Mixing
1D Wave equation mixed boundary conditions and I.C.
$begingroup$
I have been searching for a solution online, but cannot find one that fits the B.C. and I.C. for this wave equation. I read through this PDF, page 7; although I had similar conditions I just obtained trivial solutions.
Now, the system is
$left{begin{array}{ll}u_{tt}(x,t)-c^2 u_{xx}(x,t)=0,quad 0<x<L,quad t>0\u(0,t)=0,quad u_{x}(L,t)=Acos(Omega t),quad t>0\u(x,0)=0,quad u_{t}(x,0)quad 0<x<Lend{array}right.$
As usual I use separation of variables and then obtain the solutions for $X(x)$ and $T(t)$
$left{begin{array}{ll}X(x)=Bcos(omega _{1}x)+Csin(omega _{1}x)\T(t)=Dcos(omega _{2}t)+Esin(omega _{2}t)\end{array}right.$
$omega ^{2} _{1}=lambda / c^2,quad omega ^{2}_{2}=lambda,quad lambda>0$.
I use the I.C. and I obtain $D=E=0$. This is wrong. Have I used the wrong method solving this system?
Best regards//
pde boundary-value-problem wave-equation initial-value-problems
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
$endgroup$
|
show 2 more comments
$begingroup$
I have been searching for a solution online, but cannot find one that fits the B.C. and I.C. for this wave equation. I read through this PDF, page 7; although I had similar conditions I just obtained trivial solutions.
Now, the system is
$left{begin{array}{ll}u_{tt}(x,t)-c^2 u_{xx}(x,t)=0,quad 0<x<L,quad t>0\u(0,t)=0,quad u_{x}(L,t)=Acos(Omega t),quad t>0\u(x,0)=0,quad u_{t}(x,0)quad 0<x<Lend{array}right.$
As usual I use separation of variables and then obtain the solutions for $X(x)$ and $T(t)$
$left{begin{array}{ll}X(x)=Bcos(omega _{1}x)+Csin(omega _{1}x)\T(t)=Dcos(omega _{2}t)+Esin(omega _{2}t)\end{array}right.$
$omega ^{2} _{1}=lambda / c^2,quad omega ^{2}_{2}=lambda,quad lambda>0$.
I use the I.C. and I obtain $D=E=0$. This is wrong. Have I used the wrong method solving this system?
Best regards//
pde boundary-value-problem wave-equation initial-value-problems
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
$endgroup$
$begingroup$
separation of variables won't work unless the boundary conditions are homogeneous
$endgroup$
– Dylan
Feb 1 at 18:52
$begingroup$
Thanks again Dylan, any advice on what method to use?
$endgroup$
– SimpleProgrammer
Feb 1 at 18:54
1
$begingroup$
I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions".
$endgroup$
– Dylan
Feb 1 at 18:57
1
$begingroup$
Here's a different method using the characteristic lines that maybe easier computationally.
$endgroup$
– Dylan
Feb 1 at 19:03
$begingroup$
I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous?
$endgroup$
– SimpleProgrammer
Feb 1 at 19:53
|
show 2 more comments
$begingroup$
I have been searching for a solution online, but cannot find one that fits the B.C. and I.C. for this wave equation. I read through this PDF, page 7; although I had similar conditions I just obtained trivial solutions.
Now, the system is
$left{begin{array}{ll}u_{tt}(x,t)-c^2 u_{xx}(x,t)=0,quad 0<x<L,quad t>0\u(0,t)=0,quad u_{x}(L,t)=Acos(Omega t),quad t>0\u(x,0)=0,quad u_{t}(x,0)quad 0<x<Lend{array}right.$
As usual I use separation of variables and then obtain the solutions for $X(x)$ and $T(t)$
$left{begin{array}{ll}X(x)=Bcos(omega _{1}x)+Csin(omega _{1}x)\T(t)=Dcos(omega _{2}t)+Esin(omega _{2}t)\end{array}right.$
$omega ^{2} _{1}=lambda / c^2,quad omega ^{2}_{2}=lambda,quad lambda>0$.
I use the I.C. and I obtain $D=E=0$. This is wrong. Have I used the wrong method solving this system?
Best regards//
pde boundary-value-problem wave-equation initial-value-problems
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
$endgroup$
I have been searching for a solution online, but cannot find one that fits the B.C. and I.C. for this wave equation. I read through this PDF, page 7; although I had similar conditions I just obtained trivial solutions.
Now, the system is
$left{begin{array}{ll}u_{tt}(x,t)-c^2 u_{xx}(x,t)=0,quad 0<x<L,quad t>0\u(0,t)=0,quad u_{x}(L,t)=Acos(Omega t),quad t>0\u(x,0)=0,quad u_{t}(x,0)quad 0<x<Lend{array}right.$
As usual I use separation of variables and then obtain the solutions for $X(x)$ and $T(t)$
$left{begin{array}{ll}X(x)=Bcos(omega _{1}x)+Csin(omega _{1}x)\T(t)=Dcos(omega _{2}t)+Esin(omega _{2}t)\end{array}right.$
$omega ^{2} _{1}=lambda / c^2,quad omega ^{2}_{2}=lambda,quad lambda>0$.
I use the I.C. and I obtain $D=E=0$. This is wrong. Have I used the wrong method solving this system?
Best regards//
pde boundary-value-problem wave-equation initial-value-problems
pde boundary-value-problem wave-equation initial-value-problems
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
edited Feb 1 at 18:36
SimpleProgrammer
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
asked Feb 1 at 17:57
SimpleProgrammer SimpleProgrammer
719
719
$begingroup$
separation of variables won't work unless the boundary conditions are homogeneous
$endgroup$
– Dylan
Feb 1 at 18:52
$begingroup$
Thanks again Dylan, any advice on what method to use?
$endgroup$
– SimpleProgrammer
Feb 1 at 18:54
1
$begingroup$
I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions".
$endgroup$
– Dylan
Feb 1 at 18:57
1
$begingroup$
Here's a different method using the characteristic lines that maybe easier computationally.
$endgroup$
– Dylan
Feb 1 at 19:03
$begingroup$
I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous?
$endgroup$
– SimpleProgrammer
Feb 1 at 19:53
|
show 2 more comments
$begingroup$
separation of variables won't work unless the boundary conditions are homogeneous
$endgroup$
– Dylan
Feb 1 at 18:52
$begingroup$
Thanks again Dylan, any advice on what method to use?
$endgroup$
– SimpleProgrammer
Feb 1 at 18:54
1
$begingroup$
I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions".
$endgroup$
– Dylan
Feb 1 at 18:57
1
$begingroup$
Here's a different method using the characteristic lines that maybe easier computationally.
$endgroup$
– Dylan
Feb 1 at 19:03
$begingroup$
I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous?
$endgroup$
– SimpleProgrammer
Feb 1 at 19:53
$begingroup$
separation of variables won't work unless the boundary conditions are homogeneous
$endgroup$
– Dylan
Feb 1 at 18:52
$begingroup$
separation of variables won't work unless the boundary conditions are homogeneous
$endgroup$
– Dylan
Feb 1 at 18:52
$begingroup$
Thanks again Dylan, any advice on what method to use?
$endgroup$
– SimpleProgrammer
Feb 1 at 18:54
$begingroup$
Thanks again Dylan, any advice on what method to use?
$endgroup$
– SimpleProgrammer
Feb 1 at 18:54
1
1
$begingroup$
I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions".
$endgroup$
– Dylan
Feb 1 at 18:57
$begingroup$
I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions".
$endgroup$
– Dylan
Feb 1 at 18:57
1
1
$begingroup$
Here's a different method using the characteristic lines that maybe easier computationally.
$endgroup$
– Dylan
Feb 1 at 19:03
$begingroup$
Here's a different method using the characteristic lines that maybe easier computationally.
$endgroup$
– Dylan
Feb 1 at 19:03
$begingroup$
I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous?
$endgroup$
– SimpleProgrammer
Feb 1 at 19:53
$begingroup$
I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous?
$endgroup$
– SimpleProgrammer
Feb 1 at 19:53
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form
$$ u(x,t) = Axcos(Omega t) + v(x,t) $$
where the boundary function was obtained from $f(x)Acos(Omega t)$ such that $f(0)=0$ and $f'(L) = 1$
Then $v(x,t)$ satisfies
begin{cases}
v_{tt} - c^2v_{xx} = Omega^2 Axcos(Omega t)\
v(0,t) = v_x(L,t) = 0 \
v(x,0) = -Ax \
v_t(x,0) = 0
end{cases}
The equation is no longer homogeneous but you can still decompose into eigenfunctions by solving
begin{cases}
X''(x) + lambda^2 X(x) = 0 \
X(0) = X'(L) = 0
end{cases}
Then we have
$$ v(x,t) = sum_{n=0}^infty T_n(t) sinleft(frac{(2n+1)pi}{2L} xright) $$
Plugging into the equation gives
$$ sum_{n=0}^infty left[ T_n''(t) + frac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) right] sinleft(frac{(2n+1)pi}{2L} xright) = Omega^2 Ax cos(Omega t) $$
Decompose the RHS (and also the initial condition) into it's corresponding series
$$ x = sum_{n=0}^infty b_n sinleft(frac{(2n+1)pi}{2L} xright) $$
where
$$ b_n = frac{int_0^L x sinleft(frac{(2n+1)pi}{2L} xright) dx}{int_0^L sin^2left(frac{(2n+1)pi}{2L} xright) dx} $$
You'll get a family of IVPs in $T_n(t)$
begin{cases}
T_n'' + dfrac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) = b_nOmega^2 A cos(Omega t) \
T_n(0) = -b_nA \
T_n'(0) = 0
end{cases}
answered Feb 2 at 4:21
DylanDylan
14.3k31127
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
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oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form
$$ u(x,t) = Axcos(Omega t) + v(x,t) $$
where the boundary function was obtained from $f(x)Acos(Omega t)$ such that $f(0)=0$ and $f'(L) = 1$
Then $v(x,t)$ satisfies
begin{cases}
v_{tt} - c^2v_{xx} = Omega^2 Axcos(Omega t)\
v(0,t) = v_x(L,t) = 0 \
v(x,0) = -Ax \
v_t(x,0) = 0
end{cases}
The equation is no longer homogeneous but you can still decompose into eigenfunctions by solving
begin{cases}
X''(x) + lambda^2 X(x) = 0 \
X(0) = X'(L) = 0
end{cases}
Then we have
$$ v(x,t) = sum_{n=0}^infty T_n(t) sinleft(frac{(2n+1)pi}{2L} xright) $$
Plugging into the equation gives
$$ sum_{n=0}^infty left[ T_n''(t) + frac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) right] sinleft(frac{(2n+1)pi}{2L} xright) = Omega^2 Ax cos(Omega t) $$
Decompose the RHS (and also the initial condition) into it's corresponding series
$$ x = sum_{n=0}^infty b_n sinleft(frac{(2n+1)pi}{2L} xright) $$
where
$$ b_n = frac{int_0^L x sinleft(frac{(2n+1)pi}{2L} xright) dx}{int_0^L sin^2left(frac{(2n+1)pi}{2L} xright) dx} $$
You'll get a family of IVPs in $T_n(t)$
begin{cases}
T_n'' + dfrac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) = b_nOmega^2 A cos(Omega t) \
T_n(0) = -b_nA \
T_n'(0) = 0
end{cases}
answered Feb 2 at 4:21
DylanDylan
14.3k31127
$endgroup$
add a comment |
$begingroup$
You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form
$$ u(x,t) = Axcos(Omega t) + v(x,t) $$
where the boundary function was obtained from $f(x)Acos(Omega t)$ such that $f(0)=0$ and $f'(L) = 1$
Then $v(x,t)$ satisfies
begin{cases}
v_{tt} - c^2v_{xx} = Omega^2 Axcos(Omega t)\
v(0,t) = v_x(L,t) = 0 \
v(x,0) = -Ax \
v_t(x,0) = 0
end{cases}
The equation is no longer homogeneous but you can still decompose into eigenfunctions by solving
begin{cases}
X''(x) + lambda^2 X(x) = 0 \
X(0) = X'(L) = 0
end{cases}
Then we have
$$ v(x,t) = sum_{n=0}^infty T_n(t) sinleft(frac{(2n+1)pi}{2L} xright) $$
Plugging into the equation gives
$$ sum_{n=0}^infty left[ T_n''(t) + frac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) right] sinleft(frac{(2n+1)pi}{2L} xright) = Omega^2 Ax cos(Omega t) $$
Decompose the RHS (and also the initial condition) into it's corresponding series
$$ x = sum_{n=0}^infty b_n sinleft(frac{(2n+1)pi}{2L} xright) $$
where
$$ b_n = frac{int_0^L x sinleft(frac{(2n+1)pi}{2L} xright) dx}{int_0^L sin^2left(frac{(2n+1)pi}{2L} xright) dx} $$
You'll get a family of IVPs in $T_n(t)$
begin{cases}
T_n'' + dfrac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) = b_nOmega^2 A cos(Omega t) \
T_n(0) = -b_nA \
T_n'(0) = 0
end{cases}
answered Feb 2 at 4:21
DylanDylan
14.3k31127
$endgroup$
add a comment |
$begingroup$
You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form
$$ u(x,t) = Axcos(Omega t) + v(x,t) $$
where the boundary function was obtained from $f(x)Acos(Omega t)$ such that $f(0)=0$ and $f'(L) = 1$
Then $v(x,t)$ satisfies
begin{cases}
v_{tt} - c^2v_{xx} = Omega^2 Axcos(Omega t)\
v(0,t) = v_x(L,t) = 0 \
v(x,0) = -Ax \
v_t(x,0) = 0
end{cases}
The equation is no longer homogeneous but you can still decompose into eigenfunctions by solving
begin{cases}
X''(x) + lambda^2 X(x) = 0 \
X(0) = X'(L) = 0
end{cases}
Then we have
$$ v(x,t) = sum_{n=0}^infty T_n(t) sinleft(frac{(2n+1)pi}{2L} xright) $$
Plugging into the equation gives
$$ sum_{n=0}^infty left[ T_n''(t) + frac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) right] sinleft(frac{(2n+1)pi}{2L} xright) = Omega^2 Ax cos(Omega t) $$
Decompose the RHS (and also the initial condition) into it's corresponding series
$$ x = sum_{n=0}^infty b_n sinleft(frac{(2n+1)pi}{2L} xright) $$
where
$$ b_n = frac{int_0^L x sinleft(frac{(2n+1)pi}{2L} xright) dx}{int_0^L sin^2left(frac{(2n+1)pi}{2L} xright) dx} $$
You'll get a family of IVPs in $T_n(t)$
begin{cases}
T_n'' + dfrac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) = b_nOmega^2 A cos(Omega t) \
T_n(0) = -b_nA \
T_n'(0) = 0
end{cases}
answered Feb 2 at 4:21
DylanDylan
14.3k31127
$endgroup$
You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form
$$ u(x,t) = Axcos(Omega t) + v(x,t) $$
where the boundary function was obtained from $f(x)Acos(Omega t)$ such that $f(0)=0$ and $f'(L) = 1$
Then $v(x,t)$ satisfies
begin{cases}
v_{tt} - c^2v_{xx} = Omega^2 Axcos(Omega t)\
v(0,t) = v_x(L,t) = 0 \
v(x,0) = -Ax \
v_t(x,0) = 0
end{cases}
The equation is no longer homogeneous but you can still decompose into eigenfunctions by solving
begin{cases}
X''(x) + lambda^2 X(x) = 0 \
X(0) = X'(L) = 0
end{cases}
Then we have
$$ v(x,t) = sum_{n=0}^infty T_n(t) sinleft(frac{(2n+1)pi}{2L} xright) $$
Plugging into the equation gives
$$ sum_{n=0}^infty left[ T_n''(t) + frac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) right] sinleft(frac{(2n+1)pi}{2L} xright) = Omega^2 Ax cos(Omega t) $$
Decompose the RHS (and also the initial condition) into it's corresponding series
$$ x = sum_{n=0}^infty b_n sinleft(frac{(2n+1)pi}{2L} xright) $$
where
$$ b_n = frac{int_0^L x sinleft(frac{(2n+1)pi}{2L} xright) dx}{int_0^L sin^2left(frac{(2n+1)pi}{2L} xright) dx} $$
You'll get a family of IVPs in $T_n(t)$
begin{cases}
T_n'' + dfrac{c^2(2n+1)^2pi^2}{4L^2} T_n(t) = b_nOmega^2 A cos(Omega t) \
T_n(0) = -b_nA \
T_n'(0) = 0
end{cases}
answered Feb 2 at 4:21
DylanDylan
14.3k31127
edited Feb 2 at 14:34
answered Feb 2 at 4:21
DylanDylan
14.3k31127
answered Feb 2 at 4:21
DylanDylan
14.3k31127
answered Feb 2 at 4:21
DylanDylan
14.3k31127
14.3k31127
add a comment |
add a comment |
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$begingroup$
separation of variables won't work unless the boundary conditions are homogeneous
$endgroup$
– Dylan
Feb 1 at 18:52
$begingroup$
Thanks again Dylan, any advice on what method to use?
$endgroup$
– SimpleProgrammer
Feb 1 at 18:54
1
$begingroup$
I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions".
$endgroup$
– Dylan
Feb 1 at 18:57
1
$begingroup$
Here's a different method using the characteristic lines that maybe easier computationally.
$endgroup$
– Dylan
Feb 1 at 19:03
$begingroup$
I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous?
$endgroup$
– SimpleProgrammer
Feb 1 at 19:53