking you could substitute $u=x e^t$ to get
begin{aligned}
lim_{t to infty} int_{mathbb{R}} phi(x) p(t,x) dx &= lim_{t to infty} int_{mathbb{R}} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} lim_{t to infty} phi(u/e^t) p_0(u) dx \
&= int_{mathbb{R}} phi (0) p_0(u)
end{aligned}
but I am not sure if it is finished or even correct.
I also need help with iii), I know that $p$ is the density and that the integral over the density can give you the volume, but I am not sure how to physically interpret the formulas in i) and ii).
pde hyperbolic-equations transport-equation
pde hyperbolic-equations transport-equation
edited Dec 6 '18 at 13:45
Harry49
6,17331132
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
edited Dec 6 '18 at 13:45
Harry49
6,17331132
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
edited Dec 6 '18 at 13:45
Harry49
6,17331132
edited Dec 6 '18 at 13:45
Harry49
6,17331132
edited Dec 6 '18 at 13:45
Harry49
6,17331132
6,17331132
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
asked Oct 15 '18 at 15:11
Tierra TelderTierra Telder
234
234
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The PDE $p_t−(xp)_x=0$ in (i) rewrites as $p_t−xp_x=p$. Its solution obtained via the method of characteristics is indeed $p(t,x)=e^t, p_0(x e^t)$. Setting $u = x e^t $, the integral in (ii) rewrites as
$$
Phi(t) = int_{Bbb R}phi(x), p(t,x),text d x = int_{Bbb R}phi(ue^{-t}), p_0(u),text d u
$$
which limit at infinity is
$$
lim_{tto {+infty}}Phi(t) = phi(0)int_{Bbb R}p_0(u),text d u , .
$$
The PDE in (i) is a conservation law. If $p$ is the density of a given substance, then the total amount of substance $P(t) = int_{Bbb R} p(t,x),text d x$ evolves as
$$
P'(t) = int_{Bbb R} partial_t p(t,x),text d x = [x, p(t,x)]_{x=-infty}^{x=+infty} = 0, ,
$$
i.e. $P(t)= P(0)= int_{Bbb R} p_0$. The product $-xp$ is the flux of this substance, where $-x$ is analogous to a transport velocity (cf. conservation of mass). When $t$ goes to infinity, the substance flows towards $x=0$ where the speed is zero, and it stays concentrated there. This can be illustrated by a plot of the characteristic curves $x=x_0 e^{-t}$ in the $x$-$t$ plane:
The quantity $Phi$ in (ii) expresses the total amount of $phi$ related to the substance over the whole domain $Bbb R$. According to the flow of the substance, the quantity $Phi$ equals $phi(0)$ multiplied by the total amount of substance $P (0)$ as $t$ goes to infinity.
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
$endgroup$
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
1
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
add a comment |
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$begingroup$
The PDE $p_t−(xp)_x=0$ in (i) rewrites as $p_t−xp_x=p$. Its solution obtained via the method of characteristics is indeed $p(t,x)=e^t, p_0(x e^t)$. Setting $u = x e^t $, the integral in (ii) rewrites as
$$
Phi(t) = int_{Bbb R}phi(x), p(t,x),text d x = int_{Bbb R}phi(ue^{-t}), p_0(u),text d u
$$
which limit at infinity is
$$
lim_{tto {+infty}}Phi(t) = phi(0)int_{Bbb R}p_0(u),text d u , .
$$
The PDE in (i) is a conservation law. If $p$ is the density of a given substance, then the total amount of substance $P(t) = int_{Bbb R} p(t,x),text d x$ evolves as
$$
P'(t) = int_{Bbb R} partial_t p(t,x),text d x = [x, p(t,x)]_{x=-infty}^{x=+infty} = 0, ,
$$
i.e. $P(t)= P(0)= int_{Bbb R} p_0$. The product $-xp$ is the flux of this substance, where $-x$ is analogous to a transport velocity (cf. conservation of mass). When $t$ goes to infinity, the substance flows towards $x=0$ where the speed is zero, and it stays concentrated there. This can be illustrated by a plot of the characteristic curves $x=x_0 e^{-t}$ in the $x$-$t$ plane:
The quantity $Phi$ in (ii) expresses the total amount of $phi$ related to the substance over the whole domain $Bbb R$. According to the flow of the substance, the quantity $Phi$ equals $phi(0)$ multiplied by the total amount of substance $P (0)$ as $t$ goes to infinity.
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
$endgroup$
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
1
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
add a comment |
$begingroup$
The PDE $p_t−(xp)_x=0$ in (i) rewrites as $p_t−xp_x=p$. Its solution obtained via the method of characteristics is indeed $p(t,x)=e^t, p_0(x e^t)$. Setting $u = x e^t $, the integral in (ii) rewrites as
$$
Phi(t) = int_{Bbb R}phi(x), p(t,x),text d x = int_{Bbb R}phi(ue^{-t}), p_0(u),text d u
$$
which limit at infinity is
$$
lim_{tto {+infty}}Phi(t) = phi(0)int_{Bbb R}p_0(u),text d u , .
$$
The PDE in (i) is a conservation law. If $p$ is the density of a given substance, then the total amount of substance $P(t) = int_{Bbb R} p(t,x),text d x$ evolves as
$$
P'(t) = int_{Bbb R} partial_t p(t,x),text d x = [x, p(t,x)]_{x=-infty}^{x=+infty} = 0, ,
$$
i.e. $P(t)= P(0)= int_{Bbb R} p_0$. The product $-xp$ is the flux of this substance, where $-x$ is analogous to a transport velocity (cf. conservation of mass). When $t$ goes to infinity, the substance flows towards $x=0$ where the speed is zero, and it stays concentrated there. This can be illustrated by a plot of the characteristic curves $x=x_0 e^{-t}$ in the $x$-$t$ plane:
The quantity $Phi$ in (ii) expresses the total amount of $phi$ related to the substance over the whole domain $Bbb R$. According to the flow of the substance, the quantity $Phi$ equals $phi(0)$ multiplied by the total amount of substance $P (0)$ as $t$ goes to infinity.
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
$endgroup$
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
1
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
add a comment |
$begingroup$
The PDE $p_t−(xp)_x=0$ in (i) rewrites as $p_t−xp_x=p$. Its solution obtained via the method of characteristics is indeed $p(t,x)=e^t, p_0(x e^t)$. Setting $u = x e^t $, the integral in (ii) rewrites as
$$
Phi(t) = int_{Bbb R}phi(x), p(t,x),text d x = int_{Bbb R}phi(ue^{-t}), p_0(u),text d u
$$
which limit at infinity is
$$
lim_{tto {+infty}}Phi(t) = phi(0)int_{Bbb R}p_0(u),text d u , .
$$
The PDE in (i) is a conservation law. If $p$ is the density of a given substance, then the total amount of substance $P(t) = int_{Bbb R} p(t,x),text d x$ evolves as
$$
P'(t) = int_{Bbb R} partial_t p(t,x),text d x = [x, p(t,x)]_{x=-infty}^{x=+infty} = 0, ,
$$
i.e. $P(t)= P(0)= int_{Bbb R} p_0$. The product $-xp$ is the flux of this substance, where $-x$ is analogous to a transport velocity (cf. conservation of mass). When $t$ goes to infinity, the substance flows towards $x=0$ where the speed is zero, and it stays concentrated there. This can be illustrated by a plot of the characteristic curves $x=x_0 e^{-t}$ in the $x$-$t$ plane:
The quantity $Phi$ in (ii) expresses the total amount of $phi$ related to the substance over the whole domain $Bbb R$. According to the flow of the substance, the quantity $Phi$ equals $phi(0)$ multiplied by the total amount of substance $P (0)$ as $t$ goes to infinity.
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
$endgroup$
The PDE $p_t−(xp)_x=0$ in (i) rewrites as $p_t−xp_x=p$. Its solution obtained via the method of characteristics is indeed $p(t,x)=e^t, p_0(x e^t)$. Setting $u = x e^t $, the integral in (ii) rewrites as
$$
Phi(t) = int_{Bbb R}phi(x), p(t,x),text d x = int_{Bbb R}phi(ue^{-t}), p_0(u),text d u
$$
which limit at infinity is
$$
lim_{tto {+infty}}Phi(t) = phi(0)int_{Bbb R}p_0(u),text d u , .
$$
The PDE in (i) is a conservation law. If $p$ is the density of a given substance, then the total amount of substance $P(t) = int_{Bbb R} p(t,x),text d x$ evolves as
$$
P'(t) = int_{Bbb R} partial_t p(t,x),text d x = [x, p(t,x)]_{x=-infty}^{x=+infty} = 0, ,
$$
i.e. $P(t)= P(0)= int_{Bbb R} p_0$. The product $-xp$ is the flux of this substance, where $-x$ is analogous to a transport velocity (cf. conservation of mass). When $t$ goes to infinity, the substance flows towards $x=0$ where the speed is zero, and it stays concentrated there. This can be illustrated by a plot of the characteristic curves $x=x_0 e^{-t}$ in the $x$-$t$ plane:
The quantity $Phi$ in (ii) expresses the total amount of $phi$ related to the substance over the whole domain $Bbb R$. According to the flow of the substance, the quantity $Phi$ equals $phi(0)$ multiplied by the total amount of substance $P (0)$ as $t$ goes to infinity.
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
edited Oct 17 '18 at 13:52
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
answered Oct 15 '18 at 17:41
Harry49Harry49
6,17331132
6,17331132
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
1
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
add a comment |
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
1
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
$begingroup$
Thank you so much for your explanation, the physical perspective makes a lot more sense now. One thing I'm still not sure about is in ii) where the limit and integral switch places. I know you can use the Dominated Convergence Theorem, but for that there needs to be an integrable function g so that $|f_n| leq g$. Why is this the case here?
$endgroup$
– Tierra Telder
Oct 17 '18 at 13:40
1
1
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
@TierraTelder Assuming that $phi$ is positive and bounded $0 leq phi leq M$, we have $0 leq phi(ue^{-t}) p_0(u) leq M p_0(u)$ for all $t$. Since $p_0$ is integrable ($int_{Bbb R} p_0 = P(0)$), we can pass to the limit.
$endgroup$
– Harry49
Oct 17 '18 at 13:47
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
$begingroup$
Thank you very much!
$endgroup$
– Tierra Telder
Oct 18 '18 at 11:15
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
