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Finding the constants $A$ and $B$ for the general series solution of the wave equation with initial condition $g(x) = max 0, 1 − $
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I am trying to find the constants $A$ and $B$ for the general series solution of the wave equation
$$u(x, t) = sum_n = 1^infty left( A_n cos(c n pi t) + B_n sin(c n pi t) right) sin(pi n x)$$
with initial condition $g(x) = max $.
I'm assuming this means $u(x, 0) = g(x) = max $? Would I be correct?
I know that we can use the orthogonality of $sin(n pi x)$ to find the explicit form of the coefficients $A_n$ and $B_n$, by multiplying each equation by $sin(m pi x)$ and integrating.
It says the solution is
$$A_n = left( frac8pi n right)^2 left( sinleft( fracn pi16 right) right)^2 left( sinleft( fracn pi2 right) right)$$
for odd $n$ and $A_n = 0$ for even $n$.
And apparently the second condition gives us $B_n = 0$.
However, I'm not sure how to use this initial condition.
The first thing I calculated was when $g(x) = 1 − |8x − 4|$ and when $g(x) = 0$:
$$1 - |8x - 4| > 0$$
$$rightarrow |8x - 4| < 1$$
By my reasoning, we have two possibilities:
$$8x - 4 < 1$$ (If $8x - 4 > 0$)
$$rightarrow 8x < 5$$
$$x < frac58$$
$$-(8x - 4) < 1$$ (If $8x - 4 < 0$)
$$rightarrow 8x - 4 > -1$$
$$rightarrow 8x > 3$$
$$x > frac38$$
This means that $dfrac58 > x > dfrac38$ to have $1 - |8x - 4| > 0$.
I am not sure how to proceed from here?
Thank you for any help.
pde
asked Jul 26 at 9:25
Wyuw
1526
add a comment |Â
up vote
1
down vote
favorite
I am trying to find the constants $A$ and $B$ for the general series solution of the wave equation
$$u(x, t) = sum_n = 1^infty left( A_n cos(c n pi t) + B_n sin(c n pi t) right) sin(pi n x)$$
with initial condition $g(x) = max $.
I'm assuming this means $u(x, 0) = g(x) = max $? Would I be correct?
I know that we can use the orthogonality of $sin(n pi x)$ to find the explicit form of the coefficients $A_n$ and $B_n$, by multiplying each equation by $sin(m pi x)$ and integrating.
It says the solution is
$$A_n = left( frac8pi n right)^2 left( sinleft( fracn pi16 right) right)^2 left( sinleft( fracn pi2 right) right)$$
for odd $n$ and $A_n = 0$ for even $n$.
And apparently the second condition gives us $B_n = 0$.
However, I'm not sure how to use this initial condition.
The first thing I calculated was when $g(x) = 1 − |8x − 4|$ and when $g(x) = 0$:
$$1 - |8x - 4| > 0$$
$$rightarrow |8x - 4| < 1$$
By my reasoning, we have two possibilities:
$$8x - 4 < 1$$ (If $8x - 4 > 0$)
$$rightarrow 8x < 5$$
$$x < frac58$$
$$-(8x - 4) < 1$$ (If $8x - 4 < 0$)
$$rightarrow 8x - 4 > -1$$
$$rightarrow 8x > 3$$
$$x > frac38$$
This means that $dfrac58 > x > dfrac38$ to have $1 - |8x - 4| > 0$.
I am not sure how to proceed from here?
Thank you for any help.
pde
asked Jul 26 at 9:25
Wyuw
1526
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to find the constants $A$ and $B$ for the general series solution of the wave equation
$$u(x, t) = sum_n = 1^infty left( A_n cos(c n pi t) + B_n sin(c n pi t) right) sin(pi n x)$$
with initial condition $g(x) = max $.
I'm assuming this means $u(x, 0) = g(x) = max $? Would I be correct?
I know that we can use the orthogonality of $sin(n pi x)$ to find the explicit form of the coefficients $A_n$ and $B_n$, by multiplying each equation by $sin(m pi x)$ and integrating.
It says the solution is
$$A_n = left( frac8pi n right)^2 left( sinleft( fracn pi16 right) right)^2 left( sinleft( fracn pi2 right) right)$$
for odd $n$ and $A_n = 0$ for even $n$.
And apparently the second condition gives us $B_n = 0$.
However, I'm not sure how to use this initial condition.
The first thing I calculated was when $g(x) = 1 − |8x − 4|$ and when $g(x) = 0$:
$$1 - |8x - 4| > 0$$
$$rightarrow |8x - 4| < 1$$
By my reasoning, we have two possibilities:
$$8x - 4 < 1$$ (If $8x - 4 > 0$)
$$rightarrow 8x < 5$$
$$x < frac58$$
$$-(8x - 4) < 1$$ (If $8x - 4 < 0$)
$$rightarrow 8x - 4 > -1$$
$$rightarrow 8x > 3$$
$$x > frac38$$
This means that $dfrac58 > x > dfrac38$ to have $1 - |8x - 4| > 0$.
I am not sure how to proceed from here?
Thank you for any help.
pde
asked Jul 26 at 9:25
Wyuw
1526
I am trying to find the constants $A$ and $B$ for the general series solution of the wave equation
$$u(x, t) = sum_n = 1^infty left( A_n cos(c n pi t) + B_n sin(c n pi t) right) sin(pi n x)$$
with initial condition $g(x) = max $.
I'm assuming this means $u(x, 0) = g(x) = max $? Would I be correct?
I know that we can use the orthogonality of $sin(n pi x)$ to find the explicit form of the coefficients $A_n$ and $B_n$, by multiplying each equation by $sin(m pi x)$ and integrating.
It says the solution is
$$A_n = left( frac8pi n right)^2 left( sinleft( fracn pi16 right) right)^2 left( sinleft( fracn pi2 right) right)$$
for odd $n$ and $A_n = 0$ for even $n$.
And apparently the second condition gives us $B_n = 0$.
However, I'm not sure how to use this initial condition.
The first thing I calculated was when $g(x) = 1 − |8x − 4|$ and when $g(x) = 0$:
$$1 - |8x - 4| > 0$$
$$rightarrow |8x - 4| < 1$$
By my reasoning, we have two possibilities:
$$8x - 4 < 1$$ (If $8x - 4 > 0$)
$$rightarrow 8x < 5$$
$$x < frac58$$
$$-(8x - 4) < 1$$ (If $8x - 4 < 0$)
$$rightarrow 8x - 4 > -1$$
$$rightarrow 8x > 3$$
$$x > frac38$$
This means that $dfrac58 > x > dfrac38$ to have $1 - |8x - 4| > 0$.
I am not sure how to proceed from here?
Thank you for any help.
pde
asked Jul 26 at 9:25
Wyuw
1526
asked Jul 26 at 9:25
Wyuw
1526
asked Jul 26 at 9:25
Wyuw
1526
asked Jul 26 at 9:25
Wyuw
1526
1526
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
1
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Break up the absolute value
$$ |8x-4| = begincases 8x-4, & textif & x > frac12 \ -(8x-4), & textif& x < frac12 endcases $$
Hence
$$ 1-|8x-4| = begincases 5-8x, & textif & x > frac12 \ 8x-3, & textif& x < frac12 endcases $$
You already deduced that the function has to be zero outside $(frac38, frac58)$, therefore
$$ g(x) = begincases 0, & textif & 0 < x < frac38 \ 8x-3, & textif & frac38 < x < frac12 \ 5-8x, & textif & frac12 < x < frac58 \ 0, & textif & frac58 < x < 1 endcases $$
I assume the the domain is $[0,1]$ from the Fourier series expression.
You can integrate piece-wise to find $A_n$
You need a second initial condition to determine $B_n$. Check the problem again to see what it might be. If $B_n=0$ then it's likely $u_t(x,0)=0$
answered Jul 26 at 14:52
Dylan
11.4k31026
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Break up the absolute value
$$ |8x-4| = begincases 8x-4, & textif & x > frac12 \ -(8x-4), & textif& x < frac12 endcases $$
Hence
$$ 1-|8x-4| = begincases 5-8x, & textif & x > frac12 \ 8x-3, & textif& x < frac12 endcases $$
You already deduced that the function has to be zero outside $(frac38, frac58)$, therefore
$$ g(x) = begincases 0, & textif & 0 < x < frac38 \ 8x-3, & textif & frac38 < x < frac12 \ 5-8x, & textif & frac12 < x < frac58 \ 0, & textif & frac58 < x < 1 endcases $$
I assume the the domain is $[0,1]$ from the Fourier series expression.
You can integrate piece-wise to find $A_n$
You need a second initial condition to determine $B_n$. Check the problem again to see what it might be. If $B_n=0$ then it's likely $u_t(x,0)=0$
answered Jul 26 at 14:52
Dylan
11.4k31026
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
add a comment |Â
up vote
1
down vote
Break up the absolute value
$$ |8x-4| = begincases 8x-4, & textif & x > frac12 \ -(8x-4), & textif& x < frac12 endcases $$
Hence
$$ 1-|8x-4| = begincases 5-8x, & textif & x > frac12 \ 8x-3, & textif& x < frac12 endcases $$
You already deduced that the function has to be zero outside $(frac38, frac58)$, therefore
$$ g(x) = begincases 0, & textif & 0 < x < frac38 \ 8x-3, & textif & frac38 < x < frac12 \ 5-8x, & textif & frac12 < x < frac58 \ 0, & textif & frac58 < x < 1 endcases $$
I assume the the domain is $[0,1]$ from the Fourier series expression.
You can integrate piece-wise to find $A_n$
You need a second initial condition to determine $B_n$. Check the problem again to see what it might be. If $B_n=0$ then it's likely $u_t(x,0)=0$
answered Jul 26 at 14:52
Dylan
11.4k31026
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Break up the absolute value
$$ |8x-4| = begincases 8x-4, & textif & x > frac12 \ -(8x-4), & textif& x < frac12 endcases $$
Hence
$$ 1-|8x-4| = begincases 5-8x, & textif & x > frac12 \ 8x-3, & textif& x < frac12 endcases $$
You already deduced that the function has to be zero outside $(frac38, frac58)$, therefore
$$ g(x) = begincases 0, & textif & 0 < x < frac38 \ 8x-3, & textif & frac38 < x < frac12 \ 5-8x, & textif & frac12 < x < frac58 \ 0, & textif & frac58 < x < 1 endcases $$
I assume the the domain is $[0,1]$ from the Fourier series expression.
You can integrate piece-wise to find $A_n$
You need a second initial condition to determine $B_n$. Check the problem again to see what it might be. If $B_n=0$ then it's likely $u_t(x,0)=0$
answered Jul 26 at 14:52
Dylan
11.4k31026
Break up the absolute value
$$ |8x-4| = begincases 8x-4, & textif & x > frac12 \ -(8x-4), & textif& x < frac12 endcases $$
Hence
$$ 1-|8x-4| = begincases 5-8x, & textif & x > frac12 \ 8x-3, & textif& x < frac12 endcases $$
You already deduced that the function has to be zero outside $(frac38, frac58)$, therefore
$$ g(x) = begincases 0, & textif & 0 < x < frac38 \ 8x-3, & textif & frac38 < x < frac12 \ 5-8x, & textif & frac12 < x < frac58 \ 0, & textif & frac58 < x < 1 endcases $$
I assume the the domain is $[0,1]$ from the Fourier series expression.
You can integrate piece-wise to find $A_n$
You need a second initial condition to determine $B_n$. Check the problem again to see what it might be. If $B_n=0$ then it's likely $u_t(x,0)=0$
answered Jul 26 at 14:52
Dylan
11.4k31026
edited Jul 26 at 15:04
answered Jul 26 at 14:52
Dylan
11.4k31026
answered Jul 26 at 14:52
Dylan
11.4k31026
answered Jul 26 at 14:52
Dylan
11.4k31026
11.4k31026
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
add a comment |Â
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
Yes sorry about that. We have $c = 1/2$ and $u_t(x, 0) = 0$.
– Wyuw
Jul 26 at 18:48
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
What's $c$? Define your variables
– Dylan
Jul 27 at 12:43
add a comment |Â
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