Linear PDEs

  • #1
Somefantastik
230
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?
 

Answers and Replies

  • #2
Defconist
7
0
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
 
  • #3
Somefantastik
230
0
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.
 

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