Linear Partial Differential Equations Explained

Click For Summary
A linear partial differential equation (PDE) involves a function u and its partial derivatives appearing linearly, potentially with coefficients that are functions of independent variables. The equation can be classified as homogeneous if it equals zero, or nonhomogeneous if it includes a function f dependent on the independent variables. An example provided demonstrates that the PDE u_x + u_yy = exp(x^2) is linear, while (u_x)^2 = 0 is not linear due to the squared term. The operator L is defined to facilitate understanding linearity, where L(u + v) = Lu + Lv and L(cu) = c*Lu. The discussion emphasizes the importance of recognizing these properties in studying PDEs, particularly in preparation for further study.
Somefantastik
Messages
226
Reaction score
0
"A PDE satisfied by a suitably differentiable function u of the independent variables x1, ... ,xn, is said to be linear if u and its partial derivatives only occur linearly and, posibly, with coefficients that are functions of independent variables. This PDE may, or may not, contain a function f that depends only on the independent variables. The PDE is said to be homogeneous when the function f(x1,...,xn) = 0; otherwise it is nonhomogeneous."

Is there a pictorial way to expound on this for me?
 
Physics news on Phys.org
I don't know of any pictorial way, but this is an interpretation I use. It is useful mainly because it is easily verifiable in a routine "not thinking" way. Of course as you gain confidence studying you will eventually just see this. This from Differential equations, an intoduction by Strauss.

"Write the equation in the form Lu = 0,
where L is an operator. That is, if v is any function, Lv is a new function. For
instance,L = d/dx is the operator that takes v into its derivative dv/dx.

The definition we want for linearity is: L(u + v) = Lu + Lv and L(cu) = c*Lu
for any functions u, v and any constant c."

Is PDE u_x + u_yy = exp(x^2) linear ? Yes, because (u+v)_x + (u+v)_yy = u_x + v_x + u_yy + v_yy and what about (u_x)^2 = 0 ? No, it is not linear because (u_x + v_x)^2 is NOT equal to (u_x)^2 + (u_y)^2

First one of theese was not homogenous because the right side is 0 the other is though.
That homogenous part is some function of independent variables free from derivatives.
cheers,

Adam
 
THANK YOU, that really helps.
I have Strauss's PDE book for next semester. I've looked through it and he looks pretty clear and concise.

Thanks again, it really helped.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K