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Meaning of quasi-linear PDE (Where is linearity in quasi-linear PDE?)
My book says, "PDE is said to be quasi-linear if it is linear in
the highest-ordered derivative of the unknown function." For example,
$$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.
differential-equations pde
asked Nov 20 '18 at 21:10
ramanujan
714713
add a comment |
My book says, "PDE is said to be quasi-linear if it is linear in
the highest-ordered derivative of the unknown function." For example,
$$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.
differential-equations pde
asked Nov 20 '18 at 21:10
ramanujan
714713
add a comment |
My book says, "PDE is said to be quasi-linear if it is linear in
the highest-ordered derivative of the unknown function." For example,
$$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.
differential-equations pde
asked Nov 20 '18 at 21:10
ramanujan
714713
My book says, "PDE is said to be quasi-linear if it is linear in
the highest-ordered derivative of the unknown function." For example,
$$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.
differential-equations pde
differential-equations pde
asked Nov 20 '18 at 21:10
ramanujan
714713
asked Nov 20 '18 at 21:10
ramanujan
714713
asked Nov 20 '18 at 21:10
ramanujan
714713
asked Nov 20 '18 at 21:10
ramanujan
714713
asked Nov 20 '18 at 21:10
ramanujan
714713
714713
add a comment |
add a comment |
1 Answer
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oldest
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In short, you have
begin{align}
u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
which is linear in the highest derivative.
On the other hand
begin{align}
u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
is not quasi-linear.
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
2
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
2
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
|
show 1 more comment
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
In short, you have
begin{align}
u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
which is linear in the highest derivative.
On the other hand
begin{align}
u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
is not quasi-linear.
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
2
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
2
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
|
show 1 more comment
In short, you have
begin{align}
u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
which is linear in the highest derivative.
On the other hand
begin{align}
u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
is not quasi-linear.
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
2
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
2
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
|
show 1 more comment
In short, you have
begin{align}
u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
which is linear in the highest derivative.
On the other hand
begin{align}
u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
is not quasi-linear.
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
In short, you have
begin{align}
u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
which is linear in the highest derivative.
On the other hand
begin{align}
u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
end{align}
is not quasi-linear.
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:16
Jacky Chong
17.8k21128
17.8k21128
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
2
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
2
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
|
show 1 more comment
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
2
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
2
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
– ramanujan
Nov 20 '18 at 21:20
2
2
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
@ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
– Jacky Chong
Nov 20 '18 at 21:22
2
2
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
– AlexanderJ93
Nov 20 '18 at 21:25
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
@Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
– ramanujan
Nov 20 '18 at 21:29
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
Yes. You are correct.
– Jacky Chong
Nov 20 '18 at 21:30
|
show 1 more comment
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