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Approximate Solutions to the Inverse Heat Problem

Armando I. Perez

Texas A&M University

Abstract. A simple one-dimensional heat equation is reviewed. This paper will discuss its derivation along with some techniques for analysis. The techniques to be discussed include familiar ones used in the analysis of differential equations. These initial attempts to solve the heat equation will later be applied to the task of finding possible solutions to the inverse heat problem. We will show that a solution to this problems exists if certain conditions are met.

Introduction

In mathematical physics, one of the classical equations that arises is that of the one-dimensional heat equation. This equation occurs in the theory of heat flow, i.e., heat is transferred by conduction in a rod or thin wire. We are concerned with finding solutions to the heat equation using methods such as separation of variables, boundary value problems, and Fourier series. This will lead us to the task of trying to solve the inverse heat problem. In other words, given some temperature distribution in a wire at time we want to solve the heat equation backward in time for $t\leq T$ and find the initial values which produce the given temperature distribution at time This problem is ``ill-posed'' because, as it will be shown, the solution does not always exist.

Derivation of Heat Equation

We begin by deriving the one-dimensional heat equation. Consider the wire along the interval

$\qquad$ Need to label graph.

Let be the temperature at a point at a time where $0\leq x\leq L$ and $t\geq 0.$ The principle of heat conduction states that heat flow is proportional to and in the opposite direction of the temperature gradient

Heat flow

where is a positive constant.

Let be a small segment of the wire

in out $\Rightarrow$

If we assume the wire is insulated then the rate of change of the total amount of heat in the segment is the amount of heat entering at $x_{1}$ minus the amount of heat leaving at $x_{2}:$

Rate of change of heat in a segment

By the Fundamental Theorem of Calculus, we have

(1) insert label $\qquad$

$\vspace{1pt}$

Also, the total amount of heat in the segment is given by

$\vspace{1pt}$

where is a constant related to the density and heat capacity of the wire.

$\vspace{1pt}$

Thus, the rate of change of heat in the segment is

$\vspace{1pt}$

(2) insert label $\qquad$

Equating $\left( 1\right)$ and $\left( 2\right)$ gives

Moving the derivative in $\left( 2\right)$ through the integral sign and dividing through by yields

and since is an arbitrary segment, we have

(3) insert label

This is the one-dimensional heat equation. For simplicity we will assume $\frac{c}{k}=1.$

Fourier Series

To aid in solving $\left( 3\right) ,$ we review some facts about Fourier series. We make the definitions for the Fourier coefficients as follows:

where $f\left( x\right)$ represents an integrable function.

A series of the form

is called a Fourier series.

We can evaluate the Fourier coefficients more easily by observing whether the function under the integral is even or odd.

We will make use of the following:

Definition

A function is said to be even if and is said to be odd if for all in the domain of

Observe that the cosines in a Fourier expansion are even functions and the sine terms are odd functions. Then, if is an even function $f(x)\sin (kx)$ is an odd function. Thus, the Fourier sine coefficient $b_{k}=0.$ Similarly, if is an odd function $f(x)\cos (kx)$ is also an odd function and $a_{k}=0.$ It follows that an even function has only cosine terms in its Fourier series and an odd function has only sine terms in its Fourier series. It is not difficult to prove that

if is odd

if is even

For example, consider the function $-\pi <x<\pi .$ Then, using the fact that is an odd function and $x\cos (x)$ is an odd function

Also, since $x\sin (kx)$ is an even function

Integration by parts gives us

Notice we used the observation in the last step.

Thus, the Fourier series is

Now, consider the function MATH

Then,

Then, the Fourier series is

A question to consider is whether the Fourier series has anything to do with the function. We have the following:

Definition

Let be defined on the interval Then the Fourier series for converges to if the sequence converges to That is, we have We use to represent the limit

Theorem

Let be bounded and piecewise-monotone on Then, the $2\pi$ periodic Fourier series of converges at every point to a periodic extension of In particular, if is continuous at then the series converges to If is discontinuous at $x_{0,}$ the series converges to the average value

In the examples above, you can see how partial sums of a Fourier series converge. In each of the examples, the function satisfies the conditions of boundedness and piecewise-monotonicity; thus the series converges to the values claimed in the theorem. The above theorem is useful provided we know the function will converge.

Since the partial sums of a Fourier series are periodic functions, a function to which they converge must also be periodic. This is called a periodic extension of .

For example, consider the function We have shown that the Fourier series is:

Since is continuous on we expect the partial sums to converge to on the interval

$\vdots$

This is depicted by the following

Insert picture of partial sums.

Notice that at the endpoints, i.e., $x=-\pi ,x=\pi ,$ the series converges to the average value

Solving the Heat Equation

To solve the heat equation with boundary conditions

(4) insert label

and initial condition

(5) insert label

we use separation of variables. We guess the solution can be written

Substitution in the equation yields:

$^{\prime }(t)$

$^{\prime }(x)T(t)$

We then have

$^{\prime }(x)=X$

dividing through by gives

In order for this to be true, there must be a constant $-\lambda ^{2}$ such that

This gives the equations

(6) need to label

(7) need to label

The solution to $\left( 4\right)$ and $\left( 5\right) ,$ respectively, are

(8) need to label

(9) need to label

Since we need thus, $c_{1}=0.$ Similarly, $u(\pi ,t)=0$ requiring $X(\pi )=0$ and

This can be achieved by letting $\lambda =k=int.$ Then, we have

$T(t)=ce^{-k^{2}t}$

We will solve and using a series of the form

(10) need to label

To satisfy the initial condition we let in and get

We then find a Fourier sine series for

Let us now turn to the problem of explicitly solving the heat equation, i.e., subject to the following boundary and initial conditions:

$u(0,t)=u(\pi ,t)=0$

$0<x<\pi$

Insert graph of

We can extend on the interval in such a way that the cosine terms in the expansion of will all be zero, leaving only the sine terms to be computed. We do this by extending the graph of asymmetrically, about the vertical axis so as to represent an odd function. Then, $a_{k}=0$ because $h(x)\cos (kx)$ is an odd function on

This follows from the theorem known as the superposition principle.

Theorem

If are solutions of a homogeneous linear partial differential equation, then the linear combination where $c_{i},$ are constants, is also a solution.

We then need only to find the Fourier series for $b_{k}$ thus,

then,

MATH if $k\neq 1$ .

Thus,

Let

$cc_{2}=1$

then,

and

which satisfies the conditions.

Let us solve the heat equation, i.e.,

subject to the following boundary and initial conditions:

$u(0,t)=u(\pi ,t)=0$

$0<x<\pi$

Insert Graph

By extending as before, we obtain $a_{k}=0.$

$-\pi <x<\pi$

Solving for $b_{k}$ we use the fact that the Fourier coefficient is

Thus,

Letting $cc_{2}=b_{k}$ we have

Thus,

and

Since $b_{k}$ are the Fourier coefficients of we have

Thus, the conditions are satisfied.

The Inverse Heat Equation

Insert picture for inverse heat equation.

Let be the temperature distribution in a wire at a time Then, if we hold the temperature at the endpoints fixed, the temperature distribution in the wire at time is given by and it satisfies the following condition:

MATH (11) need to label

Notice we have assumed the temperature at the endpoints is zero. Now suppose is the temperature distribution in a wire at a fixed time Our goal is to find an initial value such that if we solve with as the initial heat distribution then This is called the inverse heat problem.

Let us solve the heat equation subject to the following initial and boundary conditions:

$u(x,T)=c\sin (kx)$

$u(0,t)=u(\pi ,t)=0$

$t\geq T$

The Fourier series of $u(x,T)=c\sin (kx).$ From separation of variables we obtain

Then,

is a solution to the inverse heat problem.

The inverse heat problem does not always have a solution.

Consider solving the heat equation subject to the following conditions:

$u(x,T)=c\sin (kx)$ for some fixed

MATH for

We guess:

where

since

we need

thus

This will satisfy

with and

Then, the Fourier series for each is

thus,

From separation of variables we obtain

We want:

If we let $c_{0}=-\alpha ,$ notice

where

MATH

Then,

So, we want

We need

$c_{0}=-\alpha$

Thus,

So,

But,

and this will not converge as $e^{Tj^{2}}$ grows without bound!

This leads us to the task of making further assumptions and eventually arriving at a possible solution to the inverse heat problem. Remember, we want to converge. In this manner we can find a solution. Let's find where satisfies the following

$u(0,t)=u(\pi ,t)=0$

The following new theorem (will call Mando's Theorem or Perez Theorem) is proposed and proved.

Theorem

Suppose Then, there exists a solution to the inverse heat problem.

Proof:

From separation of variables we guess

Then, the Fourier series for is

We want

Let's define

Then

So,

and this is our solution to the inverse heat problem.

Conclusion (under construction)

In conclusion, we have shown that starting with a simple one-dimensional heat equation we can further investigate its properties and arrive at a "backward" heat equation. This is called the Inverse Heat Problem. Solutions to this problem do not always exist, but we have shown that under certain assumptions we can indeed find a solution. The methods and results used in this paper can be readily applied to other more complex problems. These include, but are not limited to, the wave equation, Laplace's equation in two dimensions, and the telegraph equation. Further study leads to useful insights on electron beam lithography, an area of particular concern in industrial mathematics. Need to expand paragraphs, etc...Include electron beam lit...

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