Parallelogram identity in the wave equation
Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:
$$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
$$u(0,t) = h(t)$$
$$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
where $f, g, h ∈ C_2{[0, ∞)}$
I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.
Edit: The parallelogram identity is
$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$
pde
edited May 24 '18 at 10:32
Dylan
12.4k31026
asked Oct 29 '12 at 22:39
MariaMaria
262
add a comment |
Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:
$$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
$$u(0,t) = h(t)$$
$$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
where $f, g, h ∈ C_2{[0, ∞)}$
I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.
Edit: The parallelogram identity is
$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$
pde
edited May 24 '18 at 10:32
Dylan
12.4k31026
asked Oct 29 '12 at 22:39
MariaMaria
262
add a comment |
Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:
$$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
$$u(0,t) = h(t)$$
$$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
where $f, g, h ∈ C_2{[0, ∞)}$
I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.
Edit: The parallelogram identity is
$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$
pde
edited May 24 '18 at 10:32
Dylan
12.4k31026
asked Oct 29 '12 at 22:39
MariaMaria
262
Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:
$$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
$$u(0,t) = h(t)$$
$$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
where $f, g, h ∈ C_2{[0, ∞)}$
I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.
Edit: The parallelogram identity is
$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$
pde
pde
edited May 24 '18 at 10:32
Dylan
12.4k31026
asked Oct 29 '12 at 22:39
MariaMaria
262
edited May 24 '18 at 10:32
Dylan
12.4k31026
asked Oct 29 '12 at 22:39
MariaMaria
262
edited May 24 '18 at 10:32
Dylan
12.4k31026
edited May 24 '18 at 10:32
Dylan
12.4k31026
edited May 24 '18 at 10:32
Dylan
12.4k31026
12.4k31026
asked Oct 29 '12 at 22:39
MariaMaria
262
asked Oct 29 '12 at 22:39
MariaMaria
262
asked Oct 29 '12 at 22:39
MariaMaria
262
262
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.
Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$
$u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$
Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$
$u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.
Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$
$u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$
Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$
$u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
add a comment |
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.
Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$
$u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$
Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$
$u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
add a comment |
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.
Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$
$u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$
Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$
$u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.
Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$
$u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$
Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$
$u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
edited Jan 5 '13 at 22:02
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
answered Oct 29 '12 at 23:28
doraemonpauldoraemonpaul
12.5k31660
12.5k31660
add a comment |
add a comment |
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