General Solution for Partial Differential Equation
$begingroup$
To be honest I'm a bit lost on this, and I would like to get a hint or something that can help me, thanks.
I need to find the general solution $U(x,y,z)$ of the next equation:
$$U_{xx}+U_{yy}+4U_{zz}-2U_{xy}+4U_{xz}-4U_{yz}=xyz$$
I know that most sure exists a change of variable that would help to solve the equation, but I don't know how to find it, our teacher of Multivariable Calculus asked to solve this.
Thanks!
multivariable-calculus partial-derivative
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
asked Dec 18 '18 at 19:04
GingerGinger
475
$endgroup$
add a comment |
$begingroup$
To be honest I'm a bit lost on this, and I would like to get a hint or something that can help me, thanks.
I need to find the general solution $U(x,y,z)$ of the next equation:
$$U_{xx}+U_{yy}+4U_{zz}-2U_{xy}+4U_{xz}-4U_{yz}=xyz$$
I know that most sure exists a change of variable that would help to solve the equation, but I don't know how to find it, our teacher of Multivariable Calculus asked to solve this.
Thanks!
multivariable-calculus partial-derivative
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
asked Dec 18 '18 at 19:04
GingerGinger
475
$endgroup$
add a comment |
$begingroup$
To be honest I'm a bit lost on this, and I would like to get a hint or something that can help me, thanks.
I need to find the general solution $U(x,y,z)$ of the next equation:
$$U_{xx}+U_{yy}+4U_{zz}-2U_{xy}+4U_{xz}-4U_{yz}=xyz$$
I know that most sure exists a change of variable that would help to solve the equation, but I don't know how to find it, our teacher of Multivariable Calculus asked to solve this.
Thanks!
multivariable-calculus partial-derivative
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
asked Dec 18 '18 at 19:04
GingerGinger
475
$endgroup$
To be honest I'm a bit lost on this, and I would like to get a hint or something that can help me, thanks.
I need to find the general solution $U(x,y,z)$ of the next equation:
$$U_{xx}+U_{yy}+4U_{zz}-2U_{xy}+4U_{xz}-4U_{yz}=xyz$$
I know that most sure exists a change of variable that would help to solve the equation, but I don't know how to find it, our teacher of Multivariable Calculus asked to solve this.
Thanks!
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
asked Dec 18 '18 at 19:04
GingerGinger
475
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
asked Dec 18 '18 at 19:04
GingerGinger
475
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
edited Dec 18 '18 at 19:27
mathreadler
15.3k72263
15.3k72263
asked Dec 18 '18 at 19:04
GingerGinger
475
asked Dec 18 '18 at 19:04
GingerGinger
475
asked Dec 18 '18 at 19:04
GingerGinger
475
475
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This PDE is linear so it's solution can be obtained as
$$
U = U_h + U_p
$$
We have
$$
left(partial_{xx}^2+partial_{yy}^2+4partial_{zz}^2-2partial_xpartial_y + 4partial_xpartial_z-4partial_ypartial_zright)U = left(partial_x-partial_y+2partial_zright)^2U = x y z
$$
so calling
$$
V = left(partial_x-partial_y+2partial_zright)U
$$
we have
$$
left(partial_x-partial_y+2partial_zright)V = x y z
$$
using the characteristic method, we have
$$
frac{dx}{1}=frac{dy}{-1} = frac{dz}{2}
$$
giving the characteristics
$$
x+y = eta\
2y+z = xi\
2x-z = mu
$$
Choosing instead
$$
x-y = eta\
2y+z = xi\
2x-z = mu
$$
because $xi + mu = 2(x+y)$ and introducing now this change of variables into the full PDE we will obtain
$$
V_{eta}(eta ,xi ,mu )=frac{1}{64} (2 eta -mu +xi ) (-2 eta +mu +xi ) (2 eta +mu +xi )
$$
now considering $V = V_h + V_p$ this PDE can be easily solved.
$$
V_h(eta ,xi ,mu) = C_1+f(xi,mu)
$$
In this case $V_p(eta,xi,mu)$ is a polynomial form.
Finally after solving $V$ we will solve
$$
left(partial_x-partial_y+2partial_zright)U = V
$$
with the same process.
NOTE
Here
$$
V_p(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right)
$$
then
$$
V(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right) + f(xi,mu)
$$
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
$endgroup$
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
add a comment |
Your Answer
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This PDE is linear so it's solution can be obtained as
$$
U = U_h + U_p
$$
We have
$$
left(partial_{xx}^2+partial_{yy}^2+4partial_{zz}^2-2partial_xpartial_y + 4partial_xpartial_z-4partial_ypartial_zright)U = left(partial_x-partial_y+2partial_zright)^2U = x y z
$$
so calling
$$
V = left(partial_x-partial_y+2partial_zright)U
$$
we have
$$
left(partial_x-partial_y+2partial_zright)V = x y z
$$
using the characteristic method, we have
$$
frac{dx}{1}=frac{dy}{-1} = frac{dz}{2}
$$
giving the characteristics
$$
x+y = eta\
2y+z = xi\
2x-z = mu
$$
Choosing instead
$$
x-y = eta\
2y+z = xi\
2x-z = mu
$$
because $xi + mu = 2(x+y)$ and introducing now this change of variables into the full PDE we will obtain
$$
V_{eta}(eta ,xi ,mu )=frac{1}{64} (2 eta -mu +xi ) (-2 eta +mu +xi ) (2 eta +mu +xi )
$$
now considering $V = V_h + V_p$ this PDE can be easily solved.
$$
V_h(eta ,xi ,mu) = C_1+f(xi,mu)
$$
In this case $V_p(eta,xi,mu)$ is a polynomial form.
Finally after solving $V$ we will solve
$$
left(partial_x-partial_y+2partial_zright)U = V
$$
with the same process.
NOTE
Here
$$
V_p(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right)
$$
then
$$
V(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right) + f(xi,mu)
$$
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
$endgroup$
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
add a comment |
$begingroup$
This PDE is linear so it's solution can be obtained as
$$
U = U_h + U_p
$$
We have
$$
left(partial_{xx}^2+partial_{yy}^2+4partial_{zz}^2-2partial_xpartial_y + 4partial_xpartial_z-4partial_ypartial_zright)U = left(partial_x-partial_y+2partial_zright)^2U = x y z
$$
so calling
$$
V = left(partial_x-partial_y+2partial_zright)U
$$
we have
$$
left(partial_x-partial_y+2partial_zright)V = x y z
$$
using the characteristic method, we have
$$
frac{dx}{1}=frac{dy}{-1} = frac{dz}{2}
$$
giving the characteristics
$$
x+y = eta\
2y+z = xi\
2x-z = mu
$$
Choosing instead
$$
x-y = eta\
2y+z = xi\
2x-z = mu
$$
because $xi + mu = 2(x+y)$ and introducing now this change of variables into the full PDE we will obtain
$$
V_{eta}(eta ,xi ,mu )=frac{1}{64} (2 eta -mu +xi ) (-2 eta +mu +xi ) (2 eta +mu +xi )
$$
now considering $V = V_h + V_p$ this PDE can be easily solved.
$$
V_h(eta ,xi ,mu) = C_1+f(xi,mu)
$$
In this case $V_p(eta,xi,mu)$ is a polynomial form.
Finally after solving $V$ we will solve
$$
left(partial_x-partial_y+2partial_zright)U = V
$$
with the same process.
NOTE
Here
$$
V_p(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right)
$$
then
$$
V(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right) + f(xi,mu)
$$
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
$endgroup$
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
add a comment |
$begingroup$
This PDE is linear so it's solution can be obtained as
$$
U = U_h + U_p
$$
We have
$$
left(partial_{xx}^2+partial_{yy}^2+4partial_{zz}^2-2partial_xpartial_y + 4partial_xpartial_z-4partial_ypartial_zright)U = left(partial_x-partial_y+2partial_zright)^2U = x y z
$$
so calling
$$
V = left(partial_x-partial_y+2partial_zright)U
$$
we have
$$
left(partial_x-partial_y+2partial_zright)V = x y z
$$
using the characteristic method, we have
$$
frac{dx}{1}=frac{dy}{-1} = frac{dz}{2}
$$
giving the characteristics
$$
x+y = eta\
2y+z = xi\
2x-z = mu
$$
Choosing instead
$$
x-y = eta\
2y+z = xi\
2x-z = mu
$$
because $xi + mu = 2(x+y)$ and introducing now this change of variables into the full PDE we will obtain
$$
V_{eta}(eta ,xi ,mu )=frac{1}{64} (2 eta -mu +xi ) (-2 eta +mu +xi ) (2 eta +mu +xi )
$$
now considering $V = V_h + V_p$ this PDE can be easily solved.
$$
V_h(eta ,xi ,mu) = C_1+f(xi,mu)
$$
In this case $V_p(eta,xi,mu)$ is a polynomial form.
Finally after solving $V$ we will solve
$$
left(partial_x-partial_y+2partial_zright)U = V
$$
with the same process.
NOTE
Here
$$
V_p(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right)
$$
then
$$
V(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right) + f(xi,mu)
$$
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
$endgroup$
This PDE is linear so it's solution can be obtained as
$$
U = U_h + U_p
$$
We have
$$
left(partial_{xx}^2+partial_{yy}^2+4partial_{zz}^2-2partial_xpartial_y + 4partial_xpartial_z-4partial_ypartial_zright)U = left(partial_x-partial_y+2partial_zright)^2U = x y z
$$
so calling
$$
V = left(partial_x-partial_y+2partial_zright)U
$$
we have
$$
left(partial_x-partial_y+2partial_zright)V = x y z
$$
using the characteristic method, we have
$$
frac{dx}{1}=frac{dy}{-1} = frac{dz}{2}
$$
giving the characteristics
$$
x+y = eta\
2y+z = xi\
2x-z = mu
$$
Choosing instead
$$
x-y = eta\
2y+z = xi\
2x-z = mu
$$
because $xi + mu = 2(x+y)$ and introducing now this change of variables into the full PDE we will obtain
$$
V_{eta}(eta ,xi ,mu )=frac{1}{64} (2 eta -mu +xi ) (-2 eta +mu +xi ) (2 eta +mu +xi )
$$
now considering $V = V_h + V_p$ this PDE can be easily solved.
$$
V_h(eta ,xi ,mu) = C_1+f(xi,mu)
$$
In this case $V_p(eta,xi,mu)$ is a polynomial form.
Finally after solving $V$ we will solve
$$
left(partial_x-partial_y+2partial_zright)U = V
$$
with the same process.
NOTE
Here
$$
V_p(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right)
$$
then
$$
V(eta ,xi ,mu) = frac{1}{192} eta left(-6 eta ^3+4 eta ^2 (mu -xi )+3 eta (mu +xi )^2-3 (mu -xi ) (mu +xi )^2right) + f(xi,mu)
$$
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
edited Dec 20 '18 at 9:19
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
answered Dec 18 '18 at 19:17
CesareoCesareo
9,4923517
9,4923517
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
add a comment |
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
So how do you turn $xyz$ into that RHS expression?
$endgroup$
– mathreadler
Dec 18 '18 at 19:20
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
@mathreadler Linearity is a great ally!
$endgroup$
– Cesareo
Dec 18 '18 at 19:25
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
I doubt someone just trying to start to learn it can get it if I can't get it.
$endgroup$
– mathreadler
Dec 18 '18 at 19:26
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Try to walk with shorter steps.
$endgroup$
– Cesareo
Dec 18 '18 at 19:29
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
$begingroup$
@mathreadler Some additional information was included. I hope it helps now.
$endgroup$
– Cesareo
Dec 18 '18 at 19:52
add a comment |
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