- #1

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[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.

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- Thread starter chaotixmonjuish
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- #1

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- 0

[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.

- #2

tiny-tim

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… u(2-x,0)=(2-x) but off by a scalar.

I'm confused …

u(2-x,0) = (2-x)(2-(2-x)) = … ?

- #3

HallsofIvy

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- #4

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Oh geeze, so how do I verify this?

- #5

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

- #6

tiny-tim

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- #7

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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.

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

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 u

- #9

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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).

- #11

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

- #12

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- #13

HallsofIvy

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

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