Proving linear operator for partial derivatives.

In summary: 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?
  • #1
NINHARDCOREFAN
118
0
How do I go about proving the following partial derivitive is a linear operator?
d/dx[k(x)du/dx)]
 
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  • #2
What you've written is not a differential operator, it's a function.

Or did you mean d/dx[k(x)d/dx)]?
 
  • #3
Nope, what I originally wrote is how it is in the book. If it matters, K is with a subscript of o.
 
  • #4
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.
 
  • #5
But where is the partial derivative ?

Daniel.
 
  • #6
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, [tex]\frac{\partial }{\partial x}[/tex] is the operator, [tex]\frac{\partial u}{\partial x}[/tex] 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 [tex]\frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial}{\partial x}\right)[/tex] is a linear operator
JUST DO IT!

What is [tex]\frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial (au+ bv}{\partial x}\right)[/tex]?
 
  • #7
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.

HallsofIvy said:
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, [tex]\frac{\partial }{\partial x}[/tex] is the operator, [tex]\frac{\partial u}{\partial x}[/tex] 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 [tex]\frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial}{\partial x}\right)[/tex] is a linear operator
JUST DO IT!

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

Related to Proving linear operator for partial derivatives.

What is a linear operator for partial derivatives?

A linear operator for partial derivatives is a mathematical function that takes in a multi-variable function as input and produces another function as output. It is used to calculate how a function changes with respect to different variables.

Why is it important to prove a linear operator for partial derivatives?

Proving a linear operator for partial derivatives is important because it allows us to understand the behavior of a function and how it changes with respect to its input variables. This information is useful in many fields, including physics, engineering, and economics.

What are the steps involved in proving a linear operator for partial derivatives?

The steps involved in proving a linear operator for partial derivatives include defining the operator, applying it to a function, using the product rule and chain rule to simplify the resulting expression, and finally showing that the resulting expression satisfies the properties of a linear operator.

What are the properties of a linear operator for partial derivatives?

The properties of a linear operator for partial derivatives include linearity, meaning that it follows the rules of addition and multiplication by a constant; product rule, meaning that it can be applied to the product of two functions; and chain rule, meaning that it can be applied to composite functions.

How is a linear operator for partial derivatives used in real-world applications?

A linear operator for partial derivatives is used in many real-world applications, including optimization problems, modeling physical systems, and predicting the behavior of economic systems. It allows us to analyze the sensitivity of a function to changes in its input variables and make predictions about its behavior.

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