LIBRARY

PRODUCTS

FORUMS

CART

Join CodeCogs

Or login with:

Maths › Differential Equations ›

Partial Differential Equations

An Introduction to Partial Differential Equations

View version details

Contents

Overview
The Solution Of Partial Differential Equations.
Page Comments

Overview

Equations which connect three or more variables and their partial derivatives are called Partial Differential Equations. They can be formed by in the same way as ordinary differential equations, That is by eliminating one or more arbitrary c $\displaystyle e^{-\,ipx\,-\,p^2a^2t}$ $\displaystyle e^{-\,ipx\,-\,p^2a^2t}$ constants, but they can also be formed by eliminating one or more arbitrary functions.

The equations which will be considered occur frequently in engineering problems covering the conduction of heat; vibration of strings; gravitation; electro-magnetic waves and the diffusion of solvents.

Example 1:

$z\;=\;(x\;-\;A)^2\;+\;(y\;-\;B)^2$

(1)

$\therefore\;\;\;\;\;\;\frac{\partial z}{\partial x}\;=\;2(x\;-\;A)\;\;\;\;and\;\;\;\;\frac{\partial z}{\partial y}\;=\;2(y\;-\;B)$

(2)

This equation is of the first order although the equation from which it is derived has two arbitrary constants.

Example 2:

Eliminate A and p from $\displaystyle z\;=\;A\,e^{pt}\;sin\;px$

From ths equation it can be seen that:-

$\frac{\partial ^2z}{\partial x^2}\;=\;-\;p^2\,Ae^{pt}\;sin\;px$

(3)

$And\;\;\;\frac{\partial ^2z}{\partial t^2}\;=\;p^2\,Ae^{pt}\;sin\;px$

(4)

$therefore\;\;\;\;\;\;\frac{\partial ^2z}{\partial x^2}\;+\;\frac{\partial ^2z}{\partial t^2}\;=\;0$

(5)

Example 3:

Eliminate a; b; and c from the following equation:-

$z\;=\;a(x\;+\;y)\;+\;b(x\;-\;y)\;+\;abt\;+c$

(6)

Differentiating we get:-

$\frac{\partial z}{\partial x}\;=\;a\;+\;b\;\;\;\;\;\frac{\partial z}{\partial y}\;=\;a\;-\;b\;\;\;\;and\;\;\;\frac{\partial z}{\partial t}\;=\;ab$

(7)

$But\;\;\;\;\;\;\;(a\;+\;b)^2\;-\;(a\;-\;b)^2\;=\;4\,ab$

(8)

$\therefore\;\;\;\;\;\left(\frac{\partial z}{\partial x} \right)^2\;-\;\left(\frac{\partial z}{\partial y} \right)^2\;=\;4\,\frac{\partial z}{\partial t}$

(9)

Example 4:

$If\;\;\;\;\;\;z\;=\;f(x^2\;+\;y^2)$

(10)

$Then\;\;\;\;\;\frac{\partial z}{\partial x}\;=\;2xf'(x^2\;-\;y^2)\;\;\;\;and\;\;\;\;\frac{\partial z}{\partial y}\;=\;-\;2yf'(x^2\;-\;y^2)$

(11)

$\therefore\;\;\;\;\;x\frac{\partial z}{\partial y}\;+\;y\frac{\partial z}{\partial x}\;=\;0$

(12)

Example 5:

$If\;\;\;\;\;\;y\;=\;f(x\;-\;at)\;+\;F(x\;+at)$

(13)

$Then\;\;\;\;\;\;\frac{\partial y}{\partial x}\;=\;f'(x\;-\;at)\;+\;F'(x\;+at)$

(14)

$And\;\;\;\;\;\;\frac{\partial^2 y}{\partial x^2}\;=\;f''(x\;-\;at)\;+\;F''(x\;+at)$

(15)

$Similarly\;\;\;\;\;\;\frac{\partial y}{\partial t}\;=\;-\;af'(x\;-\;at)\;+\;aF'(x\;+at)$

(16)

$And\;\;\;\;\;\;\frac{\partial^2 y}{\partial t^2}\;=\;a^2f''(x\;-\;at)\;+\;a^2F''(x\;+at)$

(17)

From equations (15) and (17)

$\frac{\partial ^2y}{\partial x^2}\;=\;\frac{1}{a^2}\;\frac{\partial ^2y}{\partial t^2}$

(18)

$\phi \left(\frac{y}{x}\;\frac{z}{x} \right)\;=\;0$

(19)

Example 6:

If it is required to eliminate the Arbitrary Function from :-

$\phi \left(\frac{y}{x}\;\frac{z}{x} \right)\;=\;0$

(20)

Then suppose the following that $\displaystyle \xi \;=\;\frac{y}{x}\;\;\;\;and\;\;\;\;\eta \;=\;\frac{z}{x}$

$i.e.\;\;\;\;\;\;\phi (\xi \,,\,\eta )\;=\;0$

(21)

Now assume that y is kept constant and that z is a function of x only, differentiate with respect to x.

$Thus\;\;\;\;\;\frac{\partial \phi }{\partial \xi }\;\frac{\partial \xi }{\partial x}\;+\;\frac{\partial \phi }{\partial \eta }\;\frac{\partial \eta }{\partial x}\;=\;0$

(22)

$\displaystyle \therefore\;\;\;\;\;\;x\,\frac{\partial z }{\partial x}\;+\;y\,\frac{\partial z}{\partial y}\;=\;z$

$\therefore\;\;\;\;\;\;\frac{\partial \phi }{\partial \xi }\left(-\,\frac{y}{x^2} \right)\;+\;\frac{\partial \phi }{\partial \eta }\left(\frac{1}{x}\,\frac{\partial z}{\partial x}\;-\;\frac{z}{x^2} \right)\;=\;0$

(23)

Similarly if we keep x constant and differentiate with respect to y

$\frac{\partial \phi }{\partial \xi }\left(\frac{1}{x} \right)\;+\;\frac{\partial \phi }{\partial \eta }\left(\frac{1}{x}\,\frac{\partial z}{\partial y} \right)\;=\;0$

(24)

$\displaystyle Eliminating\;\;\;\;\;\;\frac{\partial \phi }{\partial \xi }/\frac{\partial \phi }{\partial \eta }$

$\frac{1}{x}\left(\frac{1}{x}\,\frac{\partial z}{\partial x}\;-\;\frac{z}{x^2} \right)\;+\;\frac{y}{x^2}\,\frac{1}{x}\frac{\partial z}{\partial y}\;=\;0$

(25)

The Solution Of Partial Differential Equations.

It w3ill be clear from these examples that the methods used for the solution of ordinary differential equations will not apply to Partial Differential Equations without considerable modification. A general discussion of partial differential equations is both difficult and lengthy. The objective in the following examples is to show some of the substitutions which may be used in the solution of the types of equation which occur in Scientific and engineering applications.

Example 7:

The following linear equation gives the conduction of heat in one direction.

$\frac{\partial ^2z}{\partial x^2}\;=\;\frac{1}{a^2}\;\frac{\partial z}{\partial t}$

(26)

In the treatment of ordinary linear equations it was found that the use of exponentials was useful. This suggests $\displaystyle z\;=\;e^{mx+nt}$ as a trial solution. Substituting in the differential equation:-

$m^2\;e^{mx+nt}\;=\;\frac{1}{a^2}\;ne^{mx+nt}$

(27)

Which is true if $\displaystyle n\;=\;m^2\,a^2$

Thus $\displaystyle e^{mx+m^2a^2t}$ is a solution. Changing the sign of m gives $\displaystyle e^{-mx+m^2a^2t}$ which is also a solution.

Example 8:

Find a solution to the same differential equation as in example 7 that vanishes when $\displaystyle t\;=\;+\,\infty$ . In the previous solutions t occurs in $\displaystyle e^{m^2a^2t}$ . This increases with t since $\displaystyle m^2a^2$ is positive if m and a are real. To make it decrease it is necessary to put $\displaystyle m\;=\;i\,p$ so that $\displaystyle m^2\;a^2\;=\;-\;p^2\;a^2$ .

As in the above example this gives $\displaystyle e^{ipx\,-\,p^2a^2t}$ as a solution. Similarly $\displaystyle e^{-\,ipx\,-\,p^2a^2t}$ is also a solution.

Hence as the differential equation is linear the following solution can be written:-

$e^{-p^2a^2t}\;(A\,e^{-ipx}\;+\;B\,e^{ipx})$

(28)

As usual this can be replaced by:-

$e^{-p^2a^2t}(E\;cos\;px\;+\;F\;sin\;px)$

(29)

Example 9:

Find a solution of $\displaystyle \frac{\partial ^2z}{\partial x^2}\;+\;\frac{\partial ^2z}{\partial y^2}\;=\;0$ which will vanish when $\displaystyle y\;=\;+\,\infty$ and also when x = 0.

The condition, when $\displaystyle y\;=\;+\,\infty$ demands that n should be both real and negative say n = -p.

$Then\;\;\;\;\;\m\;=\;\pm \;ip$

(30)

$Hence\;\;\;\;\;e^{-py}(A\,e^{ipx}\;+\;B\,e^{ipx})\;\;\;\;is\;a\;solution$

(31)

As in previous examples this can be re-written as:-

$e^{-py}\;(E\;cos\,px\;+\;F\;sin\;px)$

(32)

But z = 0 if x = 0 and so E = 0

Thus the solution required is therefore:-

$F\;e^{-py}\;\;sin\;px$

(33)

Example 10:

Find a solution to the following equation which occurs in hydrodynamics:-

$\frac{\partial }{\partial r}\left(r^2\frac{\partial \phi }{\partial r} \right)\;+\;\cosec\,\theta \;\frac{\partial }{\partial \theta }\left(sin\,\theta \;\frac{\partial \phi }{\partial \theta } \right)\;=\;0$

(34)

The form of the second term suggests the following substitution:-

$\phi \,(r,\theta )\;=\;cos\,\theta f(r)$

(35)

$Then\;\;\;\;\;\frac{\partial \phi }{\partial r}\;=\;\cos\;\frac{df}{dr}\,,\;\;\;\;\;\frac{\partial \phi }{\partial \theta }\;=\;-\;\sin\,\theta f(r)$

(36)

The original equation now becomes:-

$\frac{\partial }{\partial r}\left(r^2\;\cos\,\theta \;\frac{df}{dr} \right)\;+\;cosec\:\theta\;\frac{\partial }{\partial \theta }\{-\;s\sin^2\theta \;f\;(r)\}\;=\;0$

(37)

$\therefore\;\;\;\;\;\;\cos\,\theta \;\frac{d}{dr}\left(r^2\;\frac{df}{dr} \right)\;-\;2f(r)\;\cos\;\theta \;=\;0$

(38)

$\therefore\;\;\;\;\;r^2\;\frac{d^2f}{dr^2}\;+\;2r\;\frac{df}{dr}\;-\;2f\;=\;0$

(39)

$Putting\;\;\;\;r\;=\;e^t,\;\;\;\;Then\;\;\;\;\frac{d^2f}{dt^2}\;+\;\frac{df}{dt}\;-\;2f\;=\;0$

(40)

$\therefore\;\;\;\;\;\;f\;=\;A\,e^t\;+\;B\,e^{-2t}$

(41)

A Solution to the given equation is therefore given by:-

$\phi (r,\,\theta )\;=\;\cos\,\theta \left(Ar\;+\;\frac{B}{r^2} \right)$

(42)

Example 11:

Find a Solution of the following equation:-

$\left(\frac{\partial z}{\partial x} \right)^2\;+\;\left(\frac{\partial z}{\partial y} \right)^2\;=\;1$

(43)

Assume that $\displaystyle \frac{\partial z}{\partial x}\;\;\;\;\;and\;\;\;\;\;\frac{\partial z}{\partial y}\;\;\;\;are\;Constants$

$\displaystyle Let\;\;\;\;\;\frac{\partial z}{\partial x}\;=\;a\;\;\;\;and\;\;\;\;\;\frac{\partial z}{\partial y}\;=\;b$

$Then\;\;\;\;\;\;dz\;=\;\frac{\partial z}{\partial x}\;dx\;+\;\frac{\partial z}{\partial y}\;dy$

(44)

$\therefore\;\;\;\;\;dz\;=\;a\,dx\;+b\,dy$

(45)

$or\;\;\;\;\;z\;=\;ax\;+\;by\;+\;c\;\;\;\;where\;\;a^2\;+\;b^2\;=\;1$

(46)

In General, If $\displaystyle If\;\phi \left(\frac{\partial z}{\partial x}\;,\;\frac{\partial z}{\partial y}\right)\;=\;0$ then we assume that $\displaystyle \frac{\partial z}{\partial x}\;=\;a\;\;\;\;\;and\;\;\;\;\frac{\partial z}{\partial y}\;=\;b$ where $\displaystyle \phi \;(\;a,b)\;=\;0$

The Solution is then given by:-