# PDE question

Given this equation:

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

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
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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
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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)

Oh geeze, so how do I verify this?

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

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

uhh?

2 - (2 - x) = x.

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.

Last edited:
tiny-tim
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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 ut - uxx = 0 ?

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

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).

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

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

HallsofIvy