4.1 Example 1

Let \(\,L=2\pi ,0<x<2\pi ,D=\frac{1}{10},\)\[ u\left ( x,0\right ) =f\left ( x\right ) =\sin x \] And boundary conditions \(u\left ( 0,t\right ) =u\left ( 2\pi ,0\right ) =0\). From above, after carrying the forward transformation, the PDE to solve is found to be

\[ \psi _{t}=D\psi _{xx}\]

With transformed boundary conditions

\begin{align*} \psi _{x}\left ( 0,t\right ) & =0\qquad t>0\\ \psi _{x}\left ( 2\pi ,t\right ) & =0\qquad t>0 \end{align*}

And transformed initial conditions

\[ \psi \left ( x,0\right ) =e^{-\frac{1}{2D}\int \sin \left ( x\right ) \ dx}=e^{\frac{1}{2D}\cos \left ( x\right ) }\]

This heat PDE is standard and has known solution by separation of variables which is

\begin{align*} \psi \left ( x,t\right ) & =c_{0}+\sum _{n=1}^{\infty }c_{n}e^{-D\lambda _{n}t}\cos \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}

Or, since \(L=2\pi \),

\begin{align*} \psi \left ( x,t\right ) & =c_{0}+\sum _{n=1}^{\infty }c_{n}e^{-D\frac{n^{2}}{4}t}\cos \left ( \frac{n}{2}x\right ) \\ \lambda _{n} & =\frac{n^{2}}{4}\qquad n=1,2,3,\cdots \end{align*}

Where \begin{align*} c_{0} & =\frac{1}{L}\int _{0}^{L}\psi \left ( x,0\right ) dx\\ & =\frac{1}{2\pi }\int _{0}^{2\pi }e^{\frac{\cos \left ( x\right ) }{2D}}dx\\ & =\operatorname{BesselI}\left ( 0,\frac{1}{2D}\right ) \end{align*}

And

\begin{align*} c_{n} & =\frac{2}{L}\int _{0}^{L}\psi \left ( x,0\right ) \cos \left ( \sqrt{\lambda _{n}}x\right ) dx\\ & =\frac{1}{\pi }\int _{0}^{2\pi }e^{\frac{\cos \left ( x\right ) }{2D}}\cos \left ( \frac{n}{2}x\right ) dx \end{align*}

The after integral has no closed form solution. Hence the solution is

\[ \psi \left ( x,t\right ) =\operatorname{BesselI}\left ( 0,\frac{1}{2D}\right ) +\frac{1}{\pi }\sum _{n=1}^{\infty }\left ( \int _{0}^{2\pi }e^{\frac{\cos \left ( x\right ) }{2D}}\cos \left ( \frac{n}{2}x\right ) dx\right ) e^{-D\frac{n^{2}}{4}t}\cos \left ( \frac{n}{2}x\right ) \]

But \(\psi =\phi e^{-\int G\left ( t\right ) dt}\), therefore

\[ \phi \left ( x,t\right ) =\psi \left ( x,t\right ) e^{\int G\left ( t\right ) dt}\]

But I do not know what \(G\left ( t\right ) \) is. This was used during the forward transformation only and was eliminated. So how to find \(\phi \left ( x,t\right ) \)? Need to find \(\phi \left ( x,t\right ) \) to be able to find \(u\left ( x,t\right ) \) from \(u\left ( x,t\right ) =-2D\frac{\phi _{x}}{\phi }\).