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Is the Laplace equation quasi linear?
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Assume we have the Laplace equation $u_x_1 x_1+...+u_x_n x_n=0$. Then I would say that this equation is quasi-linear since the highest order terms are linear. But I cannot confirm my answer if I search on the internet. So I am probably wrong. Can someone explain to me why the equation is not quasi-linear?
differential-equations pde laplace-transform
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
asked Aug 3 at 17:01
esmo
255
add a comment |Â
up vote
0
down vote
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Assume we have the Laplace equation $u_x_1 x_1+...+u_x_n x_n=0$. Then I would say that this equation is quasi-linear since the highest order terms are linear. But I cannot confirm my answer if I search on the internet. So I am probably wrong. Can someone explain to me why the equation is not quasi-linear?
differential-equations pde laplace-transform
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
asked Aug 3 at 17:01
esmo
255
1
I put "$" around your equation to make it render properly. When using $LaTeX$, remember to surround your math with "$" signs or it won't work. Cheers!
– Robert Lewis
Aug 3 at 17:03
2
I'd just call it linear.
– Lord Shark the Unknown
Aug 3 at 17:03
But can you also argue why it is not quasi-linear?
– esmo
Aug 3 at 17:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume we have the Laplace equation $u_x_1 x_1+...+u_x_n x_n=0$. Then I would say that this equation is quasi-linear since the highest order terms are linear. But I cannot confirm my answer if I search on the internet. So I am probably wrong. Can someone explain to me why the equation is not quasi-linear?
differential-equations pde laplace-transform
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
asked Aug 3 at 17:01
esmo
255
Assume we have the Laplace equation $u_x_1 x_1+...+u_x_n x_n=0$. Then I would say that this equation is quasi-linear since the highest order terms are linear. But I cannot confirm my answer if I search on the internet. So I am probably wrong. Can someone explain to me why the equation is not quasi-linear?
differential-equations pde laplace-transform
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
asked Aug 3 at 17:01
esmo
255
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
edited Aug 3 at 17:03
Lord Shark the Unknown
84k949111
84k949111
asked Aug 3 at 17:01
esmo
255
asked Aug 3 at 17:01
esmo
255
asked Aug 3 at 17:01
esmo
255
255
1
I put "$" around your equation to make it render properly. When using $LaTeX$, remember to surround your math with "$" signs or it won't work. Cheers!
– Robert Lewis
Aug 3 at 17:03
2
I'd just call it linear.
– Lord Shark the Unknown
Aug 3 at 17:03
But can you also argue why it is not quasi-linear?
– esmo
Aug 3 at 17:07
add a comment |Â
1
I put "$" around your equation to make it render properly. When using $LaTeX$, remember to surround your math with "$" signs or it won't work. Cheers!
– Robert Lewis
Aug 3 at 17:03
2
I'd just call it linear.
– Lord Shark the Unknown
Aug 3 at 17:03
But can you also argue why it is not quasi-linear?
– esmo
Aug 3 at 17:07
1
1
I put "$" around your equation to make it render properly. When using $LaTeX$, remember to surround your math with "$" signs or it won't work. Cheers!
– Robert Lewis
Aug 3 at 17:03
I put "$" around your equation to make it render properly. When using $LaTeX$, remember to surround your math with "$" signs or it won't work. Cheers!
– Robert Lewis
Aug 3 at 17:03
2
2
I'd just call it linear.
– Lord Shark the Unknown
Aug 3 at 17:03
I'd just call it linear.
– Lord Shark the Unknown
Aug 3 at 17:03
But can you also argue why it is not quasi-linear?
– esmo
Aug 3 at 17:07
But can you also argue why it is not quasi-linear?
– esmo
Aug 3 at 17:07
add a comment |Â
1 Answer
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Usually the terminology comes from operator theory. i.e. If you have an operator $L$, we say it's linear if $x,y$ and s (functions wherever they are defined and a scalar) satisfy
$$ L(x+y) = L(x) + L(y) quad & quad L(sx) = sL(x)$$
Quasi-linear in the context of differential equations means that your differential operator can be represented as
$$ Q(x+y) = L(x+y) + f = L(x) + L(y) + f = Q(x) + Q(y) - f$$
where $f$ is independent of $x,y$. i.e. a linear and with some extra term as a cost. Notice every linear operator is quasi-linear with $f= 0$. If $f$ depends on $x,y$, we say the operator is non-linear.
In your case you have
$$Delta (x+y) = Delta(x) + Delta (y) quad & quad Delta(sx) = sDelta(x)$$
So we'd call it linear.
answered Aug 3 at 19:12
Jeb
3,8541612
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Usually the terminology comes from operator theory. i.e. If you have an operator $L$, we say it's linear if $x,y$ and s (functions wherever they are defined and a scalar) satisfy
$$ L(x+y) = L(x) + L(y) quad & quad L(sx) = sL(x)$$
Quasi-linear in the context of differential equations means that your differential operator can be represented as
$$ Q(x+y) = L(x+y) + f = L(x) + L(y) + f = Q(x) + Q(y) - f$$
where $f$ is independent of $x,y$. i.e. a linear and with some extra term as a cost. Notice every linear operator is quasi-linear with $f= 0$. If $f$ depends on $x,y$, we say the operator is non-linear.
In your case you have
$$Delta (x+y) = Delta(x) + Delta (y) quad & quad Delta(sx) = sDelta(x)$$
So we'd call it linear.
answered Aug 3 at 19:12
Jeb
3,8541612
add a comment |Â
up vote
0
down vote
Usually the terminology comes from operator theory. i.e. If you have an operator $L$, we say it's linear if $x,y$ and s (functions wherever they are defined and a scalar) satisfy
$$ L(x+y) = L(x) + L(y) quad & quad L(sx) = sL(x)$$
Quasi-linear in the context of differential equations means that your differential operator can be represented as
$$ Q(x+y) = L(x+y) + f = L(x) + L(y) + f = Q(x) + Q(y) - f$$
where $f$ is independent of $x,y$. i.e. a linear and with some extra term as a cost. Notice every linear operator is quasi-linear with $f= 0$. If $f$ depends on $x,y$, we say the operator is non-linear.
In your case you have
$$Delta (x+y) = Delta(x) + Delta (y) quad & quad Delta(sx) = sDelta(x)$$
So we'd call it linear.
answered Aug 3 at 19:12
Jeb
3,8541612
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Usually the terminology comes from operator theory. i.e. If you have an operator $L$, we say it's linear if $x,y$ and s (functions wherever they are defined and a scalar) satisfy
$$ L(x+y) = L(x) + L(y) quad & quad L(sx) = sL(x)$$
Quasi-linear in the context of differential equations means that your differential operator can be represented as
$$ Q(x+y) = L(x+y) + f = L(x) + L(y) + f = Q(x) + Q(y) - f$$
where $f$ is independent of $x,y$. i.e. a linear and with some extra term as a cost. Notice every linear operator is quasi-linear with $f= 0$. If $f$ depends on $x,y$, we say the operator is non-linear.
In your case you have
$$Delta (x+y) = Delta(x) + Delta (y) quad & quad Delta(sx) = sDelta(x)$$
So we'd call it linear.
answered Aug 3 at 19:12
Jeb
3,8541612
Usually the terminology comes from operator theory. i.e. If you have an operator $L$, we say it's linear if $x,y$ and s (functions wherever they are defined and a scalar) satisfy
$$ L(x+y) = L(x) + L(y) quad & quad L(sx) = sL(x)$$
Quasi-linear in the context of differential equations means that your differential operator can be represented as
$$ Q(x+y) = L(x+y) + f = L(x) + L(y) + f = Q(x) + Q(y) - f$$
where $f$ is independent of $x,y$. i.e. a linear and with some extra term as a cost. Notice every linear operator is quasi-linear with $f= 0$. If $f$ depends on $x,y$, we say the operator is non-linear.
In your case you have
$$Delta (x+y) = Delta(x) + Delta (y) quad & quad Delta(sx) = sDelta(x)$$
So we'd call it linear.
answered Aug 3 at 19:12
Jeb
3,8541612
answered Aug 3 at 19:12
Jeb
3,8541612
answered Aug 3 at 19:12
Jeb
3,8541612
answered Aug 3 at 19:12
Jeb
3,8541612
3,8541612
add a comment |Â
add a comment |Â
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1
I put "$" around your equation to make it render properly. When using $LaTeX$, remember to surround your math with "$" signs or it won't work. Cheers!
– Robert Lewis
Aug 3 at 17:03
2
I'd just call it linear.
– Lord Shark the Unknown
Aug 3 at 17:03
But can you also argue why it is not quasi-linear?
– esmo
Aug 3 at 17:07