### 9 Heat PDE inside disk

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#### 9.1 circumference ﬁxed at zero temperature, no $$\theta$$ dependency, No source

problem number 75

Taken from Mathematica DSolve help pages

Solve the heat equation in polar coordinates for $$u(r,t)$$

$\frac{ \partial u}{\partial t}= \frac{ \partial ^2 u}{\partial r^2} + \frac{1}{r} \frac{ \partial u}{\partial r}$

For $$0<r<1$$ and $$t>0$$. The boundary conditions are \begin{align*} u(1,t) &= 0 \end{align*}

Initial condition is $$u(r,0)=1-r$$.

Mathematica

$\text{DSolve}\left [\left \{u^{(0,1)}(r,t)=f(r,t)+k \left (\frac{u^{(1,0)}(r,t)}{r}+u^{(2,0)}(r,t)\right ),u(r,0)=0,u(a,t)=0\right \},u(r,t),\{r,t\},\text{Assumptions}\to r<a\right ]$

Maple

$u \left ( r,t \right ) ={\it invlaplace} \left ( \BesselJ \left ( 0,\sqrt{-s}r \right ){\it \_F2} \left ( s \right ) ,s,t \right ) -{\it invlaplace} \left ({\frac{\BesselY \left ( 0,\sqrt{-s}r \right ) \BesselJ \left ( 0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ){\it \_F2} \left ( s \right ) }{\BesselY \left ( 0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ) }},s,t \right ) -1/2\,\pi \,{\it invlaplace} \left ({\frac{\BesselY \left ( 0,\sqrt{-s}r \right ) \StruveH \left ( 0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ) \LommelS 1 \left ( 2,0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ) }{\BesselY \left ( 0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ){s}^{2}}},s,t \right ) -{\it invlaplace} \left ({\frac{\BesselY \left ( 0,\sqrt{-s}r \right ) \left ( \LommelS 1 \left ( 2,0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ) \right ) ^{2}}{\BesselY \left ( 0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ){s}^{2}}},s,t \right ) -{\it invlaplace} \left ({\frac{\BesselY \left ( 0,\sqrt{-s}r \right ) }{\BesselY \left ( 0,1/2\,\pi \,\StruveH \left ( 0,\sqrt{-s} \right ) +\LommelS 1 \left ( 2,0,\sqrt{-s} \right ) \right ) s}},s,t \right ) +{\it invlaplace} \left ({\frac{\LommelS 1 \left ( 2,0,\sqrt{-s}r \right ) }{ \left ( -s \right ) ^{3/2}}},s,t \right ) +1$ But has unresolved inverse Laplace transforms

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#### 9.2 circumference ﬁxed at zero temperature, no $$\theta$$ dependency, zero initial conditions, source that depends on time and radius (Haberman 8.3.5)

problem number 76

Added Nov 24, 2018.

Problem 8.3.5 from Richard Haberman applied partial diﬀerential equations book, 5th edition

Solve for $$u(r,t)$$

$\frac{ \partial u}{\partial t}= k \nabla ^2 u + f(r,t)$

Inside the circle ($$r<a$$) with $$u=0$$ at $$r=a$$ and initially $$u=0$$.

See my hand solution for this in HW 9, Math 322, UW Madison. One of the problem here, is how to tell CAS the implicit condition when solving this which is that $$u(0,t)<\infty$$.

Mathematica

$\text{DSolve}\left [\left \{u^{(0,0,1)}(r,\theta ,t)=\frac{1}{50} \left (\frac{u^{(0,2,0)}(r,\theta ,t)}{r^2}+\frac{u^{(1,0,0)}(r,\theta ,t)}{r}+u^{(2,0,0)}(r,\theta ,t)\right ),u^{(1,0,0)}(1,\theta ,t)=0,\left \{u(r,0,t)=0,u\left (r,\frac{\pi }{2},t\right )=0\right \},u(r,\theta ,0)=\left (r-\frac{r^3}{3}\right ) \sin (\theta )\right \},u(r,\theta ,t),\{r,\theta ,t\}\right ]$

Maple

$\text{ sol=() }$

Hand solution

Since this problem has homogeneous B.C. but has time dependent source (i.e. non-homogenous in the PDE itself), then we will use the method of eigenfunction expansion. In this method, we ﬁrst ﬁnd the eigenfunctions $$\phi _{n}\left ( x\right )$$ of the associated homogenous PDE without the source being present. Then use these $$\phi _{n}\left ( x\right )$$ to expand the source $$f\left ( x,t\right )$$ as generalized Fourier series. We now switch to the associated homogenous PDE in order to ﬁnd the eigenfunctions. $$u\equiv u\left ( r,t\right )$$. There is no $$\theta$$. Hence\begin{align} \frac{\partial u\left ( r,t\right ) }{\partial t} & =k\left ( \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}\right ) \tag{1}\\ u\left ( a,t\right ) & =0\nonumber \\ \left \vert u\left ( 0,t\right ) \right \vert & <\infty \nonumber \\ u\left ( r,0\right ) & =0\nonumber \end{align}

We need to solve the above in order to ﬁnd the eigenfunctions $$\phi _{n}\left ( r\right )$$. Let $$u=R\left ( r\right ) T\left ( t\right )$$. Substituting this back into (1) gives$T^{\prime }R=k\left ( R^{\prime \prime }T+\frac{1}{r}R^{\prime }T\right )$ Dividing by $$RT$$$\frac{1}{k}\frac{T^{\prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}$ Let separation constant be $$-\lambda$$. We obtain$T^{\prime }+k\lambda T=0$ And\begin{align*} \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R} & =-\lambda \\ R^{\prime \prime }+\frac{1}{r}R^{\prime } & =-\lambda R\\ rR^{\prime \prime }+R^{\prime }+\lambda rR & =0 \end{align*}

This is a singular Sturm-Liouville ODE. Standard form is$\left ( rR^{\prime }\right ) ^{\prime }=-\lambda rR$ Hence\begin{align*} p & =r\\ q & =0\\ \sigma & =r \end{align*}

We solved $$R^{\prime \prime }+\frac{1}{r}R^{\prime }+\lambda R=0$$ before. The solution is$R_{n}\left ( r\right ) =J_{0}\left ( \sqrt{\lambda _{n}}r\right )$ Where $$\sqrt{\lambda _{n}}$$ is found by solving $$J_{0}\left ( \sqrt{\lambda _{n}}a\right ) =0$$. Now that we know what the eigenfunctions are, then we write$$u\left ( r,t\right ) =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \tag{2}$$ Where $$a_{n}\left ( t\right )$$ is function of time since it includes the time solution in it. Now we use the above in the original PDE with the source in it$$\frac{\partial u\left ( r,t\right ) }{\partial t}=k\nabla ^{2}u+f\left ( r,t\right ) \tag{3}$$ Where $$\nabla ^{2}u=-\lambda u$$. Substituting (2) into (3), and using $$f\left ( r,t\right ) =\sum _{n=1}^{\infty }f_{n}\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right )$$ gives\begin{align*} \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) & =-k\sum _{n=1}^{\infty }\lambda _{n}a_{n}\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) +\sum _{n=1}^{\infty }f_{n}\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \\ \sum _{n=1}^{\infty }\left ( a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) \right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) & =\sum _{n=1}^{\infty }f_{n}\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) \end{align*}

Applying orthogonality, the above simpliﬁes to$a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) =f_{n}\left ( t\right )$ The solution is$a_{n}\left ( t\right ) =e^{-k\lambda _{n}t}\int _{0}^{t}f_{n}\left ( s\right ) e^{k\lambda _{n}s}ds+ce^{-k\lambda _{n}t}$ To ﬁnd constant of integration $$c$$ in the above, we use initial conditions. At $$t=0$$$c=a_{n}\left ( 0\right )$ Hence the solution becomes\begin{align*} a_{n}\left ( t\right ) & =e^{-k\lambda _{n}t}\int _{0}^{t}f_{n}\left ( s\right ) e^{k\lambda _{n}s}ds+a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}\\ & =e^{-k\lambda _{n}t}\left ( a_{n}\left ( 0\right ) +\int _{0}^{t}f_{n}\left ( s\right ) e^{k\lambda _{n}s}ds\right ) \end{align*}

To ﬁnd $$a_{n}\left ( 0\right )$$, from (2), putting $$t=0$$ gives$0=\sum _{n=1}^{\infty }a_{n}\left ( 0\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right )$ Hence $$a_{n}\left ( 0\right ) =0$$. Therefore $$a_{n}\left ( t\right )$$ becomes.$a_{n}\left ( t\right ) =e^{-k\lambda _{n}t}\int _{0}^{t}f_{n}\left ( s\right ) e^{k\lambda _{n}s}ds$ And ﬁnally, to ﬁnd $$f_{n}\left ( t\right )$$, which is the generalized Fourier coeﬃcient of the expansion of the source in (3) above, we also use orthogonality\begin{align*} \int _{0}^{a}f\left ( r,t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr & =f_{n}\left ( t\right ) \int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr\\ f_{n}\left ( t\right ) & =\frac{\int _{0}^{a}f\left ( r,t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr} \end{align*}

Summary of solution

The solution to $$\frac{\partial u\left ( r,t\right ) }{\partial t}=k\left ( \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}\right ) +f\left ( r,t\right )$$ is given by $u\left ( r,t\right ) =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right )$ Where $$a_{n}\left ( t\right )$$ is the solution to $a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) =f_{n}\left ( t\right )$ Given by$a_{n}\left ( t\right ) =e^{-k\lambda _{n}t}\int _{0}^{t}f_{n}\left ( s\right ) e^{k\lambda _{n}s}ds$ Where$f_{n}\left ( t\right ) =\frac{\int _{0}^{a}f\left ( r,t\right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt{\lambda _{n}}r\right ) rdr}$

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#### 9.3 Articolo example 6.9.1. On quarter disk, homogeneous BC (Articolo 6.9.1)

problem number 77

Added December 20, 2018.

Example 6.9.1 from Partial diﬀerential equations and boundary value problems with Maple/George A. Articolo, 2nd ed :

We seek the temperature distribution $$u(r,\theta ,t)$$ in a thin circular plate over the two-dimensional domain $$D=\{(r,\theta ) \text{s.t.} 0<r<1, 0< \theta < \frac{\pi }{2}\}$$.

The lateral surfaces of the plate are insulated. The edges $$r = 1$$ and $$\theta = 0$$ are at a ﬁxed temperature of $$0$$, and the edge $$\theta = \frac{\pi }{2}$$ is insulated. The initial temperature distribution $$u(r, \theta , 0) = f(r, \theta )$$ is $$u(r,\theta ,0)=(r-r^3)\sin (\theta )$$.

The thermal diﬀusivity is $$k = \frac{1}{50}$$.

Solve for $$u(r,\theta ,t)$$ the heat PDE

$\frac{ \partial u}{\partial t}= k \left ( \frac{ \partial ^2 u}{\partial r^2} + \frac{1}{r} \frac{ \partial u}{\partial r} + \frac{1}{r^2} \frac{ \partial u^2}{\partial \theta ^2} \right )$

With boundary conditions \begin{align*} |u(0,\theta ,t)| &< \infty \\ u(1,\theta ,t) &= 0\\ u(r,0,t) &=0\\ u(r,\frac{\pi }{2},t) & 0 \end{align*}

Mathematica

$\text{DSolve}\left [\left \{u^{(0,0,1)}(r,\theta ,t)=\frac{1}{25} \left (\frac{u^{(0,2,0)}(r,\theta ,t)}{r^2}+\frac{u^{(1,0,0)}(r,\theta ,t)}{r}+u^{(2,0,0)}(r,\theta ,t)\right ),u^{(1,0,0)}(1,\theta ,t)=0,\{u(r,0,t)=0,u(r,\pi ,t)=0\},u(r,\theta ,0)=\left (r-\frac{r^3}{3}\right ) \sin (\theta )\right \},u(r,\theta ,t),\{r,\theta ,t\}\right ]$

Maple

$u \left ( r,\theta ,t \right ) =\mbox{{\tt casesplit/ans}} \left ( \sum _{{\it n1}=0}^{\infty } \left ( \sum _{n=0}^{\infty }-128\,{\frac{{\lambda _{{{\it n1}}}}^{-1-4\,n}\Gamma \left ( 3+4\,n \right ) \sin \left ( \left ( 2+4\,n \right ) \theta \right ) \left ( \left ( -8\,n-2 \right ) \BesselJ \left ( 1+4\,n,\lambda _{{{\it n1}}}r \right ) +\BesselJ \left ( 4\,n,\lambda _{{{\it n1}}}r \right ) \lambda _{{{\it n1}}}r \right ){16}^{-n}{{\rm e}^{-{\frac{1}{50}}\,{\lambda _{{{\it n1}}}}^{2}t}}}{\pi \,r \left ( 5+4\,n \right ) \left ( 7+4\,n \right ) \left ( 3+4\,n \right ) \left ( 1+4\,n \right ) ^{2} \left ({\mbox{_2F_3}(2+4\,n,5/2+4\,n;\,3+4\,n,3+4\,n,4+8\,n;\,-{\lambda _{{{\it n1}}}}^{2})}-{\mbox{_3F_4}(2+4\,n,2+4\,n,3/2+4\,n;\,3+4\,n,3+4\,n,1+4\,n,8\,n+3;\,-{\lambda _{{{\it n1}}}}^{2})} \right ) } \left ( \left ( n+1/2 \right ){256}^{n} \left ( 7/4+n \right ){\mbox{_1F_2}(5/2+2\,n;\,3+4\,n,7/2+2\,n;\,-1/4\,{\lambda _{{{\it n1}}}}^{2})}-{\frac{{\mbox{_1F_2}(7/2+2\,n;\,3+4\,n,9/2+2\,n;\,-1/4\,{\lambda _{{{\it n1}}}}^{2})} \left ({2}^{8+8\,n}{n}^{2}+ \left ( 448\,n+160 \right ){256}^{n} \right ) }{768}} \right ) } \right ) , \left \{{\it And} \left ( -4\,\BesselJ \left ( 2+4\,n,\lambda _{{{\it n1}}} \right ) n+\BesselJ \left ( 3+4\,n,\lambda _{{{\it n1}}} \right ) \lambda _{{{\it n1}}}-2\,\BesselJ \left ( 2+4\,n,\lambda _{{{\it n1}}} \right ) =0,0<\lambda _{{{\it n1}}} \right ) \right \} \right )$

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#### 9.4 On half disk, homogeneous BC (Articolo 6.9.2)

problem number 78

Added December 20, 2018.

Example 6.9.2 from Partial diﬀerential equations and boundary value problems with Maple/George A. Articolo, 2nd ed :

We seek the temperature distribution in a thin circular plate over the two-dimensional domain $$D = \{(r,\theta ) \text{s.t.} 0 < r < 1, 0 < \theta < \pi \}$$. The lateral surfaces of the plate are insulated. The sides $$\theta = 0$$ and $$\theta = \pi$$ are at a ﬁxed temperature of $$0$$, and the edge $$r = 1$$ is insulated. The initial temperature distribution is $$u(r, \theta , 0) = \left ( r - \frac{r^3}{3} \right )\sin \theta$$.

The thermal diﬀusivity is $$k = \frac{1}{25}$$.

Solve for $$u(r,\theta ,t)$$ the heat PDE

$\frac{ \partial u}{\partial t}= k \left ( \frac{ \partial ^2 u}{\partial r^2} + \frac{1}{r} \frac{ \partial u}{\partial r} + \frac{1}{r^2} \frac{ \partial u^2}{\partial \theta ^2} \right )$

With boundary conditions \begin{align*} |u(0,\theta ,t) &< \infty \\ u(1,\theta ,t) &= 0\\ u(r,0,t) &=0\\ u(r,\pi ,t) & 0 \end{align*}

Mathematica

$\text{DSolve}\left [\left \{u^{(0,0,1)}(x,y,t)=\frac{1}{50} \left (u^{(0,2,0)}(x,y,t)+u^{(2,0,0)}(x,y,t)\right ),\left \{u^{(1,0,0)}(0,y,t)=0,u^{(1,0,0)}(1,y,t)+u(1,y,t)=0,u(x,0,t)=0,u(x,1,t)=0\right \},u(x,y,0)=\left (1-\frac{x^2}{3}\right ) (1-y) y\right \},u(x,y,t),\{x,y,t\}\right ]$

Maple

$u \left ( r,\theta ,t \right ) =\mbox{{\tt casesplit/ans}} \left ( \sum _{n=0}^{\infty }-4/3\,{\frac{\BesselJ \left ( 1,\lambda _{{n}}r \right ) \sin \left ( \theta \right ){{\rm e}^{-1/25\,{\lambda _{{n}}}^{2}t}} \left ( \BesselJ \left ( 0,\lambda _{{n}} \right ){\lambda _{{n}}}^{3}-\BesselJ \left ( 1,\lambda _{{n}} \right ){\lambda _{{n}}}^{2}+4\,\lambda _{{n}}\BesselJ \left ( 0,\lambda _{{n}} \right ) -8\,\BesselJ \left ( 1,\lambda _{{n}} \right ) \right ) }{{\lambda _{{n}}}^{3} \left ( \left ( \BesselJ \left ( 0,\lambda _{{n}} \right ) \right ) ^{2}\lambda _{{n}}+ \left ( \BesselJ \left ( 1,\lambda _{{n}} \right ) \right ) ^{2}\lambda _{{n}}-2\,\BesselJ \left ( 0,\lambda _{{n}} \right ) \BesselJ \left ( 1,\lambda _{{n}} \right ) \right ) }}, \left \{{\it And} \left ( -\BesselJ \left ( 1,\lambda _{{n}} \right ) +\BesselJ \left ( 2,\lambda _{{n}} \right ) \lambda _{{n}}=0,0<\lambda _{{n}} \right ) \right \} \right )$

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#### 9.5 Inside disk with time-independent boundary conditions and $$\theta$$ dependency (Haberman 8.2.5)

problem number 79

Added Feb 24, 2019.

Problem 8.2.5 from from Richard Haberman applied partial diﬀerential equations book, 5th edition.

Solve the initial value problem for a two-dimensional heat equation inside a circle (of radius $$a$$) with time-independent boundary conditions: \begin{align*} u_t &= k \triangledown ^2 u \\ u(a,\theta ) &= g(\theta ) \\ u(r,\theta ,0) &=f(r,\theta ) \end{align*}

Mathematica

$\text{DSolve}\left [\left \{u^{(0,0,1)}(r,\theta ,t)=\frac{u^{(0,2,0)}(r,\theta ,t)}{r^2}+\frac{u^{(1,0,0)}(r,\theta ,t)}{r}+u^{(2,0,0)}(r,\theta ,t),u(a,\theta ,t)=g(\theta ),u(r,\theta ,0)=g(r,\theta )\right \},u(r,\theta ,t),\{r,\theta ,t\},\text{Assumptions}\to \{a>0,a<r\}\right ]$

Maple

$\text{ sol=() }$

Hand solution

\begin{align*} \frac{\partial u\left ( r,\theta ,t\right ) }{\partial t} & =k\left ( \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}}\right ) \\ \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u\left ( a,\theta ,t\right ) & =g\left ( \theta \right ) \\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac{\partial u}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac{\partial u}{\partial \theta }\left ( r,\pi ,t\right ) \end{align*}

With initial conditions $$u\left ( r,\theta ,0\right ) =f\left ( r,\theta \right )$$.

Since the boundary conditions are not homogenous, and since there are no time dependent sources, then in this case we look for $$u_{E}\left ( r,\theta \right )$$ which is solution at steady state which needs to satisfy the nonhomogeneous B.C., where $$u\left ( r,\theta ,t\right ) =v\left ( r,\theta ,t\right ) +u_{E}\left ( r,\theta \right )$$ and $$v\left ( r,\theta ,t\right )$$ solves the PDE but with homogenous B.C. Therefore, we need to ﬁnd equilibrium solution for Laplace PDE on disk, that only needs to satisfy the nonhomogeneous B.C.\begin{align*} \nabla ^{2}u_{E} & =0\\ \frac{\partial ^{2}u_{E}}{\partial r^{2}}+\frac{1}{r}\frac{\partial u_{E}}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}u_{E}}{\partial \theta ^{2}} & =0 \end{align*}

With boundary condition\begin{align*} \left \vert u_{E}\left ( 0,\theta \right ) \right \vert & <\theta \\ u_{E}\left ( a,\theta \right ) & =g\left ( \theta \right ) \\ u_{E}\left ( r,-\pi \right ) & =u_{E}\left ( r,\pi \right ) \\ \frac{\partial u_{E}}{\partial \theta }\left ( r,-\pi \right ) & =\frac{\partial u_{E}}{\partial \theta }\left ( r,\pi \right ) \end{align*}

But this PDE we have already solved before. But to practice, will solve it again. Let $u_{E}\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right )$ Where $$R\left ( r\right )$$ is the solution in radial dimension and $$\Theta \left ( \theta \right )$$ is solution in angular dimension. Substituting $$u_{E}\left ( r,\theta \right )$$ in the PDE gives$R^{\prime \prime }\Theta +\frac{1}{r}R^{\prime }\Theta +\frac{1}{r^{2}}\Theta ^{\prime \prime }R=0$ Dividing by $$R\left ( r\right ) \Phi \left ( \theta \right )$$\begin{align*} \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta } & =0\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R} & =-\frac{\Theta ^{\prime \prime }}{\Theta } \end{align*}

Hence each side is equal to constant, say $$\lambda$$ and we obtain\begin{align*} r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R} & =\lambda \\ -\frac{\Theta ^{\prime \prime }}{\Theta } & =\lambda \end{align*}

Or\begin{align} r^{2}R^{\prime \prime }+rR^{\prime }-\lambda R & =0\tag{1}\\ \Theta ^{\prime \prime }+\lambda \Theta & =0\tag{2} \end{align}

We start with $$\Phi$$ ODE. The boundary conditions on (3) are \begin{align*} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \\ \frac{\partial \Theta }{\partial \theta }\left ( -\pi \right ) & =\frac{\partial \Theta }{\partial \theta }\left ( \pi \right ) \end{align*}

case $$\lambda =0$$ The solution is $$\Phi =c_{1}\theta +c_{2}$$. Hence we obtain, from ﬁrst initial conditions\begin{align*} -\pi c_{1}+c_{2} & =\pi c_{1}+c_{2}\\ c_{1} & =0 \end{align*}

Second boundary conditions just says that $$c_{2}=c_{2}$$, so any constant will do. Hence $$\lambda =0$$ is an eigenvalue with constant being eigenfunction.

case $$\lambda >0$$ The solution is $\Theta \left ( \theta \right ) =c_{1}\cos \sqrt{\lambda }\theta +c_{2}\sin \sqrt{\lambda }\theta$ The ﬁrst boundary conditions gives\begin{align} c_{1}\cos \left ( -\sqrt{\lambda }\pi \right ) +c_{2}\sin \left ( -\sqrt{\lambda }\pi \right ) & =c_{1}\cos \left ( \sqrt{\lambda }\pi \right ) +c_{2}\sin \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ c_{1}\cos \left ( \sqrt{\lambda }\pi \right ) -c_{2}\sin \left ( \sqrt{\lambda }\pi \right ) & =c_{1}\cos \left ( \sqrt{\lambda }\pi \right ) +c_{2}\sin \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ 2c_{2}\sin \left ( \sqrt{\lambda }\pi \right ) & =0 \tag{3} \end{align}

From second boundary conditions we obtain$\Theta ^{\prime }\left ( \theta \right ) =-\sqrt{\lambda }c_{1}\sin \sqrt{\lambda }\theta +c_{2}\sqrt{\lambda }\cos \sqrt{\lambda }\theta$ Therefore\begin{align} -\sqrt{\lambda }c_{1}\sin \left ( -\sqrt{\lambda }\pi \right ) +c_{2}\sqrt{\lambda }\cos \left ( -\sqrt{\lambda }\pi \right ) & =-\sqrt{\lambda }c_{1}\sin \left ( \sqrt{\lambda }\pi \right ) +c_{2}\sqrt{\lambda }\cos \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ \sqrt{\lambda }c_{1}\sin \left ( \sqrt{\lambda }\pi \right ) +c_{2}\sqrt{\lambda }\cos \left ( \sqrt{\lambda }\pi \right ) & =-\sqrt{\lambda }c_{1}\sin \left ( \sqrt{\lambda }\pi \right ) +c_{2}\sqrt{\lambda }\cos \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ \sqrt{\lambda }c_{1}\sin \left ( \sqrt{\lambda }\pi \right ) & =-\sqrt{\lambda }c_{1}\sin \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ 2c_{1}\sin \left ( \sqrt{\lambda }\pi \right ) & =0 \tag{4} \end{align}

Both (3) and (4) are satisﬁed if\begin{align*} \sqrt{\lambda }\pi & =n\pi \qquad n=1,2,3,\cdots \\ \lambda & =n^{2}\qquad n=1,2,3,\cdots \end{align*}

Therefore$$\Theta _{n}\left ( \theta \right ) =\overset{\lambda =0}{\overbrace{\tilde{A}_{0}}}+\sum _{n=1}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) +\tilde{B}_{n}\sin \left ( n\theta \right ) \tag{5}$$ I put tilde on top of these constants, so not confuse them with constants used for $$v\left ( r,\theta ,t\right )$$ found later below. Now we go back to the $$R$$ ODE (2) given by $$r^{2}R^{\prime \prime }+rR^{\prime }-\lambda _{n}R=0$$ and solve it. This is Euler PDE whose solution is found by substituting $$R\left ( r\right ) =r^{\alpha }$$. The solution comes out to be (Lecture 9)$$R_{n}\left ( r\right ) =c_{0}+\sum _{n=1}^{\infty }c_{n}r^{n} \tag{6}$$ Combining (5,6) we now ﬁnd $$u_{E}$$ as\begin{align} u_{E_{n}}\left ( r,\theta \right ) & =R_{n}\left ( r\right ) \Theta _{n}\left ( \theta \right ) \nonumber \\ u_{E}\left ( r,\theta \right ) & =\tilde{A}_{0}+\sum _{n=1}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) r^{n}+\tilde{B}_{n}\sin \left ( n\theta \right ) r^{n}\nonumber \\ & =\sum _{n=0}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) r^{n}+\sum _{n=1}^{\infty }\tilde{B}_{n}\sin \left ( n\theta \right ) r^{n} \tag{7} \end{align}

Where $$c_{0}$$ was combined with $$A_{0}$$. Now the above equilibrium solution needs to satisfy the non-homogenous B.C. $$u_{E}\left ( a,\theta \right ) =g\left ( \theta \right )$$. Using orthogonality on (7) to ﬁnd $$A_{n},B_{n}$$ gives\begin{align*} g\left ( \theta \right ) & =\sum _{n=0}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) a^{n}+\sum _{n=1}^{\infty }\tilde{B}_{n}\sin \left ( n\theta \right ) a^{n}\\ \int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n^{\prime }\theta \right ) d\theta & =\int _{0}^{2\pi }\sum _{n=0}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) a^{n}d\theta +\int _{0}^{2\pi }\sum _{n=1}^{\infty }\tilde{B}_{n}\sin \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) a^{n}d\theta \\ & =\sum _{n=0}^{\infty }\int _{0}^{2\pi }\tilde{A}_{n}\cos \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) a^{n}d\theta +\sum _{n=0}^{\infty }\overset{0}{\overbrace{\int _{0}^{2\pi }\tilde{B}_{n}\sin \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) a^{n}d\theta }}\\ & =\tilde{A}_{n^{\prime }}\int _{0}^{2\pi }\cos ^{2}\left ( n^{\prime }\theta \right ) a^{n}d\theta \end{align*}

For $$n=0$$\begin{align*} \int _{0}^{2\pi }g\left ( \theta \right ) d\theta & =\tilde{A}_{0}\int _{0}^{2\pi }d\theta \\ \tilde{A}_{0} & =\frac{1}{2\pi }\int _{0}^{2\pi }g\left ( \theta \right ) d\theta \end{align*}

For $$n>0$$\begin{align*} \int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta & =\tilde{A}_{n}\int _{0}^{2\pi }\cos ^{2}\left ( n\theta \right ) a^{n}d\theta \\ \tilde{A}_{n} & =\frac{1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \end{align*}

Similarly, we apply orthogonality to ﬁnd $$\tilde{B}_{n}\,$$ which gives (for $$n>0$$ only)$\tilde{B}_{n}=\frac{1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta$ Therefore, we have found $$u_{E}\left ( r,\theta \right )$$ completely now. It is given by\begin{align*} u_{E}\left ( r,\theta \right ) & =\tilde{A}_{0}+\sum _{n=1}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) r^{n}+\tilde{B}_{n}\sin \left ( n\theta \right ) r^{n}\\ \tilde{A}_{0} & =\frac{1}{2\pi }\int _{0}^{2\pi }g\left ( \theta \right ) d\theta \\ \tilde{A}_{n} & =\frac{1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \\ \tilde{B}_{n} & =\frac{1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \end{align*}

The above satisﬁes the non-homogenous B.C. $$u_{E}\left ( a,\theta \right ) =g\left ( \theta \right )$$. Now, since $$u\left ( r,\theta ,t\right ) =v\left ( r,\theta ,t\right ) +u_{E}\left ( r,\theta \right )$$, then we need to solve now for $$v\left ( r,\theta ,t\right )$$ speciﬁed by\begin{align} \frac{\partial v\left ( r,\theta ,t\right ) }{\partial t} & =k\left ( \frac{\partial ^{2}v}{\partial r^{2}}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}v}{\partial \theta ^{2}}\right ) \tag{8}\\ \left \vert v\left ( 0,\theta ,t\right ) \right \vert & <\theta \nonumber \\ v\left ( a,\theta ,t\right ) & =0\nonumber \\ v\left ( r,-\pi ,t\right ) & =v\left ( r,\pi ,t\right ) \nonumber \\ \frac{\partial v}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac{\partial v}{\partial \theta }\left ( r,\pi ,t\right ) \nonumber \end{align}

Let $$v\left ( r,\theta ,t\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right )$$. Substituting into (8) gives$T^{\prime }R\Theta =k\left ( R^{\prime \prime }T\Theta +\frac{1}{r}R^{\prime }T\Theta +\frac{1}{r^{2}}\Theta ^{\prime \prime }RT\right )$ Dividing by $$R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) \neq 0$$ gives$\frac{1}{k}\frac{T^{\prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta }$ Let ﬁrst separation constant be $$-\lambda$$, hence the above becomes\begin{align*} \frac{1}{k}\frac{T^{\prime }}{T} & =-\lambda \\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta } & =-\lambda \end{align*}

Or\begin{align*} T^{\prime }+\lambda kT & =0\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda & =-\frac{\Theta ^{\prime \prime }}{\Theta } \end{align*}

We now separate the second equation above using $$\mu$$ giving\begin{align*} r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda & =\mu \\ -\frac{\Theta ^{\prime \prime }}{\Theta } & =\mu \end{align*}

Or\begin{align} R^{\prime \prime }+\frac{1}{r}R^{\prime }+R\left ( \lambda -\frac{\mu }{r^{2}}\right ) & =0\tag{9}\\ \Theta ^{\prime \prime }+\mu \Theta & =0 \tag{10} \end{align}

Equation (9) is Sturm-Liouville ODE with boundary conditions $$R\left ( a\right ) =0$$ and bounded at $$r=0$$ and (10) has periodic boundary conditions as was solved above. The solution to (10) is given in (5) above, no change for this part.\begin{align} \Theta _{n}\left ( \theta \right ) & =\overset{\lambda =0}{\overbrace{A_{0}}}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \nonumber \\ & =\sum _{n=0}^{\infty }A_{n}\cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( n\theta \right ) \tag{11} \end{align}

Therefore (9) becomes $$R^{\prime \prime }+\frac{1}{r}R^{\prime }+R\left ( \lambda -\frac{n^{2}}{r^{2}}\right ) =0$$ with $$n=0,1,2,\cdots$$. We found the solution to this Sturm-Liouville before, it is given by$$R_{nm}\left ( r\right ) =J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \tag{12}$$ Where $$\sqrt{\lambda _{nm}}=\frac{a}{z_{nm}}$$ where $$a$$ is the radius of the disk and $$z_{nm}$$ is the $$m^{th}$$ zero of the Bessel function of order $$n$$. This is found numerically. We now just need to ﬁnd the time solution from $$T^{\prime }+\lambda _{nm}kT=0$$. This has solution $$T_{nm}\left ( t\right ) =e^{-\sqrt{k\lambda _{nm}}t} \tag{13}$$ Now we combine (11,12,13) to ﬁnd solution for $$v\left ( r,\theta ,t\right )$$\begin{align} v_{nm}\left ( r,\theta ,t\right ) & =\Theta _{n}\left ( \theta \right ) R_{nm}\left ( r\right ) T_{nm}\left ( t\right ) \nonumber \\ v\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) e^{-\sqrt{k\lambda _{nm}}t}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) e^{-\sqrt{k\lambda _{nm}}t} \tag{14} \end{align}

We now need to ﬁnd $$A_{n},B_{n}$$, which is found from initial conditions on $$v\left ( r,\theta ,0\right )$$ which is given by\begin{align*} v\left ( r,\theta ,0\right ) & =u\left ( r,\theta ,0\right ) -u_{E}\left ( r,\theta \right ) \\ & =f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \end{align*}

Hence from (14), at $$t=0$$$$f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \tag{15}$$ For each $$n$$, inside the $$m$$ sum, $$\cos \left ( n\theta \right )$$ and $$\sin \left ( n\theta \right )$$ will be constant. So we need to apply orthogonality twice in order to remove both sums. Multiplying (15) by $$\cos \left ( n^{\prime }\theta \right )$$ and integrating gives\begin{align*} \int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n^{\prime }\theta \right ) d\theta & =\int _{-\pi }^{\pi }\sum _{n=0}^{\infty }\left ( \sum _{m=1}^{\infty }A_{n}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \right ) \cos \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) d\theta \\ & +\int _{-\pi }^{\pi }\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{n}J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) \right ) \sin \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) \end{align*}

The second sum in the RHS above goes to zero due to $$\int _{-\pi }^{\pi }\sin \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) d\theta$$ and we end up with$\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) d\theta =A_{n}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) \sum _{m=1}^{\infty }J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) d\theta$ We now apply orthogonality again, but on Bessel functions and remember to add the weight $$r$$. The above becomes\begin{align*} \int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm^{\prime }}}r\right ) rd\theta dr & =A_{n}\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) \sum _{m=1}^{\infty }J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) J_{n}\left ( \sqrt{\lambda _{nm^{\prime }}}r\right ) rd\theta dr\\ & =A_{n}\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt{\lambda _{nm^{\prime }}}r\right ) rd\theta dr \end{align*}

Hence$A_{n}=\frac{\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots$ We will repeat the same thing to ﬁnd $$B_{n}$$. The only diﬀerence now is to use $$\sin n\theta$$. repeating these steps gives$B_{n}=\frac{\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \sin \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots$ This complete the solution.

Summary of solution

\begin{align*} u\left ( r,\theta ,t\right ) & =v\left ( r,\theta ,t\right ) +u_{E}\left ( r,\theta \right ) \\ & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) e^{-\sqrt{k\lambda _{nm}}t}+\\ & \sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) e^{-\sqrt{k\lambda _{nm}}t}+u_{E}\left ( r,\theta \right ) \end{align*}

Where\begin{align*} u_{E}\left ( r,\theta \right ) & =\tilde{A}_{0}+\sum _{n=1}^{\infty }\tilde{A}_{n}\cos \left ( n\theta \right ) r^{n}+\tilde{B}_{n}\sin \left ( n\theta \right ) r^{n}\\ \tilde{A}_{0} & =\frac{1}{2\pi }\int _{0}^{2\pi }g\left ( \theta \right ) d\theta \\ \tilde{A}_{n} & =\frac{1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \\ \tilde{B}_{n} & =\frac{1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \end{align*}

And$A_{n}=\frac{\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots$ And$B_{n}=\frac{\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \sin \left ( n\theta \right ) J_{n}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt{\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots$ Where $$\sqrt{\lambda _{nm}}=\frac{a}{z_{nm}}$$ where $$a$$ is the radius of the disk and $$z_{nm}$$ is the $$m^{th}$$ zero of the Bessel function of order $$n$$.