BIEN 501Partial Differential Equations
Physiological ModelingJanuary 17, 2006
Review of Partial Differential Equations
Partial differential equations arise whenever the function to be found varies with more than one independent variable. In fluid mechanics, solid mechanics, heat transfer and mass transfer, the variables of interest (velocity, stress, temperature, and concentration, respectively) will vary with three spatial dimensions and with time. Take, for example, a one-dimensional time-dependent mass transfer problem. The concentration, depends on both , the spatial dimension, and , time. The differential equation is written in terms of partial derivatives () as opposed to total derivatives (). The operator means that the derivative with respect to is taken while all other variables are held constant.
Partial Differentiation:
If a variable, such as concentration, depends on both space and time, one must evaluate the total derivative to find a rate of change. If and are the variables of interest, then the total derivative is:
The rounded operator () is a necessary element of this equation and cannot be substituted for the straight operator (), which would otherwise mean ”the rate of change of along a given pathway.” The second term on the right hand side of Equation 1 cannot be evaluated until the pathway is described. Thus, the curved operator () is used to denote that the derivative is to be taken with respect to the variable while the variable is held constant.
To gain physical insight into the concept of partial differentiation, consider the temperature to which a particle is exposed as it moves in the x-direction. Assume that the ambient temperature changes with both the x position and time, as in Figure 1. The electric field changes along the x-direction, but also changes with time, so that . One wishes to find the rate at which the temperature surrounding the particle changes as the particle moves through a path .
As a specific example, let the temperature field increase linearly with and sinusoidally in time so that . Let the particle move with constant velocity in the direction, so that . The correct expression for the rate of change of temperature is:
The first term on the right hand side is the rate of change of temperature at a given point caused by the changing temperature field. The second term is the rate of change in temperature that occurs because the particle moves along the direction and temperature increases with . To demonstrate clearly that the second term is necessary, consider what happens when there is no change in the temperature field with time, i.e. . The partial derivative of this field with respect to is zero, so the first term on the right hand side of Eq. 2 is zero. However, there must be a change in temperature because the particle is moving into a region of higher as increases. The rate of that change is the second term on the right hand side of Eq. 2, which becomes:
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As expected, the temperature changes at a rate proportional to the velocity of the particle.
A sample partial differential equation for mass transfer is:
Solutions to Partial Differential Equations:
To solve an ordinary differential equation, one seeks specific functions, such as or that satisfy the equation. In contrast, solutions to partial differential equations are determined by the argument of the functional form. For example, in Equation 3, any function whose argument is is a solution to the equation since, regardless of the form for ,
;
so
; ;
The first objective in solving a partial differential equation is to find a way to convert the partial derivatives to total derivatives, which means that the derivatives must operate on a function of a single variable only. In this course, three methods will be used for solving partial differential equations. These are:
- Separation of Variables
- Transform Methods
- Similarity Solutions
In all three cases, the ultimate goal is to reduce the partial differential equation to one or more ordinary differential equations. The choice as to which method to use depends more on the boundary conditions of the problem than on the form of the partial differential equation.
Separation of Variables:
For separation of variables, the function to be found is written as a product of two or more functions, each of which depends on only one variable. Thus, if one seeks the solution for a velocity , where is a velocity component, one writes that . This assumption is restrictive in that many functions cannot be written in this form. For example, while the function is in such a form (with , , and ), a function such as cannot be written in such a manner.
Exercise 1: Which of the following forms is not separable?
BIEN 501Partial Differential Equations
Physiological ModelingJanuary 17, 2006
a.
b.
c.
d.
e.
f.
g.
h.
BIEN 501Partial Differential Equations
Physiological ModelingJanuary 17, 2006
Exercise 2: For each of the separable forms in Exercise 1, what are the separated functions? (Example: for , the separated functions are and ).
If a solution to a partial differential equation is separable, then the boundary conditions must be separable.
To illustrate the method of separation of variables, the following equation will be considered:
with the boundary conditions:
→
→
→
→
The function will not yet be specified, but in general it is given an explicit form in the statement of the problem. The boundary conditions are separable. The substitution is made in the differential equation so that:
Since is not a function of , it can be treated as a constant with respect to the derivative in . Similarly, since is not a function of , it can be treated as a constant with respect to the derivative in . Therefore:
.
Next, the equation is divided by the product to yield.
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The left hand side of this equation is a function of only, and the right hand side of the equation is a function of only. But if the left hand side changes as changes, it is not possible for the right hand side to change because it does not vary with . Similarly, it is not possible for the left hand side to change as the right hand side changes with . Thus, the only way the two sides can be equal is if they are both independent of and . They must therefore be equal to some constant . It follows that:
In addition, since the functions and are functions of only and only, respectively, it is now possible to replace the partial derivative with an ordinary derivative so that:
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Each equation can now be multiplied by or , as appropriate and the two ordinary differential equations to be solved are:
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The solution to the equation in is
,
and the solution to the equation in is
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The boundary conditions at and require that . Furthermore, the boundary condition for requires that , so that , where is any integer value from to . These are the eigenvalues of the equation. With these eigenvalues,
,
where the two constants and have been combined into the single constant . The subscript in designates that this is the solution corresponding to a specific eigenvalue .
Now the boundary condition for must be satisfied. In the simplest case, where has the form (with some specified constants and ), the solution is readily obtained with and . It is not likely, however, that we will be lucky enough to enjoy this result. Therefore, we must consider the more general case of an arbitrary . Note that must satisfy the boundary conditions at and or else it will not be a valid boundary condition for this problem.
Through the process of separation of variables, we have found a family of solutions to the differential equations. Because the differential equation is linear, any linear combination of these solutions is also a solution to the equation. Specifically, we can write:
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This is then evaluated at , using the last boundary condition to evaluate the coefficients . The following equation is obtained:
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Students who are familiar with the techniques used to derive Fourier series will immediately recognize the method required to determine the value of for a given value of . First, multiply the equation by . Then integrate both sides from to. One obtains:
.
The integral and summation on the right hand side can be interchanged, and since is independent of ,
But the sine wave is an orthogonal function so that the integral is zero unless , in which case it has the value of . Therefore:
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If we substitute this result back into the functional form for , we find that
Exercise 3: Evaluate the coefficients for the case in which has a triangular shape at (i.e. it is zero at and and has a maximum at ).
Transform Methods:
The invocation of Fourier series at the end of the above discussion suggests that the various transforms may be valuable in solving partial differential equations. In illustration of the transform method, a solution to equations 3 and 4 will be sought with via the Fourier transform method.
First, one chooses the variable to be transformed, and then the differential equation and boundary conditions are transformed with respect to that variable. In this case, the variable will be transformed. The differential equation becomes:
where is the Fourier transform of with respect to . The boundary conditions become:
The solution to equation 5 is immediately recognized as:
Again, the first boundary condition leads to . The second boundary condition leads to:
,
so that:
.
This result and Equation 7 combine to yield:
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It follows that is the inverse Fouirer transform of the equation above.
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Once this inverse Fourier transform is evaluated, the boundary conditions forand can be applied.
Note that other transform methods can be used in place of the Fourier transform. The most obvious of these is the Laplace transform. However, transforms can be based on any group of functions that are orthogonal to one another. The Fourier transform is based on sines and cosines, whereas the Laplace transforms are based on exponentials. Other types of transforms include Legendre transformes (based on the orthgonality of the Legendre polynomials), and Fourier Bessel transforms (based on the orthogonality of the Bessel functions).
Similarity Solutions:
Similaritiy solutions are sometimes referred to as “Combination of Variables.” One seeks to combine the variables in the problem in such a way that the solution depends only on that combination. Similarity solutions are often useful in mass transfer problems, and can also be used in many nonlinear problems. The example to be used here is the differential equation
,
with the boundary conditions:
First, we seek a reasonable combination of the governing parameter and the independent variables and . Because the units of are cm2/s, the combination is non dimensional. Next, we must write the differential equation in terms of . We make use of the following relationships:
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With these substitutions, the differential equation becomes:
This can be divided by to yield:
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The combination can now be replaced with to give:
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This form of the equation depends on only. Thus, the partial differential equation has been converted to the ordinary differential equation:
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This equation can be solved by seeking a solution for . The equation is rewritten as:
which is separated as:
.
Thus:
.
So:
where . The concentration becomes:
.
The boundary condition at means that for and . Hence, the limits of the integration and the constant were selected to satisfy this condition.
This integral is the well-known error function and cannot be evaluated in closed form. However, the error function is easy to calculate numerically. The lower limit is taken as since we know that for the concentration must go to zero (because is large and/or because ).