expressions for no growth, linear growth, exponential growth

growing_domain/notes.tex

@@ -88,22 +88,45 @@ Finally, we express the second term on the RHS as a function of $T$ to write:
 The above equation can be solved easily by separation of variables. Writing $C(\xi, T) = u(\xi) v(T)$,
 The above equation can be solved easily by separation of variables. Writing $C(\xi, T) = u(\xi) v(T)$,
 \eqn{\frac{1}{v} \frac{dv}{dt} = \frac{1}{u \xi^2} \frac{d}{d \xi} \left( \xi^2 \frac{du}{d \xi} \right) + f(T)}
 \eqn{\frac{1}{v} \frac{dv}{dt} = \frac{1}{u \xi^2} \frac{d}{d \xi} \left( \xi^2 \frac{du}{d \xi} \right) + f(T)}
 Thus, 
 Thus, 
-\eqn{\frac{1}{v} \frac{dv}{dt} - f(T) = \frac{1}{u \xi^2} \frac{d}{d \xi} \left( \xi^2 \frac{du}{d \xi} \right) = -k^2}
+\eqn{\frac{1}{v} \frac{dv}{dT} - f(T) = \frac{1}{u \xi^2} \frac{d}{d \xi} \left( \xi^2 \frac{du}{d \xi} \right) = -n^2}
 Solving the $v$ equation first:
 Solving the $v$ equation first:
-\eqn{v(T) = v(0) \exp\left[-k^2 T + \int_0^T f(T^{\star}) dT^\star \right]}
+\eqn{v(T) = v(0) \exp\left[-n^2 T + \int_0^T f(T^{\star}) dT^\star \right]}
 And the $u$ equation reads:
 And the $u$ equation reads:
-\eqn{\frac{d^2 u}{d \xi^2} + \frac{2}{\xi} \frac{du}{d \xi} + k^2 u = 0}
+\eqn{\frac{d^2 u}{d \xi^2} + \frac{2}{\xi} \frac{du}{d \xi} + n^2 u = 0}
 This is the Bessel equation of zeroth order, having solutions
 This is the Bessel equation of zeroth order, having solutions
-\eqn{u(\xi) = c_1 j_0 (k \xi) + c_2 y_0 (k \xi)}
+\eqn{u(\xi) = c_1 j_0 (n \xi) + c_2 y_0 (n \xi)}
 where $j_0, y_0$ are spherical bessel functions given by 
 where $j_0, y_0$ are spherical bessel functions given by 
-\eq{j_0 (k \xi) = \frac{\sin(k \xi)}{k \xi}}
-\eq{y_0 (k \xi) = -\frac{\cos(k \xi)}{k \xi}}
+\eq{j_0 (n \xi) = \frac{\sin(n \xi)}{n \xi}}
+\eq{y_0 (n \xi) = -\frac{\cos(n \xi)}{n \xi}}
 Thus, the exact solution in the $(\xi, T)$ coordinates is :
 Thus, the exact solution in the $(\xi, T)$ coordinates is :
-\eqn{C(\xi, T) = \left[ a_1 j_0 (k \xi) + a_2 y_0 (k \xi)\right] \exp\left[-k^2 T + \int_0^T f(T^\star) dT^\star \right]}
+\eqn{C(\xi, T) = \left[ a_1 j_0 (n \xi) + a_2 y_0 (n \xi)\right] \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}
 Since we want the solution to be bounded at $\xi = 0$, we keep only the $j_0$ term:
 Since we want the solution to be bounded at $\xi = 0$, we keep only the $j_0$ term:
-\eqn{C(\xi, T) = a_1 j_0 (k \xi) \exp\left[-k^2 T + \int_0^T f(T^\star) dT^\star \right]}  
+\eqn{C(\xi, T) = a_1 j_0 (n \xi) \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}  
 And
 And
-\eqn{\frac{\partial C}{\partial \xi} = a_1 \left(\frac{1}{\xi} \cos(k \xi) - \frac{1}{k \xi^2} \sin(k \xi) \right) \exp\left[-k^2 T + \int_0^T f(T^\star) dT^\star \right].\label{neumann}}
-Demanding Eq(\ref{neumann}) to be zero at $\xi = 1$ (Neumann boundary conditions at $\xi = 1$), one can get $k$ to be the solution of:
-\eqn{\tan (k)= k} 
+\eqn{\frac{\partial C}{\partial \xi} = a_1 \left(\frac{1}{\xi} \cos(n \xi) - \frac{1}{n \xi^2} \sin(n \xi) \right) \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right].\label{neumann}}
+Demanding Eq(\ref{neumann}) to be zero at $\xi = 1$ (Neumann boundary conditions at $\xi = 1$), one can get $n$ to be the solution of:
+\eqn{\tan (n)= n} 
+Thus, the most general solution is:
+\eq{C(\xi, T) = \sum_{n} a_n j_0(n \xi) \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}
+where the sum is over solutions of $\tan(n) = n$ and $a_n$ is determined by the initial conditions.
+\section{No growth}
+In this case, $R(t) = R, \sigma(t) = 0, f(T^\star) = R^2 k/D$.
+This gives \eq{C(\xi, T) = \sum_n a_n \left( \frac{\sin (n \xi)}{n \xi}\right) e^{-n^2 T} e^{R^2 k T/D}}
+Plugging in $T = D t/R^2 $ and $\xi = r/R$,
+\eqn{C(r,t) = \sum_n a_n \left(\frac{\sin(nr/R)}{nr/R}\right) e^{-\frac{n^2 D t}{R^2}}e^{kt}}
+\section{Exponential domain growth}
+In this case, $R(t) = R_0 e^{\alpha t}, \sigma(t)=\alpha$
+\eq{ T = \frac{D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2}}
+And \eq{f(	T^\star) = \frac{R_0^2 (k - 3 \alpha)}{D - 2 \alpha R_0^2 T^{\star}}} 
+Then,
+\eq{C(\xi, T) = \sum_n a_n j_0(n \xi) e^{-n^2 T} \frac{k - 3 \alpha}{2 \alpha} \left(1 - \frac{2 \alpha R_0^2 T}{D}\right)}
+and \eqn{C(r,t) = \frac{k - 3 \alpha}{2 \alpha} e^{- 2 \alpha t} \sum_n a_n j_0 \left( \frac{n r}{R(t)}\right) e^{-\frac{n^2 D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2}}}
+\section{Linear domain growth}
+In this case, $R(t)=R_0 + bt, \sigma(t) = b/R(t)$.\\
+Further, \eq{T = \frac{D t}{R(t) R_0}}, or, 
+\eq{t = \frac{R_0^2 T}{D - R_0 b T}}
+And \eq{f(T^\star) = \frac{R_0}{(D - R_0 bT^\star)^2} (k D R_0 - 3 b D + 3bR_0T^\star)}
+Then
+\eq{C(\xi, T) = e^{kt} \left(\frac{R_0}{R(t)}\right)^3 \sum_n a_n j_0(n \xi) e^{-n^2 T} e^{\frac{k R_0^2 T}{D - R_0 b T}}\left(\frac{D - R_0 b T}{D}\right)^3}
+\eqn{C(r, t) = \sum_n a_n j_0\left(\frac{n r}{R(t)}\right)e^{-\frac{n^2 D t}{R(t) R_0}} }
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