Partial differential equations represented as operators

[kingwinner]

Partial differential equations represented as "operators"

Homework Statement

Partial differential equations (PDEs) can be represented in the form Lu=f(x,y) where L is an operator.
Example:
Input: u(x,y)
Operator: L=∂_xy + cos(x) + (∂_y)²
=> Output: Lu = u_xy+cos(x) u + (u_y)²

Homework Equations

N/A

The Attempt at a Solution

I don't understand the parts in red.
What does (∂_y)² mean? Shouldn't it mean (∂_y)(∂_y) which (I believe) means taking the second partial derivative with respect to y. But the above example seems to imply that (∂_y)² means take the partial derivative with respect to y and then square the result. But I just can't possibly understand how (∂_y)²u would be equal to (∂_yu)²


I don't have a lot of experience working with operators, so I am feeling really confused. Could someone please kindly explain?
Thank you very much!

 

[Dick]

I think you are right. It should be u_yy, the second derivative. Not (u_y)^2. It's a linear operator. Is it a typo?

 

[kingwinner]

I think you are right. It should be u_yy, the second derivative. Not (u_y)^2. It's a linear operator. Is it a typo?

But if that is the case, why don't they write ∂_yy (instead of (∂_y)²? (I am not sure whether it's a typo, but it seems to me that the context of it highly suggests it isn't)

Also, if they really meant (u_y)², how could it be expressed as an "operator"?

Thanks!

 

[Dick]

This is getting a little overly semantic. del_yy=(del_y)^2. If they mean O(u)=(u_y)^2, that is an operator. But it's not a linear operator. And I wouldn't write it as del_yy or (del_y)^2.

 

[kingwinner]

For example, if we have the PDE:
u_xy+cos(x) u + (u_y)² = x

How can we express it in "operator" form Lu=f(x,y)?

OK, in this case, f(x,y)=x, but what would L be?
L=∂_xy + cos(x) + ?[/color]

 

[Dick]

For example, if we have the PDE:
u_xy+cos(x) u + (u_y)² = x

How can we express it in "operator" form Lu=f(x,y)?

OK, in this case, f(x,y)=x, but what would L be?
L=∂_xy + cos(x) + ?[/color]

I told you, that's not a linear operator. Why do you think you should be able to express it using the notation of linear operators?

 

[kingwinner]

I think I may be missing something...so whenever we write L= (...) + (...) + (...), is L necessarily always a LINEAR operator? Can non-linear operators be expressed this way?


Also, there is something in general that I don't understand about operators...(I'm just guessing that I can do things like the following)

If L=∂_xy + cos(x) + (∂_y)², to find Lu, should we just imagine multiplying L with u (just like multiplying real numbers?), so
Lu = [∂_xy + cos(x) + (∂_y)²] u
=∂_xy u + cos(x) u + (∂_y)² u (by distributive law of multiplication) ?

 

[Dick]

If L is just a sum of differential operators and functions as you are writing then every term in Lu will only have one u in it. So you are going to have a hard time getting (u_y)^2. And yes, you just distribute the operator, so in your example you get Lu=u_xy+cos(x)u+u_yy. Or u_yx+cos(x)u+u_yy depending on how you define ∂xy.

 

[kingwinner]

If L is just a sum of differential operators and functions as you are writing then every term in Lu will only have one u in it. So you are going to have a hard time getting (u_y)^2. And yes, you just distribute the operator, so in your example you get Lu=u_xy+cos(x)u+u_yy. Or u_yx+cos(x)u+u_yy depending on how you define ∂xy.

Can we distribute it that way for any operator in general, or just for linear operators?

 

[Dick]

Can we distribute it that way for any operator in general, or just for linear operators?

If you write L=(L1+L2+L3) then, yes, Lu=L1u+L2u+L3u. That's what addition of operators means, any operators.

 

[kingwinner]

If you write L=(L1+L2+L3) then, yes, Lu=L1u+L2u+L3u. That's what addition of operators means, any operators.

Are "operators" the same as "functions"?

 

[Dick]

Are "operators" the same as "functions"?

What do YOU think? Can't you look up the definitions?