Non-linear partial differential equation of order 1
$begingroup$
Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$
$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$
Using Charpit's method
$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.
pde
edited Jan 22 at 17:25
El borito
714216
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
$endgroup$
add a comment |
$begingroup$
Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$
$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$
Using Charpit's method
$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.
pde
edited Jan 22 at 17:25
El borito
714216
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
$endgroup$
2
$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54
1
$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11
add a comment |
$begingroup$
Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$
$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$
Using Charpit's method
$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.
pde
edited Jan 22 at 17:25
El borito
714216
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
$endgroup$
Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $displaystyle p = frac{partial z}{partial x}$, $,displaystyle q = frac{partial z}{partial y}$
$f_p=2q+2x\ f_q=2p+2x\ f_z=0\ f_x=2q+2x\ f_y=2p+2y$
Using Charpit's method
$$frac{dp}{2q+2x}=frac{dq}{2p+2y}=frac{dz}{-4pq-2py-2qx}=frac{dx}{-2q-2y}=frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.
pde
pde
edited Jan 22 at 17:25
El borito
714216
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
edited Jan 22 at 17:25
El borito
714216
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
edited Jan 22 at 17:25
El borito
714216
edited Jan 22 at 17:25
El borito
714216
edited Jan 22 at 17:25
El borito
714216
714216
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
asked Jan 15 at 7:17
Piyush DivyanakarPiyush Divyanakar
3,338325
3,338325
2
$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54
1
$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11
add a comment |
2
$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54
1
$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11
2
2
$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54
$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54
1
1
$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11
$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.
HINT :
Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.
For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.
Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.
$z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$
Determine $F(y)$ and $G(x)$ to make them consistent.
The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :
$$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$
$p=z_x=u_x-x=P-x$
$q=z_y=u_y-y=Q-y$
$$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
After simplification :
$$PQ+frac12(x-y)^2=0$$
$$u_xu_y+frac12(x-y)^2=0$$
$$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
$$begin{cases}
P+Q=c_1 \
PQ=-frac12(x-y)^2
end{cases}$$
etc.
Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .
Note: It is a pity that no boundary condition be specified.
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
$endgroup$
add a comment |
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$begingroup$
$$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.
HINT :
Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.
For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.
Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.
$z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$
Determine $F(y)$ and $G(x)$ to make them consistent.
The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :
$$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$
$p=z_x=u_x-x=P-x$
$q=z_y=u_y-y=Q-y$
$$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
After simplification :
$$PQ+frac12(x-y)^2=0$$
$$u_xu_y+frac12(x-y)^2=0$$
$$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
$$begin{cases}
P+Q=c_1 \
PQ=-frac12(x-y)^2
end{cases}$$
etc.
Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .
Note: It is a pity that no boundary condition be specified.
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
$endgroup$
add a comment |
$begingroup$
$$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.
HINT :
Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.
For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.
Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.
$z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$
Determine $F(y)$ and $G(x)$ to make them consistent.
The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :
$$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$
$p=z_x=u_x-x=P-x$
$q=z_y=u_y-y=Q-y$
$$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
After simplification :
$$PQ+frac12(x-y)^2=0$$
$$u_xu_y+frac12(x-y)^2=0$$
$$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
$$begin{cases}
P+Q=c_1 \
PQ=-frac12(x-y)^2
end{cases}$$
etc.
Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .
Note: It is a pity that no boundary condition be specified.
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
$endgroup$
add a comment |
$begingroup$
$$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.
HINT :
Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.
For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.
Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.
$z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$
Determine $F(y)$ and $G(x)$ to make them consistent.
The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :
$$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$
$p=z_x=u_x-x=P-x$
$q=z_y=u_y-y=Q-y$
$$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
After simplification :
$$PQ+frac12(x-y)^2=0$$
$$u_xu_y+frac12(x-y)^2=0$$
$$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
$$begin{cases}
P+Q=c_1 \
PQ=-frac12(x-y)^2
end{cases}$$
etc.
Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .
Note: It is a pity that no boundary condition be specified.
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
$endgroup$
$$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.
HINT :
Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$.
For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$.
Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.
$z=int p(x,y)dx+F(y)$ and $z=int q(x,y)dy+G(x)$
Determine $F(y)$ and $G(x)$ to make them consistent.
The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :
$$z(x,y)=u(x,y)-frac12 x^2-frac12 y^2$$
$p=z_x=u_x-x=P-x$
$q=z_y=u_y-y=Q-y$
$$2left((P-x)(Q-y)+(P-x)y+(Q-y)xright)+x^2+y^2=0$$
After simplification :
$$PQ+frac12(x-y)^2=0$$
$$u_xu_y+frac12(x-y)^2=0$$
$$frac{dx}{Q}=frac{dy}{P}=frac{du}{2PQ}=frac{dP}{y-x}=frac{dQ}{x-y}=ds$$
$$begin{cases}
P+Q=c_1 \
PQ=-frac12(x-y)^2
end{cases}$$
etc.
Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .
Note: It is a pity that no boundary condition be specified.
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
edited Jan 20 at 11:18
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
answered Jan 20 at 11:07
JJacquelinJJacquelin
43.5k21853
43.5k21853
add a comment |
add a comment |
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$begingroup$
Just so you know, writing $p$ and $q$ for derivatives aren't really common in other countries, so you should specify the notation in every question.
$endgroup$
– Dylan
Jan 15 at 15:54
1
$begingroup$
I will keep that in mind in future.
$endgroup$
– Piyush Divyanakar
Jan 15 at 16:11