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Confusion regarding the notation of associated operator of a PDE.
Numerical Analysis and Differential equations book recommendations focusing on the given topics.Solve the 3 Dimension first order PDE $xu_x+yu_y+zu_z=0$Energy associated with PDESome clarification on nonlinear PDEsConfusion in college book on the introduction of PDE (method of characteristics)Explaining Characteristics of a PDEexample of classical solution of elliptic PDEFinding the integrating factor for an ODE through the PDE characterisitc methodWhat precisely does the notation $frac partial h partial z$ mean in this context?Using the Multivariable Chain Rule to Show a Solution to a PDESolving the Hyperbolic PDE $u_xx-2u_xy-3u_yy=0$
$begingroup$
I'm given the following PDE: $$u_tt-u_xx+u^3=0$$ My source says that the associated operator is $$L:=fracpartial^2partial t^2-fracpartial^2partial x^2+u^2$$ which is arrived at simply by factoring the common term $u$ and thinking of the operator as left multiplying the function.$$fracpartial^2partial t^2u-fracpartial^2partial x^2u+u^3=left (fracpartial^2partial t^2-fracpartial^2partial x^2+u^2 right )u=Lu=0$$
Is this correct? In general how do I find and write the associated operator of a PDE? Do I simply do as my source does and just factor out the $u$? For eg. is the operator in $$Lu=u_x+u_y+1$$ given by $$L=
fracpartialpartial x+fracpartialpartial y+frac1u$$ as given here?
ordinary-differential-equations pde notation
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
$endgroup$
|
show 3 more comments
$begingroup$
I'm given the following PDE: $$u_tt-u_xx+u^3=0$$ My source says that the associated operator is $$L:=fracpartial^2partial t^2-fracpartial^2partial x^2+u^2$$ which is arrived at simply by factoring the common term $u$ and thinking of the operator as left multiplying the function.$$fracpartial^2partial t^2u-fracpartial^2partial x^2u+u^3=left (fracpartial^2partial t^2-fracpartial^2partial x^2+u^2 right )u=Lu=0$$
Is this correct? In general how do I find and write the associated operator of a PDE? Do I simply do as my source does and just factor out the $u$? For eg. is the operator in $$Lu=u_x+u_y+1$$ given by $$L=
fracpartialpartial x+fracpartialpartial y+frac1u$$ as given here?
ordinary-differential-equations pde notation
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
$endgroup$
1
$begingroup$
It's not linear because of the $u^3$ term
$endgroup$
– Dylan
Mar 14 at 7:24
$begingroup$
@Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question.
$endgroup$
– Rhaldryn
Mar 14 at 7:32
1
$begingroup$
The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $partial^2_t - partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$.
$endgroup$
– bgsk
yesterday
1
$begingroup$
Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (partial^2_t - partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments).
$endgroup$
– bgsk
yesterday
1
$begingroup$
@bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books?
$endgroup$
– Rhaldryn
yesterday
|
show 3 more comments
$begingroup$
I'm given the following PDE: $$u_tt-u_xx+u^3=0$$ My source says that the associated operator is $$L:=fracpartial^2partial t^2-fracpartial^2partial x^2+u^2$$ which is arrived at simply by factoring the common term $u$ and thinking of the operator as left multiplying the function.$$fracpartial^2partial t^2u-fracpartial^2partial x^2u+u^3=left (fracpartial^2partial t^2-fracpartial^2partial x^2+u^2 right )u=Lu=0$$
Is this correct? In general how do I find and write the associated operator of a PDE? Do I simply do as my source does and just factor out the $u$? For eg. is the operator in $$Lu=u_x+u_y+1$$ given by $$L=
fracpartialpartial x+fracpartialpartial y+frac1u$$ as given here?
ordinary-differential-equations pde notation
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
$endgroup$
I'm given the following PDE: $$u_tt-u_xx+u^3=0$$ My source says that the associated operator is $$L:=fracpartial^2partial t^2-fracpartial^2partial x^2+u^2$$ which is arrived at simply by factoring the common term $u$ and thinking of the operator as left multiplying the function.$$fracpartial^2partial t^2u-fracpartial^2partial x^2u+u^3=left (fracpartial^2partial t^2-fracpartial^2partial x^2+u^2 right )u=Lu=0$$
Is this correct? In general how do I find and write the associated operator of a PDE? Do I simply do as my source does and just factor out the $u$? For eg. is the operator in $$Lu=u_x+u_y+1$$ given by $$L=
fracpartialpartial x+fracpartialpartial y+frac1u$$ as given here?
ordinary-differential-equations pde notation
ordinary-differential-equations pde notation
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
edited Mar 14 at 7:30
Rhaldryn
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
asked Mar 14 at 6:29
RhaldrynRhaldryn
352416
352416
1
$begingroup$
It's not linear because of the $u^3$ term
$endgroup$
– Dylan
Mar 14 at 7:24
$begingroup$
@Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question.
$endgroup$
– Rhaldryn
Mar 14 at 7:32
1
$begingroup$
The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $partial^2_t - partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$.
$endgroup$
– bgsk
yesterday
1
$begingroup$
Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (partial^2_t - partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments).
$endgroup$
– bgsk
yesterday
1
$begingroup$
@bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books?
$endgroup$
– Rhaldryn
yesterday
|
show 3 more comments
1
$begingroup$
It's not linear because of the $u^3$ term
$endgroup$
– Dylan
Mar 14 at 7:24
$begingroup$
@Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question.
$endgroup$
– Rhaldryn
Mar 14 at 7:32
1
$begingroup$
The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $partial^2_t - partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$.
$endgroup$
– bgsk
yesterday
1
$begingroup$
Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (partial^2_t - partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments).
$endgroup$
– bgsk
yesterday
1
$begingroup$
@bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books?
$endgroup$
– Rhaldryn
yesterday
1
1
$begingroup$
It's not linear because of the $u^3$ term
$endgroup$
– Dylan
Mar 14 at 7:24
$begingroup$
It's not linear because of the $u^3$ term
$endgroup$
– Dylan
Mar 14 at 7:24
$begingroup$
@Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question.
$endgroup$
– Rhaldryn
Mar 14 at 7:32
$begingroup$
@Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question.
$endgroup$
– Rhaldryn
Mar 14 at 7:32
1
1
$begingroup$
The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $partial^2_t - partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$.
$endgroup$
– bgsk
yesterday
$begingroup$
The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $partial^2_t - partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$.
$endgroup$
– bgsk
yesterday
1
1
$begingroup$
Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (partial^2_t - partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments).
$endgroup$
– bgsk
yesterday
$begingroup$
Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (partial^2_t - partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments).
$endgroup$
– bgsk
yesterday
1
1
$begingroup$
@bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books?
$endgroup$
– Rhaldryn
yesterday
$begingroup$
@bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books?
$endgroup$
– Rhaldryn
yesterday
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
As indicated by the OP, I will regroup my comments as an answer.
The source notes you cite make use of rather ambiguous/confusing notation. Indeed, one cannot simply "factor out" the term $partial_t^2 - partial_x^2 + u^2$ and call it "the operator", as this object clearly depends on the function $u$. The operator should be an object which is a priori independent of the function: it takes a "function" $u$ as input and yields another "function" $g$ as output.
In relation to your actual question. I believe it is better to consider the "operator" $Nu = (partial_t^2 - partial_x^2)u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. It is now clear that the operator $N$ is nonlinear, and the linear and nonlinear parts are moreover explicitly split.
For futher reference, the linear operator $L = partial_t^2 - partial_x^2$ is called the wave operator (also called the d'Alembertian), as it is the governing linear operator in the wave equation
beginequation
partial_t^2 u - partial_t^2 u = 0.
endequation
Hence the equation associated to the "operator" you were considering is the nonlinear wave equation
beginequation
partial_t^2 u - partial_t^2 u + u^3 = 0.
endequation
answered 22 hours ago
bgskbgsk
175111
$endgroup$
add a comment |
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$begingroup$
As indicated by the OP, I will regroup my comments as an answer.
The source notes you cite make use of rather ambiguous/confusing notation. Indeed, one cannot simply "factor out" the term $partial_t^2 - partial_x^2 + u^2$ and call it "the operator", as this object clearly depends on the function $u$. The operator should be an object which is a priori independent of the function: it takes a "function" $u$ as input and yields another "function" $g$ as output.
In relation to your actual question. I believe it is better to consider the "operator" $Nu = (partial_t^2 - partial_x^2)u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. It is now clear that the operator $N$ is nonlinear, and the linear and nonlinear parts are moreover explicitly split.
For futher reference, the linear operator $L = partial_t^2 - partial_x^2$ is called the wave operator (also called the d'Alembertian), as it is the governing linear operator in the wave equation
beginequation
partial_t^2 u - partial_t^2 u = 0.
endequation
Hence the equation associated to the "operator" you were considering is the nonlinear wave equation
beginequation
partial_t^2 u - partial_t^2 u + u^3 = 0.
endequation
answered 22 hours ago
bgskbgsk
175111
$endgroup$
add a comment |
$begingroup$
As indicated by the OP, I will regroup my comments as an answer.
The source notes you cite make use of rather ambiguous/confusing notation. Indeed, one cannot simply "factor out" the term $partial_t^2 - partial_x^2 + u^2$ and call it "the operator", as this object clearly depends on the function $u$. The operator should be an object which is a priori independent of the function: it takes a "function" $u$ as input and yields another "function" $g$ as output.
In relation to your actual question. I believe it is better to consider the "operator" $Nu = (partial_t^2 - partial_x^2)u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. It is now clear that the operator $N$ is nonlinear, and the linear and nonlinear parts are moreover explicitly split.
For futher reference, the linear operator $L = partial_t^2 - partial_x^2$ is called the wave operator (also called the d'Alembertian), as it is the governing linear operator in the wave equation
beginequation
partial_t^2 u - partial_t^2 u = 0.
endequation
Hence the equation associated to the "operator" you were considering is the nonlinear wave equation
beginequation
partial_t^2 u - partial_t^2 u + u^3 = 0.
endequation
answered 22 hours ago
bgskbgsk
175111
$endgroup$
add a comment |
$begingroup$
As indicated by the OP, I will regroup my comments as an answer.
The source notes you cite make use of rather ambiguous/confusing notation. Indeed, one cannot simply "factor out" the term $partial_t^2 - partial_x^2 + u^2$ and call it "the operator", as this object clearly depends on the function $u$. The operator should be an object which is a priori independent of the function: it takes a "function" $u$ as input and yields another "function" $g$ as output.
In relation to your actual question. I believe it is better to consider the "operator" $Nu = (partial_t^2 - partial_x^2)u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. It is now clear that the operator $N$ is nonlinear, and the linear and nonlinear parts are moreover explicitly split.
For futher reference, the linear operator $L = partial_t^2 - partial_x^2$ is called the wave operator (also called the d'Alembertian), as it is the governing linear operator in the wave equation
beginequation
partial_t^2 u - partial_t^2 u = 0.
endequation
Hence the equation associated to the "operator" you were considering is the nonlinear wave equation
beginequation
partial_t^2 u - partial_t^2 u + u^3 = 0.
endequation
answered 22 hours ago
bgskbgsk
175111
$endgroup$
As indicated by the OP, I will regroup my comments as an answer.
The source notes you cite make use of rather ambiguous/confusing notation. Indeed, one cannot simply "factor out" the term $partial_t^2 - partial_x^2 + u^2$ and call it "the operator", as this object clearly depends on the function $u$. The operator should be an object which is a priori independent of the function: it takes a "function" $u$ as input and yields another "function" $g$ as output.
In relation to your actual question. I believe it is better to consider the "operator" $Nu = (partial_t^2 - partial_x^2)u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. It is now clear that the operator $N$ is nonlinear, and the linear and nonlinear parts are moreover explicitly split.
For futher reference, the linear operator $L = partial_t^2 - partial_x^2$ is called the wave operator (also called the d'Alembertian), as it is the governing linear operator in the wave equation
beginequation
partial_t^2 u - partial_t^2 u = 0.
endequation
Hence the equation associated to the "operator" you were considering is the nonlinear wave equation
beginequation
partial_t^2 u - partial_t^2 u + u^3 = 0.
endequation
answered 22 hours ago
bgskbgsk
175111
answered 22 hours ago
bgskbgsk
175111
answered 22 hours ago
bgskbgsk
175111
answered 22 hours ago
bgskbgsk
175111
175111
add a comment |
add a comment |
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1
$begingroup$
It's not linear because of the $u^3$ term
$endgroup$
– Dylan
Mar 14 at 7:24
$begingroup$
@Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question.
$endgroup$
– Rhaldryn
Mar 14 at 7:32
1
$begingroup$
The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $partial^2_t - partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$.
$endgroup$
– bgsk
yesterday
1
$begingroup$
Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (partial^2_t - partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x in mathbbR$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments).
$endgroup$
– bgsk
yesterday
1
$begingroup$
@bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books?
$endgroup$
– Rhaldryn
yesterday