User:Egm6322.s09.mafia/HW5

=Comments on R3=

The clarifications for the Report 3 asked by the prof. Dr. Loc Vu-Quoc are showed in the respective position on Report 3.

=Comments on R4=

Egm6322.s09.bit.sahin 16:34, 24 April 2009 (UTC)

The clarifications for the Report 4 asked by the Prof. Dr. Loc Vu-Quoc are showed in the respective position on Report 4.

=Solution of unsteady heat equation with no heat source=

Separation of Variables
Egm6322.s09.bit.sahin 16:35, 24 April 2009 (UTC)

$$\frac {1}{rR} \frac {\partial} {\partial r} (r \frac {dR} {dr})$$ $$+ \frac {1}{r^2 \Theta} \frac {d^2 \Theta}{d \theta^2 \theta}$$ $$- \frac {1}{T} \frac {dT}{dt}=0$$

Since $$\left (r,\theta ,t \right )$$ are independent varibles,

$$f\left (r,\theta \right )+g\left (t \right )=0$$  (1)

$$f\left (r,\theta \right )=-g\left (t \right )=\lambda =constant$$   (2)

Therefore,

$$\frac{1}{T}\frac{dT}{dt}=-\lambda$$

$$\frac{dT}{T}=-\lambda dt$$

$$logT=-\lambda t+k_{1}$$

$$T\left (t \right )=exp\left (-\lambda t+K_{1} \right )$$

$$T\left (t \right )=exp\left (-\lambda t \right )expK_{1}$$

Defining $$K_{2}=expK_{1}$$, we have

$$T\left (t \right )= K_{2}exp\left (-\lambda t \right )$$

It is easily seen that as $$t\rightarrow +\infty $$, $$exp\left (-\lambda t \right )$$ approaching to zero. Therefore we choose $$\lambda >0$$ at the beginning.

Egm6322.s09.bit.gk 20:46, 24 April 2009 (UTC)

$$f\left (r,\theta \right )=\frac{1}{rR}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\frac{1}{r^2\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}=-\lambda (const)$$

$$\Rightarrow \frac{r}{R}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\frac{1}{\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}+\lambda r^2 =0 $$ --(1)

let,

$$m(\theta)=\frac{1}{\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}$$

and

$$l(r)=\frac{r}{R}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\lambda r^2$$

$$such\ that\ l(r)+m(\theta)=0$$

$$\Rightarrow l(r)=-m(\theta)=\rho$$ (2)

$$\Rightarrow \frac{1}{\Theta }\frac{\mathrm{d^2\Theta } }{\mathrm{d} \theta ^2}=-\rho$$

$$\Rightarrow {\Theta}''+\rho\Theta=0$$ ---(3)

Assume,

$$ \Theta(\theta)=e^{\alpha\theta}$$, Where $$\alpha$$ is unknown

$$\Rightarrow {\Theta }'\left (\theta \right )=\frac{\mathrm{d} \Theta}{\mathrm{d} \theta}=\alpha e^{\alpha\theta}=\alpha\Theta$$

and,

$$ {\Theta }''\left (\theta \right )=\frac{\mathrm{d^2} \Theta}{\mathrm{d} \theta^2}=\alpha^2 e^{\alpha\theta}=\alpha^2 \Theta$$

Making the above substitutions in the Eqn.(3)

We have,

$$\alpha^2\Theta +\rho\Theta=0$$

$$\Rightarrow \alpha^2+\rho=0$$

$$\Rightarrow \alpha=\pm i\sqrt{\rho}$$

$$\Rightarrow \Theta \left (\theta \right )=k_3e^{i\sqrt{\rho}}+k_4e^{-i\sqrt{\rho}} $$

$$\underline V=v_i \underline e_i=\overline v_i \underline \bar e_i$$

$$\underline e_i$$ linearly independent

$$\left \{\underline \bar e_i \right \}$$ linearly independent

Similarly for functions:linearly independent function. $$ \left \{ e^{i\sqrt\rho},e^{-i\sqrt\rho} \right \}$$ linearly independent.

Method One
--Egm6322.s09.xyz 17:16, 24 April 2009 (UTC)

Egm6322.s09.bit.gk 21:01, 24 April 2009 (UTC) We have from De Moivre's theorem

$$e^{i\theta}=cos(\theta)+i sin(\theta)$$

and $$ e^{-i\theta}=cos(\theta)-i sin(\theta)$$

$$\Rightarrow cos(\theta)=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)$$

$$\Rightarrow sin(\theta)=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)$$

$$\therefore$$ We have

$$ \Theta(\theta)=k_3 \left [cos(\sqrt\rho\theta)+i sin(\sqrt\rho\theta ) \right ]+k_4 \left [cos(\sqrt\rho\theta)-i sin(\sqrt\rho\theta ) \right ]$$

$$ =  \left [(k_3+k_4)cos(\sqrt\rho\theta)+\left( i(k_3+k_4)sin(\sqrt\rho\theta) \right) \right]$$

$$ =  \left [B(\rho)cos(\sqrt\rho\theta)+C(\rho)sin(\sqrt\rho\theta \right]$$

Recall solution for T: $$T(t) = A(\lambda)exp(-\lambda t)$$

As derived above, solution for $$\theta$$: $$ \Theta (\theta)=B(\rho)cos(\sqrt{\rho}\theta) + C(\rho)sin(\sqrt{\rho}\theta)$$ Note: in these equations, A,B,C are all constants.B and C are obtained by clubbing k_3 and k_4 into single constants. These constants are also a function of constant $$\lambda$$ and $$\rho$$, which are determined by 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 $$k = T_o = 1$$
 * $$\therefore A(\lambda)=exp(\lambda t_o)$$

$$\bar u$$ $$(r, \theta)=$$ $$R(r) \Theta (\theta)= \frac {1}{k} R(r)\Theta (\theta)$$

Since $$k$$ is arbitrary, select $$k=1 $$

Now why $$B=B(\rho), C=C(\rho)$$?

Hint:

General initial condition : $$u(r, \theta, t_0)= {\bar u} (r, \theta)$$

Simplify $$u(r, \theta ,t_0)={\bar u (\theta)}$$

$$u(r, \theta_1 ,t_0)={\bar u (\theta_1)}$$

$$u(r, \theta_2 ,t_0)={\bar u (\theta_2)}$$

$${\bar u} (\theta)= \Theta (\theta)= \frac {1}{k} \Theta (\theta) k$$

Let $$\frac {1}{k} \Theta (\theta) = \Theta (\theta)$$,

$$k=T(t_0)$$

$$\begin{cases} {\bar u} (\theta_1)= Bsin(\sqrt{\rho} \theta_1)+Ccos(\sqrt{\rho} \theta_1)\\ {\bar u} (\theta_2)= Bsin(\sqrt{\rho} \theta_2)+Ccos(\sqrt{\rho} \theta_2) \end{cases}$$

Then we can solve B, C and finally we will find that B and C are functions of $$\rho$$

$$B=B(\rho, \theta_1, \theta_2)$$

$$C=C(\rho, \theta_1, \theta_2)$$

--EGM6322.S09.TIAN 17:19, 24 April 2009 (UTC)

Method Two
There is another means of determining $$B$$ and $$C$$ for the general solution. For Case 2, we must multiply the equation:

$$\bar u (\theta)=\Theta (\theta)$$ by $$sin (\sqrt \rho \theta )$$ and $$cos (\sqrt \rho \theta )$$, and evaluate the integral over the domain from $$\theta_1$$ and $$\theta_2$$:

$$\int\limits_{{\theta}_1}^{{\theta}_2}\bar{u}(\theta) sin (\sqrt \rho \theta )\, d\theta =\int\limits_{{\theta}_1}^{{\theta}_2}\Theta(\theta) sin (\sqrt \rho \theta )\, d\theta $$

$$\int\limits_{{\theta}_1}^{{\theta}_2}\bar{u}(\theta) cos (\sqrt \rho \theta )\, d\theta =\int\limits_{{\theta}_1}^{{\theta}_2}\Theta(\theta) cos (\sqrt \rho \theta )\, d\theta $$.

This leads to the following equations:

$$\int\limits_{{\theta}_1}^{{\theta}_2}\bar{u}(\theta) sin (\sqrt \rho \theta ) d \theta = \left [ \int\limits_{{\theta}_1}^{{\theta}_2}B(\rho) cos( \sqrt \rho \theta)+ C(\rho) sin( \sqrt \rho \theta)\right ]sin (\sqrt \rho)\, d\theta $$

$$\int\limits_{{\theta}_1}^{{\theta}_2}\bar{u}(\theta) cos (\sqrt \rho \theta ) d \theta = \left [ \int\limits_{{\theta}_1}^{{\theta}_2}B(\rho) cos( \sqrt \rho \theta)+ C(\rho) sin( \sqrt \rho \theta) \right] cos (\sqrt \rho)\, d\theta $$

This can be rearranged into the following matrix form:

$$ \begin{Bmatrix} \int\limits_{{\theta}_1}^{{\theta}_2}\bar{u}(\theta) cos (\sqrt \rho \theta ) d \theta \\ \int\limits_{{\theta}_1}^{{\theta}_2}\bar{u}(\theta) sin (\sqrt \rho \theta ) d \theta \end{Bmatrix}= \begin{bmatrix} \int\limits_{{\theta}_1}^{{\theta}_2}cos^2( \sqrt \rho \theta) d \theta& \int\limits_{{\theta}_1}^{{\theta}_2}sin( \sqrt \rho \theta) cos( \sqrt \rho \theta)d \theta \\ \int\limits_{{\theta}_1}^{{\theta}_2}sin( \sqrt \rho \theta) cos( \sqrt \rho \theta)d \theta & \int\limits_{{\theta}_1}^{{\theta}_2}sin^2( \sqrt \rho \theta) d \theta \end{bmatrix}

\begin{Bmatrix} B(\rho)  \\ C(\rho) \end{Bmatrix}

$$

Note here that $$\sqrt \rho$$ is not an integer, but is in the set of real numbers, and therefore can be irrational.

However, there is a problem with both methods.

$$\bar{u}$$ is an arbitrary function, and $$\bar{u} \ne \Theta (\theta)$$ in general. Solved with either the first or second method, it has only two terms, both of which are sinusoidal functions. Even if $$\bar{u}$$ is as simple as a constant, it cannot be expressed with only two sinusoidal functions

It is, instead, a superposition of many sine and cosine functions, expressed as a Fourier Series:

$$\bar{u} =\sum_\rho B(\rho) sin( \sqrt \rho \theta)+ C(\rho) cos( \sqrt \rho \theta) $$

From Eqn.(2), solving for l(r), we have

l(r)= $$l(r)=\frac{r}{R}\frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\lambda r^2 = \rho$$

Multiplying through by $$R(r)$$, we are left with the following equation:

$$l(r)= r \frac{\mathrm{d} }{\mathrm{d} r}\left (r\frac{\mathrm{d} R}{\mathrm{d} r} \right )+\lambda r^2 R= \rho R$$

Simplifying the differential term and rearranging

$$ l(r)= r^2 \left( \frac{\mathrm{d^2}R }{\mathrm{d} r^2} \right)+ r \left (\frac{\mathrm{d}R }{\mathrm{d} r} \right)+\left ((\sqrt\lambda r)^2-\rho) \right)R= 0 $$

The solution is a Bessel Function of the first and second type:

$$R(r)=D(\rho, \lambda){J}_{\sqrt \rho}(\sqrt \lambda r) +E(\rho, \lambda) {Y}_{\sqrt \rho} (\sqrt \lambda r)$$

Where $$J$$ and $$Y$$ represent Bessel functions of the first and second type, respectively.

Recall that the Bessel functions $${J}_{\nu}$$ and $${Y}_{\nu}$$ are solutions to the Ordinary Differential Equation:

$$z^2 \frac{d^2 w}{d z^2}+ z \frac{dw}{dz} + (z^2-{\nu}^2)w=0$$.

The solution is in the form $$ w(z)= \begin{cases}

{J}_{\nu}(z)\\

{Y}_{\nu}(z) \end{cases}$$

Here, in general, $$z$$ and $$\nu$$ are complex numbers of the form $$z=x+iy$$. For the particular case where $$\nu$$ is a non-negative integer, the equation for which the Bessel functions are the solution becomes:

$$z \frac{d}{d z} \left ( z \frac{dw}{dz} \right ) + (z^2-{\nu}^2)w=0$$

Bessel Functions
Homework:Plot Bessel Functions





Matlab Script Code Matlab Script Code

Verify Solution to Bessel Equation of the First Kind

--Egm6322.s09.lapetina 02:07, 17 April 2009 (UTC)

Major aid from Advanced Engineering Mathematics by E. Kreyszig

Verify that the integral expression of :

$$w=J_n (z)= \frac{1}{\pi} \int_0^{\pi} cos(z sin(\theta)-n \theta) d \theta$$, where $$n$$ is the set of non-negative integers, is in fact a solution to the equation:

$$z^2 w'' +zw'+(z^2-{n}^2)w=0$$

It would be convenient to simply take derivatives of the solution with respect to $$z$$ to the Bessel equation, which leaves us with:

$$w'=-\frac{1}{\pi} \int_0^{\pi} sin(\theta)sin(z sin(\theta)-n \theta) d \theta$$ and

$$w''=-\frac{1}{\pi} \int_0^{\pi} sin^2(\theta)cos(z sin(\theta)-n \theta) d \theta$$.

This leaves us with:

$$-\frac{1}{\pi} \int_0^{\pi} sin^2(\theta)cos(z sin(\theta)-n \theta) d \theta - \frac{1}{z}\frac{1}{\pi} \int_0^{\pi} sin(\theta)sin(z sin(\theta)-n \theta) d \theta + (1-\frac{n^2}{z^2})\frac{1}{\pi} \int_0^{\pi} cos(z sin(\theta)-n \theta) d \theta$$

It is extremely difficult to solve these integrals analytically.

Another option would be to re-express the integral form of the Bessel solution as a summation using the Method of Frobenius, which applies for this type of equation, because it is of the form:

$$z^2 w'' +z b(z) w' +c(z) w=0$$

where

$$b=1$$

and

$$c=(z^2-{n}^2)$$.

We will assume:

$$w= z^k \sum_{n}^{\infty} a_{n}z^{n}$$

where $$k$$ is any number such that $$a_0 \neq 0$$ making

$$w'= (n+k)z^k\sum_{n}^{\infty} a_{n} z^{n-1}$$

and

$$w''= (n+k)(n+k-1)z^k \sum_n^{\infty} a_n z^{n-2}$$.

If we take these summations and functions and plug them into the original differential equation, we are left with:

$$z^2((n+k)(n+k-1)\sum_n^{\infty} a_n z^{n+k-2}) +z((n+k)\sum_{n}^{\infty} a_{n} z^{n+k-1})  + (z^2-{n}^2) (\sum_{n}^{\infty} a_{n}z^{n+k}) =0$$

which simplifies to:

$$((n+k)(n+k-1)\sum_n^{\infty} a_n z^{n+k}) +((n+k)\sum_{n}^{\infty} a_{n} z^{n+k})  + (\sum_{n}^{\infty} a_{n}z^{n+k+2}) +n^2 (\sum_{n}^{\infty} a_{n}z^{n+k})=0$$

It should be noted that all the exponents within the summations are equivalent. Because this entire equation equals zero, and none of the exponential terms can be zero, the sum of the coefficients for each power must be zero. Let us assume $$n+k=r$$

For the first series, this leaves us with the polynomial:

$$(r)(r-1)a_0+(r)a_0-n^2 a_0=0$$

for the second series, with:

$$(r+1)(r)a_1+(r+1)a_1-n^2 a_1=0$$

and for subsequent series $$s=2,3,...,\infty$$, with:

$$(s+r)(s+r-1)a_s+(s+r)a_s+a_{s-2}-n^2 a_s=0$$.

For the first series, it is clear that the roots of the equation can be found from:

$$r^2-n^2=0$$

and the roots of this equation are then clearly $$n$$ and $$-n$$.

Using the root $$n$$the second series becomes:

$$(n+1)(n)a_1+(n+1)a_1-n^2 a_1=0$$

$$n^2 a_1+2n a_1 +a_1-n^2 a_1=0$$,

indicating $$a_1=0$$.

For higher series, the series becomes:

$$(s+n)(s+n-1)a_s+(s+n)a_s+a_{s-2}-n^2 a_s=0$$ which simplifies to:

$$((s+n)(s+n-1)+(s+n)-n^2)a_s+a_{s-2}=0$$

$$(s^2+2ns)a_s+a_{s-2}=0$$.

Interestingly, because $$a_1=0$$, and $$n$$ is non-negative, all odd series must have coefficients of 0.

Thus, we only have to work with even series $$m$$. This results in the coefficients of the even series $$m$$ being represented as:

$$(2m+2n)2ma_{2m} + a_{2m-2}=0$$.

This results in:

$$a_{2m}=-\frac{-1}{2^2 m (n+m)}a_{2m-2}$$.

This is recursive, which has helpful properties. The sign oscillates, and terms become factorials.

For series $$m$$, this leaves us with:

$$a_{2m}=\frac{-1^m a_0}{2^{2m} m! (n+1)(n+2)...(n+m)}$$.

Choosing $$a_0 =frac{1}{2^n n!}$$ simplifies the problem such that:

$$a_{2m}=\frac{-1^m}{2^{2m+n} m!(n+m)!}$$.

Finally, we are left with the Bessel function of the first kind in summation form:

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

The derivatives of this are then:

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

and

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

Plugging these derivatives into the original equation yields:

$$z^n \sum_{m=2}^\infin \frac{(2m)(2m-1) (-1)^m z^{2m}}{2^{2m+n} m! (n+m)!} + nz^{n}\sum_{m=1}^\infin \frac{2m (-1)^m z^{2m}}{2^{2m+n} m! (n+m)!}  +  nz^{n} \sum_{m=1}^\infin \frac{(2m)(-1)^m z^{2m}}{2^{2m+n} m! (n+m)!}  +  $$ $$ n(n-1)z^{n-2} \sum_{m=0}^\infin \frac{(-1)^m z^{2m}}{2^{2m+n} m! (n+m)!} +   z^{n-1} \sum_{m=1}^\infin \frac{2m (-1)^m z^{2m-1}}{2^{2m+n} m! (n+m)!}  +  nz^{n} \sum_{m=0}^\infin \frac{(-1)^m z^{2m}}{2^{2m+n} m! (n+m)!} + $$

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

These summations equate to zero.

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

--Egm6322.s09.xyz 17:17, 24 April 2009 (UTC)

(source: Advanced Engineering Mathematics by E. Kreyszig )

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

Superposition: 2 separation of variables, constant $$\rho$$, $$\lambda$$

$$\sum_{\lambda}\sum_{\rho}[A(\lambda)exp(-\lambda t)][B(\rho)sin(\sqrt{\rho}\theta)+C(\rho)cos(\sqrt{\rho}\theta)] [D(\rho,\lambda)J_{\sqrt{\rho}}(\sqrt{\lambda} r)+E(\rho,\lambda)Y_{\sqrt{\rho}}(\sqrt{\lambda} r)]$$

Up to now, no restriction on $$\rho$$ and $$\lambda$$, i.e. they can be continuous real numbers (as opposed to discrete integral numbers as in Fourier Series: think of Fourier integral or Fourier Transformation and Laplace Transformation)

$$\Rightarrow$$ Replace discrete sum $$\sum$$ by continuous sum of $$\int$$.

Define:

$$\bar{B}(\rho,\lambda):=A(\lambda)B(\rho)$$

$$\bar{C}(\rho,\lambda):=A(\lambda)C(\rho)$$

General solution:

$$u(r,\theta,t)=\int d\lambda \int d\rho[exp(-\lambda t)][\bar{B}(\rho,\lambda)sin(\sqrt{\rho}\theta)+\bar{C}(\rho,\lambda)cos(\sqrt{\rho}\theta)][D(\rho,\lambda)J_{\sqrt{\rho}}(\sqrt{\lambda} r)+E(\rho,\lambda)Y_{\sqrt{\lambda}}r] $$

Egm6322.s09.three.liu 16:29, 24 April 2009 (UTC)

Egm6322.s09.Three.nav 13:44, 24 April 2009 (UTC)

Particularizing the general solution:

Consider a Domain $$ \mho $$ that is a full circle i.e. $$\theta \ \epsilon \ [0,2\pi]$$

$$ \Rightarrow$$ Solution has to be periodic w.r.t $$\theta$$

$$ \Rightarrow u(r,\theta+2k\pi, t) = u(r,\theta,t) $$, for any non-negative integer 'k'.

This implies that solution for $$\theta$$($$\theta$$), which consists of sines and cosines (periodic functions) is only possible only for integers n= 1,2,...

$$ \Rightarrow \sqrt{\rho}= n $$

$$ \Rightarrow \rho= n^{2} $$

Also replace the continuous integral $$ \int d\rho$$ by a discrete sum $$\sum_{0}^{n}$$ in Eqn.(2)

Now the equation becomes

$$ u(r,\theta,t)= \int_{0}^{\infty} d\lambda \left [\sum_{0}^{n} exp(-\lambda t)\left (\bar{B}(\rho, \lambda)sin(\sqrt{\rho}\theta) + \bar{C}(\rho,\lambda)cos(\sqrt{\rho}\theta) \right)\left (D(\rho, \lambda)J_{\sqrt{\rho}}(\sqrt{\lambda}r)+(E(\rho, \lambda)Y_{\sqrt{\rho}}(\sqrt{\lambda}r \right) \right] $$

An important point to remember here is that the above solution, obtained as a result of assuming periodicity, is applicable to the annulus domain.

Q: What if the domain is a solid disk, instead of an annulus?

In the plots of $$ J_{n}(z)$$ and $$Y_{n}(z)$$ Vs z, it was seen that $$Y_{n}(z) \to -\infty \ as\ z\ \to \ 0^{+} $$

For the solution to remain meaningful, this cannot be allowed to happen. Hence we force $$E(\rho, \lambda)$$ = 0.

Hence the equation reduces to

$$ u(r,\theta,t)= \int_{0}^{\infty} d\lambda \left [\sum_{0}^{n} exp(-\lambda t)\left (\bar{\bar{B}}(\rho, \lambda)sin(\sqrt{\rho}\theta) + \bar{\bar{C}}(\rho,\lambda)cos(\sqrt{\rho}\theta) \right)J_{\sqrt{\rho}}(\sqrt{\lambda}r)\right] $$

where $$\bar{\bar{B}}= \bar{B}(\rho,\lambda)D(\rho,\lambda)$$ and

$$\bar{\bar{C}}= \bar{C}(\rho,\lambda)D(\rho,\lambda)$$

Remembering that $$\sqrt{\rho} = n$$, the above equation can be rewritten as

$$ u(r,\theta,t)= \int_{0}^{\infty} d\lambda \left [\sum_{0}^{n} exp(-\lambda t)\left (\bar{\bar{B}}(\rho, \lambda)sin(n\theta) + \bar{\bar{C}}(\rho,\lambda)cos(n\theta) \right)J_{n}(\sqrt{\lambda}r)\right] $$

Example Problem: Solve the above given form of heat equation using the same solution method given above but with different initial conditions and boundary conditions as listed below:

Initial Condition: $$u(r,\theta,t=0)= u_{0}\left (1-(\frac{r}{R})^2 \right)$$

Boundary Condition:

$$ For \ \theta \neq [-\frac{\pi}{4}, \frac{\pi}{4}], u(R,\theta,t)=0$$

$$ For \ \theta \epsilon [-\frac{\pi}{4}, \frac{\pi}{4}], u= \begin{Bmatrix} u(\frac{u_{1}}{t_{1}}), tt_{1} \end{Bmatrix}$$

Note: Was unable to solve this problem. Egm6322.s09.Three.nav 16:13, 24 April 2009 (UTC)

=References=

=Signatures=

--Egm6322.s09.xyz 19:22, 26 March 2009 (UTC)

Egm6322.s09.bit.la 21:03, 26 March 2009 (UTC)

Egm6322.s09.Three.nav 05:28, 27 March 2009 (UTC)

--EGM6322.S09.TIAN 14:22, 27 March 2009 (UTC)

Egm6322.s09.Three.ge 16:04, 27 March 2009 (UTC)

Egm6322.s09.bit.sahin 17:59, 27 March 2009 (UTC)

Egm6322.s09.three.liu 19:56, 27 March 2009 (UTC)

--Egm6322.s09.lapetina 20:28, 27 March 2009 (UTC)

Egm6322.s09.bit.gk 20:53, 27 March 2009 (UTC)