Toronto Math Forum
Welcome,
Guest
. Please
login
or
register
.
1 Hour
1 Day
1 Week
1 Month
Forever
Login with username, password and session length
News:
Home
Help
Search
Calendar
Login
Register
Toronto Math Forum
»
APM346-2016F
»
APM346--Tests
»
Q1
»
Q1-P1
« previous
next »
Print
Pages: [
1
]
Author
Topic: Q1-P1 (Read 4941 times)
Victor Ivrii
Administrator
Elder Member
Posts: 2607
Karma: 0
Q1-P1
«
on:
September 29, 2016, 09:29:03 PM »
Consider first order equations and determine if they are linear homogeneous, linear inhomogeneous, quasilinear or non-linear ($u$ is an unknown function):
\begin{align}
&u_t+xu_x-u= 0,\label{eq-1}\\[5pt]
&u_x^2+u_y^2-1= 0. \label{eq-2}
\end{align}
Logged
Shentao YANG
Full Member
Posts: 24
Karma: 0
Re: Q1-P1
«
Reply #1 on:
September 29, 2016, 09:43:40 PM »
Below is my solution:
$$u_t+xu_x-u= 0\text{ : linear homogeneous}$$
$$u_x^2+u_y^2-1= 0\text{ : nonlinear}$$
«
Last Edit: September 29, 2016, 09:50:40 PM by Shentao YANG
»
Logged
John Menacherry
Newbie
Posts: 1
Karma: 0
Re: Q1-P1
«
Reply #2 on:
September 29, 2016, 10:16:46 PM »
Aren't they both linear inhomogeneous?
Logged
Jaisen
Newbie
Posts: 1
Karma: 0
Re: Q1-P1
«
Reply #3 on:
September 30, 2016, 10:03:37 AM »
John, I think because of the minuses you are right they are both inhomogeneous. But (1) is Semi linear since F=u but for Linear F has to be a function of (x,y). As for (2) it is fully non-linear (not quasi-linear) because of the squares.
Logged
Victor Ivrii
Administrator
Elder Member
Posts: 2607
Karma: 0
Re: Q1-P1
«
Reply #4 on:
September 30, 2016, 10:12:46 AM »
Shentao YANG
is correct, it is linear homogeneous since $f(x,t,u)= c(x,y)u$.
The second equation is non-linear
Logged
Print
Pages: [
1
]
« previous
next »
Toronto Math Forum
»
APM346-2016F
»
APM346--Tests
»
Q1
»
Q1-P1