$u_{xx} + u_{xy} + u_{yy} = 0$ in canonical form
How do i put $u_{xx} + u_{xy} + u_{yy} = 0$ in canonical form?
$a=1, b=1/2, c=1 $ implies that it is elliptic as $b^2 - ac <0$
$dy/dx = lambda$ where $alambda^2-2blambda+c=0$ gives $lambda = pmsqrt{frac{-3}{4}} + 1/2$
As these roots are complex I am not sure how to proceed?
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
differential-equations pde partial-derivative canonical-transformation
asked Nov 26 at 20:20
pablo_mathscobar
836
add a comment |
How do i put $u_{xx} + u_{xy} + u_{yy} = 0$ in canonical form?
$a=1, b=1/2, c=1 $ implies that it is elliptic as $b^2 - ac <0$
$dy/dx = lambda$ where $alambda^2-2blambda+c=0$ gives $lambda = pmsqrt{frac{-3}{4}} + 1/2$
As these roots are complex I am not sure how to proceed?
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
differential-equations pde partial-derivative canonical-transformation
asked Nov 26 at 20:20
pablo_mathscobar
836
add a comment |
How do i put $u_{xx} + u_{xy} + u_{yy} = 0$ in canonical form?
$a=1, b=1/2, c=1 $ implies that it is elliptic as $b^2 - ac <0$
$dy/dx = lambda$ where $alambda^2-2blambda+c=0$ gives $lambda = pmsqrt{frac{-3}{4}} + 1/2$
As these roots are complex I am not sure how to proceed?
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
differential-equations pde partial-derivative canonical-transformation
asked Nov 26 at 20:20
pablo_mathscobar
836
How do i put $u_{xx} + u_{xy} + u_{yy} = 0$ in canonical form?
$a=1, b=1/2, c=1 $ implies that it is elliptic as $b^2 - ac <0$
$dy/dx = lambda$ where $alambda^2-2blambda+c=0$ gives $lambda = pmsqrt{frac{-3}{4}} + 1/2$
As these roots are complex I am not sure how to proceed?
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
differential-equations pde partial-derivative canonical-transformation
differential-equations pde partial-derivative canonical-transformation
asked Nov 26 at 20:20
pablo_mathscobar
836
asked Nov 26 at 20:20
pablo_mathscobar
836
edited Nov 26 at 21:05
asked Nov 26 at 20:20
pablo_mathscobar
836
asked Nov 26 at 20:20
pablo_mathscobar
836
asked Nov 26 at 20:20
pablo_mathscobar
836
836
add a comment |
add a comment |
1 Answer
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$lambda_1=frac{1+isqrt{3}}{2}, lambda_2=overline{lambda_1}=frac{1-isqrt{3}}{2}\
implies varepsilon=y-lambda_1x, eta=y-overline{lambda_1}x$
The standard procedure to deal with complex roots involves finding the canonical form by making the substitution $alpha=frac{varepsilon+eta}{2}=y-frac{lambda_1+overlinelambda_1}{2}x=y-Re(lambda_1)x, beta=frac{eta-varepsilon}{2i}=frac{lambda_1-barlambda_1}{2i}x=Im(lambda_1)x$.
The canonical form so obtained will not contain complex constants.
answered Nov 26 at 21:05
Shubham Johri
3,826716
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
|
show 1 more comment
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$lambda_1=frac{1+isqrt{3}}{2}, lambda_2=overline{lambda_1}=frac{1-isqrt{3}}{2}\
implies varepsilon=y-lambda_1x, eta=y-overline{lambda_1}x$
The standard procedure to deal with complex roots involves finding the canonical form by making the substitution $alpha=frac{varepsilon+eta}{2}=y-frac{lambda_1+overlinelambda_1}{2}x=y-Re(lambda_1)x, beta=frac{eta-varepsilon}{2i}=frac{lambda_1-barlambda_1}{2i}x=Im(lambda_1)x$.
The canonical form so obtained will not contain complex constants.
answered Nov 26 at 21:05
Shubham Johri
3,826716
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
|
show 1 more comment
$lambda_1=frac{1+isqrt{3}}{2}, lambda_2=overline{lambda_1}=frac{1-isqrt{3}}{2}\
implies varepsilon=y-lambda_1x, eta=y-overline{lambda_1}x$
The standard procedure to deal with complex roots involves finding the canonical form by making the substitution $alpha=frac{varepsilon+eta}{2}=y-frac{lambda_1+overlinelambda_1}{2}x=y-Re(lambda_1)x, beta=frac{eta-varepsilon}{2i}=frac{lambda_1-barlambda_1}{2i}x=Im(lambda_1)x$.
The canonical form so obtained will not contain complex constants.
answered Nov 26 at 21:05
Shubham Johri
3,826716
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
|
show 1 more comment
$lambda_1=frac{1+isqrt{3}}{2}, lambda_2=overline{lambda_1}=frac{1-isqrt{3}}{2}\
implies varepsilon=y-lambda_1x, eta=y-overline{lambda_1}x$
The standard procedure to deal with complex roots involves finding the canonical form by making the substitution $alpha=frac{varepsilon+eta}{2}=y-frac{lambda_1+overlinelambda_1}{2}x=y-Re(lambda_1)x, beta=frac{eta-varepsilon}{2i}=frac{lambda_1-barlambda_1}{2i}x=Im(lambda_1)x$.
The canonical form so obtained will not contain complex constants.
answered Nov 26 at 21:05
Shubham Johri
3,826716
$lambda_1=frac{1+isqrt{3}}{2}, lambda_2=overline{lambda_1}=frac{1-isqrt{3}}{2}\
implies varepsilon=y-lambda_1x, eta=y-overline{lambda_1}x$
The standard procedure to deal with complex roots involves finding the canonical form by making the substitution $alpha=frac{varepsilon+eta}{2}=y-frac{lambda_1+overlinelambda_1}{2}x=y-Re(lambda_1)x, beta=frac{eta-varepsilon}{2i}=frac{lambda_1-barlambda_1}{2i}x=Im(lambda_1)x$.
The canonical form so obtained will not contain complex constants.
answered Nov 26 at 21:05
Shubham Johri
3,826716
answered Nov 26 at 21:05
Shubham Johri
3,826716
answered Nov 26 at 21:05
Shubham Johri
3,826716
answered Nov 26 at 21:05
Shubham Johri
3,826716
3,826716
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
|
show 1 more comment
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
Can i take $ξ=y-x$ and $η=x$ to give me $u_{ξξ} + u_{ηη} - u_{ηξ} =0$ which is in canonical form?
– pablo_mathscobar
Nov 26 at 21:06
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
This is not in canonical form. It involves all the second order partial derivatives of $u$, whilst the aim of reducing an equation to its canonical form is to remove such derivatives to make the analysis of the partial differential equation easier. The equation you got is not very different from the original equation.
– Shubham Johri
Nov 26 at 21:11
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
using your method i still get second order partials of u, i cant seem to get it into canonical form?
– pablo_mathscobar
Nov 26 at 21:23
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
You may not be able to remove all the second order partial derivatives. The canonical forms of parabolic and hyperbolic equations also contain second order partial derivatives. The point is to be able to remove at least some of them. Do you still get 3 second order partial derivatives?
– Shubham Johri
Nov 26 at 21:27
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
no, i get 2: $u_{αα} + u_{ββ} = 0$ thanks
– pablo_mathscobar
Nov 26 at 21:29
|
show 1 more comment
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