Wave equation with inhomogeneous boundary conditions

Click For Summary
SUMMARY

The discussion focuses on solving the wave equation given by the partial differential equation (1) \(\frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial^2 u}{\partial x^2}\) on the interval \(0 PREREQUISITES

  • Understanding of wave equations and their properties
  • Familiarity with separation of variables technique
  • Knowledge of boundary value problems and initial conditions
  • Proficiency in manipulating partial differential equations (PDEs)
NEXT STEPS
  • Study the method of separation of variables in depth
  • Learn about homogeneous and inhomogeneous boundary conditions
  • Explore the derivation of ordinary differential equations from partial differential equations
  • Investigate the implications of initial conditions on wave equations
USEFUL FOR

Students and professionals in applied mathematics, physics, and engineering who are working on wave equations and boundary value problems, particularly those interested in advanced PDE techniques.

josh146
Messages
4
Reaction score
0
I'm reposting this because there was a problem with the title/LaTeX last time.

Homework Statement



Solve the wave equation (1) on the region 0<x<2 subject to the boundary conditions (2) and the initial condition (3) by separation of variables.

Homework Equations



(1) \frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial^2 u}{\partial x^2}

(2) \frac{\partial u}{\partial x}(0,t)=1 ; \frac{\partial u }{\partial x}(2,t)=1

(3) \frac{\partial u}{\partial t}(x,0)=0

The Attempt at a Solution



I've defined \theta(x,t)=u(x,t)-u_{st}(x) = u(x,t)-x-h(t) where u_st is the steady state solution (4). I've used this to create a new PDE with homogeneous boundary conditions.

The PDE is:

\frac{\partial^2 \theta}{\partial t^2} + h&#039;&#039;(t)=c^2 \frac{\partial^2 \theta}{\partial x^2}.

By subbing in \theta=f(t)g(x) I get:

f&#039;&#039;(t)g(x)+h&#039;&#039;(t)=c^2 f(t) g&#039;&#039;(x)

I'm not sure how to transform this into two ODEs. Can someone help?

(4): The solution to \frac{\partial^2 u}{\partial x^2} = 0 subject to (2).
 
Physics news on Phys.org
I think you just need to introduce new function v: u(x,t) = v(x,t) + x. The equation for v(x,t) will be the same but boundary conditions will become homogeneous =) and btw, you are missing one more initial condition like u(x,t=0) = ...
 
Im also stuck on this one, for me its the initial conditions and boundary conditions being partial derivitives that throws me.
 
Last edited:
I think you just need to introduce new function v: u(x,t) = v(x,t) + x. The equation for v(x,t) will be the same but boundary conditions will become homogeneous =) and btw, you are missing one more initial condition like u(x,t=0) = ...

The eqn. given for boundary conditions is a partial derivative so do i intergrate it to find u(x,t)
in which case is it that; u(o,t) = x and u(2,t) = x. so that.
u(x,t) = v(x,t) + x so that in both cases... umm then i get stuck.
 
No-one at all?
 
So far I've got that:
U_{st} (x) = x
(a) Find the steady state solution, ust(x), by solving \frac{\delta ^{2} U_{st} }{\delta x^{2}} = 0 and applying the boundary
conditions. (You will only be able to determine one of the two arbitrary constants).

confused here, surely U_{st} (x) = x + c (as boundary conditions given are derivitives)

then initial condition:

<br /> \frac{\partial u}{\partial t}(x,0)=0<br />

given means that c = 0? :S
 
Latex is fail :(

So far I've got that:
U_st (x) = x
(a) Find the steady state solution, u_st(x), by solving (partial second derivative u by x = 0) and applying the boundary
conditions. (You will only be able to determine one of the two arbitrary constants).

confused here, surely U_st (x) = x + c (as boundary conditions given are derivitives)

then initial condition:

(as above... in first post)

given means that c = 0? :S
 
Can no one point me in the right direction?
 
Instead of integrating the BC, you may differentiate the solution you get and then apply the BC.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
3
Views
2K
Replies
4
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K