Partial Differential Equations PowerTool

by Dr. Jim Herod

Section 4.1: The Simple Heat Equation

We consider the simple heat equation. This is a standard linear realization for diffusion in one dimension. Derivations for this model can be found in most texts in this subject. The equation typically has the following form:   u  is a function of t  and x  and is written as u ( t , x ). We suppose that u  is differentiable as a function of t  and twice differentiable as a function of x . In this model, these derivatives are related by

              =    

In the heat equation, there are boundary conditions and initial conditions. In this Section, we will take the boundary conditions to be specified at x  = 0 and at x  = 1:

        u (t, 0) = 0 and u (t, 1) = 0.

We suppose that the value of u  when t  = 0 is specified as, say, f ( x ). Thus, we have an initial condition:

       u (0, x) = f ( x ).

A physical realization of the above equation is generally taken to be a thin, uniform bar over the interval [0, 1]. In this simple model, one takes the lateral surface of the rod to be insulated. Both ends of the rod are held at a fixed temperature. There are no internal or external heat sources. The rod has an initial heat distribution given by f ( x ). The heat diffuses from the places where it is warm toward the places where it is cooler. The rate of diffusion is proportional to the curvature of the present heat distribution. This statement about curvature takes the form of the second derivative. Thus, if the current distribution is concave down, the second derivative is negative and the temperature at that point will be decreasing. If the distribution is concave up, the temperature is increasing.

We will make the model more complicated later. For now, let's analyze this situation with the experience and knowledge we have. To this point, the principle idea we have used is the Fourier Idea . Here, we introduce the second idea: We suppose that the solution can be separate into a function of t  alone and of x  alone, so that

           u ( t , x ) = X( x ) T( t ).

This notion for approaching the problem is called the Method of Separation of Variables .

As we will see, this assumption leads to ordinary differential equations.

>	u:=(t,x)->X(x)*T(t);

>	diff(u(t,x),t)=diff(u(t,x),x,x);

>

If we bring all the functions of t  to one side and all the functions of x  to the other side, we have

          T ' / T  = X '' / X.

Since the left side is independent of x , then as x  changes, X '' / X does not change. This quotient is constant. In a similar manner, the quotient T ' / T is constant, and, from the equality, is the same constant. Call it .

The boundary conditions give that T( t ) X(0) = 0. If there is a single t  so that T( t ) is not zero, we can conclude that X(0) = 0. In a similar manner, X(1) = 0. Thus, we have a differential equation

          X '' =  X, with X(0) = 0 = X(1).

We have already examined this differential equation. We found that there are an infinite number of such  's and they are all of the form

and corresponding to each such constant, there is the solution

          X  ( x ) = sin( n   x ).

What about the possible T solutions? The quotient T '/ T is the same quotient, so that

          T ' =  T.

Solutions for this equation are

          T  ( t ) = exp(     t ).

Consequently, for each integer n, we have a solution

          u( t , x ) = exp(     t ) sin( n   x ).

But, since this is a linear problem, constant multiples of these solutions are also solutions, and (infinite) sums of solutions are solutions. Hence, a general solution can be written in the form of

          u( t , x ) =  .

There is one more piece of information we have not used. We have not used the initial value. This condition determines the coefficients:

           f ( x ) = u(0, x ) = .

This looks like a job for Fourier Series. We know that  can be written as a quotient of integrals which resolve as

           =    .

It will be valuable at this point to take a particular f , solve the equation, and draw some pictures. We take
           f ( x ) =  .

To visualize the initial value, we draw a graph.

>	f:=x->x*(1-x)^3;

>	plot(f(x),x=0..1);

>

We form the Fourier representation for this f  and compare the graph of f  and of our series representation. First, note that the function is continuous and its periodic extension is continuous, so that the Fourier series will converge uniformly.

>	for n from 1 to 7 do    c[n]:=int(f(x)*sin(n*Pi*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1): end do: n:='n';

>	approx:=x->sum(c[n]*sin(n*Pi*x),n=1..7):

>	plot([f(x),approx(x)],x=0..1,color=[BLACK,RED]);

As you can see, this is a very good fit. Now, we form the truncated series expansion for u.

>	u:=(t,x)->sum(c[n]*exp(-n^2*Pi^2*t)*sin(n*Pi*x),n=1..7);

     In this worksheet, we have solved a simple heat equation with zero boundary conditions and with an initial distribution. Here are the main steps:

Step 1:  By separation of variables, give two differential equations with boundary conditions whose solutions will determine the solution for the partial differential equation.

Step 2:  Solve the two ordinary differential equations with boundary conditions.

Step 3:  Construct the general solution for the partial differential equation with boundary condition.

Step 4:  Find the particular solution determined by the initial conditions.

EMAIL: herod@math.gatech.edu   or   jherod@tds.net

URL: http://www.math.gatech.edu/~herod

Copyright ©  2003  by James V. Herod

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