Verifying Solutions of u_t-u_{xx}=0

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The discussion focuses on verifying the solution of the partial differential equation \( u_t - u_{xx} = 0 \) with the initial condition \( u(x,0) = x(2-x) \) for \( x \in [0,2] \). Participants confirm that the function \( u(x,t) = u(2-x,t) \) satisfies the boundary conditions \( u(0,t) = u(2,t) = 0 \) and the initial condition. The verification process involves substituting \( u(2-x,0) \) and demonstrating that it equals \( u(x,0) \). The discussion highlights the importance of understanding variable substitution in function evaluation.

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chaotixmonjuish
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Given this equation:

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

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.
 
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Hi chaotixmonjuish! :wink:
chaotixmonjuish said:
… 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:
 
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)
 
Oh geeze, so how do I verify this?
 
I got u(2-x(2-x),0)=...=u(x^2,0)
 
chaotixmonjuish said:
I got u(2-x(2-x),0)=...=u(x^2,0)

uhh? :confused:

2 - (2 - x) = x.
 
Whoops, I meant this:

<br /> u(x,t)=u(2-x,t)<br />
<br /> u(x,0)=u(2-x)<br />
<br /> u(x,0)=(2-x)(2-(2-x))<br />
<br /> u(x,0)=x(2-x)<br />
This satisifies the initial condition.
 
Last edited:
Hi chaotixmonjuish! :smile:

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

<br /> u(x,t)=u(2-x,t)<br />
<br /> u(x,0)=u(2-x)<br />
<br /> u(x,0)=(2-x)(2-(2-x))<br />
<br /> u(x,0)=x(2-x)<br />
This satisifies the initial condition.

Yup … that nails the middle condition …

and the third one is easy …

now how about ut - uxx = 0 ? :smile:
 
I'm having problems just getting why we only used 2-x in the substitution
 
  • #10
chaotixmonjuish said:
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
hi chaotixmonjuish ,, can you tell me what is the chapter name of these questions ? they look cool :)
 
  • #13
chaotixmonjuish said:
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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