Mixing
Wave PDE with Neumann and Dirichlet initial condition
$begingroup$
So I'm given the following PDE problem, which I try to solve using variable separation:
$dfrac{d^2u(x,t)}{dt^2}-c^2dfrac{d^2u(x,t)}{dx^2}=0$
$0<x<L,enspace t>0$
$u(0,t)=0, dfrac{du(L,t)}{dx}=Acos(Omega t)$
$u(x,0)=dfrac{du(x,0)}{dt}=0$
I really don't get how this can hold up since the ODE in terms of $T(t)$ has a solution in terms of $T(t) = c_1cos(k_nx)+c_2sin(k_nx)$, and both $c_1$ and $c_2$ must be $0$ for the initial conditions to hold. Since it's evaluated at $t=0$ I can't set $k_n$ to fit these conditions either. I find it highly unlikely that there are no non-trivial solutions. What am I missing here?
pde wave-equation sturm-liouville
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
asked Jan 30 at 21:14
AndreasAndreas
1
$endgroup$
add a comment |
$begingroup$
So I'm given the following PDE problem, which I try to solve using variable separation:
$dfrac{d^2u(x,t)}{dt^2}-c^2dfrac{d^2u(x,t)}{dx^2}=0$
$0<x<L,enspace t>0$
$u(0,t)=0, dfrac{du(L,t)}{dx}=Acos(Omega t)$
$u(x,0)=dfrac{du(x,0)}{dt}=0$
I really don't get how this can hold up since the ODE in terms of $T(t)$ has a solution in terms of $T(t) = c_1cos(k_nx)+c_2sin(k_nx)$, and both $c_1$ and $c_2$ must be $0$ for the initial conditions to hold. Since it's evaluated at $t=0$ I can't set $k_n$ to fit these conditions either. I find it highly unlikely that there are no non-trivial solutions. What am I missing here?
pde wave-equation sturm-liouville
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
asked Jan 30 at 21:14
AndreasAndreas
1
$endgroup$
1
$begingroup$
By setting $u(x,t) = v(x,t) + A x cos(Omega t)$ you can transfer the non-homogeneous boundary condition for $u$ to a source term for $v$.
$endgroup$
– Christoph
Jan 31 at 5:49
add a comment |
$begingroup$
So I'm given the following PDE problem, which I try to solve using variable separation:
$dfrac{d^2u(x,t)}{dt^2}-c^2dfrac{d^2u(x,t)}{dx^2}=0$
$0<x<L,enspace t>0$
$u(0,t)=0, dfrac{du(L,t)}{dx}=Acos(Omega t)$
$u(x,0)=dfrac{du(x,0)}{dt}=0$
I really don't get how this can hold up since the ODE in terms of $T(t)$ has a solution in terms of $T(t) = c_1cos(k_nx)+c_2sin(k_nx)$, and both $c_1$ and $c_2$ must be $0$ for the initial conditions to hold. Since it's evaluated at $t=0$ I can't set $k_n$ to fit these conditions either. I find it highly unlikely that there are no non-trivial solutions. What am I missing here?
pde wave-equation sturm-liouville
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
asked Jan 30 at 21:14
AndreasAndreas
1
$endgroup$
So I'm given the following PDE problem, which I try to solve using variable separation:
$dfrac{d^2u(x,t)}{dt^2}-c^2dfrac{d^2u(x,t)}{dx^2}=0$
$0<x<L,enspace t>0$
$u(0,t)=0, dfrac{du(L,t)}{dx}=Acos(Omega t)$
$u(x,0)=dfrac{du(x,0)}{dt}=0$
I really don't get how this can hold up since the ODE in terms of $T(t)$ has a solution in terms of $T(t) = c_1cos(k_nx)+c_2sin(k_nx)$, and both $c_1$ and $c_2$ must be $0$ for the initial conditions to hold. Since it's evaluated at $t=0$ I can't set $k_n$ to fit these conditions either. I find it highly unlikely that there are no non-trivial solutions. What am I missing here?
pde wave-equation sturm-liouville
pde wave-equation sturm-liouville
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
asked Jan 30 at 21:14
AndreasAndreas
1
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
asked Jan 30 at 21:14
AndreasAndreas
1
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
edited Jan 30 at 22:47
DisintegratingByParts
60.3k42681
60.3k42681
asked Jan 30 at 21:14
AndreasAndreas
1
asked Jan 30 at 21:14
AndreasAndreas
1
asked Jan 30 at 21:14
AndreasAndreas
1
1
1
$begingroup$
By setting $u(x,t) = v(x,t) + A x cos(Omega t)$ you can transfer the non-homogeneous boundary condition for $u$ to a source term for $v$.
$endgroup$
– Christoph
Jan 31 at 5:49
add a comment |
1
$begingroup$
By setting $u(x,t) = v(x,t) + A x cos(Omega t)$ you can transfer the non-homogeneous boundary condition for $u$ to a source term for $v$.
$endgroup$
– Christoph
Jan 31 at 5:49
1
1
$begingroup$
By setting $u(x,t) = v(x,t) + A x cos(Omega t)$ you can transfer the non-homogeneous boundary condition for $u$ to a source term for $v$.
$endgroup$
– Christoph
Jan 31 at 5:49
$begingroup$
By setting $u(x,t) = v(x,t) + A x cos(Omega t)$ you can transfer the non-homogeneous boundary condition for $u$ to a source term for $v$.
$endgroup$
– Christoph
Jan 31 at 5:49
add a comment |
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By setting $u(x,t) = v(x,t) + A x cos(Omega t)$ you can transfer the non-homogeneous boundary condition for $u$ to a source term for $v$.
$endgroup$
– Christoph
Jan 31 at 5:49