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Return to Part VI of the course APMA0360
Introduction to Linear Algebra with Mathematica
Glossary
Preface
2D Heat Equation
2D Heat Equation in Polar Coordinates: Symmetry
Let us consider the heat equation in a polar coordinates Due to the special geometry of the spacial domain, it is natural to consider the initial boundary value problems using polar coordinates (r , θ) satisfying
\[
x = r\,\cos\theta , \qquad y = r\,\sin\theta .
\]
Upon differentiating these equations, we get
\begin{align*}
\cos\theta\,r_x - r \left( \sin\theta \right) \theta_x &= 1 ,
\\
\left( \cos\theta \right) r_y - r \left( \sin\theta \right) \theta_y &= 0,
\\
\left( \sin\theta \right) r_x + r \left( \cos\theta \right) \theta_x &= 0 ,
\\
\left( \sin\theta \right) r_y + r \left( \cos\theta \right) \theta_y &= 1.
\end{align*}
From latter, it follows
\begin{align*}
r_x &= \frac{x}{r} , \qquad r_y = \frac{y}{r} . \qquad r_{xx} = \frac{1}{r} - \frac{x^2}{r^3} , \qquad r_{yy} = \frac{1}{r} - \frac{y^2}{r^3} ,
\\
\theta_x &= - \frac{\sin\theta}{r} = - \frac{y}{r^2} , \qquad \theta_y = \frac{\cos\theta}{r} = \frac{x}{r^2} , \qquad \theta_{xx} = \frac{2xy}{r^4} , \qquad \theta_{yy} = - \frac{2xy}{r^4} .
\end{align*}
Therefore,
\begin{align*}
u_{xx} &= u_{rr} \frac{x^2}{r^2} - u_{r\theta} \frac{2xy}{r^3} + u_{\theta\theta} \frac{y^2}{r^4} + u_r \left( \frac{1}{r} - \frac{x^2}{r^3} \right) + u_{\theta} \frac{2xy}{r^4} ,
\\
u_{yy} &= u_{rr} \frac{y^2}{r^2} + u_{r\theta} \frac{2xy}{r^3} + u_{\theta\theta} \frac{x^2}{r^4} + u_r \left( \frac{1}{r} - \frac{y^2}{r^3} \right) + u_{\theta} \left( -\frac{2xy}{r^4} \right) .
\end{align*}
The heat equation in polar coordinates becomes
\begin{equation} \label{Eq2heat.1}
u_t = \alpha \left( u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\theta\theta} \right)
\end{equation}
Example 1: Consider the inner Dirichlet problem for the heat equation in a 2D disc
\begin{align*}
&\mbox{Heat equation:} \qquad &u_t &= \alpha \left( u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\vartheta\theta} \right) \qquad\mbox{for } 0 \le r < a \mbox{ and } t > 0,
\\
&\mbox{Boundary condition:} \qquad &u(r = a, \vartheta ,t) &= 0 , \qquad t > 0 ,
\\
&\mbox{Initial condition:} \qquad &u(r, \vartheta ,,0) &= f(r, \vartheta ) , \qquad 0 \le r < a,
\\
&\mbox{Periodic condition:} \qquad &u(r, \vartheta ,t) &= u(r, \vartheta + 2\pi ,t) , \qquad t > 0 ,
\\
&\mbox{Regularity condition:} \qquad & f(r=a, \vartheta ) &= 0 , \quad\qquad u, u_t , u_{rr} , u_{\vartheta\vartheta}\in 𝔏².
\end{align*}
We apply separation of variables to solve this equation.
First we try to find non-trivial “basic” solutions of the form
\[
u(r, \vartheta ,t) = R(r)\,\Theta (\vartheta )\,T(t).
\]
Substituting this product into the heat equation, we reach
\[
R(r)\,\Theta (\vartheta )\,T' (t) = \alpha \left( R'' (r)\,\Theta (\vartheta ) + \frac{1}{r}\, R' (r)\,\Theta (\vartheta ) + \frac{1}{r^2}\, R(r)\,\Theta '' (\vartheta ) \right) T(t) .
\]
Dividing both sides by u = R(r) Θ(ϑ) T(t), we get
\[
\frac{T' (t)}{T(t)} = \alpha \left( \frac{R'' (r)}{R(r)} + \frac{1}{r} \cdot \frac{R' (r)}{R(r)} + \frac{1}{r^2} \cdot \frac{\Theta ''(\vartheta )}{\Theta (\vartheta )} \right) .
\]
As the left-hand side only involves time variable t and the right-hand side only spacial variables r, ϑ, they can be equal only when both sides are a constant. So there exists a constant λ such that
\begin{align*}
\frac{T' (t)}{T(t)} &= - \alpha\,\lambda ,
\\
\frac{R'' (r)}{R(r)} + \frac{1}{r} cdot \frac{R' (r)}{R(r)} + \frac{1}{r^2} \cdot \frac{\Theta ''(\vartheta )}{\Theta (\vartheta )} &= -\lambda .
\end{align*}
Multiply both sides by r², we have
\[
\frac{r^2 R'' (r)}{R(r)} + \frac{r\,R' (r)}{R(r)} + r^2 \lambda = \frac{\Theta ''(\vartheta )}{\Theta (\vartheta )} .
\]
The left-hand side only involves r and the right-hand side only ϑ. Thus, there is a constant μ such that
\[
\frac{\Theta ''(\vartheta )}{\Theta (\vartheta )} = \mu , \qquad
\frac{r^2 R'' (r)}{R(r)} + \frac{r\,R' (r)}{R(r)} + r^2 \lambda = \mu .
\]
As Θ(ϑ) is obviously 2π periodic, we obtain
\[
\mu = n^2 , \qquad n=1,2,3,\ldots .
\]
and
\[
\Theta_n (n \vartheta ) = a_n \cos (n\vartheta ) + b_n \sin (n \vartheta ), \qquad n=1,2,3,\ldots .
\tag{1.1}
\]
On the other hand, the equation for R now becomes
\[
r^2 R'' (r) + r\,R' (r) + \left( r^2 \lambda - n^2 \right) R(r) = 0 , \qquad 0 \le r < a.
\tag{1.2}
\]
This function must also satisfy the boundary condition
\[
R(a) = 0 .
\tag{1.3}
\]
Differential equation (1.2) is a Bessel equation. For each n, we have
λn, k eigenvalues, such that the boundary value problem (1.2), (1.3)
has a solution Rn, k. Then the required solution is represented by a double sum:
\[
u(r, \vartheta , t) = \sum_{n,k} \left[ a_{n, k} R_{n, k}(r)\,\cos (n\vartheta ) + b_{n, k} R_{n, k}(r)\,\sin (n\vartheta ) \right] e^{-\alpha\,\lambda_{n, k}t} .
\tag{1.4}
\]
The coefficients are determined from the expansion
\[
f(r, \vartheta ) = \sum_{n,k} \left[ a_{n, k} R_{n, k}(r)\,\cos (n\vartheta ) + b_{n, k} R_{n, k}(r)\,\sin (n\vartheta ) \right] .
\tag{1.5}
\]
ϑ
ϑ
ϑ
2D Heat Equation in Polar coordinate: General Case
Fokas Method for 2D Heat Equation in Polar coordinate
\begin{align*}
&\mbox{Heat equation:} \qquad &u_t &= \alpha \left( u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\theta\theta} \right) \qquad\mbox{for } 0 &le r < a \mbox{ and } t > 0, \\
&\mbox{Initial condition:} \qquad &u(r, \theta ,,0) &= f(x) , \qquad 0 \le r > a, \\
&\mbox{Boundary condition:} \qquad &u(0,t) &= g(t) , \qquad t > 0 , \\
&\mbox{Regularity condition:} \qquad & \lim_{x,t \to +\infty}\,|u(x,t)| &= 0 , \quad\qquad u, u_t , u_{xx} \in 𝔏²(R).
\end{align*}
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