Verifying Solutions of u_t-u_{xx}=0

[chaotixmonjuish]

Given this equation:

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

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.

 

[tiny-tim]

Hi chaotixmonjuish!

… u(2-x,0)=(2-x) but off by a scalar.

I'm confused …

u(2-x,0) = (2-x)(2-(2-x)) = … ?

 

[HallsofIvy]

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)

 

[chaotixmonjuish]

Oh geeze, so how do I verify this?

 

[chaotixmonjuish]

I got u(2-x(2-x),0)=...=u(x^2,0)

 

[tiny-tim]

I got u(2-x(2-x),0)=...=u(x^2,0)

uhh?

2 - (2 - x) = x.

 

[chaotixmonjuish]

Whoops, I meant this:

$$u(x,t)=u(2-x,t)$$
$$u(x,0)=u(2-x)$$
$$u(x,0)=(2-x)(2-(2-x))$$
$$u(x,0)=x(2-x)$$
This satisifies the initial condition.

 

[tiny-tim]

Hi chaotixmonjuish!

(erm … why bother with LaTeX when you could just have typed it as text? )

Whoops, I meant this:

$$u(x,t)=u(2-x,t)$$
$$u(x,0)=u(2-x)$$
$$u(x,0)=(2-x)(2-(2-x))$$
$$u(x,0)=x(2-x)$$
This satisifies the initial condition.

Yup … that nails the middle condition …

and the third one is easy …

now how about u_t - u_xx = 0 ?

 

[chaotixmonjuish]

I'm having problems just getting why we only used 2-x in the substitution

 

[tiny-tim]

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).

 

[chaotixmonjuish]

So it that why we can pop a 2-x in for u?

 

[Lord Dark]

hi chaotixmonjuish ,, can you tell me what is the chapter name of these questions ? they look cool :)

 

[HallsofIvy]

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!