Lecture 8: Differential Equations Philip Moriarty, philip.moriarty@nottingham.ac.uk NB Notes based heavily on lecture slides prepared by DE Rourke for the F32SMS module, 2006
8.1 Overview In this final lecture, we shall cover the application of Fourier transforms to the solution of differential equations. It was in deriving the solution to a particular differential equation the heat equation that Fourier invented the use of Fourier series. He was interested in determining just what happens in terms of heat flow when a long metal bar is touched at the middle with an intense source of heat energy. Initially, of course, the middle of the bar will become very hot (possibly white hot). Before long, however, the heat will flow in both directions, causing the middle to cool down and the rest of the bar to heat up. This is intuitively obvious, of course. What is less obvious is the equation which describes the heat flow. The heat equation (in one dimension) may be written as : where u(x,t) represents temperature and λ is the thermal diffusivity. This is a partial differential equation which describes how the temperature varies in both position and time. The larger the value of λ, the faster heat diffuses out of a hot spot. For a finite length bar Fourier used the separation of variables technique (see last semester s Elements of Mathematical Physics notes) and a Fourier series approach to solve the heat equation. In many cases, however, it is reasonable to assume that the bar is infinitely long. For this case, Fourier invented the Fourier transform to solve the problem. In the following, we ll cover how to solve not only the heat equation but a broad range of differential equations using Fourier transforms. 8.2 A selection of differential equations Alan Turing stated that Science is a differential equation (with an interesting qualifying statement which I ll not go into here!). You will know that differential equations are at the heart of our description of Nature. Just some of the differential equations you have encountered thus far are listed below:
NOTES The final equation on the previous page is that for a damped, driven oscillator. You ll have seen a form of this equation in both Vibrations and Waves last year and in the 2 nd Year Lab. module. The solution of this equation will form the basis of this week s problems class. In this lecture we will instead focus on the solution of the heat equation. 8.3 Fourier transforms of derivatives The key to using Fourier transforms (FTs) to solve differential equations is that the FT of a derivative is very simple. We need to be careful, however, when using FTs to solve differential equations as the variables may be functions of more than one variable. In the case of the heat equation, for example, the temperature of the bar is a function of both position and time. That means the Fourier transform could be with respect to either x or t. Throughout this lecture, however, the transforms will always be taken with respect to x. That is: To describe differential equations we need of course to deal with derivatives and, therefore, Fourier transforms of derivatives. In this case, let s consider the Fourier transform of the (partial) derivative of u with respect to x:
This is just our standard expression for the Fourier transform of a function with f(x) replaced by the partial derivative. We can solve the integral using integration by parts: If we now make the (very reasonable) assumption that the temperature goes to zero as x ±, then we re left with the following: Remarkably, we can therefore write the FT of a derivative in terms of the FT of the original function. (This holds true for any function, not just the temperature example here): We can then generalise this result to higher order derivatives: NOTES
8.4 Solving the 1D heat equation Using FTs, the heat equation can be solved in four steps. The first step is to reduce the heat equation, a partial differential equation (PDE), to an ordinary differential equation (ODE). We can do this by taking Fourier transforms of both sides:? From what we ve discussed in Section 8.3, can you write the FT of the second derivative of u wrt x above in terms of FT{u}? We can also write the left hand side of the equation in terms of FT{u}: However, the RHS of this expression is nothing but the FT of u, hence:
The key difference is that we now have an equation which involves differentiation in only t (as compared to both x and t). That is, we have converted the PDE to an ODE. Step 2 The next step is to solve the ODE. You should be able to show that the equation: has the general solution: (Make sure that you can do this!).? How might we determine the value of the constant A? NOTES The area under the φ(x) curve between, for example, a and b gives the probability, p, that the mass takes a value between a and b. That is, This of course means that the total area under the φ(x) curve, if the function is appropriately normalised, must be 1, since the probability of x lying between - and + must be 1.? Given a normalised probability density function φ(x), how would you find the mean value of x, i.e. <x>? Similarly, how would you find the variance, σ 2, where σ 2 = <(x-<x>)> 2? We can then write down the solution for U(k,t) as follows: Now, remember that what we re trying to find is not U(k,t) but the real space solution to the differential equation. This is related to U(k,t) as follows:
To clarify the reasoning here, let s focus on an example. Suppose we have a metal bar which has a temperature of zero everywhere except for a region around x=0 where the temperature equals 1. (To keep things as simple as possible, I m not going to explicitly include units for the moment. Position should of course have units of metres; time, units of seconds; and diffusivity units of m 2 s -1 ). It s a hard problem to work out directly from this initial temperature distribution what the temperature distribution will be at a later time. On the other hand, we can FT the temperature distribution to get U(k,t=0) which will of course give us the sinc function shown below. Let s say we want to know the temperature along the bar at a later time, say t = 0.05 (assuming a thermal diffusivity of 1). To do this we simply multiply U(k,0) by exp(-λk 2 (0.05)) (compare this to the solution for U(k,t) given on the previous page). However, exp(-λk 2 (0.05)) is just a Gaussian function in k (see below). Having evaluated U(k, t =0.05) we then need to get the inverse Fourier transform to find u(x, t =0.05). Graphically: This type of procedure, where we solve a hard problem by transferring to a different space (in this case k-space), solve a simpler problem in that space and then transform back to the original space is very common in physics and maths. For the heat equation, however, we can do even better and solve the hard problem directly by making use of the heat equation.
NOTES 8.5 Solution via convolution Importantly, the Fourier transform of a Gaussian function is another Gaussian function. We can therefore write the solution for U(k,t) as follows: Hence, we have the product (in k-space) of two Fourier transforms which is equivalent to a convolution of the corresponding functions in x : Or, writing out the convolution integral in full:
Some examples: 1. Delta-function temperature distribution Let s take the simplest example first a delta function temperature distribution at x=0. (This is equivalent to touching the bar with a very intense point source of heat at t = 0). Let s substitute the expression for our initial temperature distribution into the convolution integral on the previous page: Using the sifting property of the delta-function we get: Note that what we ve done is apply a point of heat to the bar and seen how this point spreads out. The Gaussian function that represents the solution to the heat equation tells us how the point of heat spreads out and is so called the point spread function for the system. (It is also often called the Green s function, which is why g(x,t) has been used to represent it in the formula above). The constant in front of the Gaussian tells us the height of the Gaussian. The width of the Gaussian is contained in the exponent of the exponential function (as highlighted in the equation above). If we calculate what the curve looks like for λ =1 and t = 0.05, we get a sharply peaked Gaussian, showing that the heat energy is still fairly concentrated at x = 0. If we wait a bit longer and look at the bar at t = 0.5, we see that the Gaussian is lower and wider (see below). (Hopefully the analogy with the solution to the diffusion equation covered in last year s Thermal and Kinetic module is clear!).
NOTES Some examples: 2. Unit step function As a second example, let the initial temperature be zero for x < 0 and 1 for x>0, i.e. a unit step. To find the temperature at a later time we need to convolve the initial temperature function with the PSF, g(x,t), which from Example 1, looks like a Gaussian at a fixed value of t. Unfortunately, it s rather difficult to calculate the convolution of a unit step with a Gaussian, so let s use a graphical method to roughly see what the temperature distribution will be. (Refer to Problems Class 3 for a description of how to graphically convolve two functions). The graphical process and the resulting form of the temperature distribution is described in the figure below:
8.6 Conclusions Fourier transforms represent a powerful method of solving both partial and ordinary differential equations. The key to solving, for example, the heat equation lies in the relationship of the Fourier transform of a derivative of a function to the FT of the original function. We ve covered two methods of solving the heat equation: (i) via the use of FTs and IFTs, and (ii) directly via the convolution integral. The signature behaviour of the system is represented by the point spread function (or Green s function) and describes, for the case of the application of the heat equation we ve described, how the initial point of heat spreads out across the bar. All these ideas can, of course, be applied to other differential equations. NOTES