MHB -b.1.3.21-24 Order and linear or nonlinear

[karush]

5000
In each of Problems 21 through 24,determine the order of the given partial differential equation;also state whether the equation is linear or nonlinear. Partial derivatives are denoted by subscripts.
21. $u_{xx} + u_{yy} + u_{zz}= 0$
23. $u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0$
22. $u_{xx} + u_{yy} + uu_{x} + uu_{y} + u = 0$
24. $ut + uu_{x} = 1 + u_{xx} $

Answers
21. 2nd, linear 22. 2nd, nonlinear 23. 4th,linear 24. 2nd, nonlinearThis isn't HW, just trying to go thru the book on my own
Apparently the subscripts give the answer but ?

 

[Ackbach]

It's not entirely clear what your question is. The subscripts tell you what you're taking the partial derivative w.r.t. For it to be linear, then you can't have any products or functions of the dependent variable or any of its derivatives.

 

[karush]

https://www.dropbox.com/s/y6ashh8hg... Boundary Value Problems [10th].1-52.pdf?dl=0

OK here is a link to chp 1 of the book it's exercise 1.3
https://www.physicsforums.com/attachments/5082

 

[MarkFL]

As Adrian stated you cannot have any products in a linear equation, so look at 22.) and 24.)...notice both have the product $uu_x$. So they are non-linear whereas the others are linear. :)

 

[karush]

OK, wonder why the goofy subscribes or is that just an destraction

Btw, having a good latex writer on MHB is really a plus.

I quit going to other math sites because the latex is such a pain

 

[MarkFL]

OK, wonder why the goofy subscribes or is that just an destraction

Btw, having a good latex writer on MHB is really a plus.

I quit going to other math sites because the latex is such a pain

The subscript notation tells you with respect to which variable(s) the partial is taken. For example:

$$\pd{u}{x}$$ is equivalent to $$u_x$$

$$\pd{u^2}{^2x}$$ is equivalent to $$u_{xx}$$

$$\frac{\partial u^2}{\partial x\partial y}$$ is equivalent to $$u_{xy}$$

And yes, we are very pleased with our $\LaTeX$ tools here at MHB. (Yes)

 

[karush]

I try to continue to finish this page but will start a new thread to do so.