#### 1.1.11 Diﬀusion-Convection. Both ends at zero temperature, no source. (also called advection)

$$u_{t}=ku_{xx}+au_{x} \tag{1}$$

Where $$k>0,a>0$$ and $$t>0,0<x<\pi$$. BC are\begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( \pi ,t\right ) & =0 \end{align*}

Initial conditions$u\left ( x,0\right ) =\sin \left ( x\right ) \qquad 0<x<\pi$ This problem can not currently be solved in Maple nor by Mathematica analytically.

solution

Trying separation of variables. Let $$u=XT$$, then the PDE becomes\begin{align*} T^{\prime }X & =kX^{\prime \prime }T+aX^{\prime }T\\ \frac{1}{k}\frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}+\frac{a}{k}\frac{X^{\prime }}{X}=-\lambda \end{align*}

Where $$\lambda$$ is separation constant. Hence$T^{\prime }+k\lambda T=0$ Which has solution $$T\left ( t\right ) =e^{-k\lambda t}$$ (the constant of integration is not needed, it will be combined with constant coming from the spatial ODE), and the spatial ODE is\begin{align*} \frac{X^{\prime \prime }}{X}+\frac{a}{k}\frac{X^{\prime }}{X} & =-\lambda \\ X^{\prime \prime }+\frac{a}{k}X^{\prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( \pi \right ) & =0 \end{align*}

The characteristic equation is $$r^{2}+\frac{a}{k}r+\lambda =0$$.

Case $$\lambda =0$$ Then $$r^{2}+\frac{a}{k}r=0$$ or $$r\left ( r+\frac{a}{k}\right ) =0$$. Hence $$r=0$$ or $$r=-\frac{a}{k}$$. So the solution is$X\left ( x\right ) =c_{1}+c_{2}e^{-\frac{a}{k}x}$ At $$X\left ( 0\right ) =0$$ the above gives $$0=c_{1}+c_{2}$$. Therefore the solution is $$X\left ( x\right ) =c_{1}\left ( 1-e^{\frac{-a}{k}x}\right )$$. At $$X\left ( \pi \right ) =0$$ then $$c_{1}\left ( 1-e^{\frac{a}{k}\pi }\right ) =0$$. This means this for non-trivial solution $$e^{\frac{-a}{k}\pi }=1$$ or $$\frac{-a}{k}\pi =0$$. Hence $$\lambda =0$$ is not an eigenvalue.

Case $$\lambda <0$$ Then $$r^{2}+\frac{a}{k}r+\lambda =0$$ or $$r=\frac{-b}{2}\pm \frac{\sqrt{b^{2}-4ac}}{2a}=\frac{-a}{2k}\pm \frac{\sqrt{\frac{a^{2}}{k^{2}}-4\lambda }}{2}$$. Since $$\lambda <0$$ then the term under the sqrt root is positive. Hence the solution will have real roots. Not complex conjugate. Hence the solution is pure exponentials.  Let $$\frac{a^{2}}{k^{2}}-4\lambda =\alpha$$ and $$\frac{a}{2k}=\beta$$ then\begin{align*} r_{1} & =-\beta +\sqrt{\alpha }\\ r_{2} & =-\beta -\sqrt{\alpha } \end{align*}

Then $$X\left ( x\right ) =c_{1}\cosh \left ( \left ( -\beta +\alpha \right ) x\right ) +c_{2}\sinh \left ( \left ( -\beta -\alpha \right ) x\right )$$. At $$X\left ( 0\right ) =0$$ this gives $$0=c_{1}$$. Hence solution is $$X\left ( x\right ) =c_{2}\sinh \left ( \left ( -\beta -\alpha \right ) x\right )$$. At $$X\left ( \pi \right ) =0$$, this gives$0=c_{2}\sinh \left ( \left ( -\beta -\alpha \right ) \pi \right )$ For non-trivial solution, $$\sinh \left ( \left ( -\beta -\alpha \right ) \pi \right ) =0$$. But this is zero only when $$\left ( -\beta -\alpha \right ) \pi$$ or $$\beta =-\alpha$$. Or $$\frac{-a}{2k}=-\sqrt{\frac{a^{2}}{k^{2}}-4\lambda }$$. Or $$\frac{a^{2}}{4k^{2}}=\frac{a^{2}}{k^{2}}-4\lambda$$ or\begin{align*} 4\lambda & =\frac{a^{2}}{k^{2}}-\frac{a^{2}}{4k^{2}}\\ \lambda & =\frac{3}{16}\frac{a^{2}}{k^{2}} \end{align*}

But $$\lambda$$ was assumed negative. Hence this is not possible. $$\lambda <0$$ is not eigenvalue.

Case $$\lambda >0$$ Then $$r^{2}+\frac{a}{k}r+\lambda =0$$ or $$r=\frac{-a}{2k}\pm \frac{1}{2}\sqrt{\frac{a^{2}}{k^{2}}-4\lambda }=\frac{-a}{2k}\pm \sqrt{\frac{a^{2}}{4k^{2}}-\lambda }$$. Since $$\lambda >0$$ then the term under the sqrt root can be negative. Then only when $$\lambda >\frac{a^{2}}{4k^{2}}$$ will there be complex roots. Therefore assuming $$\lambda >\frac{a^{2}}{4k^{2}}$$, then\begin{align*} r_{1} & =\frac{-a}{2k}+i\sqrt{\lambda -\frac{a^{2}}{4k^{2}}}\\ r_{2} & =\frac{-a}{2k}-i\sqrt{\lambda -\frac{a^{2}}{4k^{2}}} \end{align*}

The solution is$X\left ( x\right ) =e^{\frac{-a}{2k}x}\left ( c_{1}\cos \left ( \sqrt{\lambda -\frac{a^{2}}{4k^{2}}}x\right ) +c_{2}\sin \left ( \sqrt{\lambda -\frac{a^{2}}{4k^{2}}}x\right ) \right )$ At $$X\left ( 0\right ) =0$$ this gives $$0=c_{1}$$. Hence solution is$X\left ( x\right ) =e^{\frac{-a}{2k}x}c_{2}\sin \left ( \sqrt{\lambda -\frac{a^{2}}{4k^{2}}}x\right )$ At $$X\left ( \pi \right ) =0$$, this gives$0=e^{\frac{-a}{2k}\pi }c_{2}\sin \left ( \sqrt{\lambda -\frac{a^{2}}{4k^{2}}}\pi \right )$ For non-trivial solution, $$\sin \left ( \sqrt{\lambda -\frac{a^{2}}{4k^{2}}}\pi \right ) =0$$ or $$\sqrt{\lambda -\frac{a^{2}}{4k^{2}}}\pi =n\pi$$. Hence\begin{align*} \lambda -\frac{a^{2}}{4k^{2}} & =n^{2}\qquad n=1,2,3,\cdots \\ \lambda _{n} & =n^{2}+\frac{a^{2}}{4k^{2}}\qquad n=1,2,3,\cdots \end{align*}

And the corresponding eigenfunction is \begin{align*} X_{n}\left ( x\right ) & =e^{\frac{-a}{2k}x}c_{n}\sin \left ( \sqrt{\lambda -\frac{a^{2}}{4k^{2}}}x\right ) \qquad n=1,2,3,\cdots \\ & =e^{\frac{-a}{2k}x}c_{n}\sin \left ( nx\right ) \end{align*}

Therefore the solution is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }c_{n}e^{-k\lambda _{n}t}e^{\frac{-a}{2k}x}\sin \left ( nx\right ) \\ & =\sum _{n=1}^{\infty }c_{n}e^{-k\left ( n^{2}+\frac{a^{2}}{4k^{2}}\right ) t}e^{\frac{-a}{2k}x}\sin \left ( nx\right ) \end{align*}

$$c_{n}$$ is found from initial conditions. At $$t=0$$\begin{align*} \sin \left ( x\right ) & =\sum _{n=1}^{\infty }c_{n}e^{\frac{-a}{2k}x}\sin \left ( nx\right ) \\ e^{\frac{a}{2k}x}\sin \left ( x\right ) & =\sum _{n=1}^{\infty }c_{n}\sin \left ( nx\right ) \end{align*}

Therefore, by orthogonality\begin{align*} \int _{0}^{\pi }e^{\frac{a}{2k}x}\sin \left ( x\right ) \sin \left ( nx\right ) dx & =c_{n}\frac{\pi }{2}\\ \frac{-16nak^{3}\left ( 1+(-1)^{n}e^{\frac{\pi a}{2k}}\right ) }{a^{4}+16k^{4}\left ( n^{2}-1\right ) ^{2}+8a^{2}k^{2}\left ( 1+n^{2}\right ) } & =c_{n}\frac{\pi }{2}\\ c_{n} & =-\frac{1}{\pi }\frac{32nak^{3}\left ( 1+(-1)^{n}e^{\frac{\pi a}{2k}}\right ) }{a^{4}+16k^{4}\left ( n^{2}-1\right ) ^{2}+8a^{2}k^{2}\left ( 1+n^{2}\right ) } \end{align*}

Hence the solution becomes\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }c_{n}e^{-k\left ( n^{2}+\frac{a^{2}}{4k^{2}}\right ) t}e^{\frac{-a}{2k}x}\sin \left ( nx\right ) \\ & =\frac{-32ak^{3}}{\pi }\sum _{n=1}^{\infty }\left ( \frac{n\left ( 1+(-1)^{n}e^{\frac{\pi a}{2k}}\right ) }{a^{4}+16k^{4}\left ( n^{2}-1\right ) ^{2}+8a^{2}k^{2}\left ( 1+n^{2}\right ) }\right ) e^{-k\left ( n^{2}+\frac{a^{2}}{4k^{2}}\right ) t}e^{\frac{-a}{2k}x}\sin \left ( nx\right ) \end{align*}

The ﬁrst animation is for $$a=1,k=1$$ and the second for $$a=1,k=0.1$$.

The following ﬁrst animation is for $$a=5,k=1$$ and the second for $$a=10,k=1$$. This shows the eﬀect of increasing the convection.