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

[Mark_Twain_MO]

Hello,

Could anyone help explain why

\partial/\partialx[K(x,u)\partialu/\partialu]

is not a linear operator?

 

[Coto]

Are you sure you have that written down right?

\frac{\partial u}{\partial u} = 1

 

[Mark_Twain_MO]

Sorry, I meant:

\partial/\partialx[K(x,u)\partialu/\partialx]

I understand that

\partial/\partialx[K(x)\partialu/\partialx]

is a linear operator, but I do not get why making K a function of both x and u should non-linearize this operator.

 

[Coto]

The best way to test quickly for a linear operator is just to plug in the definition... that is put in

\lambda _1 u_1 + \lambda _2 u_2

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.

 

[g_edgar]

If K(x,u) = u, then what is your operator? Is it linear?

 

[HallsofIvy]

It's not linear because you have a function of the dependent variable, u, K(x,u), multiplying the derivative of u, \partial u/\partial x.

(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?)