Just plug u + v into the pde and use the linearity of everything. Do the same for ku.
Dear Folks,
Can anyone get me started on how to approach the following question:
Consider the following 1st order linear PDE
If u and v are 2 solutions shown that a) u + v and b) ku for any constant k are also solutions?
Just a tip would be good, I want to tackle it myself.
Thanks
No need to put them on different sides of the equation. Just group them differently. You can separate out all the u terms, and shove them to the left. Group the v terms, and shove them to the right, but all terms are still on the LHS of the equation. Write this out. Then what can you say?
The other proof works in a very similar vein. Yeah, it can sometimes be tough knowing when something simple is perfectly adequate, as opposed to the latest complicated proof technique. Incidentally, can you see how this proof would fail if, say, you had u^2 multiplying the c(x,y)?
Well, because of the unknown function u is squared, the 1st order PDE becomes nonlinear. But not sure how to prove it mathematically...
I could multiply out the terms and get something like the 'partial terms' + c(x,y)u^2+c(x,y)v^2=0
I dont see how it would fail....
Also, I have attempted the ku problem...see attached jpg (part 2).
Thanks
No, no. Your solution for ku is not correct. You DON'T know that all those partial derivatives are zero. In fact, they are probably nonzero. However, you do know that k, a constant, behaves in a certain fashion under derivative signs, right? I mean, how could you simplify the expression
Your right, they cant be 0, poor judgement on my part. I guess you can pull out the k's and divide across by k. And you could do this for any constant k. See attached... Hope im right!
I also replied to your query about what if u was squared?... Thanks
Attachment 19464