Burgers' equation $u_y + uu_x = 0$ with $u(x,0)=-x$
$begingroup$
I couldn't solve this problem, can you help me please?
The Burgers' equation
$$ u_y + uu_x = 0 $$ $ - infty < x < infty $ , $ y > 0 $ , $ u(x,0)=f(x) $
My question;
is there any solution $ f(x)= -x $ for all $ y>0 $ , if not, why?
Thank you for your help.
pde hyperbolic-equations
edited Dec 7 '18 at 13:20
Harry49
6,22331132
asked Jul 29 '13 at 11:29
JohnJohn
132
$endgroup$
add a comment |
$begingroup$
I couldn't solve this problem, can you help me please?
The Burgers' equation
$$ u_y + uu_x = 0 $$ $ - infty < x < infty $ , $ y > 0 $ , $ u(x,0)=f(x) $
My question;
is there any solution $ f(x)= -x $ for all $ y>0 $ , if not, why?
Thank you for your help.
pde hyperbolic-equations
edited Dec 7 '18 at 13:20
Harry49
6,22331132
asked Jul 29 '13 at 11:29
JohnJohn
132
$endgroup$
$begingroup$
You can use the method of characteristics to solve it.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 12:11
$begingroup$
Hi Dear, I tried but it doesn't work.
$endgroup$
– John
Jul 29 '13 at 12:34
$begingroup$
See here.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 13:18
add a comment |
$begingroup$
I couldn't solve this problem, can you help me please?
The Burgers' equation
$$ u_y + uu_x = 0 $$ $ - infty < x < infty $ , $ y > 0 $ , $ u(x,0)=f(x) $
My question;
is there any solution $ f(x)= -x $ for all $ y>0 $ , if not, why?
Thank you for your help.
pde hyperbolic-equations
edited Dec 7 '18 at 13:20
Harry49
6,22331132
asked Jul 29 '13 at 11:29
JohnJohn
132
$endgroup$
I couldn't solve this problem, can you help me please?
The Burgers' equation
$$ u_y + uu_x = 0 $$ $ - infty < x < infty $ , $ y > 0 $ , $ u(x,0)=f(x) $
My question;
is there any solution $ f(x)= -x $ for all $ y>0 $ , if not, why?
Thank you for your help.
pde hyperbolic-equations
pde hyperbolic-equations
edited Dec 7 '18 at 13:20
Harry49
6,22331132
asked Jul 29 '13 at 11:29
JohnJohn
132
edited Dec 7 '18 at 13:20
Harry49
6,22331132
asked Jul 29 '13 at 11:29
JohnJohn
132
edited Dec 7 '18 at 13:20
Harry49
6,22331132
edited Dec 7 '18 at 13:20
Harry49
6,22331132
edited Dec 7 '18 at 13:20
Harry49
6,22331132
6,22331132
asked Jul 29 '13 at 11:29
JohnJohn
132
asked Jul 29 '13 at 11:29
JohnJohn
132
asked Jul 29 '13 at 11:29
JohnJohn
132
132
$begingroup$
You can use the method of characteristics to solve it.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 12:11
$begingroup$
Hi Dear, I tried but it doesn't work.
$endgroup$
– John
Jul 29 '13 at 12:34
$begingroup$
See here.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 13:18
add a comment |
$begingroup$
You can use the method of characteristics to solve it.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 12:11
$begingroup$
Hi Dear, I tried but it doesn't work.
$endgroup$
– John
Jul 29 '13 at 12:34
$begingroup$
See here.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 13:18
$begingroup$
You can use the method of characteristics to solve it.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 12:11
$begingroup$
You can use the method of characteristics to solve it.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 12:11
$begingroup$
Hi Dear, I tried but it doesn't work.
$endgroup$
– John
Jul 29 '13 at 12:34
$begingroup$
Hi Dear, I tried but it doesn't work.
$endgroup$
– John
Jul 29 '13 at 12:34
$begingroup$
See here.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 13:18
$begingroup$
See here.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 13:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I rewrite equation to common view
$$frac{partial u}{partial t}+ufrac{partial u}{partial x}=0$$
There is a solution with each initial value, but this solution exists only for $tle t_{r}$, where $t_{r}$ is a point where the phenomenon of "rollover" occurs. This phenomenon occurs if some characteristics of this equations intersect. Here is a picture of characteristics:
It shows that characteristics of this equations intersect on the moment $t=1$
Before the moment $t=1$ the solution is:
$$u(x,t)=f(x-ut),$$ where $$f(x)=-x $$
After the point $t=1$ the solution does not exist. There you should change your equation to conservation law:
$$frac{partial u}{partial t}+frac{partial}{partial x}left(frac{u^2}{2}right)=0$$
and define the order of this law.
answered Jul 29 '13 at 12:54
coolcool
901618
$endgroup$
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{dt}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0t+f(u_0)=uy+f(u)$ , i.e. $u=F(x-uy)$
$u(x,0)=-x$ :
$F(x)=-x$
$therefore u=-(x-uy)=-x+uy$
Since $u(x,y)$ has perfect explicit form $u(x,y)=dfrac{x}{y-1}$ , of course it has solution for all $y>0$ .
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
1
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
I rewrite equation to common view
$$frac{partial u}{partial t}+ufrac{partial u}{partial x}=0$$
There is a solution with each initial value, but this solution exists only for $tle t_{r}$, where $t_{r}$ is a point where the phenomenon of "rollover" occurs. This phenomenon occurs if some characteristics of this equations intersect. Here is a picture of characteristics:
It shows that characteristics of this equations intersect on the moment $t=1$
Before the moment $t=1$ the solution is:
$$u(x,t)=f(x-ut),$$ where $$f(x)=-x $$
After the point $t=1$ the solution does not exist. There you should change your equation to conservation law:
$$frac{partial u}{partial t}+frac{partial}{partial x}left(frac{u^2}{2}right)=0$$
and define the order of this law.
answered Jul 29 '13 at 12:54
coolcool
901618
$endgroup$
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
add a comment |
$begingroup$
I rewrite equation to common view
$$frac{partial u}{partial t}+ufrac{partial u}{partial x}=0$$
There is a solution with each initial value, but this solution exists only for $tle t_{r}$, where $t_{r}$ is a point where the phenomenon of "rollover" occurs. This phenomenon occurs if some characteristics of this equations intersect. Here is a picture of characteristics:
It shows that characteristics of this equations intersect on the moment $t=1$
Before the moment $t=1$ the solution is:
$$u(x,t)=f(x-ut),$$ where $$f(x)=-x $$
After the point $t=1$ the solution does not exist. There you should change your equation to conservation law:
$$frac{partial u}{partial t}+frac{partial}{partial x}left(frac{u^2}{2}right)=0$$
and define the order of this law.
answered Jul 29 '13 at 12:54
coolcool
901618
$endgroup$
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
add a comment |
$begingroup$
I rewrite equation to common view
$$frac{partial u}{partial t}+ufrac{partial u}{partial x}=0$$
There is a solution with each initial value, but this solution exists only for $tle t_{r}$, where $t_{r}$ is a point where the phenomenon of "rollover" occurs. This phenomenon occurs if some characteristics of this equations intersect. Here is a picture of characteristics:
It shows that characteristics of this equations intersect on the moment $t=1$
Before the moment $t=1$ the solution is:
$$u(x,t)=f(x-ut),$$ where $$f(x)=-x $$
After the point $t=1$ the solution does not exist. There you should change your equation to conservation law:
$$frac{partial u}{partial t}+frac{partial}{partial x}left(frac{u^2}{2}right)=0$$
and define the order of this law.
answered Jul 29 '13 at 12:54
coolcool
901618
$endgroup$
I rewrite equation to common view
$$frac{partial u}{partial t}+ufrac{partial u}{partial x}=0$$
There is a solution with each initial value, but this solution exists only for $tle t_{r}$, where $t_{r}$ is a point where the phenomenon of "rollover" occurs. This phenomenon occurs if some characteristics of this equations intersect. Here is a picture of characteristics:
It shows that characteristics of this equations intersect on the moment $t=1$
Before the moment $t=1$ the solution is:
$$u(x,t)=f(x-ut),$$ where $$f(x)=-x $$
After the point $t=1$ the solution does not exist. There you should change your equation to conservation law:
$$frac{partial u}{partial t}+frac{partial}{partial x}left(frac{u^2}{2}right)=0$$
and define the order of this law.
answered Jul 29 '13 at 12:54
coolcool
901618
edited Jul 29 '13 at 14:08
answered Jul 29 '13 at 12:54
coolcool
901618
answered Jul 29 '13 at 12:54
coolcool
901618
answered Jul 29 '13 at 12:54
coolcool
901618
901618
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
add a comment |
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Here is a mistake. Wait for 1 minute..
$endgroup$
– cool
Jul 29 '13 at 14:01
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Please,look at the answer,@John
$endgroup$
– cool
Jul 29 '13 at 14:11
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect.
$endgroup$
– cool
Jul 29 '13 at 14:19
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
$begingroup$
@John "There you should change your equation to conservation law"... I would rather say that the classical solution breaks down at $t=1$ (it then equals simultaneously all values from $-infty$ to $+infty$). And there is no weak solution to start after this blow-up happened.
$endgroup$
– Harry49
Dec 5 '18 at 17:54
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{dt}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0t+f(u_0)=uy+f(u)$ , i.e. $u=F(x-uy)$
$u(x,0)=-x$ :
$F(x)=-x$
$therefore u=-(x-uy)=-x+uy$
Since $u(x,y)$ has perfect explicit form $u(x,y)=dfrac{x}{y-1}$ , of course it has solution for all $y>0$ .
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
1
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{dt}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0t+f(u_0)=uy+f(u)$ , i.e. $u=F(x-uy)$
$u(x,0)=-x$ :
$F(x)=-x$
$therefore u=-(x-uy)=-x+uy$
Since $u(x,y)$ has perfect explicit form $u(x,y)=dfrac{x}{y-1}$ , of course it has solution for all $y>0$ .
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
1
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
add a comment |
$begingroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{dt}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0t+f(u_0)=uy+f(u)$ , i.e. $u=F(x-uy)$
$u(x,0)=-x$ :
$F(x)=-x$
$therefore u=-(x-uy)=-x+uy$
Since $u(x,y)$ has perfect explicit form $u(x,y)=dfrac{x}{y-1}$ , of course it has solution for all $y>0$ .
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
$endgroup$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$dfrac{dx}{dt}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0t+f(u_0)=uy+f(u)$ , i.e. $u=F(x-uy)$
$u(x,0)=-x$ :
$F(x)=-x$
$therefore u=-(x-uy)=-x+uy$
Since $u(x,y)$ has perfect explicit form $u(x,y)=dfrac{x}{y-1}$ , of course it has solution for all $y>0$ .
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
answered Jul 30 '13 at 2:07
doraemonpauldoraemonpaul
12.6k31660
12.6k31660
1
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
add a comment |
1
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
1
1
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
$begingroup$
I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist.
$endgroup$
– cool
Jul 30 '13 at 13:09
add a comment |
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$begingroup$
You can use the method of characteristics to solve it.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 12:11
$begingroup$
Hi Dear, I tried but it doesn't work.
$endgroup$
– John
Jul 29 '13 at 12:34
$begingroup$
See here.
$endgroup$
– Mhenni Benghorbal
Jul 29 '13 at 13:18