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My final Diff Eq question for now

  • #1
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Here u(x,t) refers to D'ALembert's solution for the one dimensional wave. L represnets the legnth of the string and c, the speed of the wave.

Find u(3/4,2) where l=c=1 f(x) = x(1-x), g(x) = x^2 (1-x)
Ok i know that from earlier that i ahve to extend g as an even periodic function over this interval

But what about f? Does f have to be extended ?

For teh extension of g... Do i simply move that function to right?
so [tex] g_{1} (r) = r^2 (1-r) [/tex]

for [tex] g_{2} (r) = (-r-\frac{1}{2})^2)[1-(-r-\frac{1}{2})] [/tex] over the interval [5/4,7/4]

and finally [tex] g_{3} (r) = (r-1)^2)[1-(r-1)] [/tex] over the interval [7/4,2]

am i going in the right direction??

Simialr question i have is
If [itex] f(x) = x^2 (1-x)^2, g(x) = 1, \partial u/ \partial x (0,t)=0, c=1, find u(3/4,7/2)[/itex]

does f(x) have to be extended
if so would it be extended by a polynomial of degree 4? in which case
for [1,2] (x-1.5)^4
for [2,3.5] (2.5-x)^4 [/tex]

your help would be greatly appreciated!
p.s. the other question was not required to be answered tahts why you dont see that one, thanks!
 
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Answers and Replies

  • #2
saltydog
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Stunner, the extensions depend on the boundary conditions: if the boundary conditions are of a form u(0,t)=0 or u(L,t)=0 then odd-extend about those points. If the boundary conditions are of the form [itex]u_x(0,t)=0[/itex] or [itex]u_x(L,t)=0[/itex], then even-extend about the points. If the left boundary is u(0,t) then odd-extend about that point and if the right boundary is [itex]u_x(L,t)=0[/itex], then even-extend about that point.

You don't state the boundary conditions above.
 
  • #3
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the first one is just like this question from earlier https://www.physicsforums.com/showthread.php?t=90027

the second one has the following conditions
[itex] f(x) = x^2 (1-x)^2, g(x) = 1, \partial u/ \partial x (0,t)=\partial u/ \partial x (1,t)0, c=1 [/itex] find u(3/4,7/2)

ok to odd extend is for a function to follow this rules f(-x) = -f(x)
and to even extend is fora function to follow f(-x) = f(x)

i nthe first case i need to odd extend - but am right in what i have done?

in the second case i need to even extend

now heres the stupid part - how do i even or odd extend?
 
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  • #4
saltydog
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stunner5000pt said:
the first one is just like this question from earlier https://www.physicsforums.com/showthread.php?t=90027

the second one has the following conditions
[itex] f(x) = x^2 (1-x)^2, g(x) = 1, \partial u/ \partial x (0,t)=\partial u/ \partial x (1,t)0, c=1 [/itex] find u(3/4,7/2)
Ok, the first one has boundary conditions:

u(0,t)=0 and u(L,0)=0

So then both f(x) and g(x) would be odd-extended about 0 and L. The second one has derivative boundary conditions so f(x) and g(x) for that one would be even-extended. Can you draw plots of both of these extensions?
 
  • #5
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the one on the left applies to the first one

te one of hte right applies to the second one
 

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  • #6
saltydog
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Not exactly Stunner. You need to extend both f(x) and g(x) for each problem. Plots below are the odd-extensions for the first problem. First is odd-extension of f(x)=x(1-x), second one is the odd-extension for g(x)=x^2(1-x). Both are extended in the range of -5/4 to 11/4 which corresponds to the intervals needed in order to calculate the value for u(3/4, 2). Can you come up with the equations for each interval for each plot in order to plug it into D'Alembert's formula?
 

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  • #7
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ok so then for the first one
For f(x)
x(1-x) [0,1]
(x-1)(x-2) [1,2]
(2-x)(x-3) [2,3]

x(x+1) [-1,0]
(x+1)(x+2) [-2,-1]

for g(x)
x^2 (1-x) [0,1]
(x-2)^2 (-1-x) [1,2]
-x(x+1)^2 [-1,-2]
 
  • #8
saltydog
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stunner5000pt said:
ok so then for the first one
For f(x)
x(1-x) [0,1]
(x-1)(x-2) [1,2]
(2-x)(x-3) [2,3]

x(x+1) [-1,0]
(x+1)(x+2) [-2,-1]

for g(x)
x^2 (1-x) [0,1]
(x-2)^2 (-1-x) [1,2]
-x(x+1)^2 [-1,-2]
This is what I get:

For f(x):

[tex](-1,0)\Rightarrow \quad x(1+x)[/tex]
[tex](0,1)\Rightarrow \quad x(1-x)[/tex]
[tex](1,2)\Rightarrow \quad -(x-1)+(x-1)^2[/tex]
[tex](2,3)\Rightarrow \quad (x-2)-(x-2)^2 [/tex]

for g(x):

[tex](-1,0)\Rightarrow \quad -x^2(1+x)[/tex]
[tex](0,1)\Rightarrow \quad x^2(1-x)[/tex]
[tex](1,2)\Rightarrow \quad -(2-x)^2[1-(2-x)][/tex]
[tex](2,3)\Rightarrow \quad (x-2)^2[1-(x-2)][/tex]
 

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