Domain of definition for $u_x + uu_y = 1$
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How do i find the domain of definition for $u_x + uu_y = 1$
with $u = x/2$ on $y=x$ , $0 leq x leq 1$
I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 leq s leq 1$ at $t=0$
The characteristic equations are:
$dx/dt = 1$, $dy/dt = u$, $du/dt = 1$
Solving $dx/dt = 1$ gives $x=t +s$
Solving $du/dt = 1$ gives $u=t+ s/2$
Solving $dy/dt = u$ gives $y = (1/2)t^2 + st/2 + s$
This is where I get stuck,
$t=x-s$ so $y = (1/2)(x-s)^2 + (1/2)s(x-s) + s$
and $0 leq s leq 1$ so am i correct in saying that the domain of definition is the region between $y = (1/2)x^2 $ and $y = (1/2)(x-1)^2 + (1/2)(x-1) + 1$
pde characteristics
edited Dec 10 at 9:41
Harry49
5,87421030
asked Nov 19 at 22:24
pablo_mathscobar
696
add a comment |
up vote
2
down vote
favorite
How do i find the domain of definition for $u_x + uu_y = 1$
with $u = x/2$ on $y=x$ , $0 leq x leq 1$
I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 leq s leq 1$ at $t=0$
The characteristic equations are:
$dx/dt = 1$, $dy/dt = u$, $du/dt = 1$
Solving $dx/dt = 1$ gives $x=t +s$
Solving $du/dt = 1$ gives $u=t+ s/2$
Solving $dy/dt = u$ gives $y = (1/2)t^2 + st/2 + s$
This is where I get stuck,
$t=x-s$ so $y = (1/2)(x-s)^2 + (1/2)s(x-s) + s$
and $0 leq s leq 1$ so am i correct in saying that the domain of definition is the region between $y = (1/2)x^2 $ and $y = (1/2)(x-1)^2 + (1/2)(x-1) + 1$
pde characteristics
edited Dec 10 at 9:41
Harry49
5,87421030
asked Nov 19 at 22:24
pablo_mathscobar
696
I did it by means of another procedure and I get the same result. Maybe you want a full answer...
– Rafa Budría
Nov 20 at 11:33
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How do i find the domain of definition for $u_x + uu_y = 1$
with $u = x/2$ on $y=x$ , $0 leq x leq 1$
I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 leq s leq 1$ at $t=0$
The characteristic equations are:
$dx/dt = 1$, $dy/dt = u$, $du/dt = 1$
Solving $dx/dt = 1$ gives $x=t +s$
Solving $du/dt = 1$ gives $u=t+ s/2$
Solving $dy/dt = u$ gives $y = (1/2)t^2 + st/2 + s$
This is where I get stuck,
$t=x-s$ so $y = (1/2)(x-s)^2 + (1/2)s(x-s) + s$
and $0 leq s leq 1$ so am i correct in saying that the domain of definition is the region between $y = (1/2)x^2 $ and $y = (1/2)(x-1)^2 + (1/2)(x-1) + 1$
pde characteristics
edited Dec 10 at 9:41
Harry49
5,87421030
asked Nov 19 at 22:24
pablo_mathscobar
696
How do i find the domain of definition for $u_x + uu_y = 1$
with $u = x/2$ on $y=x$ , $0 leq x leq 1$
I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 leq s leq 1$ at $t=0$
The characteristic equations are:
$dx/dt = 1$, $dy/dt = u$, $du/dt = 1$
Solving $dx/dt = 1$ gives $x=t +s$
Solving $du/dt = 1$ gives $u=t+ s/2$
Solving $dy/dt = u$ gives $y = (1/2)t^2 + st/2 + s$
This is where I get stuck,
$t=x-s$ so $y = (1/2)(x-s)^2 + (1/2)s(x-s) + s$
and $0 leq s leq 1$ so am i correct in saying that the domain of definition is the region between $y = (1/2)x^2 $ and $y = (1/2)(x-1)^2 + (1/2)(x-1) + 1$
pde characteristics
pde characteristics
edited Dec 10 at 9:41
Harry49
5,87421030
asked Nov 19 at 22:24
pablo_mathscobar
696
edited Dec 10 at 9:41
Harry49
5,87421030
asked Nov 19 at 22:24
pablo_mathscobar
696
edited Dec 10 at 9:41
Harry49
5,87421030
edited Dec 10 at 9:41
Harry49
5,87421030
edited Dec 10 at 9:41
Harry49
5,87421030
5,87421030
asked Nov 19 at 22:24
pablo_mathscobar
696
asked Nov 19 at 22:24
pablo_mathscobar
696
asked Nov 19 at 22:24
pablo_mathscobar
696
696
I did it by means of another procedure and I get the same result. Maybe you want a full answer...
– Rafa Budría
Nov 20 at 11:33
add a comment |
I did it by means of another procedure and I get the same result. Maybe you want a full answer...
– Rafa Budría
Nov 20 at 11:33
I did it by means of another procedure and I get the same result. Maybe you want a full answer...
– Rafa Budría
Nov 20 at 11:33
I did it by means of another procedure and I get the same result. Maybe you want a full answer...
– Rafa Budría
Nov 20 at 11:33
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
$$u_x+uu_y= tag 1$$
Your three equations written on a equivalent form:
$$frac{dx}{1}=frac{dy}{u}=frac{du}{1}=dt$$
A first family of characteristic equations comes from $frac{dx}{1}=frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $frac{dy}{u}=frac{du}{1}$
$$frac{u^2}{2}-y=c_2$$
The general solution of the PDE expressed on the form of implicite equation is :
$$frac{u^2}{2}-y=F(u-x) tag 2$$
where $F$ is an arbitrary equation, to be determined according to boundary condition :
$$u(x,x)=frac{x}{2}quadimpliesquadfrac{x^2}{8}-x=Fleft(frac{x}{2}-xright)$$
$X=-frac{x}{2}quad;quad x=-2X$
$$F(X)=frac{(-2X)^2}{8}-(-2X)=frac{X^2}{2}+2X$$
So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$
$$frac{u^2}{2}-y=frac{(u-x)^2}{2}+2(u-x)$$
After simplification :
$$u(x,y)=frac{y+frac{x^2}{2}-2x}{x-2}$$
You can find the domain of definition.
answered Dec 7 at 10:03
JJacquelin
42.2k21750
1
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
add a comment |
up vote
0
down vote
This PDE is a non-homogeneous inviscid Burgers' equation, for which the same question has been asked here (see previous link for details). The method of characteristics introduces the characteristic curves $y = x(x-s)/2 + s$ for $0<s<1$, along which $u$ is given by (cf. answer by JJacquelin)
$$
u(x,y) = frac{x^2/2 - 2x + y}{x-2} , .
$$
This expression is not defined at $x=2$, where the denominator vanishes.
On the one hand, if we start following a characteristic curve at some point $(s,s)$ with $0<s<1$, we can decrease $x$ without restriction. On the other hand, we cannot increase $x$ further than $x=2^-$. At the point $(x,y) = (2,2)$, all characteristics coming from $lbrace(s,s), 0<s<1rbrace$ intersect: the classical solution collapses.
Finally, the solution is only valid over the domain in yellow
$$
(x,y) in biglbrace big(x, x(x-s)/2+sbig),; 0<s<1,; x<2bigrbrace
$$
answered Dec 10 at 9:37
Harry49
5,87421030
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$$u_x+uu_y= tag 1$$
Your three equations written on a equivalent form:
$$frac{dx}{1}=frac{dy}{u}=frac{du}{1}=dt$$
A first family of characteristic equations comes from $frac{dx}{1}=frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $frac{dy}{u}=frac{du}{1}$
$$frac{u^2}{2}-y=c_2$$
The general solution of the PDE expressed on the form of implicite equation is :
$$frac{u^2}{2}-y=F(u-x) tag 2$$
where $F$ is an arbitrary equation, to be determined according to boundary condition :
$$u(x,x)=frac{x}{2}quadimpliesquadfrac{x^2}{8}-x=Fleft(frac{x}{2}-xright)$$
$X=-frac{x}{2}quad;quad x=-2X$
$$F(X)=frac{(-2X)^2}{8}-(-2X)=frac{X^2}{2}+2X$$
So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$
$$frac{u^2}{2}-y=frac{(u-x)^2}{2}+2(u-x)$$
After simplification :
$$u(x,y)=frac{y+frac{x^2}{2}-2x}{x-2}$$
You can find the domain of definition.
answered Dec 7 at 10:03
JJacquelin
42.2k21750
1
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
add a comment |
up vote
0
down vote
$$u_x+uu_y= tag 1$$
Your three equations written on a equivalent form:
$$frac{dx}{1}=frac{dy}{u}=frac{du}{1}=dt$$
A first family of characteristic equations comes from $frac{dx}{1}=frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $frac{dy}{u}=frac{du}{1}$
$$frac{u^2}{2}-y=c_2$$
The general solution of the PDE expressed on the form of implicite equation is :
$$frac{u^2}{2}-y=F(u-x) tag 2$$
where $F$ is an arbitrary equation, to be determined according to boundary condition :
$$u(x,x)=frac{x}{2}quadimpliesquadfrac{x^2}{8}-x=Fleft(frac{x}{2}-xright)$$
$X=-frac{x}{2}quad;quad x=-2X$
$$F(X)=frac{(-2X)^2}{8}-(-2X)=frac{X^2}{2}+2X$$
So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$
$$frac{u^2}{2}-y=frac{(u-x)^2}{2}+2(u-x)$$
After simplification :
$$u(x,y)=frac{y+frac{x^2}{2}-2x}{x-2}$$
You can find the domain of definition.
answered Dec 7 at 10:03
JJacquelin
42.2k21750
1
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
add a comment |
up vote
0
down vote
up vote
0
down vote
$$u_x+uu_y= tag 1$$
Your three equations written on a equivalent form:
$$frac{dx}{1}=frac{dy}{u}=frac{du}{1}=dt$$
A first family of characteristic equations comes from $frac{dx}{1}=frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $frac{dy}{u}=frac{du}{1}$
$$frac{u^2}{2}-y=c_2$$
The general solution of the PDE expressed on the form of implicite equation is :
$$frac{u^2}{2}-y=F(u-x) tag 2$$
where $F$ is an arbitrary equation, to be determined according to boundary condition :
$$u(x,x)=frac{x}{2}quadimpliesquadfrac{x^2}{8}-x=Fleft(frac{x}{2}-xright)$$
$X=-frac{x}{2}quad;quad x=-2X$
$$F(X)=frac{(-2X)^2}{8}-(-2X)=frac{X^2}{2}+2X$$
So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$
$$frac{u^2}{2}-y=frac{(u-x)^2}{2}+2(u-x)$$
After simplification :
$$u(x,y)=frac{y+frac{x^2}{2}-2x}{x-2}$$
You can find the domain of definition.
answered Dec 7 at 10:03
JJacquelin
42.2k21750
$$u_x+uu_y= tag 1$$
Your three equations written on a equivalent form:
$$frac{dx}{1}=frac{dy}{u}=frac{du}{1}=dt$$
A first family of characteristic equations comes from $frac{dx}{1}=frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $frac{dy}{u}=frac{du}{1}$
$$frac{u^2}{2}-y=c_2$$
The general solution of the PDE expressed on the form of implicite equation is :
$$frac{u^2}{2}-y=F(u-x) tag 2$$
where $F$ is an arbitrary equation, to be determined according to boundary condition :
$$u(x,x)=frac{x}{2}quadimpliesquadfrac{x^2}{8}-x=Fleft(frac{x}{2}-xright)$$
$X=-frac{x}{2}quad;quad x=-2X$
$$F(X)=frac{(-2X)^2}{8}-(-2X)=frac{X^2}{2}+2X$$
So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$
$$frac{u^2}{2}-y=frac{(u-x)^2}{2}+2(u-x)$$
After simplification :
$$u(x,y)=frac{y+frac{x^2}{2}-2x}{x-2}$$
You can find the domain of definition.
answered Dec 7 at 10:03
JJacquelin
42.2k21750
answered Dec 7 at 10:03
JJacquelin
42.2k21750
answered Dec 7 at 10:03
JJacquelin
42.2k21750
answered Dec 7 at 10:03
JJacquelin
42.2k21750
42.2k21750
1
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
add a comment |
1
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
1
1
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
I understand your solution but how would I then find the domain of definition? i am confused
– pablo_mathscobar
Dec 7 at 12:59
add a comment |
up vote
0
down vote
This PDE is a non-homogeneous inviscid Burgers' equation, for which the same question has been asked here (see previous link for details). The method of characteristics introduces the characteristic curves $y = x(x-s)/2 + s$ for $0<s<1$, along which $u$ is given by (cf. answer by JJacquelin)
$$
u(x,y) = frac{x^2/2 - 2x + y}{x-2} , .
$$
This expression is not defined at $x=2$, where the denominator vanishes.
On the one hand, if we start following a characteristic curve at some point $(s,s)$ with $0<s<1$, we can decrease $x$ without restriction. On the other hand, we cannot increase $x$ further than $x=2^-$. At the point $(x,y) = (2,2)$, all characteristics coming from $lbrace(s,s), 0<s<1rbrace$ intersect: the classical solution collapses.
Finally, the solution is only valid over the domain in yellow
$$
(x,y) in biglbrace big(x, x(x-s)/2+sbig),; 0<s<1,; x<2bigrbrace
$$
answered Dec 10 at 9:37
Harry49
5,87421030
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
add a comment |
up vote
0
down vote
This PDE is a non-homogeneous inviscid Burgers' equation, for which the same question has been asked here (see previous link for details). The method of characteristics introduces the characteristic curves $y = x(x-s)/2 + s$ for $0<s<1$, along which $u$ is given by (cf. answer by JJacquelin)
$$
u(x,y) = frac{x^2/2 - 2x + y}{x-2} , .
$$
This expression is not defined at $x=2$, where the denominator vanishes.
On the one hand, if we start following a characteristic curve at some point $(s,s)$ with $0<s<1$, we can decrease $x$ without restriction. On the other hand, we cannot increase $x$ further than $x=2^-$. At the point $(x,y) = (2,2)$, all characteristics coming from $lbrace(s,s), 0<s<1rbrace$ intersect: the classical solution collapses.
Finally, the solution is only valid over the domain in yellow
$$
(x,y) in biglbrace big(x, x(x-s)/2+sbig),; 0<s<1,; x<2bigrbrace
$$
answered Dec 10 at 9:37
Harry49
5,87421030
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
add a comment |
up vote
0
down vote
up vote
0
down vote
This PDE is a non-homogeneous inviscid Burgers' equation, for which the same question has been asked here (see previous link for details). The method of characteristics introduces the characteristic curves $y = x(x-s)/2 + s$ for $0<s<1$, along which $u$ is given by (cf. answer by JJacquelin)
$$
u(x,y) = frac{x^2/2 - 2x + y}{x-2} , .
$$
This expression is not defined at $x=2$, where the denominator vanishes.
On the one hand, if we start following a characteristic curve at some point $(s,s)$ with $0<s<1$, we can decrease $x$ without restriction. On the other hand, we cannot increase $x$ further than $x=2^-$. At the point $(x,y) = (2,2)$, all characteristics coming from $lbrace(s,s), 0<s<1rbrace$ intersect: the classical solution collapses.
Finally, the solution is only valid over the domain in yellow
$$
(x,y) in biglbrace big(x, x(x-s)/2+sbig),; 0<s<1,; x<2bigrbrace
$$
answered Dec 10 at 9:37
Harry49
5,87421030
This PDE is a non-homogeneous inviscid Burgers' equation, for which the same question has been asked here (see previous link for details). The method of characteristics introduces the characteristic curves $y = x(x-s)/2 + s$ for $0<s<1$, along which $u$ is given by (cf. answer by JJacquelin)
$$
u(x,y) = frac{x^2/2 - 2x + y}{x-2} , .
$$
This expression is not defined at $x=2$, where the denominator vanishes.
On the one hand, if we start following a characteristic curve at some point $(s,s)$ with $0<s<1$, we can decrease $x$ without restriction. On the other hand, we cannot increase $x$ further than $x=2^-$. At the point $(x,y) = (2,2)$, all characteristics coming from $lbrace(s,s), 0<s<1rbrace$ intersect: the classical solution collapses.
Finally, the solution is only valid over the domain in yellow
$$
(x,y) in biglbrace big(x, x(x-s)/2+sbig),; 0<s<1,; x<2bigrbrace
$$
answered Dec 10 at 9:37
Harry49
5,87421030
edited Dec 10 at 11:43
answered Dec 10 at 9:37
Harry49
5,87421030
answered Dec 10 at 9:37
Harry49
5,87421030
answered Dec 10 at 9:37
Harry49
5,87421030
5,87421030
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
add a comment |
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Why do we get different values for y when solving the characteristics eq? I cant see why mine is wrong
– pablo_mathscobar
Dec 10 at 11:10
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
Okay thanks, but if my characteristics eq are correct, why is my domain of definition wrong? I thought all you do is sub s in to find the domain
– pablo_mathscobar
Dec 10 at 11:27
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
ah makes sense, im just having trouble solving the 3 equations to get u explicitly, any ideas?
– pablo_mathscobar
Dec 10 at 11:41
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
@pablo_mathscobar The expression of $u$ is derived in the answer by JJacquelin (resolution of a quadratic equation)
– Harry49
Dec 10 at 11:43
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
cheers mate, i appreciate it
– pablo_mathscobar
Dec 10 at 11:48
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I did it by means of another procedure and I get the same result. Maybe you want a full answer...
– Rafa Budría
Nov 20 at 11:33