Introduction to partial differential equation and solution of Laplace equation in cartesiaan co-ordinate system.

Partial differential equation:

There are various types of partial differential equation in mathematical physics which are used to analysis in various phenomena in differential branch of physics. They are;

1. One dimensional wave equations:

$$\frac{\delta^2 u}{\delta u^2}=\frac{1}{c^2}\frac{\delta^2 u}{\delta t^2}$$

Where c is the velocity of transverse wave in string 'u' is the displacement of transverse wave.

2. Laplace equation:

The form of this equation is \(\nabla \phi=0\)

Where \(\rho\) is the charge density and \(\phi\) represent scalar electric potential.

3. Poisson Equation:

\(\nabla^2 \phi= -\frac{\rho}{\epsilon_{\circ}}\)

Where, \(\rho\) is the charge density and \(\phi\) represent scalar electric potential.

4. Schodinger wave equations:

The form of this equation is,

$$i\hbar \frac{\delta \phi}{\delta t}=\frac{-\hbar^2}{2m}\nabla^2\psi+v(x,y,z)\psi$$

$$i\hbar\frac{\delta \psi}{\delta t}=\frac{-\hbar^2}{2m}\nabla\psi+v(x,y,z)\psi$$

5. Helmoltz equations;

The form of this equation is,

$$(\nabla^2+k^2)\psi=0$$

Where, \(\psi\) represents the various potential .

6. Heat conduction equations:

The form of this equation is.

$$\nabla^2 \phi=\frac{1}{h^2}\frac{\delta \theta}{\delta t}$$

Here, h is the diffusivity and \(\theta\) is the temperature distribution.

Solution of Laplace equation in Cartesian co-ordinate system:

The laplace equation in Cartesian co-ordinate is,

$$\nabla^2\psi=0$$

$$or,\;\; \frac{\delta^2\psi}{\delta x^2}+\frac{\delta^2\psi}{\delta y^2}+\frac{\delta^2\psi}{\delta z^2}=0\dotsm(1)$$

Here, \(\psi=\psi(x,y,x)\dotsm(2)\)

From equation (1) and (2)

$$yz\frac{\delta ^2 x}{\delta x^2}+ xz\frac{\delta^2y}{\delta y^2}+yx \frac{\delta^2z}{\delta z^2}=0$$

Dividing both sides by xyz, we get;

$$\frac{1}{x} \frac{\delta^2 x}{\delta x^2}+\frac1y \frac{\delta^2y}{\delta y^2}+\frac1z \frac{\delta^2z}{\delta z^2}=0$$

Since, all terms of above equation are separated so these must be equal to some constant. i.e.

Let, \(\frac1x \frac{\delta^2X}{\delta x^2}=-\alpha^2 (constant)\)

$$\frac{\delta^2 X}{\delta x^2}+ \alpha^2 x=0$$

$$\Rightarrow X=A_1 cos\alpha x+ A_2 sin\alpha x$$

Also,

$$\frac 1y \frac{\delta^2 Y}{\delta y^2}=-\beta^2$$

$$or,\;\; \frac{\delta^2 Y}{\delta y^2}+\beta^2 Y=0$$

$$\Rightarrow Y= A_2 cos\beta y+ B_2 sin\beta y$$

Now,

$$\frac1z \frac{\delta^2 Z}{\delta z^2}=\gamma^2$$

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

$$or,\;\; z(x)= A_3 e^{yz} + B_3 e^{-\gamma_2}$$

Where, \(-\alpha^2- \beta^2+\gamma^2=0\)

Now, the solution of \(\psi\) is,

$$\psi(x,y,z)= X(x) Y(y) Z(z)$$

$$= (A_1 cos\alpha x+ \beta_1 sin\alpha x)(A_2 cos\beta Y+ B_2 sin\beta y)(A_3 e^{y_2} + B_3 e^{-y_2}$$

Variable separation method:

This method which reduces a partial differential equations with 'n' independent variable to a ordinary differential equation involving (n-1) independent separation constant.

Condition needed to used variable separation method.

(i). The parameters used in variable separation method must be independent.

Eg,

In spherical polar co-ordinate

$$\psi(r,\theta,\phi)= R(r) \Phi(\phi)\oplus (\theta)\)

Here, R,\(\theta\), I bare independent.

(ii). The solution obtained by setting variable separation method must satisfy the original equation.

(iii). The separation constant used in this method are set in such a way that the boundary condition and initial condition can be freely imposed to the function.

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.