solution of standard types of partial differential equations
First-Order Linear Partial Differential Equations
1.1. Equations of the Form f(x, y)wx + g(x, y)wy = 0
- wx + [f(x)y + g(x)]wy = 0.
- wx + [f(x)y + g(x)yk]wy = 0.
- wx + [f(x)eλy + g(x)]wy = 0.
- f(x)wx + g(y)wy = 0.
- [f(y) + amxnym−1]wx − [g(x) + anxn−1ym]wy = 0.
- [eαxf(y) + cβ]wx − [eβyg(x) + cα]wy = 0.
- wx + f(ax + by + c)wy = 0.
- wx + f(y/x)wy = 0.
- xwx + yf(xnym)wy = 0.
- wx + yf(eαxym)wy = 0.
- xwx + f(xneαy)wy = 0.
1.2. Equations of the Form f(x, y)wx + g(x, y)wy = h(x, y)
- awx + bwy = f(x).
- wx + awy = f(x)yk.
- wx + awy = f(x)eλy.
- awx + bwy = f(x) + g(y).
- wx + awy = f(x)g(y).
- wx + awy = f(x, y).
- wx + [ay + f(x)]wy = g(x).
- wx + [ay + f(x)]wy = g(x)h(y).
- wx + [f(x)y + g(x)yk]wy = h(x).
- wx + [f(x) + g(x)eλy]wy = h(x).
- axwx + bywy = f(x, y).
- f(x)wx + g(y)wy = h1(x) + h2(y).
- f(x)wx + g(y)wy = h(x, y).
- f(y)wx + g(x)wy = h(x, y).
1.3. Equations of the Form f(x, y)wx + g(x, y)wy = h(x, y)w + r(x, y)
- awx + bwy = f(x)w.
- awx + bwy = f(x)w + g(x).
- awx + bwy = [f(x) + g(y)]w.
- wx + awy = f(x, y)w.
- wx + awy = f(x, y)w + g(x, y).
- axwx + bywy = f(x)w + g(x).
- axwx + bywy = f(x, y)w.
- xwx + aywy = f(x, y)w + g(x, y).
- f(x)wx + g(y)wy = [h1(x) + h2(y)]w.
- f1(x)wx + f2(y)wy = aw + g1(x) + g2(y).
- f(x)wx + g(y)wy = h(x, y)w + r(x, y).
- f(y)wx + g(x)wy = h(x, y)w + r(x, y). courtesy:eqworld.ipmnet.ru/en/solutions/fpde/fpdetoc1.htm
A partial differential equation is an equation that involves an unknown function and its partial derivatives. The order of the highest derivative defines the order of the equation. The equation is called linear if the unknown function only appears in a linear form. Finally, the equation is homogeneous if every term involves the unknown function or its partial derivatives and inhomogeneous if it does not.
The course will only consider linear partial differential equations of second order. Examples of important linear partial differential equations are:
Equation (2.1) is the one dimensional wave equation, equation (2.2) is the one dimensional heat (or diffusion) equation and equation (2.3) is the two dimensional Laplace equation. They are all examples of homogeneous, linear partial differential equations.
An example of an inhomogeneous equation is the two dimensional Poisson equation, namely
The solutions to the above equations are numerous. For example, if one considers equation (2.3) then it is easily verified that all of the functions
are solutions. So how do we determine the actual solution we are looking for? The answer, of course, lies in the application of boundary conditions. Once the initial conditions (conditions at ) and the boundary conditions (conditions at specific values of ), where appropriate, are specified, there will be a unique solution to the linear partial differential equation.
Using our knowledge of linear, ordinary differential equations, we expect that it should be possible to linearly superimpose solutions to the equations to obtain the most general solution. This is indeed the case. Thus, if and are both solutions to (2.1), then
is also a solution if and are constants. However, unlike second order ordinary differential equations where there are two linearly independent solutions and two arbitrary constants, linear partial differential equations may well have an infinte number of linearly independent solutions and we may have to add together solutions involving an infinite number of constants. We will return to this later on.
Example 2. .12Obtain a solution, , to the equation
Note that there are only derivatives with respect to and none with respect to . Thus, we can treat the equation like an ordinary differential equation and use first year work to write the solution.
However, unlike the ordinary differential equation case and are not constants but are, in fact, functions of the other independent variable. Hence, the actual solution is
as can be verified by differenitating and substituting into the differential equation.
Example 2. .13Consider the equation
To solve this we set
so that the equation becomes
Thus, the solution is
Now we must integrate with respect to to get the solution . Therefore, we get
and is an arbitrary function of integration when we integrate with respect to .
The wave equation has many physical applications from sound waves in air to magnetic waves in the Sun's atmosphere. However, one of the simplest systems to visualise and describe are waves on a stretched elastic string.
Initially the string is horizontal. Then we distort it by displacing it in the vertical direction and at some time, say , we release it and the string starts to oscillate. The aim is to try and determine the vertical displacement of the string, , as a function of space, , and time . The typical situation is illustrated in Figure 2.1 at a typical time .
To derive the wave equation we need to make some simplifying assumptions
(i) The density of the string, , is constant so that the mass of the string between and is simply times the length of the string between and , namely . Thus,
(ii) The displacement, , and its derivatives are assumed small so that
and the mass of the portion of the string is
(iii) The only forces acting on this portion of the string are the tensions, at and at . The gravitational force is neglected.
(iv) The motions are purely vertical. There is no horizontal motion of any portion of the string.
Now we consider the forces acting on the typical string portion shown in Figure 2.1. Tension acts tangential to the string and the gradient of the tangent is, of course, the slope of or simply
. Now we resolve the forces into their horizontal and vertical components.
Horizontal: At the tension force is
and it acts to the left, whereas at the force is
, acting to the right. Since there is no horizontal motion, these forces must balance and so
Vertical: From Figure 2.1 it is clear that the force at is
and at is
. Then Newton's law of motion gives
Divide by and use (2.5) to give
is the gradient of the tangent to at and this is just
. In a similar manner,
Thus, we get
Using the definition of the derivative
we let tend to zero to obtain
, the tension divided by the density, is the wave speed squared.
A particularly neat solution to the wave equation, that is valid when the string is so long that it may be approximated by one of infinite length, was obtained by d'Alembert. The idea is to change coordinates from and to and in order to simplify the equation. Anticipating the final result, we choose the following linear transformation
and we must use the chain rule to express derivatives in terms of and as derivatives in terms of and . Hence,
The second derivatives require a bit of care.
and in a similar manner
Thus, (2.6) becomes
This equation is much simpler and can be solved by direct integration. First of all integrate with respect to to give
where is an arbitrary function of . Then integrate with respect to to obtain
where is an arbitrary function of and
Finally, we replace and by their expressions in terms of and to obtain
Example 2. .14Verify that
is a solution to the wave equation. Here
Note that we may use the trigonometric identities for
and we remark that this is the product of a function of with a function of . This result will be used later.
D'Alembert's solution involves two arbitrary functions that are determined (normally) by two initial conditions. If the initial conditions are
Then, from (2.7), we have
Note that is a function of but at . The tricky part is calculating
but at and . Thus, the functional form of and are obtained by replacing by and .
Example 2. .15If, for example,
The initial speed, that is
evaluated at , is then
Integrating with respect to , as both and are functions of when , we get
Hence, d'Alembert's solution that satisfies the initial conditions (2.8) is
Example 2. .16 can be thought of as a ``shape", defined by the function , moving to the right. We can see this if we consider the value . Say
Then at when
Thus, we are moving to the right to larger values of . Similarily corresponds to a wave propagating to the left.
Example 2. .17
Thus, the solution to the wave equation is
corresponding to a wave travelling to the right (increasing ).
Here is the amplitude,
is the wavenumber - ,
is the wavelength - distance from one crest to the next,
is the angular frequency ,
is the period of the wave - time from one crest to the next to pass afixed point.
Example 2. .18
This gives the solution
The solution is a standing wave as shown in Figure 2.2. The maximum and minimum displacements are at
and the nodes (where ) are at
Example 2. .19
Consider the case of the string initially at rest with the above initial displacement. Thus,
where is a constant. In Figure 2.3, .
From d'Alembert's solution we have
where is a constant. Adding the two solutions we have
Hence, the solution is
and the arbitrary constant, , does not appear in the solution. Thus, we may, in fact, take without any loss of generality. So we have 2 equal triangles of height going to the right and the left. This is shown in Figure 2.4.