Characteristic Curve of a PDE
$begingroup$
Question from a Exam:
Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.
Can someone please tell me how are these types of problems handled?
I dont want exact solution.Please tell me how to solve these type of questions.
pde characteristics
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
$endgroup$
|
show 2 more comments
$begingroup$
Question from a Exam:
Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.
Can someone please tell me how are these types of problems handled?
I dont want exact solution.Please tell me how to solve these type of questions.
pde characteristics
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
$endgroup$
$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43
$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50
$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51
$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52
$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00
|
show 2 more comments
$begingroup$
Question from a Exam:
Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.
Can someone please tell me how are these types of problems handled?
I dont want exact solution.Please tell me how to solve these type of questions.
pde characteristics
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
$endgroup$
Question from a Exam:
Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$.
Find the characteristic curves of the above.
Can someone please tell me how are these types of problems handled?
I dont want exact solution.Please tell me how to solve these type of questions.
pde characteristics
pde characteristics
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
asked Dec 3 '18 at 15:24
Join_PhDJoin_PhD
3818
3818
$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43
$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50
$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51
$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52
$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00
|
show 2 more comments
$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43
$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50
$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51
$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52
$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00
$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43
$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43
$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50
$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50
$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51
$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51
$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52
$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52
$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00
$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.
You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
$$xg'+2Ax^3=g-1$$
$$g-xg'=2Ax^3+1$$
$$g=-Ax^3+Bx+1$$
Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
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add a comment |
$begingroup$
$$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
$v=u_x$
$$xv_{x}+2x^2v_{y}=v-1 tag 2$$
Charpit-Lagrange equations:
$$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
$$y-x^2=c_1$$
Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
$$frac{v-1}{x}=c_2$$
General solution of the PDE Eq.$(2)$:
$$frac{v-1}{x}=F(y-x^2)$$
where $F$ is an arbitrary function.
$$v(x,y)=1+xF(y-x^2)$$
$u=int left(1+xF(y-x^2)right)dx$
$$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
$F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.
Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
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add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.
You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
$$xg'+2Ax^3=g-1$$
$$g-xg'=2Ax^3+1$$
$$g=-Ax^3+Bx+1$$
Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.
You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
$$xg'+2Ax^3=g-1$$
$$g-xg'=2Ax^3+1$$
$$g=-Ax^3+Bx+1$$
Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.
You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
$$xg'+2Ax^3=g-1$$
$$g-xg'=2Ax^3+1$$
$$g=-Ax^3+Bx+1$$
Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
$endgroup$
Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.
You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives:
$$xg'+2Ax^3=g-1$$
$$g-xg'=2Ax^3+1$$
$$g=-Ax^3+Bx+1$$
Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=frac{A}{2}x^2y-frac{A}{4}x^4+frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
answered Dec 3 '18 at 16:07
MoKo19MoKo19
1914
1914
add a comment |
add a comment |
$begingroup$
$$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
$v=u_x$
$$xv_{x}+2x^2v_{y}=v-1 tag 2$$
Charpit-Lagrange equations:
$$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
$$y-x^2=c_1$$
Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
$$frac{v-1}{x}=c_2$$
General solution of the PDE Eq.$(2)$:
$$frac{v-1}{x}=F(y-x^2)$$
where $F$ is an arbitrary function.
$$v(x,y)=1+xF(y-x^2)$$
$u=int left(1+xF(y-x^2)right)dx$
$$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
$F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.
Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
$endgroup$
add a comment |
$begingroup$
$$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
$v=u_x$
$$xv_{x}+2x^2v_{y}=v-1 tag 2$$
Charpit-Lagrange equations:
$$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
$$y-x^2=c_1$$
Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
$$frac{v-1}{x}=c_2$$
General solution of the PDE Eq.$(2)$:
$$frac{v-1}{x}=F(y-x^2)$$
where $F$ is an arbitrary function.
$$v(x,y)=1+xF(y-x^2)$$
$u=int left(1+xF(y-x^2)right)dx$
$$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
$F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.
Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
$endgroup$
add a comment |
$begingroup$
$$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
$v=u_x$
$$xv_{x}+2x^2v_{y}=v-1 tag 2$$
Charpit-Lagrange equations:
$$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
$$y-x^2=c_1$$
Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
$$frac{v-1}{x}=c_2$$
General solution of the PDE Eq.$(2)$:
$$frac{v-1}{x}=F(y-x^2)$$
where $F$ is an arbitrary function.
$$v(x,y)=1+xF(y-x^2)$$
$u=int left(1+xF(y-x^2)right)dx$
$$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
$F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.
Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
$endgroup$
$$xu_{xx}+2x^2u_{xy}=u_x-1 tag 1$$
$v=u_x$
$$xv_{x}+2x^2v_{y}=v-1 tag 2$$
Charpit-Lagrange equations:
$$frac{dx}{x}=frac{dy}{2x^2}=frac{dv}{v-1}$$
First family of characteristic curves from $frac{dx}{x}=frac{dy}{2x^2}$ :
$$y-x^2=c_1$$
Second family of characteristic curves from $frac{dx}{x}=frac{dv}{v-1}$
$$frac{v-1}{x}=c_2$$
General solution of the PDE Eq.$(2)$:
$$frac{v-1}{x}=F(y-x^2)$$
where $F$ is an arbitrary function.
$$v(x,y)=1+xF(y-x^2)$$
$u=int left(1+xF(y-x^2)right)dx$
$$u(x,y)=x+int xF(y-x^2) dx+G(y)$$
$F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.
Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
answered Dec 8 '18 at 8:50
JJacquelinJJacquelin
43.3k21853
43.3k21853
add a comment |
add a comment |
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$begingroup$
You can change the variable to $v=u_x$ and apply the characteristic curves method.
$endgroup$
– Dante Grevino
Dec 3 '18 at 15:43
$begingroup$
@DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:50
$begingroup$
then i have if $a=x,b=2x^2,c=v-1implies frac{dx}{a}=frac{dy}{2a^2}=frac{dv}{v-1}$
$endgroup$
– Join_PhD
Dec 3 '18 at 15:51
$begingroup$
@DanteGrevino;how to solve it now
$endgroup$
– Join_PhD
Dec 3 '18 at 15:52
$begingroup$
Do you have boundary conditions?
$endgroup$
– Dante Grevino
Dec 3 '18 at 16:00