Wave PDE with Neumann and Dirichlet initial condition












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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







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  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 22:47









DisintegratingByParts

60.3k42681




60.3k42681










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














  • 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










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