- #1

- 287

- 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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter chaotixmonjuish
- Start date

Hi chaotixmonjuish, can you tell me what is the chapter name of these questions ? they look cool :)There is no chapter name for these questions, they are just questions.f

- #1

- 287

- 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

Science Advisor

Homework Helper

- 25,833

- 256

… u(2-x,0)=(2-x) but off by a scalar.

I'm confused …

u(2-x,0) = (2-x)(2-(2-x)) = … ?

- #3

Science Advisor

Homework Helper

- 43,008

- 974

- #4

- 287

- 0

Oh geeze, so how do I verify this?

- #5

- 287

- 0

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

- #6

Science Advisor

Homework Helper

- 25,833

- 256

- #7

- 287

- 0

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

Science Advisor

Homework Helper

- 25,833

- 256

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

[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

- 287

- 0

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

- #10

Science Advisor

Homework Helper

- 25,833

- 256

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

- 287

- 0

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

- #12

- 121

- 0

- #13

Science Advisor

Homework Helper

- 43,008

- 974

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

Share:

- Replies
- 7

- Views
- 436

- Replies
- 6

- Views
- 363

- Replies
- 7

- Views
- 505

- Replies
- 5

- Views
- 561

- Replies
- 5

- Views
- 1K

- Replies
- 11

- Views
- 392

- Replies
- 1

- Views
- 546

- Replies
- 5

- Views
- 716

- Replies
- 7

- Views
- 594

- Replies
- 7

- Views
- 592