# PDE Linearity Textbook Solution: $\mathscr{L} = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}$

I derived the operator $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}$$ from the PDE $$u_x + uu_y = 0$$ in order to figure out whether it is linear.

The textbook solutions take the following steps in finding whether the operator is linear:

$$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u + v)\dfrac{\partial{(u + v)}}{\partial{y}} = \dots = \mathscr{L}u + \mathscr{L}v + uv_y + vu_y,$$

which is obviously nonlinear. But I proceeded as follows:

$$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u)\dfrac{\partial{(u + v)}}{\partial{y}} = \dots = \mathscr{L}u + \mathscr{L}v,$$

which obviously is linear.

I don't understand why the author let $$u$$ in the operator become $$u = u + v$$? It seemed to me that, since $$u$$ is part of the operator, it should remain as $$u$$ and not $$u + v$$?

I would greatly appreciate it if people could please take the time to clarify this. Why is it wrong? My point is that $$u$$ is part of the operator, and the operator is acting upon $$(u + v)$$, so why is it that $$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u + v)\dfrac{\partial{(u + v)}}{\partial{y}}$$ instead of $$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u)\dfrac{\partial{(u + v)}}{\partial{y}}$$?

EDIT: For future reference, I would also like to present another interesting example.

Take the operator $$\mathscr{L} u = u_x + u_y + 1$$. Therefore, $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + \dfrac{1}{u}$$. And so,

$$\mathscr{L} (u + v) = \left( \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + \dfrac{1}{u + v} \right) (u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + \dfrac{\partial{(u + v)}}{\partial{y}} + 1$$

EDIT2:

I think the derivation of $$\mathscr{L}$$ in the above edit is incorrect. Here is what I think the correct derivation is:

Take the operator $$\mathscr{L} u = u_x + u_y + 1$$. Therefore, $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + 1$$. And so,

$$\mathscr{L} (u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + \dfrac{\partial{(u + v)}}{\partial{y}} + 1$$

• $\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u+v)\dfrac{\partial{(u + v)}}{\partial{y}}$ Apr 15, 2019 at 5:57
• You wrote $u$ in the second term which is obviously wrong. Apr 15, 2019 at 5:59
• @AnubhabGhosal Yes, but why is it wrong? My point is that $u$ is part of the operator, and the operator is acting upon $(u + v)$, so why is it that $\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u + v)\dfrac{\partial{(u + v)}}{\partial{y}}$ instead of $\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u)\dfrac{\partial{(u + v)}}{\partial{y}}$? Apr 15, 2019 at 5:59
• Your calculation is like showing $f(u) = u^2$ is linear by first writing $f(u+v) = u(u+v) = \cdots$. Apr 15, 2019 at 6:07
• @ArcticChar Ahh, yes, seems like it confused others too. Apr 15, 2019 at 6:13

Obviously the notation $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}$$ meant $$\mathscr{L}[u] = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}.$$ So the $$u$$ should be replaced too.

• Ok, thanks. I'm not too familiar with operators, and the authors never write it explicitly like $\mathscr{L}[u]$, so it becomes confusing. Apr 15, 2019 at 6:08
• Take a look at my edit. In light of this new example, I don't think your notational answer is correct. Apr 15, 2019 at 8:18
• By making that deriviation, you are assuming $\mathscr L$ is linear. You can't have $$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + \dfrac{1}{u}$$ by division. Apr 15, 2019 at 8:37
• Hmm, what do you mean we are assuming it is linear? The author of the textbook asks whether the operator $\mathscr{L} u = u_x + u_y + 1$ is linear, so I don't see how there is such an assumption? See the bottom of page 3 for the solution: stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%201/… Apr 15, 2019 at 8:45
• Furthermore, if we follow through with the calculations for $\mathscr{L} (u + v) = \left( \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + \dfrac{1}{u + v} \right) (u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + \dfrac{\partial{(u + v)}}{\partial{y}} + 1$, we find that the operator is not linear, so I don't see how we are assuming linearity here? Apr 15, 2019 at 9:27

Let me just say that

$$\mathcal L = \frac{\partial}{\partial x} + u\frac{\partial }{\partial y}$$

is NOT the correct notation. The above implies that $$u$$ is a fixed function, and the operator acts as

$$\mathcal L (f) = \frac{\partial f}{\partial x} + u \frac{\partial f}{\partial y}$$

for all differentiable function $$f$$.

To avoid confusion, your operator should be written as

$$\mathcal L' (f) = \frac{\partial f}{\partial x} + f \frac{\partial f}{\partial y}$$

and now it is clear why we have

$$\mathcal L' (f_1+ f_2) = \frac{\partial}{\partial x} (f_1+f_2) + (f_1+f_2) \frac{\partial }{\partial y} (f_1+f_2).$$

• Thank you for the clarification. Apr 16, 2019 at 0:49