Solution of Laplace equation in cylindrical co-ordinate system and Heat conduction equation

Laplace equation in cylindrical co-ordinate system:

$$\nabla^2\psi=\frac{1}{\rho} \frac{\delta}{\delta \rho}\biggl(\frac{\rho \delta \psi}{\delta \rho}\biggr)+ \frac{1}{\rho^2} \frac{\delta^2\psi}{\delta \phi^2}+\frac{\delta^2 \psi}{\delta z^2}=0\dotsm(1)$$

Here, \(\psi=\psi(\rho,\phi,z)\)

Using variable separation method

$$\psi(\rho,\phi,z)=p(y)\cdot \Phi(\phi)Z(z)\dotsm(2)$$

Using equation (2) in equation (1) we get,

$$\frac{\phi_z}{\rho}\cdot \frac{\delta}{\delta \rho}\biggl(\rho \frac{\delta \rho}{\delta \rho}\biggr)+\frac{pZ}{\rho^2}\frac{\delta^2\phi}{\delta \phi^2}+P\Phi \frac{delta^2 Z}{\delta z^2}=0$$

Dividing both sides by \(P\Phi z\) we get,

$$\frac{1}{p\rho}\cdot \frac{\delta}{\delta \rho}(\rho \frac{\delta p}{\delta \rho}) +\frac{1}{\rho^2\Phi}\frac{\delta \Phi}{\delta \phi^2}+\frac1z \frac{\delta^2Z}{\delta z^2}=0\dotsm(2)$$

The lost term of above equation is function of z only. So it is equal to some constant \((\lambda^2\))

$$\frac{1}{z}\frac{d^2 Z}{dz^2}=\lambda^2$$

$$or,\;\; \frac{\delta^2 Z}{\delta z^2}-\lambda^2 z=0$$

$$\Rightarrow Z(z)= A_1 e^{\lambda z}+B_1 e^{-\lambda z}$$

Now equation (2) can be written as,

$$ \frac{1}{p \rho} \cdot \frac{\delta}{\delta \rho}(\rho \frac{\delta P}{\delta s})+\frac{1}{\rho^2\phi}\frac{\delta^2\phi}{\delta \phi^2}+\lambda^2=0$$

Multiplying above equation by \(\rho^2\) we get,

$$\frac{\rho}{P} \frac{\delta}{\delta \rho}(\rho \frac{\delta p}{\delta \rho})+\frac{1}{\Phi}\cdot\frac{\delta^2\phi}{\delta \phi^2}+\hat e^2 \rho^2=0\dotsm(4)$$

The second term of above equation is function of only \(\Phi\) so it can be set to some constant i.e \(-m^2\).

$$\frac{1}{\phi}\frac{\delta^2\phi}{\delta \phi^2}= -mn$$

$$\frac{\delta^2 \Phi}{\delta \phi^2}+m^2\Phi=0$$

$$\Rightarrow \Phi(\phi)= A_2 cos m\phi+ B_2 sin m\phi$$

The equation (4) reduces to

$$\frac{\rho}{P}\frac{\delta}{\delta \rho}(\rho \frac{\delta p}{\delta \rho})+\lambda^2 \rho^2-m^2=0$$

Multiplying above equation by 'P' we get,

$$\rho\cdot\frac{\delta}{\delta \rho}(\frac{\rho \delta p}{\delta \rho})+(\lambda^2 \rho^2-m^2)p=0$$

Let,

$$\lambda \rho=x$$

$$\lambda \delta \rho= \delta x$$

$$or,\;\;\lambda \frac{\delta}{\delta x}=\frac{\delta}{\delta \rho}$$

Then, $$\rho\lambda\frac{\delta}{delta x}(\rho\lambda\frac{\delta P}{\delta x})+(x^2-m^2)P=0$$

$$or,\;\;x\cdot \frac{\delta}{\delta x}(x\frac{\delta p}{\delta x})+(x^2-m^2)p=0$$

$$or,\;\;x^2\frac{\delta^2 P}{\delta x^2}+x\frac{\delta P}{\delta x}+(x^2-m^2)P=0$$

This is the Bessel differential equation whose solution is Bessel polynomials

$$P(\rho)= A_3 J_m(x)+ B_3 J_{-m}(m)$$

$$\Rightarrow P(\rho)= A_3 J_m (\lambda \rho)+ B_3 J_{-m} (\lambda \rho)$$

Now, the complete solution is

$$u(\rho,\phi,z)= P(\rho)\psi(\phi) Z(z)$$

$$[A_3 J_m (\lambda_3)]+B_3 J_{-m} (\lambda \rho)][A_2 cos m\phi + B_2 sin n\phi][A_1 e^{\lambda_2}+B_1 e^{-\lambda_2}]$$

Heat conduction equation (Heat Equation):

Let us consider a substance of mass 'm' and specific heat capacity 's' then the rate of loss of heat can be written as,

$$\frac{\delta Q}{\delta t}=-ms\frac{\delta Q}{\delta t}$$

If 'k' be the thermal conductivity of the substance then the rate of loss of heat can be written as,

$$\frac{\delta Q}{\delta t}=-k\iint_s \nabla Q\cdot \hat n da$$

From (1) and (2) we get,

$$k\iint_s \nabla Q\cdot \hat n da= ms \frac{\delta Q}{\delta t}$$

Using \(m= \iiint_v \rho dv\) and using gauss divergence theorem we get,

$$k\iiint_v \nabla\cdot \nabla Q dv= \iiint_v \rho s\frac{\delta Q}{\delta t}dv$$

$$or,\;\; \iiint_v\biggl[ k\nabla^2 Q- \rho s\frac{\delta Q}{\delta k}\biggr] dv=0$$

Since, \(dv \ne 0\). So, the integrand equal to zero.

i.e $$ k\nabla^2 Q- \rho s\frac{\delta Q}{\delta t}=0$$

$$or,\; \nabla^2 Q= \frac{\rho s}{k}\frac{\delta Q}{\delta t}$$

$$or,\;\; nabla^2 Q= \frac{1}{h^2} \frac{\delta Q}{delta t}$$

Where,

\(\frac{1}{h^2}=\frac{\rho s}{k}\),

\( h=\sqrt{\frac{k}{\rho s}}= \) coefficient of diffusivity

\(\rho=\)density

\(S=\) Specific heat capacity

\(k=\) Thermal conductivity

This is heat equation in one dimension.

$$\frac{\delta^2 Q}{\delta x^2}=\frac{1}{h^2} \frac{\delta Q}{\delta t}$$

Reference:

1. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.