Author Topic: Q1-P1  (Read 461 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
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}

Shentao YANG

  • Full Member
  • ***
  • Posts: 24
  • Karma: 0
    • View Profile
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 »

John Menacherry

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
Re: Q1-P1
« Reply #2 on: September 29, 2016, 10:16:46 PM »
Aren't they both linear inhomogeneous?

Jaisen

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
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.

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
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