2.6 Wave PDE with inhomogeneous boundary conditions, 1D (string)

Analytical solution

\begin{align} \frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}} & =c\frac{\partial ^{2}u\left ( x,t\right ) }{\partial x^{2}}\tag{1}\\ u\left ( 0,t\right ) & =0\nonumber \\ \frac{\partial u\left ( L,t\right ) }{\partial x} & =C\nonumber \\ u\left ( x,0\right ) & =0\nonumber \\ \frac{\partial u\left ( x,0\right ) }{\partial t} & =0\nonumber \end{align}

Let \begin{equation} u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \tag{2} \end{equation} \(u_{E}\left ( x\right ) \) is the steady state solution which only needs to satisfy the non-homogeneous boundary conditions. At equilibrium \(\frac{\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=0\) and the PDE becomes \(\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=0\) or the ODE \(\frac{d^{2}u_{E}\left ( x,t\right ) }{dx^{2}}=0\) with B.C. \(u_{E}\left ( 0\right ) =0,u_{E}^{\prime }\left ( L\right ) =C\). The solution to this ODE is\[ u_{E}\left ( x\right ) =c_{1}x+c_{2}\] At first B.C.\[ 0=c_{2}\] Solution becomes \(u_{E}\left ( x\right ) =c_{1}x\). At second B.C. \(u_{E}^{\prime }\left ( x\right ) =c_{1}=C\). Therefore solution is\[ u_{E}\left ( x\right ) =Cx \] Hence\[ u\left ( x,t\right ) =v\left ( x,t\right ) +Cx \] \(v\left ( x,t\right ) \) is the solution to the PDE but with homogeneous B.C. Plugging (2) into (1) gives\[ \frac{\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}}+\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=c\left ( \frac{\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}+\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial x^{2}}\right ) \] But \(\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial x^{2}}=0\) and also \(\frac{\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=0\), hence above becomes\[ \frac{\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}}=c\frac{\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}\] With \(v\left ( x,t\right ) \) having now homogeneous B.C.\begin{align*} v\left ( 0,t\right ) & =0\\ \frac{\partial v\left ( L,t\right ) }{\partial x} & =0 \end{align*}

And initial conditions given by\begin{align*} v\left ( x,0\right ) & =u\left ( x,0\right ) -u_{E}\left ( x\right ) \\ & =0-Cx\\ & =-Cx \end{align*}

And\begin{align*} \frac{\partial v\left ( x,0\right ) }{\partial t} & =\frac{\partial u\left ( x,0\right ) }{\partial t}-\frac{\partial u_{E}\left ( x\right ) }{\partial t}\\ & =0 \end{align*}

In summary, the PDE to solve for \(v\left ( x,t\right ) \) is\begin{align} \frac{\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}} & =c\frac{\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}\tag{3}\\ v\left ( 0,t\right ) & =0\nonumber \\ \frac{\partial v\left ( L,t\right ) }{\partial x} & =0\nonumber \\ v\left ( x,0\right ) & =-Cx\nonumber \\ \frac{\partial v\left ( x,0\right ) }{\partial t} & =0\nonumber \end{align}

Now we solve for PDE (3) for \(v\left ( x,t\right ) \) using separation of variables since the boundary conditions in space are now homogeneous. Let \(v\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right ) \) and the PDE becomes\[ \frac{1}{c}T^{\prime \prime }X=X^{\prime \prime }T \] Dividing by \(XT\neq 0\) gives\begin{equation} \frac{1}{c}\frac{T^{\prime \prime }}{T}=\frac{X^{\prime \prime }}{X}=-\lambda \tag{4} \end{equation} Where \(\lambda \) is some real positive constant. The space ODE becomes\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

Case \(\lambda <0\): Let The solution is\[ X\left ( x\right ) =c_{1}\cosh \left ( \sqrt{\lambda }x\right ) +c_{2}\sinh \left ( \sqrt{\lambda }x\right ) \] At \(x=0\)\[ 0=c_{1}\] Hence solution becomes \[ X\left ( x\right ) =c_{2}\sinh \left ( \sqrt{\lambda }x\right ) \] Taking derivative\[ X^{\prime }\left ( x\right ) =\sqrt{\lambda }c_{2}\cosh \left ( \sqrt{\lambda }x\right ) \] Using second boundary conditions gives\[ 0=\sqrt{\lambda }c_{2}\cosh \left ( \sqrt{\lambda }L\right ) \] Since \(\cosh \) is zero only when its argument is zero. But we assumed \(\sqrt{\lambda }\) not zero here, then \(c_{2}=0\) in only other choice. Hence this gives trivial solution. Therefore \(\lambda <0\) is not possible.

Case \(\lambda =0\)\begin{align*} X^{\prime \prime } & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

Solution is \(X\left ( x\right ) =c_{1}x+c_{2}\). First B.C. gives \(0=c_{2}\). Solution becomes \(X\left ( x\right ) =c_{1}x\). Second B.C. gives \(c_{1}=0\). This gives trivial solution again. Hence \(\lambda =0\) is not possible eigenvalue.

Case \(\lambda >0\): The solution becomes\[ X\left ( x\right ) =B_{1}\cos \left ( \sqrt{\lambda }x\right ) +B_{2}\sin \left ( \sqrt{\lambda }x\right ) \] AT first B.C.\[ 0=B_{1}\] Hence solution becomes\[ X\left ( x\right ) =B_{2}\sin \left ( \sqrt{\lambda }x\right ) \] Taking derivative\[ X^{\prime }\left ( x\right ) =\sqrt{\lambda }B_{2}\cos \left ( \sqrt{\lambda }x\right ) \] At second B.C.\[ 0=\sqrt{\lambda }B_{2}\cos \left ( \sqrt{\lambda }L\right ) \] To avoid trivial solution, take \(\cos \left ( \sqrt{\lambda }L\right ) =0\) or \(\sqrt{\lambda }L=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\cdots \) or \begin{align} \sqrt{\lambda } & =\frac{\pi }{2L},\frac{3\pi }{2L},\frac{5\pi }{2L},\cdots \nonumber \\ \sqrt{\lambda _{n}} & =\left ( \frac{n\pi }{2L}\right ) \qquad n=1,3,5,\cdots \nonumber \\ & =\frac{\left ( 2n+1\right ) \pi }{2L}\qquad n=0,1,2,\cdots \tag{5} \end{align}

Hence the space solution is\begin{equation} X_{n}\left ( x\right ) =B_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \tag{6} \end{equation} Now we solve the time ODE \(T\left ( t\right ) \) from (4), which is\[ T^{\prime \prime }+\lambda cT=0 \] The solution is\[ T_{n}\left ( t\right ) =D_{n}\cos \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt{\lambda _{n}c}t\right ) \] Therefore\begin{align*} v\left ( x,t\right ) & =\sum _{n=0}^{\infty }T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=0}^{\infty }\left ( D_{n}\cos \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt{\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}

Where constant \(B_{n}\) merged with the other constants. Now At \(t=0\)\[ -Cx=\sum _{n=0}^{\infty }D_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \] Applying orthogonality\begin{align*} -\int _{0}^{L}Cx\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx & =D_{n}\int _{0}^{L}\sin ^{2}\left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx\\ -C\int _{0}^{L}x\sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) dx & =D_{n}\frac{L}{2}\\ -C\left ( \frac{4\left ( -1\right ) ^{n}L^{2}}{\left ( \pi +2n\pi \right ) ^{2}}\right ) & =D_{n}\frac{L}{2}\\ D_{n} & =-C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}} \end{align*}

Therefore\[ v\left ( x,t\right ) =\sum _{n=0}^{\infty }\left ( -C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt{\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \] Taking time derivative\[ \frac{\partial v\left ( x,t\right ) }{\partial t}=\sum _{n=0}^{\infty }\left ( C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\frac{\left ( 2n+1\right ) \pi }{2L}\sin \left ( \sqrt{\lambda _{n}c}t\right ) +E_{n}\sqrt{\lambda _{n}c}\cos \left ( \sqrt{\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \] At \(t=0\)\[ 0=\sum _{n=0}^{\infty }E_{n}\sqrt{\lambda _{n}c}\sin \left ( \sqrt{\lambda _{n}}x\right ) \] Hence \(E_{n}=0\). Therefore solution becomes\[ v\left ( x,t\right ) =\sum _{n=0}^{\infty }-C\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \sqrt{\lambda _{n}c}t\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \] Therefore, since \(u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \) then\begin{equation} u\left ( x,t\right ) =Cx-C\sum _{n=0}^{\infty }\frac{8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \frac{\left ( 2n+1\right ) \pi }{2L}\sqrt{c}t\right ) \sin \left ( \frac{\left ( 2n+1\right ) \pi }{2L}x\right ) \tag{7} \end{equation}

Animation

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