Partial differential equations represented as operators

In summary: I'm just not sure if I'm understanding what you are trying to say.Thanks for trying to help!I'm not sure either! I'm not an expert on this either! I'm just trying to explain what I think!Thanks for trying to help!
  • #1
kingwinner
1,270
0
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 = uxy+cos(x) u + (uy)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!
 
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  • #2


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?
 
  • #3


Dick said:
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 (uy)2, how could it be expressed as an "operator"?

Thanks!
 
  • #4


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.
 
  • #5


For example, if we have the PDE:
uxy+cos(x) u + (uy)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) + ?
 
  • #6


kingwinner said:
For example, if we have the PDE:
uxy+cos(x) u + (uy)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) + ?

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?
 
  • #7


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) ?
 
  • #8


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.
 
  • #9


Dick said:
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?
 
  • #10


kingwinner said:
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.
 
  • #11


Dick said:
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"?
 
  • #12


kingwinner said:
Are "operators" the same as "functions"?

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

1. What is a partial differential equation (PDE)?

A PDE is a mathematical equation that involves multiple independent variables and their partial derivatives. It describes the relationship between a function and its partial derivatives with respect to the independent variables.

2. How are partial differential equations represented as operators?

In order to solve a PDE, it is often rewritten as a linear operator acting on the dependent variable. This allows for the use of mathematical methods such as eigenfunction expansions and Fourier transforms.

3. What is the purpose of representing PDEs as operators?

By representing PDEs as operators, we can use techniques from linear algebra to solve them. This allows for more efficient and accurate solutions compared to traditional methods.

4. What are some common applications of PDEs represented as operators?

PDEs represented as operators are used in various fields, such as physics, engineering, and economics, to model and understand complex systems. They are particularly useful in problems involving continuous systems, such as heat transfer, fluid dynamics, and quantum mechanics.

5. Are there different types of operators used to represent PDEs?

Yes, there are various types of operators used for different types of PDEs. Some common examples include differential operators, integral operators, and Fourier operators. Each type of operator has its own properties and is used for specific types of PDEs.

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