Pde problem: inspiration needed

  • Thread starter Thread starter Hyperreality
  • Start date Start date
  • Tags Tags
    Pde
Click For Summary
SUMMARY

The discussion centers on solving the PDE defined by the equation u_{a}+u_{t}=-\mu t_{u} with boundary conditions u(a,0)=u_{0}(a) and u(0,t)=b∫_0^∞ u(a,t)da. The general solution is established as u(a,t)=F(a-t)e^{-1/2{\mu}t^{2}}. Further simplification is suggested by the lecturer, emphasizing the importance of the boundary conditions on the arbitrary function F, leading to the equation F(a-t)=b∫_0^∞ F(2a-t)da. The discussion encourages exploring the value of F(0) to derive more specific information about F.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with boundary value problems
  • Knowledge of integral transforms
  • Experience with arbitrary functions in mathematical solutions
NEXT STEPS
  • Investigate the properties of arbitrary functions in PDE solutions
  • Learn about boundary conditions and their impact on function determination
  • Explore integral equations and their applications in solving PDEs
  • Study the method of characteristics for solving first-order PDEs
USEFUL FOR

Mathematicians, physics students, and anyone involved in solving partial differential equations, particularly those interested in boundary value problems and integral transforms.

Hyperreality
Messages
201
Reaction score
0
u_{a}+u_{t}=-\mu t_{u}
u(a,0)=u_{0}(a)
u(0,t)=b\int_0^\infty \left u(a,t)da

Solve u(a,t) for the region a<t

Got this question from assignment. My solution is incomplete though, need some inspirations! I have shown that the general solution is

u(a,t)=F(a-t)e^{-1/2{\mu}t^{2}}

So for u(0, t) we have

u(0,t)=F(-t)^{-1/2{\mu}t^{2}}=b\int_0^\infty \left u(a,t)da

Substitute the general solution u(a,t) we get

F(-t)=b\int_0^\infty \left F(a-t)da

Now -t -> a-t, and we obtain F(a-t) to be

F(a-t)=b \int_0^\infty \left F(2a-t)da

Hence our solution u(a,t) for a<t is

u(a,t)=be^{-1/2{\mu}t^{2}}\int_0^\infty\left F(2a-t)da

However, when I talked to my lecturer, he told me I can simplify this further. I need some inspirations:redface:
 
Last edited:
Physics news on Phys.org
Your general solution has an arbitrary function F.
But as you can see in your steps there, the boundary conditions on u have given you quite a bit of information on F. So you can restrict it further.

So here is some inspiration:
-What is F(0) equal to?
- Note you obtained one equation with F on both sides. Can you solve for F given what F(0) is?

Good luck with your homework.
 
Last edited:

Similar threads

Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K