# Partial differential equations represented as operators

 P: 1,270 1. The problem statement, all variables and given/known data 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 2. Relevant equations N/A 3. 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!
 Sci Advisor HW Helper Thanks P: 25,251 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?
P: 1,270
 Quote by 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?
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!

 Sci Advisor HW Helper Thanks P: 25,251 Partial differential equations represented as operators 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.
 P: 1,270 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) + ?????
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P: 25,251
 Quote by kingwinner 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?
 P: 1,270 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) ?
 Sci Advisor HW Helper Thanks P: 25,251 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.
P: 1,270
 Quote by 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.
Can we distribute it that way for any operator in general, or just for linear operators?
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P: 25,251
 Quote by kingwinner 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.
P: 1,270
 Quote by Dick 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"?