Proving linear operator for partial derivatives.

[NINHARDCOREFAN]

How do I go about proving the following partial derivitive is a linear operator?
d/dx[k(x)du/dx)]

 

[quasar987]

What you've written is not a differential operator, it's a function.

Or did you mean d/dx[k(x)d/dx)]?

 

[NINHARDCOREFAN]

Nope, what I originally wrote is how it is in the book. If it matters, K is with a subscript of o.

 

[matt grime]

So you're mutliplying by something, k(x)du/dx, and then you're differentiating. I hope you know that those are linear operators individually.

 

[dextercioby]

But where is the partial derivative ?

Daniel.

 

[HallsofIvy]

Don't get naughty, Daniel! He clearly intended the "d" as partial derivative, he just didn't know how to write here.

NINHARDCOREFAN, what quasar987 said is still valid. Whatever is written in your book, \frac{\partial }{\partial x} is the operator, \frac{\partial u}{\partial x} is the function that results from applying that operator to u.

Of course, before you can prove anything is a linear operator, you have to know the definition of "linear operator"! An operator, F, is "linear" if and only if it takes linear combinations into linear combinations: F(au+ bv)= a F(u)+ bF(v) where a and b are numbers (constants) and u and v are objects in the vector space (in this case functions).

So, to prove that \frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial}{\partial x}\right) is a linear operator
JUST DO IT!

What is \frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial (au+ bv}{\partial x}\right)?

 

[jasmine_b98]

Please explain this proof futher. Let d be the partial derivative of d.
So the equation is actually L(u)=d/dx[Ko(x)du/dx]. There is the du/dx at the end of the equation. Please explain further how this is interpreted: this means take the partial derivative of any function, multiply it by Ko(x) du/dx? This reads that Ko is a function of x correct?

Please outline the stept to solve this.

Don't get naughty, Daniel! He clearly intended the "d" as partial derivative, he just didn't know how to write here.

NINHARDCOREFAN, what quasar987 said is still valid. Whatever is written in your book, \frac{\partial }{\partial x} is the operator, \frac{\partial u}{\partial x} is the function that results from applying that operator to u.

Of course, before you can prove anything is a linear operator, you have to know the definition of "linear operator"! An operator, F, is "linear" if and only if it takes linear combinations into linear combinations: F(au+ bv)= a F(u)+ bF(v) where a and b are numbers (constants) and u and v are objects in the vector space (in this case functions).

So, to prove that \frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial}{\partial x}\right) is a linear operator
JUST DO IT!

What is \frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial (au+ bv}{\partial x}\right)?