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## Analytical solution to speciﬁc Stockes ﬁrst problem PDE

June 12, 2017   Compiled on May 23, 2020 at 5:39am

Solve

\begin{align} \frac{\partial u}{\partial t} & =k\frac{\partial ^{2}u}{\partial x^{2}}\tag{1}\\ 0 & <x<L\nonumber \\ t & >0\nonumber \end{align}

Initial conditions

$u\left ( 0,x\right ) =0$

Boundary conditions

\begin{align*} u\left ( 0,t\right ) & =\sin \left ( t\right ) \\ u\left ( t,L\right ) & =0 \end{align*}

Let $$u=v+u_{E} \tag{2}$$ where $$u_{E}\left ( x,t\right )$$ is steady state solution that only needs to satisfy boundary conditions and $$v\left ( x,t\right )$$ satisﬁes the PDE itself but with homogenous B.C.  At steady state, the PDE becomes

\begin{align*} 0 & =k\frac{d^{2}u_{E}}{dx^{2}}\\ u_{E}\left ( 0\right ) & =\sin \left ( t\right ) \\ u_{E}\left ( L\right ) & =0 \end{align*}

The solution is $$u_{E}\left ( t\right ) =\left ( \frac{L-x}{L}\right ) \sin \left ( t\right )$$.  Hence (2) becomes

$u\left ( x,t\right ) =v\left ( x,t\right ) +\left ( \frac{L-x}{L}\right ) \sin \left ( t\right )$

Substituting the above in (1) gives

\begin{align} \frac{\partial v}{\partial t}+\left ( \frac{L-x}{L}\right ) \cos \left ( t\right ) & =k\frac{\partial ^{2}v}{\partial x^{2}}\nonumber \\ \frac{\partial v}{\partial t} & =k\frac{\partial ^{2}v}{\partial x^{2}}+\left ( \frac{x-L}{L}\right ) \cos \left ( t\right ) \nonumber \\ \frac{\partial v}{\partial t} & =k\frac{\partial ^{2}v}{\partial x^{2}}+Q\left ( x,t\right ) \tag{3} \end{align}

With boundary conditions $$u_{0}\left ( 0,t\right ) =0,u\left ( L,t\right ) =0$$. This is now in standard form and separation of variables can be used to solve it. $Q\left ( x,t\right ) =\left ( \frac{x-L}{L}\right ) \cos \left ( t\right )$ Now acts as a source term. The eigenfunctions are known to be $$\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right )$$ where $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}$$. Hence by eigenfunction expansion, the solution to (3) is

$$v\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{3A}$$

Substituting this into (3) gives

$$\sum _{n=1}^{\infty }\frac{dB_{n}\left ( t\right ) }{dt}\Phi _{n}\left ( x\right ) =k\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +Q\left ( x,t\right ) \tag{4}$$

Expanding $$Q\left ( x,t\right )$$ using same basis (eigenfunctions) gives

$Q\left ( x,t\right ) =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right )$

Applying orthogonality

\begin{align*} \int _{0}^{L}Q\left ( x,t\right ) \Phi _{m}\left ( x\right ) dx & =\int _{0}^{L}\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\\ & =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \int _{0}^{L}\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx \end{align*}

But $$\sum _{n=1}^{\infty }\int _{0}^{L}\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx=\int _{0}^{L}\Phi _{m}^{2}\left ( x\right ) dx=\frac{L}{2}$$ since $$\Phi _{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{L}x\right )$$ and the above simpliﬁes to

\begin{align*} \int _{0}^{L}Q\left ( x,t\right ) \Phi _{n}\left ( x\right ) dx & =\frac{L}{2}q_{n}\left ( t\right ) \\ q_{n}\left ( t\right ) & =\frac{2}{L}\int _{0}^{L}Q\left ( x,t\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx \end{align*}

But $$Q\left ( x,t\right ) =\left ( \frac{x-L}{L}\right ) \cos \left ( t\right )$$, hence

\begin{align*} q_{n}\left ( t\right ) & =\frac{2}{L}\int _{0}^{L}\left ( \frac{x-L}{L}\right ) \cos \left ( t\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =\frac{-2}{n\pi }\cos \left ( t\right ) \end{align*}

Therefore $$Q\left ( x,t\right ) =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=1}^{\infty }\frac{-2}{n\pi }\cos \left ( t\right ) \sin \left ( \frac{n\pi }{L}x\right )$$ and (4) becomes

\begin{align*} \sum _{n=1}^{\infty }\frac{dB_{n}\left ( t\right ) }{dt}\Phi _{n}\left ( x\right ) & =k\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) -\sum _{n=1}^{\infty }\frac{2}{n\pi }\cos \left ( t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ \frac{dB_{n}\left ( t\right ) }{dt}\sin \left ( \frac{n\pi }{L}x\right ) & =kB_{n}\left ( t\right ) \left ( -\frac{n^{2}\pi ^{2}}{L^{2}}\sin \left ( \frac{n\pi }{L}x\right ) \right ) -\frac{2}{n\pi }\cos \left ( t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \\ \frac{dB_{n}\left ( t\right ) }{dt}+B_{n}\left ( t\right ) k\frac{n^{2}\pi ^{2}}{L^{2}} & =-\frac{2}{n\pi }\cos \left ( t\right ) \end{align*}

This is an ODE in $$B_{n}\left ( t\right )$$ whose solution is

$B_{n}\left ( t\right ) =C_{n}e^{-k\left ( \frac{n^{2}\pi ^{2}}{L^{2}}\right ) t}-\frac{2L^{2}\left ( kn^{2}\pi ^{2}\cos t+L^{2}\sin t\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }$

From (3A) $$v\left ( x,t\right )$$ now becomes

$$v\left ( x,t\right ) =\sum _{n=1}^{\infty }C_{n}e^{-k\left ( \frac{n^{2}\pi ^{2}}{L^{2}}\right ) t}\sin \left ( \frac{n\pi }{L}x\right ) -\frac{2L^{2}\left ( kn^{2}\pi ^{2}\cos t+L^{2}\sin t\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }\sin \left ( \frac{n\pi }{L}x\right ) \tag{5}$$

To ﬁnd $$C_{n}$$, from initial conditions, at $$t=0$$ the above becomes

$0=\sum _{n=1}^{\infty }C_{n}\sin \left ( \frac{n\pi }{L}x\right ) -\frac{2L^{2}\left ( kn^{2}\pi ^{2}\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }\sin \left ( \frac{n\pi }{L}x\right )$

Hence

$C_{n}=\frac{2L^{2}\left ( kn^{2}\pi ^{2}\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }$

Therefore (5) becomes

$v\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac{2L^{2}\left ( kn^{2}\pi ^{2}\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }e^{-k\left ( \frac{n^{2}\pi ^{2}}{L^{2}}\right ) t}-\frac{2L^{2}\left ( kn^{2}\pi ^{2}\cos t+L^{2}\sin t\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }\right ) \sin \left ( \frac{n\pi }{L}x\right )$

And since $$u=v+u_{E}$$ then the solution is

$u\left ( x,t\right ) =\left ( \sum _{n=1}^{\infty }\left ( \frac{2L^{2}\left ( kn^{2}\pi ^{2}\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }e^{-k\left ( \frac{n^{2}\pi ^{2}}{L^{2}}\right ) t}-\frac{2L^{2}\left ( kn^{2}\pi ^{2}\cos t+L^{2}\sin t\right ) }{n\pi \left ( L^{4}+k^{2}n^{4}\pi ^{4}\right ) }\right ) \sin \left ( \frac{n\pi }{L}x\right ) \right ) +\left ( \frac{L-x}{L}\right ) \sin \left ( t\right )$

To simulate

Here is the animation from the above

Here is the numerical solution to compare with

Here is the animation from the above