How to derive the solution of the wave equation in a finite interval with inhomogenous boundary conditions $u(0,t)=h(t),v(l,t)=k(t)$
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We have the following problem:
$$u_tt=c^2u_xx$$
$$u(x,0)=u_t(x,0)=0$$
$$u(0,t)=h(t),u(l,t)=k(t).$$
From the first two conditions one can deduce that $u(x,t)=0$ for $x>ct$ since the solution to the wave equation is of the form $f(x+ct)+g(x-ct)$ and for $x>ct$ we have that $f+g=0$ and $f'=g'.$ Combining these conditions implies that $f(y)=C$ and $g(y)=-C$ for some constant $C.$
The boundary condition data implies that $f(ct)+g(-ct)=h(t)$ and $f(l+ct)+g(l-ct)=k(t).$ Since $ct>0$ we have that $f(ct)=C$ and therefore for $y<0$
$$g(-ct)=-C+h(t)implies g(y)=-C+h(-y/c).$$ Since the solution is a sum of $f(x+ct)$ and $g(x-ct)$ the final result will therefore be for $x<ct$
$$h(t-x/c).$$
The answer is however in the form of a series
$$u(x,t)= h(t-x/c)-h(t+(x/2l)/c)+h(t-(x+2l)/c)-...+k(t+(x-l)/c)-k(t-(x-l)/c)+..$$
How do I go about deriving this expression?
pde
asked Aug 15 at 5:30
Hello_World
2,95821328
add a comment |Â
up vote
1
down vote
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We have the following problem:
$$u_tt=c^2u_xx$$
$$u(x,0)=u_t(x,0)=0$$
$$u(0,t)=h(t),u(l,t)=k(t).$$
From the first two conditions one can deduce that $u(x,t)=0$ for $x>ct$ since the solution to the wave equation is of the form $f(x+ct)+g(x-ct)$ and for $x>ct$ we have that $f+g=0$ and $f'=g'.$ Combining these conditions implies that $f(y)=C$ and $g(y)=-C$ for some constant $C.$
The boundary condition data implies that $f(ct)+g(-ct)=h(t)$ and $f(l+ct)+g(l-ct)=k(t).$ Since $ct>0$ we have that $f(ct)=C$ and therefore for $y<0$
$$g(-ct)=-C+h(t)implies g(y)=-C+h(-y/c).$$ Since the solution is a sum of $f(x+ct)$ and $g(x-ct)$ the final result will therefore be for $x<ct$
$$h(t-x/c).$$
The answer is however in the form of a series
$$u(x,t)= h(t-x/c)-h(t+(x/2l)/c)+h(t-(x+2l)/c)-...+k(t+(x-l)/c)-k(t-(x-l)/c)+..$$
How do I go about deriving this expression?
pde
asked Aug 15 at 5:30
Hello_World
2,95821328
You may have a look at this post where the sub-case $h=0$, $k=sin$ is tackled in two answers.
– Harry49
Aug 16 at 11:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We have the following problem:
$$u_tt=c^2u_xx$$
$$u(x,0)=u_t(x,0)=0$$
$$u(0,t)=h(t),u(l,t)=k(t).$$
From the first two conditions one can deduce that $u(x,t)=0$ for $x>ct$ since the solution to the wave equation is of the form $f(x+ct)+g(x-ct)$ and for $x>ct$ we have that $f+g=0$ and $f'=g'.$ Combining these conditions implies that $f(y)=C$ and $g(y)=-C$ for some constant $C.$
The boundary condition data implies that $f(ct)+g(-ct)=h(t)$ and $f(l+ct)+g(l-ct)=k(t).$ Since $ct>0$ we have that $f(ct)=C$ and therefore for $y<0$
$$g(-ct)=-C+h(t)implies g(y)=-C+h(-y/c).$$ Since the solution is a sum of $f(x+ct)$ and $g(x-ct)$ the final result will therefore be for $x<ct$
$$h(t-x/c).$$
The answer is however in the form of a series
$$u(x,t)= h(t-x/c)-h(t+(x/2l)/c)+h(t-(x+2l)/c)-...+k(t+(x-l)/c)-k(t-(x-l)/c)+..$$
How do I go about deriving this expression?
pde
asked Aug 15 at 5:30
Hello_World
2,95821328
We have the following problem:
$$u_tt=c^2u_xx$$
$$u(x,0)=u_t(x,0)=0$$
$$u(0,t)=h(t),u(l,t)=k(t).$$
From the first two conditions one can deduce that $u(x,t)=0$ for $x>ct$ since the solution to the wave equation is of the form $f(x+ct)+g(x-ct)$ and for $x>ct$ we have that $f+g=0$ and $f'=g'.$ Combining these conditions implies that $f(y)=C$ and $g(y)=-C$ for some constant $C.$
The boundary condition data implies that $f(ct)+g(-ct)=h(t)$ and $f(l+ct)+g(l-ct)=k(t).$ Since $ct>0$ we have that $f(ct)=C$ and therefore for $y<0$
$$g(-ct)=-C+h(t)implies g(y)=-C+h(-y/c).$$ Since the solution is a sum of $f(x+ct)$ and $g(x-ct)$ the final result will therefore be for $x<ct$
$$h(t-x/c).$$
The answer is however in the form of a series
$$u(x,t)= h(t-x/c)-h(t+(x/2l)/c)+h(t-(x+2l)/c)-...+k(t+(x-l)/c)-k(t-(x-l)/c)+..$$
How do I go about deriving this expression?
pde
asked Aug 15 at 5:30
Hello_World
2,95821328
asked Aug 15 at 5:30
Hello_World
2,95821328
asked Aug 15 at 5:30
Hello_World
2,95821328
asked Aug 15 at 5:30
Hello_World
2,95821328
2,95821328
You may have a look at this post where the sub-case $h=0$, $k=sin$ is tackled in two answers.
– Harry49
Aug 16 at 11:07
add a comment |Â
You may have a look at this post where the sub-case $h=0$, $k=sin$ is tackled in two answers.
– Harry49
Aug 16 at 11:07
You may have a look at this post where the sub-case $h=0$, $k=sin$ is tackled in two answers.
– Harry49
Aug 16 at 11:07
You may have a look at this post where the sub-case $h=0$, $k=sin$ is tackled in two answers.
– Harry49
Aug 16 at 11:07
add a comment |Â
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You may have a look at this post where the sub-case $h=0$, $k=sin$ is tackled in two answers.
– Harry49
Aug 16 at 11:07