Solution to Homogenous PDE

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Homework Statement



Find the solution of:

utt-uxx = sin(∏x) for 0<x<1
u(x,0)=0 for 0<=x<=1
ut(x,0)=0 for 0<=x<=1
u(0,t)=0
u(1,t)=0


Homework Equations



utt-uxx = sin(∏x) for 0<x<1
u(x,0)=0 for 0≤x≤1
ut(x,0)=0 for 0≤x≤1
u(0,t)=0
u(1,t)=0

The Attempt at a Solution



I was thinking that I would need to somehow make a change of variables, converting the PDE to a homogenous PDE and continuing from there. I don't really know how I can do this or solve it at all without some assistance!
 

Answers and Replies

  • #2
pasmith
Homework Helper
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Homework Statement



Find the solution of:

utt-uxx = sin(∏x) for 0<x<1
u(x,0)=0 for 0<=x<=1
ut(x,0)=0 for 0<=x<=1
u(0,t)=0
u(1,t)=0


Homework Equations



utt-uxx = sin(∏x) for 0<x<1
u(x,0)=0 for 0≤x≤1
ut(x,0)=0 for 0≤x≤1
u(0,t)=0
u(1,t)=0

The Attempt at a Solution



I was thinking that I would need to somehow make a change of variables, converting the PDE to a homogenous PDE and continuing from there. I don't really know how I can do this or solve it at all without some assistance!
The strategy is the same as for linear ODEs with constant coefficients: Find a particular solution which satisfies the PDE but not necessarily the boundary conditions. Then add a complementary solution (a solution of [itex]u_{tt} - u_{xx} = 0[/itex]) in order to satisfy the boundary conditions.
 
  • #3
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The strategy is the same as for linear ODEs with constant coefficients: Find a particular solution which satisfies the PDE but not necessarily the boundary conditions. Then add a complementary solution (a solution of [itex]u_{tt} - u_{xx} = 0[/itex]) in order to satisfy the boundary conditions.
So basically something like Autt-Bxx= sin∏x and Cutt-Duxx=0 and then sum the up into a solution u(x,t)?
 
  • #4
pasmith
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So basically something like Autt-Bxx= sin∏x and Cutt-Duxx=0 and then sum the up into a solution u(x,t)?
No. You have [itex]u(x,t) = u_p(x,t) + u_c(x,t)[/itex] where [itex]u_p[/itex] is any solution of
[tex]
u_{tt} - u_{xx} = \sin \pi x
[/tex]
and [itex]u_c[/itex] is the solution of
[tex]
u_{tt} - u_{xx} = 0
[/tex]
subject to [itex]u_c(x,t) = -u_p(x,t)[/itex] on the boundary.

But here it is actually simplest to try a solution of the form [itex]u(x,t) = f(t)\sin \pi x[/itex], given that [itex](\sin(\pi x))'' = -\pi^2 \sin(\pi x)[/itex]. This reduces the problem to an ODE for f(t).
 
  • #5
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No. You have [itex]u(x,t) = u_p(x,t) + u_c(x,t)[/itex] where [itex]u_p[/itex] is any solution of
[tex]
u_{tt} - u_{xx} = \sin \pi x
[/tex]
and [itex]u_c[/itex] is the solution of
[tex]
u_{tt} - u_{xx} = 0
[/tex]
subject to [itex]u_c(x,t) = -u_p(x,t)[/itex] on the boundary.

But here it is actually simplest to try a solution of the form [itex]u(x,t) = f(t)\sin \pi x[/itex], given that [itex](\sin(\pi x))'' = -\pi^2 \sin(\pi x)[/itex]. This reduces the problem to an ODE for f(t).
So you're saying that I need to set u(x,t)= u_tt-u_xx = f(t)sin(∏x)?
How do I find the function f(t)?

What I did was this:
found u_xx, u_tt and plugged them into u_tt - u_xx = sin(∏x). What do I do with the boundary conditions and the initial conditions (how do they factor in)?
 
  • #6
pasmith
Homework Helper
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599
So you're saying that I need to set u(x,t)= u_tt-u_xx = f(t)sin(∏x)?
How do I find the function f(t)?

What I did was this:
found u_xx, u_tt and plugged them into u_tt - u_xx = sin(∏x).

What do I do with the boundary conditions and the initial conditions (how do they factor in)?
The conditions [itex]u(0,t) = f(t) \sin 0 = 0[/itex] and [itex]u(1,t) = f(t) \sin \pi = 0[/itex] are satisfied automatically.

The remaining two conditions require that
[tex]u(x,0) = f(0) \sin\pi x = 0[/tex]
for all [itex]0 < x < 1[/itex] and
[tex]u_t(x,0) = f'(0) \sin \pi x = 0[/tex]
for all [itex]0 < x < 1[/itex]. This gives you initial conditions for [itex]f[/itex] and [itex]f'[/itex].
 
  • #7
43
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Great! I got it!
How would I solve this problem using a technique like changing variables instead of separation of variables?
 

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