1
$\begingroup$

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,

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

$\endgroup$
9
  • $\begingroup$ $\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u+v)\dfrac{\partial{(u + v)}}{\partial{y}}$ $\endgroup$
    – Anubhab
    Apr 15, 2019 at 5:57
  • $\begingroup$ You wrote $u$ in the second term which is obviously wrong. $\endgroup$
    – Anubhab
    Apr 15, 2019 at 5:59
  • $\begingroup$ @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 $\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u)\dfrac{\partial{(u + v)}}{\partial{y}}$? $\endgroup$ Apr 15, 2019 at 5:59
  • 2
    $\begingroup$ Your calculation is like showing $f(u) = u^2 $ is linear by first writing $f(u+v) = u(u+v) = \cdots$. $\endgroup$ Apr 15, 2019 at 6:07
  • 1
    $\begingroup$ @ArcticChar Ahh, yes, seems like it confused others too. $\endgroup$ Apr 15, 2019 at 6:13

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Ok, thanks. I'm not too familiar with operators, and the authors never write it explicitly like $\mathscr{L}[u]$, so it becomes confusing. $\endgroup$ Apr 15, 2019 at 6:08
  • $\begingroup$ Take a look at my edit. In light of this new example, I don't think your notational answer is correct. $\endgroup$ Apr 15, 2019 at 8:18
  • $\begingroup$ 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. $\endgroup$
    – Trebor
    Apr 15, 2019 at 8:37
  • $\begingroup$ 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/… $\endgroup$ Apr 15, 2019 at 8:45
  • $\begingroup$ 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? $\endgroup$ Apr 15, 2019 at 9:27
1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ Thank you for the clarification. $\endgroup$ Apr 16, 2019 at 0:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .