User:Egm6322.s09.xyz/hw5 xyz

Verify that a linear superposition of Jn (1st kind) and Yn (2nd kind) is also a solution

From the above verification of the Bessel solution of the first kind, we found that:

$$J_n(z) = z^n \sum_{m=0}^\infty \frac{(-1)^m z^{2m}}{2^{2m + n}m!\Gamma(n+m+1)}$$

A second general solution can be found by replacing n by -n to get:

$$J_{-n}(z) = z^{-n} \sum_{m=0}^\infty \frac{(-1)^m z^{2m}}{2^{2m - n}m!\Gamma(m-n+1)}$$

If n is not and integer, then $$J_n(z)$$ and $$J_{-n}(z)$$ are linearly independent because both the first terms are finite, non-zero multiples of $$x^n$$ and $$x^{-n}$$

$$\therefore R(r) = c_1J_n(z)+ c_2J_{-n}(z)$$ which is linearly independent for all $$n\not=integer$$ and $$x\not=0$$

From the verification of Bessel function of the second kind,

$$Y_n(z) = \frac{1}{sin(n\pi)}\left[ J_n(z)cos(n\pi) - J_{-n}(z)\right]$$

and $$Y_k(z) = lim_{n \rightarrow k} Y_n(z)$$

Since $$J_n(z)$$ and $$J_{-n}(z)$$ are linearly independent and solutions of Y_n(z), then it follows that $$J_n(z)$$ and $$Y_n(z)$$ are linearly independent.

$$\therefore R(r) = D J_n(z) + E Y_n(z)$$ where D, E are constants

Recall: $$T(t) = A(\lambda)exp(-\lambda t)$$ and $$ \Theta (\theta)=B(\rho)cos(\sqrt{\rho}\theta) + C(\rho)sin(\sqrt{\rho}\theta)$$ note: in these equations $$A,B,C$$, are all constant values, but also a function of constant $$\lambda$$ and $$\rho.$$ based on initial conditions and boundary conditions This is the auxiliary solution for the time $$t$$ and angle $$\theta$$ of the PDE given previously as: $$\frac{1}{r} \frac{\partial}{\partial{r}} \left( r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = \frac{\partial u}{\partial t}$$ where $$u(r,\theta, t) = R(r) \Theta(\theta) T(t)$$

The initial condition is of the form:

$$u(r,\theta, t = t_o) = \bar{u}(r, \theta)$$


 *  (Simple Case)
 * $$\blacktriangleright$$Let $$\bar{u}(r, \theta)=T_o=constant$$
 * i.e.$$u(r,\theta, t = t_o) = R(r)\Theta(\theta)T(t_o) = T_o = constant$$
 * For this simple case, $$T(t_o) = T_o = Aexp(-\lambda t_o)$$
 * $$\therefore A = A(\lambda) = T_o exp(\lambda t_o)$$


 *  (General Case)
 * $$\blacktriangleright$$Let $$u(r, \theta, t_o) = R(r)\Theta(\theta)T(t_o) = \bar{u}(r, \theta)$$
 * $$=\left[ \frac{1}{k} R(r)\Theta(\theta)\right] k = \bar{u}(r, \theta)$$
 * Select $$k = T(t_o) = 1 = exp(-\lambda t_o)$$ where $$k = T_o = 1$$
 * $$\therefore A(\lambda)=exp(\lambda t_o)$$