### Author Topic: Classification of PDEs  (Read 541 times)

#### Weihan Luo

• Jr. Member
•  • Posts: 5
• Karma: 0 ##### Classification of PDEs
« on: January 14, 2022, 12:38:03 AM »
I am a little bit confused about the classifications of PDES. Namely, I have trouble distinguishing between linear equations versus quasi-linear equations.

In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?

#### Victor Ivrii ##### Re: Classification of PDEs
« Reply #1 on: January 14, 2022, 02:45:57 AM »
In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?
First, it will be not just quasilinear, but also  semilinear. Second, it will also be linear since you can move $c(x,y)u$ to the left

Good job, you mastered some $\LaTeX$ basics. « Last Edit: January 14, 2022, 06:40:57 AM by Victor Ivrii »

#### Weihan Luo

• Jr. Member
•  • Posts: 5
• Karma: 0 ##### Re: Classification of PDEs
« Reply #2 on: January 14, 2022, 11:28:53 AM » 