Verifying if this PDE is a solution

  • Thread starter Thread starter jc2009
  • Start date Start date
  • Tags Tags
    Pde
Click For Summary
The discussion focuses on verifying whether specific functions are solutions to given nonhomogeneous PDEs using the 1D heat operator. The functions [x+1]e^(-t), e^(-2)x sin(t), and xt are confirmed as solutions for their respective equations. The challenge arises in finding a solution for the new PDE, Hu = √2 x + πe^(-2x)(4sin(t) + cos(t)) + e^(-t)(x+1). Participants suggest leveraging the linearity of the operator, indicating that the sum of verified solutions can be used to construct a solution for the new equation. The conversation emphasizes the importance of understanding the linear nonhomogeneous nature of the problem to find a suitable solution.
jc2009
Messages
12
Reaction score
0
PROBLEM: Verify that the functions [x+1]e^(-t) ; e^(-2)sint ; and xt are respectively solutions of the nonhomogeneous equations
Hu = -e^(-t)[x+1] ; Hu = e^(-2x)[4sint+cost] ; and Hu = x
where H is the 1D heat operator H = \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2}

i did this the verification part,, the problem is with the second part of the problem
Find a solution of the PDE
Hu = \sqrt{2} x + [Pi]e^(-2x) [4sint + cost] + e^(-t)[x+1]

isn't the first part a solution for this PDE? i don't understand the question

any hints how to setup this PDE?
 
Physics news on Phys.org
You have already verified some particular solutions for Hu =\phi(x,t).
Note that the DE is linear nonhomogeneous.

If u1(x,t) (resp. u2(x,t)) is a solution of Hu =\phi_1(x,t) (resp. Hu =\phi_2(x,t) )
then
u1(x,t) + u2(x,t) will be a solution of

Hu =\phi_1(x,t) + \phi_2(x,t).
 

Similar threads

  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K