Ge the canonical form of the equation The Next CEO of Stack Overflowcanonical form for hyperbolic PDE $y^2u_xx+2xyu_xy+u_yy=0$?How do you find the second constant in a parabolic pde solution?Transforming hyperbolic PDE into normal formCanonical form 2nd order PDESolving $y^2u_xx-x^2u_yy=0$.PDE - $y^2 fracpartial^2 upartial x^2=x^2 fracpartial^2 upartial y^2$ - how to derive the general solutionHyperbolic non-homogeneous 2nd order linear PDEReduce the equation to canonical form. $u_xx+2ayu_xy+e^2xu_yy-u=0$Understanding how to obtain the substitution to reduce a PDE to canonical formFind the type of the equation
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Ge the canonical form of the equation
The Next CEO of Stack Overflowcanonical form for hyperbolic PDE $y^2u_xx+2xyu_xy+u_yy=0$?How do you find the second constant in a parabolic pde solution?Transforming hyperbolic PDE into normal formCanonical form 2nd order PDESolving $y^2u_xx-x^2u_yy=0$.PDE - $y^2 fracpartial^2 upartial x^2=x^2 fracpartial^2 upartial y^2$ - how to derive the general solutionHyperbolic non-homogeneous 2nd order linear PDEReduce the equation to canonical form. $u_xx+2ayu_xy+e^2xu_yy-u=0$Understanding how to obtain the substitution to reduce a PDE to canonical formFind the type of the equation
$begingroup$
Given $$x^2u_xx + 2xyu_xy - 3y^2u_yy - 2xu_x +4yu_y +16x^4u = 0$$
I have to find the canonical form. The determinant is the following $$ D = 16x^2y^2$$
I considered cases when $x$ and $y$ can be $0$. When calculating for the remaining case (where both $x$ and $y$ is not $0$ and the type is hyperbolic), I am not getting its canonical form. I am getting this
$$(-3y^2 - 6x^3y + 9x^6)u_xixi + (-3y^2 + frac2yx + frac1x^2)u_etaeta + (-6y^2 - 6x^2 + frac 2y(1-3x^4)x)u_etaxi + (frac-4x+4y)u_eta + 4yu_xi + 16x^4u = 0$$
But the thing is, that this doesn't meet the requirements for canonical form of hyperbolic type.
Any help will be appreciated.
pde partial-derivative differential
edited Mar 28 at 8:48
jmerry
16.9k11633
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
$endgroup$
add a comment |
$begingroup$
Given $$x^2u_xx + 2xyu_xy - 3y^2u_yy - 2xu_x +4yu_y +16x^4u = 0$$
I have to find the canonical form. The determinant is the following $$ D = 16x^2y^2$$
I considered cases when $x$ and $y$ can be $0$. When calculating for the remaining case (where both $x$ and $y$ is not $0$ and the type is hyperbolic), I am not getting its canonical form. I am getting this
$$(-3y^2 - 6x^3y + 9x^6)u_xixi + (-3y^2 + frac2yx + frac1x^2)u_etaeta + (-6y^2 - 6x^2 + frac 2y(1-3x^4)x)u_etaxi + (frac-4x+4y)u_eta + 4yu_xi + 16x^4u = 0$$
But the thing is, that this doesn't meet the requirements for canonical form of hyperbolic type.
Any help will be appreciated.
pde partial-derivative differential
edited Mar 28 at 8:48
jmerry
16.9k11633
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
$endgroup$
$begingroup$
$x^2 partial_x^2+2xypartial_xpartial_y -3y^2partial_y^2 = (xpartial_x-ypartial_y)(xpartial_x+3ypartial_y)$
$endgroup$
– Cesareo
Mar 28 at 8:49
$begingroup$
What does this give me?
$endgroup$
– Vahe Karamyan
Mar 28 at 8:54
$begingroup$
It gives you the characteristic curves.
$endgroup$
– Cesareo
Mar 28 at 10:03
add a comment |
$begingroup$
Given $$x^2u_xx + 2xyu_xy - 3y^2u_yy - 2xu_x +4yu_y +16x^4u = 0$$
I have to find the canonical form. The determinant is the following $$ D = 16x^2y^2$$
I considered cases when $x$ and $y$ can be $0$. When calculating for the remaining case (where both $x$ and $y$ is not $0$ and the type is hyperbolic), I am not getting its canonical form. I am getting this
$$(-3y^2 - 6x^3y + 9x^6)u_xixi + (-3y^2 + frac2yx + frac1x^2)u_etaeta + (-6y^2 - 6x^2 + frac 2y(1-3x^4)x)u_etaxi + (frac-4x+4y)u_eta + 4yu_xi + 16x^4u = 0$$
But the thing is, that this doesn't meet the requirements for canonical form of hyperbolic type.
Any help will be appreciated.
pde partial-derivative differential
edited Mar 28 at 8:48
jmerry
16.9k11633
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
$endgroup$
Given $$x^2u_xx + 2xyu_xy - 3y^2u_yy - 2xu_x +4yu_y +16x^4u = 0$$
I have to find the canonical form. The determinant is the following $$ D = 16x^2y^2$$
I considered cases when $x$ and $y$ can be $0$. When calculating for the remaining case (where both $x$ and $y$ is not $0$ and the type is hyperbolic), I am not getting its canonical form. I am getting this
$$(-3y^2 - 6x^3y + 9x^6)u_xixi + (-3y^2 + frac2yx + frac1x^2)u_etaeta + (-6y^2 - 6x^2 + frac 2y(1-3x^4)x)u_etaxi + (frac-4x+4y)u_eta + 4yu_xi + 16x^4u = 0$$
But the thing is, that this doesn't meet the requirements for canonical form of hyperbolic type.
Any help will be appreciated.
pde partial-derivative differential
pde partial-derivative differential
edited Mar 28 at 8:48
jmerry
16.9k11633
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
edited Mar 28 at 8:48
jmerry
16.9k11633
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
edited Mar 28 at 8:48
jmerry
16.9k11633
edited Mar 28 at 8:48
jmerry
16.9k11633
edited Mar 28 at 8:48
jmerry
16.9k11633
16.9k11633
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
asked Mar 28 at 8:39
Vahe KaramyanVahe Karamyan
346
346
$begingroup$
$x^2 partial_x^2+2xypartial_xpartial_y -3y^2partial_y^2 = (xpartial_x-ypartial_y)(xpartial_x+3ypartial_y)$
$endgroup$
– Cesareo
Mar 28 at 8:49
$begingroup$
What does this give me?
$endgroup$
– Vahe Karamyan
Mar 28 at 8:54
$begingroup$
It gives you the characteristic curves.
$endgroup$
– Cesareo
Mar 28 at 10:03
add a comment |
$begingroup$
$x^2 partial_x^2+2xypartial_xpartial_y -3y^2partial_y^2 = (xpartial_x-ypartial_y)(xpartial_x+3ypartial_y)$
$endgroup$
– Cesareo
Mar 28 at 8:49
$begingroup$
What does this give me?
$endgroup$
– Vahe Karamyan
Mar 28 at 8:54
$begingroup$
It gives you the characteristic curves.
$endgroup$
– Cesareo
Mar 28 at 10:03
$begingroup$
$x^2 partial_x^2+2xypartial_xpartial_y -3y^2partial_y^2 = (xpartial_x-ypartial_y)(xpartial_x+3ypartial_y)$
$endgroup$
– Cesareo
Mar 28 at 8:49
$begingroup$
$x^2 partial_x^2+2xypartial_xpartial_y -3y^2partial_y^2 = (xpartial_x-ypartial_y)(xpartial_x+3ypartial_y)$
$endgroup$
– Cesareo
Mar 28 at 8:49
$begingroup$
What does this give me?
$endgroup$
– Vahe Karamyan
Mar 28 at 8:54
$begingroup$
What does this give me?
$endgroup$
– Vahe Karamyan
Mar 28 at 8:54
$begingroup$
It gives you the characteristic curves.
$endgroup$
– Cesareo
Mar 28 at 10:03
$begingroup$
It gives you the characteristic curves.
$endgroup$
– Cesareo
Mar 28 at 10:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let
$$a=x^2,quad b=xy,quad c=-3y^2$$
Characteristic equations:
$$y'=fracbpmsqrtb^2-aca$$
or
$$y'=frac3yx, quad y'=-fracyx$$
Solutions:
$$fracyx^3=c_1,quad xy=c_2$$
Change variables
$$xi=fracyx^3,quad eta=xy$$
gives canonical form
$$u_xieta-frac4u_xi xi^2+xieta u_eta+4eta u4etaxi^2=0$$
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
$$a=x^2,quad b=xy,quad c=-3y^2$$
Characteristic equations:
$$y'=fracbpmsqrtb^2-aca$$
or
$$y'=frac3yx, quad y'=-fracyx$$
Solutions:
$$fracyx^3=c_1,quad xy=c_2$$
Change variables
$$xi=fracyx^3,quad eta=xy$$
gives canonical form
$$u_xieta-frac4u_xi xi^2+xieta u_eta+4eta u4etaxi^2=0$$
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Let
$$a=x^2,quad b=xy,quad c=-3y^2$$
Characteristic equations:
$$y'=fracbpmsqrtb^2-aca$$
or
$$y'=frac3yx, quad y'=-fracyx$$
Solutions:
$$fracyx^3=c_1,quad xy=c_2$$
Change variables
$$xi=fracyx^3,quad eta=xy$$
gives canonical form
$$u_xieta-frac4u_xi xi^2+xieta u_eta+4eta u4etaxi^2=0$$
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Let
$$a=x^2,quad b=xy,quad c=-3y^2$$
Characteristic equations:
$$y'=fracbpmsqrtb^2-aca$$
or
$$y'=frac3yx, quad y'=-fracyx$$
Solutions:
$$fracyx^3=c_1,quad xy=c_2$$
Change variables
$$xi=fracyx^3,quad eta=xy$$
gives canonical form
$$u_xieta-frac4u_xi xi^2+xieta u_eta+4eta u4etaxi^2=0$$
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
Let
$$a=x^2,quad b=xy,quad c=-3y^2$$
Characteristic equations:
$$y'=fracbpmsqrtb^2-aca$$
or
$$y'=frac3yx, quad y'=-fracyx$$
Solutions:
$$fracyx^3=c_1,quad xy=c_2$$
Change variables
$$xi=fracyx^3,quad eta=xy$$
gives canonical form
$$u_xieta-frac4u_xi xi^2+xieta u_eta+4eta u4etaxi^2=0$$
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
answered Mar 28 at 9:19
Aleksas DomarkasAleksas Domarkas
1,62317
1,62317
add a comment |
add a comment |
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$begingroup$
$x^2 partial_x^2+2xypartial_xpartial_y -3y^2partial_y^2 = (xpartial_x-ypartial_y)(xpartial_x+3ypartial_y)$
$endgroup$
– Cesareo
Mar 28 at 8:49
$begingroup$
What does this give me?
$endgroup$
– Vahe Karamyan
Mar 28 at 8:54
$begingroup$
It gives you the characteristic curves.
$endgroup$
– Cesareo
Mar 28 at 10:03