My final Diff Eq question for now

In summary, the extensions for f(x) and g(x) for the first problem are given by the above equations. These extensions can be used to calculate the value of u(3/4,2) using D'Alembert's solution for one-dimensional wave equation with length and speed of the wave given as 1 and the functions f(x) and g(x) as defined above.
  • #1
1,459
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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 don't see that one, thanks!
 
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  • #2
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
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 here's the stupid part - how do i even or odd extend?
 
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  • #4
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
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
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
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
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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