How to solve this PDE: $psqrt x+qsqrt y =sqrt z$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How do I solve the following PDE?Analytical Solution of a PDEDiffusion-Reaction PDE - radial coordinateHow to solve this pde equation: $(p^2 + q^2)y = qz$How to solve the following parabolic pde?Lagrange's First Order PDE Special Case .Solution to a two dimensional transport PDEA first order PDE with unsolvable characteristic equationsSolving an PDE with the Fourier TransformHow to solve this parabolic PDE?
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How to solve this PDE: $psqrt x+qsqrt y =sqrt z$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How do I solve the following PDE?Analytical Solution of a PDEDiffusion-Reaction PDE - radial coordinateHow to solve this pde equation: $(p^2 + q^2)y = qz$How to solve the following parabolic pde?Lagrange's First Order PDE Special Case .Solution to a two dimensional transport PDEA first order PDE with unsolvable characteristic equationsSolving an PDE with the Fourier TransformHow to solve this parabolic PDE?
$begingroup$
I am trying to solve this PDE, using Lagrange's method:
$psqrt x+qsqrt y =sqrt z$
Lagrange's auxiliary equation is:
$fracdxsqrt x=frac dysqrt y= fracdzsqrt z$
I already know the answer to this on which is $f(sqrt x - sqrt y, sqrt y - sqrt z)$ , but I don't know the complete solution and I have tried to solve it but I can't quite get to the same answer.
pde
asked Jan 15 '18 at 17:44
House A.House A.
113
$endgroup$
|
show 3 more comments
$begingroup$
I am trying to solve this PDE, using Lagrange's method:
$psqrt x+qsqrt y =sqrt z$
Lagrange's auxiliary equation is:
$fracdxsqrt x=frac dysqrt y= fracdzsqrt z$
I already know the answer to this on which is $f(sqrt x - sqrt y, sqrt y - sqrt z)$ , but I don't know the complete solution and I have tried to solve it but I can't quite get to the same answer.
pde
asked Jan 15 '18 at 17:44
House A.House A.
113
$endgroup$
1
$begingroup$
What is it PDE?
$endgroup$
– Michael Rozenberg
Jan 15 '18 at 17:45
1
$begingroup$
What you've posted does not contain any partial derivatives... Try reading the question again?
$endgroup$
– AlkaKadri
Jan 15 '18 at 17:46
$begingroup$
This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z.
$endgroup$
– House A.
Jan 15 '18 at 17:51
$begingroup$
What are $p$ and $q$?
$endgroup$
– Bernard Massé
Jan 15 '18 at 17:53
1
$begingroup$
@MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ).
$endgroup$
– House A.
Jan 15 '18 at 17:55
|
show 3 more comments
$begingroup$
I am trying to solve this PDE, using Lagrange's method:
$psqrt x+qsqrt y =sqrt z$
Lagrange's auxiliary equation is:
$fracdxsqrt x=frac dysqrt y= fracdzsqrt z$
I already know the answer to this on which is $f(sqrt x - sqrt y, sqrt y - sqrt z)$ , but I don't know the complete solution and I have tried to solve it but I can't quite get to the same answer.
pde
asked Jan 15 '18 at 17:44
House A.House A.
113
$endgroup$
I am trying to solve this PDE, using Lagrange's method:
$psqrt x+qsqrt y =sqrt z$
Lagrange's auxiliary equation is:
$fracdxsqrt x=frac dysqrt y= fracdzsqrt z$
I already know the answer to this on which is $f(sqrt x - sqrt y, sqrt y - sqrt z)$ , but I don't know the complete solution and I have tried to solve it but I can't quite get to the same answer.
pde
pde
asked Jan 15 '18 at 17:44
House A.House A.
113
asked Jan 15 '18 at 17:44
House A.House A.
113
edited Jan 15 '18 at 18:15
House A.
asked Jan 15 '18 at 17:44
House A.House A.
113
asked Jan 15 '18 at 17:44
House A.House A.
113
asked Jan 15 '18 at 17:44
House A.House A.
113
113
1
$begingroup$
What is it PDE?
$endgroup$
– Michael Rozenberg
Jan 15 '18 at 17:45
1
$begingroup$
What you've posted does not contain any partial derivatives... Try reading the question again?
$endgroup$
– AlkaKadri
Jan 15 '18 at 17:46
$begingroup$
This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z.
$endgroup$
– House A.
Jan 15 '18 at 17:51
$begingroup$
What are $p$ and $q$?
$endgroup$
– Bernard Massé
Jan 15 '18 at 17:53
1
$begingroup$
@MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ).
$endgroup$
– House A.
Jan 15 '18 at 17:55
|
show 3 more comments
1
$begingroup$
What is it PDE?
$endgroup$
– Michael Rozenberg
Jan 15 '18 at 17:45
1
$begingroup$
What you've posted does not contain any partial derivatives... Try reading the question again?
$endgroup$
– AlkaKadri
Jan 15 '18 at 17:46
$begingroup$
This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z.
$endgroup$
– House A.
Jan 15 '18 at 17:51
$begingroup$
What are $p$ and $q$?
$endgroup$
– Bernard Massé
Jan 15 '18 at 17:53
1
$begingroup$
@MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ).
$endgroup$
– House A.
Jan 15 '18 at 17:55
1
1
$begingroup$
What is it PDE?
$endgroup$
– Michael Rozenberg
Jan 15 '18 at 17:45
$begingroup$
What is it PDE?
$endgroup$
– Michael Rozenberg
Jan 15 '18 at 17:45
1
1
$begingroup$
What you've posted does not contain any partial derivatives... Try reading the question again?
$endgroup$
– AlkaKadri
Jan 15 '18 at 17:46
$begingroup$
What you've posted does not contain any partial derivatives... Try reading the question again?
$endgroup$
– AlkaKadri
Jan 15 '18 at 17:46
$begingroup$
This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z.
$endgroup$
– House A.
Jan 15 '18 at 17:51
$begingroup$
This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z.
$endgroup$
– House A.
Jan 15 '18 at 17:51
$begingroup$
What are $p$ and $q$?
$endgroup$
– Bernard Massé
Jan 15 '18 at 17:53
$begingroup$
What are $p$ and $q$?
$endgroup$
– Bernard Massé
Jan 15 '18 at 17:53
1
1
$begingroup$
@MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ).
$endgroup$
– House A.
Jan 15 '18 at 17:55
$begingroup$
@MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ).
$endgroup$
– House A.
Jan 15 '18 at 17:55
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$$sqrtx fracpartial zpartial x+sqrty fracpartial zpartial y =sqrt z$$
Your general solution on the form of implicit equation : $quad f(sqrt x - sqrt y, sqrt y - sqrt z)=0quad$ is correct.
An equivalent explicit form is :
$$ sqrt y - sqrt z =phi( sqrt x - sqrt y)$$
where $phi$ is any differentiable function.
As a consequence :
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
IN ADDITION :
System of characteristic ODEs : $quadfracdxsqrt x=fracdysqrt y=fracdzsqrt z$
First family of characteristic curves, from $quad fracdxsqrt x=fracdysqrt yquadtoquad sqrt x-sqrt y=c_1$
Second family of characteristic curves, from $quad fracdzsqrt z=fracdysqrt yquadtoquad sqrt z-sqrt y=c_2$
General solution : $quadsqrt y - sqrt z =phi( sqrt x - sqrt y)$
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
$endgroup$
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
add a comment |
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1 Answer
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$begingroup$
$$sqrtx fracpartial zpartial x+sqrty fracpartial zpartial y =sqrt z$$
Your general solution on the form of implicit equation : $quad f(sqrt x - sqrt y, sqrt y - sqrt z)=0quad$ is correct.
An equivalent explicit form is :
$$ sqrt y - sqrt z =phi( sqrt x - sqrt y)$$
where $phi$ is any differentiable function.
As a consequence :
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
IN ADDITION :
System of characteristic ODEs : $quadfracdxsqrt x=fracdysqrt y=fracdzsqrt z$
First family of characteristic curves, from $quad fracdxsqrt x=fracdysqrt yquadtoquad sqrt x-sqrt y=c_1$
Second family of characteristic curves, from $quad fracdzsqrt z=fracdysqrt yquadtoquad sqrt z-sqrt y=c_2$
General solution : $quadsqrt y - sqrt z =phi( sqrt x - sqrt y)$
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
$endgroup$
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
add a comment |
$begingroup$
$$sqrtx fracpartial zpartial x+sqrty fracpartial zpartial y =sqrt z$$
Your general solution on the form of implicit equation : $quad f(sqrt x - sqrt y, sqrt y - sqrt z)=0quad$ is correct.
An equivalent explicit form is :
$$ sqrt y - sqrt z =phi( sqrt x - sqrt y)$$
where $phi$ is any differentiable function.
As a consequence :
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
IN ADDITION :
System of characteristic ODEs : $quadfracdxsqrt x=fracdysqrt y=fracdzsqrt z$
First family of characteristic curves, from $quad fracdxsqrt x=fracdysqrt yquadtoquad sqrt x-sqrt y=c_1$
Second family of characteristic curves, from $quad fracdzsqrt z=fracdysqrt yquadtoquad sqrt z-sqrt y=c_2$
General solution : $quadsqrt y - sqrt z =phi( sqrt x - sqrt y)$
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
$endgroup$
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
add a comment |
$begingroup$
$$sqrtx fracpartial zpartial x+sqrty fracpartial zpartial y =sqrt z$$
Your general solution on the form of implicit equation : $quad f(sqrt x - sqrt y, sqrt y - sqrt z)=0quad$ is correct.
An equivalent explicit form is :
$$ sqrt y - sqrt z =phi( sqrt x - sqrt y)$$
where $phi$ is any differentiable function.
As a consequence :
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
IN ADDITION :
System of characteristic ODEs : $quadfracdxsqrt x=fracdysqrt y=fracdzsqrt z$
First family of characteristic curves, from $quad fracdxsqrt x=fracdysqrt yquadtoquad sqrt x-sqrt y=c_1$
Second family of characteristic curves, from $quad fracdzsqrt z=fracdysqrt yquadtoquad sqrt z-sqrt y=c_2$
General solution : $quadsqrt y - sqrt z =phi( sqrt x - sqrt y)$
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
$endgroup$
$$sqrtx fracpartial zpartial x+sqrty fracpartial zpartial y =sqrt z$$
Your general solution on the form of implicit equation : $quad f(sqrt x - sqrt y, sqrt y - sqrt z)=0quad$ is correct.
An equivalent explicit form is :
$$ sqrt y - sqrt z =phi( sqrt x - sqrt y)$$
where $phi$ is any differentiable function.
As a consequence :
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
IN ADDITION :
System of characteristic ODEs : $quadfracdxsqrt x=fracdysqrt y=fracdzsqrt z$
First family of characteristic curves, from $quad fracdxsqrt x=fracdysqrt yquadtoquad sqrt x-sqrt y=c_1$
Second family of characteristic curves, from $quad fracdzsqrt z=fracdysqrt yquadtoquad sqrt z-sqrt y=c_2$
General solution : $quadsqrt y - sqrt z =phi( sqrt x - sqrt y)$
$$z(x,y)=bigg(sqrt y - phi( sqrt x - sqrt y)bigg)^2$$
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
edited Jan 17 '18 at 22:29
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
answered Jan 15 '18 at 18:05
JJacquelinJJacquelin
45.6k21857
45.6k21857
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
add a comment |
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
I know that this is the answer, but I don't know how to get to it. I have tried couple of times, and I don't get this same answer.
$endgroup$
– House A.
Jan 16 '18 at 8:48
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
$begingroup$
See the addition to my main answer.
$endgroup$
– JJacquelin
Jan 16 '18 at 9:05
add a comment |
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1
$begingroup$
What is it PDE?
$endgroup$
– Michael Rozenberg
Jan 15 '18 at 17:45
1
$begingroup$
What you've posted does not contain any partial derivatives... Try reading the question again?
$endgroup$
– AlkaKadri
Jan 15 '18 at 17:46
$begingroup$
This is a partial differential equation of the form Pp+Qq=R, where P, Q and R are functions of x, y, z.
$endgroup$
– House A.
Jan 15 '18 at 17:51
$begingroup$
What are $p$ and $q$?
$endgroup$
– Bernard Massé
Jan 15 '18 at 17:53
1
$begingroup$
@MichaelRozenberg Partial Differential Equation. This is a standard form of a linear partial differential equation of the first order ( Lagrange's linear equation ).
$endgroup$
– House A.
Jan 15 '18 at 17:55