Linear operator or nonlinear operator?

[Elbobo]

Homework Statement

Verify whether or not the operator

L(u) = u_x + u_y + 1
is linear.

Homework Equations

An operator L is linear if for any functions u, v and any constants c, the property

L(c_1 u + c_2 v) = c_1 L(u) + c_2 L(v)
holds true.

The Attempt at a Solution

I feel as though this should be a linear operator, but the "+1" throws me off as I don't know what linear operator takes any function u, v into 1.

L = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + ?

 

[STEMucator]

Homework Statement

Verify whether or not the operator

L(u) = u_x + u_y + 1
is linear.

Homework Equations

An operator L is linear if for any functions u, v and any constants c, the property

L(c_1 u + c_2 v) = c_1 L(u) + c_2 L(v)
holds true.

The Attempt at a Solution

I feel as though this should be a linear operator, but the "+1" throws me off as I don't know what linear operator takes any function u, v into 1.

Alright so, really there's two things you want to verify, but I suppose you could combine them into one condition like that if you want.

What is L(c_1 u + c_2 v) = ?

 

[Elbobo]

L = c_1 \frac{ \partial u}{\partial x} + c_1 \frac{\partial u}{\partial y} + c_2 \frac{ \partial v}{\partial x} + c_2 \frac{\partial v}{\partial y} + ?

I'm still confused by that constant term.

 

[STEMucator]

L = c_1 \frac{ \partial u}{\partial x} + c_1 \frac{\partial u}{\partial y} + c_2 \frac{ \partial u}{\partial x} + c_2 \frac{\partial u}{\partial y} + ?

I'm still confused by that constant term.

Just literally write out what the transform would give you, so :

L(c_1 u + c_2 v) = c_1u_x + c_1u_y + c_1 + c_2v_x + c_2v_y + c_2

Can you continue from there? Get it into the form c_1 L(u) + c_2 L(v)

 

[Elbobo]

Sorry for the typos everywhere earlier (edited now).

L(c_1 u + c_2 v) = c_1 u_x + c_1 u_y + c_1 + c_2 v_x + c_2 v_y + c_2
= c_1 (u_x + u_y + 1) + c_2 (v_x + v_y + 1)
= c_1 L(u) + c_2 L(v)

So is that what you were guiding me to? If I did that correctly, it makes a lot more sense now, thank you. If not...

 

[Dick]

Sorry for the typos everywhere earlier (edited now).

L(c_1 u + c_2 v) = c_1 u_x + c_1 u_y + c_1 + c_2 v_x + c_2 v_y + c_2
= c_1 (u_x + u_y + 1) + c_2 (v_x + v_y + 1)
= c_1 L(u) + c_2 L(v)

So is that what you were guiding me to? If I did that correctly, it makes a lot more sense now, thank you. If not...

No, no, no. L(c_1 u + c_2 v) = (c_1 u + c_2 v)_x + (c_1 u + c_2 v)_y + 1