Quasilinear PDE using method of characteristics
$begingroup$
The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$
I first find that
$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$
Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$
Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$
Applying the initial conditions yields
$A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$
This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,
$A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$
Is this the correct solution?
Thanks!
pde characteristics
edited Dec 11 '18 at 22:00
Harry49
8,38931344
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
$endgroup$
add a comment |
$begingroup$
The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$
I first find that
$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$
Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$
Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$
Applying the initial conditions yields
$A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$
This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,
$A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$
Is this the correct solution?
Thanks!
pde characteristics
edited Dec 11 '18 at 22:00
Harry49
8,38931344
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
$endgroup$
add a comment |
$begingroup$
The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$
I first find that
$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$
Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$
Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$
Applying the initial conditions yields
$A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$
This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,
$A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$
Is this the correct solution?
Thanks!
pde characteristics
edited Dec 11 '18 at 22:00
Harry49
8,38931344
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
$endgroup$
The equation is: $yu_x+uu_y=-xy$ with initial conditions $u=y$ on $x=0$
I first find that
$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}$
Solving $frac{dx}{y}=frac{dy}{u}$ we get, $ux=frac{1}{2}y^2+A$
Now solving $frac{dy}{u}=-frac{du}{xy}$ we get $frac{1}{2}u^2=B-frac{1}{2}xy^2$
Applying the initial conditions yields
$A=-frac{1}{2}y^2$ and $frac{1}{2}y^2=B$, so $A=-B$, or $A+B=0$
This was obtained on the line $x=0$, but contains only constants, thus holds for all characteristics intersecting $x=0$, so we have,
$A=-B implies ux-frac{1}{2}y^2=-frac{1}{2}u^2-frac{1}{2}xy^2$
Is this the correct solution?
Thanks!
pde characteristics
pde characteristics
edited Dec 11 '18 at 22:00
Harry49
8,38931344
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
edited Dec 11 '18 at 22:00
Harry49
8,38931344
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
edited Dec 11 '18 at 22:00
Harry49
8,38931344
edited Dec 11 '18 at 22:00
Harry49
8,38931344
edited Dec 11 '18 at 22:00
Harry49
8,38931344
8,38931344
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
asked Dec 11 '18 at 16:40
Brad ScottBrad Scott
1548
1548
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.
You can solve the first and last quotient as
$$
x,dx = -duimplies 2u+x^2=c_1
$$
This you can then insert into the other equations.
$$
0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
$$
Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
$$
ϕ(2u+x^2)=2xu+frac23x^3-y^2.
$$
With the initial conditions $u=y$, $x=0$ this reduces to
$$
ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
$$
so that the surface follows the equation
$$
-(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
$$
which is a quadratic equation for the value of $u$.
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
add a comment |
$begingroup$
$$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
$$u+frac{x^2}{2}=c_1$$
Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :
$c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$
$(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$
$$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
General solution on the form of implicite equation :
$$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
$F$ is an arbitrary function to be determined according to the boundary condition.
Condition $u(0,y)=y$ :
$0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$
The function $F$ is determined :
$$F(X)=-frac{X^2}{2}$$
We put it into the general solution where $X=u+frac{x^2}{2}$
$$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
Solving for $u$ :
$$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$
This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
$endgroup$
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
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
$begingroup$
No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.
You can solve the first and last quotient as
$$
x,dx = -duimplies 2u+x^2=c_1
$$
This you can then insert into the other equations.
$$
0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
$$
Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
$$
ϕ(2u+x^2)=2xu+frac23x^3-y^2.
$$
With the initial conditions $u=y$, $x=0$ this reduces to
$$
ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
$$
so that the surface follows the equation
$$
-(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
$$
which is a quadratic equation for the value of $u$.
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
add a comment |
$begingroup$
No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.
You can solve the first and last quotient as
$$
x,dx = -duimplies 2u+x^2=c_1
$$
This you can then insert into the other equations.
$$
0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
$$
Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
$$
ϕ(2u+x^2)=2xu+frac23x^3-y^2.
$$
With the initial conditions $u=y$, $x=0$ this reduces to
$$
ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
$$
so that the surface follows the equation
$$
-(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
$$
which is a quadratic equation for the value of $u$.
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
add a comment |
$begingroup$
No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.
You can solve the first and last quotient as
$$
x,dx = -duimplies 2u+x^2=c_1
$$
This you can then insert into the other equations.
$$
0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
$$
Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
$$
ϕ(2u+x^2)=2xu+frac23x^3-y^2.
$$
With the initial conditions $u=y$, $x=0$ this reduces to
$$
ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
$$
so that the surface follows the equation
$$
-(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
$$
which is a quadratic equation for the value of $u$.
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
$endgroup$
No, already the first equation you solve is wrong. As the last denominator is not zero, $u$ is not constant along characteristic curves.
You can solve the first and last quotient as
$$
x,dx = -duimplies 2u+x^2=c_1
$$
This you can then insert into the other equations.
$$
0=u,dx -y,dyimplies 0=(c_1-x^2),dx-2y,dyimplies c_2=c_1x-frac{x^3}3-y^2
$$
Because each characteristic curve only has one constant, there is some functional dependence like $c_2=phi(c_1)$, so that
$$
ϕ(2u+x^2)=2xu+frac23x^3-y^2.
$$
With the initial conditions $u=y$, $x=0$ this reduces to
$$
ϕ(2y)=-y^2implies ϕ(t)=-frac{t^2}4
$$
so that the surface follows the equation
$$
-(u^2+x^2u+frac{x^4}4)=2xu+frac23x^3-y^2
$$
which is a quadratic equation for the value of $u$.
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
edited Dec 11 '18 at 18:03
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
answered Dec 11 '18 at 17:37
LutzLLutzL
60.1k42057
60.1k42057
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
add a comment |
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
how do you know which equations to solve?
$endgroup$
– Brad Scott
Dec 11 '18 at 17:39
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You look for combinations of quotients that reduce to only contain two variables, like the first and last quotient here. This will not always work, but mostly works for textbook exercises.
$endgroup$
– LutzL
Dec 11 '18 at 17:43
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
$begingroup$
You could also just multiply all with $y$ so that $ds=frac{dx}1=frac{y,dy}u=-frac{du}x$, which reduces to the linear system $frac{d(y^2)}{dx}=2u$ and $frac{du}{dx}=-x$.
$endgroup$
– LutzL
Dec 11 '18 at 17:49
add a comment |
$begingroup$
$$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
$$u+frac{x^2}{2}=c_1$$
Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :
$c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$
$(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$
$$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
General solution on the form of implicite equation :
$$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
$F$ is an arbitrary function to be determined according to the boundary condition.
Condition $u(0,y)=y$ :
$0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$
The function $F$ is determined :
$$F(X)=-frac{X^2}{2}$$
We put it into the general solution where $X=u+frac{x^2}{2}$
$$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
Solving for $u$ :
$$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$
This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
$endgroup$
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
add a comment |
$begingroup$
$$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
$$u+frac{x^2}{2}=c_1$$
Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :
$c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$
$(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$
$$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
General solution on the form of implicite equation :
$$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
$F$ is an arbitrary function to be determined according to the boundary condition.
Condition $u(0,y)=y$ :
$0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$
The function $F$ is determined :
$$F(X)=-frac{X^2}{2}$$
We put it into the general solution where $X=u+frac{x^2}{2}$
$$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
Solving for $u$ :
$$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$
This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
$endgroup$
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
add a comment |
$begingroup$
$$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
$$u+frac{x^2}{2}=c_1$$
Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :
$c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$
$(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$
$$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
General solution on the form of implicite equation :
$$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
$F$ is an arbitrary function to be determined according to the boundary condition.
Condition $u(0,y)=y$ :
$0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$
The function $F$ is determined :
$$F(X)=-frac{X^2}{2}$$
We put it into the general solution where $X=u+frac{x^2}{2}$
$$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
Solving for $u$ :
$$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$
This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
$endgroup$
$$frac{dx}{y}=frac{dy}{u}=-frac{du}{xy}qquad text{OK.}$$
First characteristic curves from $frac{dx}{y}=-frac{du}{xy}$ :
$$u+frac{x^2}{2}=c_1$$
Second characteristic curves from $frac{dx}{y}=frac{dy}{c_1-frac{x^2}{2}}$ :
$c_1x-frac{x^3}{6}=frac{y^2}{2}+c_2$
$(u+frac{x^2}{2})x-frac{x^3}{6}-frac{y^2}{2}=c_2$
$$xu+frac{x^3}{3}-frac{y^2}{2}=c_2$$
General solution on the form of implicite equation :
$$xu+frac{x^3}{3}-frac{y^2}{2}=Fleft(u+frac{x^2}{2}right)$$
$F$ is an arbitrary function to be determined according to the boundary condition.
Condition $u(0,y)=y$ :
$0y+frac{0^3}{3}-frac{y^2}{2}=Fleft(y+frac{0^2}{2}right)=-frac{y^2}{2}=Fleft(yright)$
The function $F$ is determined :
$$F(X)=-frac{X^2}{2}$$
We put it into the general solution where $X=u+frac{x^2}{2}$
$$xu+frac{x^3}{3}-frac{y^2}{2}=-frac{left(u+frac{x^2}{2}right)^2}{2}$$
Solving for $u$ :
$$u(x,y)=-x-frac{x^2}{2}+y:sqrt{1+frac{x^2}{y^2}+frac{x^3}{3y^2}}$$
This result has been successfully checked in putting it into the PDE $yu_x+uu_y=-xy$.
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
edited Dec 11 '18 at 18:43
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
answered Dec 11 '18 at 18:34
JJacquelinJJacquelin
45.3k21856
45.3k21856
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
add a comment |
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
Thanks! I see some solutions put in an arbitrary function and some where they do not, i was wondering if you could explain when an arbitrary function, like F here, should be used.
$endgroup$
– Brad Scott
Dec 11 '18 at 19:03
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
$begingroup$
In the general solution of a PDE there is always at least an arbitrary function, It is not a matter of "should be used" or "not used". But, depending of the method of solving that you choose to use, the arbitrary function appears explicitly or not. Thus your question : "Could you explain when an arbitrary function, like F here, should be used" has no more sens than asking for : "Could you explain when a particular method of solving or another one should be used".
$endgroup$
– JJacquelin
Dec 11 '18 at 21:55
add a comment |
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