Why is \partial/\partialx[K(x,u)\partialu/\partialu] Not a Linear Operator?

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Discussion Overview

The discussion revolves around the question of why the operator \(\partial/\partial x[K(x,u)\partial u/\partial x]\) is not considered a linear operator. Participants explore the implications of having \(K\) as a function of both \(x\) and \(u\) and compare it to simpler cases.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant initially presents the operator \(\partial/\partial x[K(x,u)\partial u/\partial u]\) but later corrects it to \(\partial/\partial x[K(x,u)\partial u/\partial x]\).
  • Another participant asserts that \(\partial/\partial x[K(x)\partial u/\partial x]\) is a linear operator, raising the question of how introducing \(K\) as a function of both \(x\) and \(u\) affects linearity.
  • A suggestion is made to test for linearity by substituting a linear combination of functions into the operator and observing the outcome, indicating that conditions on \(K(x,u)\) may influence its linearity.
  • It is noted that if \(K(x,u) = u\), the operator's linearity is questioned, leading to the assertion that the presence of \(K(x,u)\) as a function of the dependent variable \(u\) contributes to its non-linearity.
  • One participant questions the use of the partial derivative with respect to \(u\) if \(u\) is solely a function of \(x\), suggesting there may be additional variables involved.

Areas of Agreement / Disagreement

Participants express differing views on the linearity of the operator, with some asserting it is non-linear due to the functional dependence of \(K\) on \(u\), while others explore specific cases and conditions that might affect this conclusion. The discussion remains unresolved regarding the general case.

Contextual Notes

There are assumptions about the nature of \(K(x,u)\) and its dependence on \(u\) that are not fully explored, as well as the implications of treating \(u\) as a function of \(x\) alone.

Mark_Twain_MO
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Hello,

Could anyone help explain why

[tex]\partial[/tex]/[tex]\partial[/tex]x[K(x,u)[tex]\partial[/tex]u/[tex]\partial[/tex]u]

is not a linear operator?
 
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Are you sure you have that written down right?

[tex]\frac{\partial u}{\partial u} = 1[/tex]
 
Sorry, I meant:

[tex]\partial[/tex]/[tex]\partial[/tex]x[K(x,u)[tex]\partial[/tex]u/[tex]\partial[/tex]x]

I understand that

[tex]\partial[/tex]/[tex]\partial[/tex]x[K(x)[tex]\partial[/tex]u/[tex]\partial[/tex]x]

is a linear operator, but I do not get why making K a function of both x and u should non-linearize this operator.
 
The best way to test quickly for a linear operator is just to plug in the definition... that is put in

[tex]\lambda _1 u_1 + \lambda _2 u_2[/tex]

to your operator above. Use the rules of differentiation and see what comes out. This would be a good exercise.

It seems there might have to be some conditions on K(x, u) for the above to be non-linear. I'm out the door myself, so hope this helps.
 
If K(x,u) = u, then what is your operator? Is it linear?
 
It's not linear because you have a function of the dependent variable, u, K(x,u), multiplying the derivative of u, [itex]\partial u/\partial x[/itex].

(But why are you using the partial derivative if u is a function only of x? Is there another term in the equation with another independent variable?)
 

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