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PDE question

  • #1
Given this equation:

[tex]
u_t-u_{xx}=0
[/tex]
[tex]
u(x,0)=x(2-x)
x\in[0,2]
[/tex]
[tex]
u(0,t)=u(2,t)=0
t\in[0,2]
[/tex]

Verify that u(x,t)=u(2-x,t) is a solution.

To do this would I just show that:

u(0,t)=u(2,t)
u(2-x,0)=(2-x) but off by a scalar.
 

Answers and Replies

  • #2
tiny-tim
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Hi chaotixmonjuish! :wink:
… u(2-x,0)=(2-x) but off by a scalar.
I'm confused :confused:

u(2-x,0) = (2-x)(2-(2-x)) = … ? :smile:
 
  • #3
HallsofIvy
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What do you mean by "u(x,t)= u(2-x,t)"? You certainly do NOT know that u(2-x,0)= 2-x, only that u(2-x,0)= u(0,x)
 
  • #4
Oh geeze, so how do I verify this?
 
  • #5
I got u(2-x(2-x),0)=...=u(x^2,0)
 
  • #6
tiny-tim
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  • #7
Whoops, I meant this:

[tex]
u(x,t)=u(2-x,t)
[/tex]
[tex]
u(x,0)=u(2-x)
[/tex]
[tex]
u(x,0)=(2-x)(2-(2-x))
[/tex]
[tex]
u(x,0)=x(2-x)
[/tex]
This satisifies the initial condition.
 
Last edited:
  • #8
tiny-tim
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Hi chaotixmonjuish! :smile:

(erm :redface: … why bother with LaTeX when you could just have typed it as text? :wink:)
Whoops, I meant this:

[tex]
u(x,t)=u(2-x,t)
[/tex]
[tex]
u(x,0)=u(2-x)
[/tex]
[tex]
u(x,0)=(2-x)(2-(2-x))
[/tex]
[tex]
u(x,0)=x(2-x)
[/tex]
This satisifies the initial condition.
Yup … that nails the middle condition …

and the third one is easy …

now how about ut - uxx = 0 ? :smile:
 
  • #9
I'm having problems just getting why we only used 2-x in the substitution
 
  • #10
tiny-tim
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I'm having problems just getting why we only used 2-x in the substitution
Because we were seeing what happens if u(x,t) = u(2-x,t),

and for that we needed to know what u(2-x,t) is for t = 0, ie u(2-x,0). :wink:
 
  • #11
So it that why we can pop a 2-x in for u?
 
  • #12
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hi chaotixmonjuish ,, can you tell me what is the chapter name of these questions ? they look cool :)
 
  • #13
HallsofIvy
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So it that why we can pop a 2-x in for u?
You aren't putting 2-x in for u, you are replacing one of the variables in u with 2-x. And, of course, you can replace a variable in a function with anything!
 

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