Then we can use the formulas for the solution of the particular problems on the real line. Thus we obtain a couple of points x1 and x2.
The solution is then determined by the initial displacements at these points, by the initial velocity in the interval x1, x2 , and also by the number of reflections. Reflection method for the wave equation on finite interval. Its application in solving the initial boundary value problems for the wave equation led to the systematic investigation of trigonometric series much later called Fourier series.
Variables x and t are thus sepa- rated from each other. First, we will solve the boundary value prob- lem 7. Now, we go back to equation 7.
The original PDE of prob- lem 7. Let us notice that, instead of An Dn and Bn Dn , we write only An and Bn, since all these real constants are arbitrary. The function given by formula 7. Obviously, conditions 7. For this reason, we look for a solution of problem 7. It can be seen that such an expansion makes sense for a sufficiently wide class of functions. To ensure that this solution of problem 7. These issues are by no means trivial and it is the role of the Theory of Fourier Series to provide the answers to these delicate questions see, e.
They lie beyond the scope of this text and we will not deal with them. In what follows, we automatically assume that the formal calculations may be performed. Remark 7. Coefficients in front of the time variable in the arguments of trigonometric functions in expression 7.
If we go back to the string which is described by problem 7. Higher aliquot tones are then exactly the integer multiples of this basic tone. The discovery that musical tones can be described in this easy mathematical way was made by Euler in Graphic illustration of the solution of problem 7. Let us proceed in the same way as in the previous example.
If we go back to equation 7. Then Figure 7. Graphic illustration of the solution of Example 7. In the case of the wave equation, these conditions correspond to a string with free ends. If we model the diffusion process, then they describe a tube with isolated ends nothing can penetrate in or out and the flow across the boundary is zero. Similarly, when modeling the heat flow, the homogeneous Neumann conditions represent a totally isolated tube again, the heat flux across the boundary is zero.
Let us consider a problem for the wave or diffusion equation in the interval 0, l. In the case of the heat flow in a bar, these boundary conditions model the heat transfer between the bar ends and the surrounding media. Let us consider the following modeling problem illustrated in Figure 7. Let us take a vertical bar of unit length, whose upper end is kept at zero temperature while the lower end is immersed into a reservoir with water of zero temperature.
Schematic illustration of problem 7. Since we look for the nontrivial solution, i. Graphic illustration of the solution u x, t of problem 7. The reader can find basic facts of Sturm-Liouville theory in Appendix A. In the Fourier method for initial boundary value problems for nonho- mogeneous PDEs we use the analogue of this approach.
We illustrate particular steps on a simple example of the diffusion equation with homogeneous Dirichlet boundary conditions. Now, we expand all data of the original problem 7. That is why we will now study the problem of transforming the initial boundary value problem with nonhomogeneous boundary conditions to a problem with homogeneous conditions.
In the following example we illustrate the simplest model situation. Let the heat-insulated bar of length l have its ends kept at constant temperatures g0 and g1. It is thus reasonable to assume that the solution of problem 7. Due to the fact that the stationary part w x is uniquely determined by the constants g0 , g1 , we can — instead of the function u x, t — look directly for the unknown function U x, t. If we insert expression 7. We again illustrate their transformation to the homoge- neous boundary conditions through an example.
In this case we have transformed the original problem with a homogeneous equa- tion and nonhomogeneous boundary conditions into a problem with nonzero right-hand side but with homogeneous boundary conditions, which can be solved by the Fourier method. We look for a substitution that would simplify the PDE of problem 7. We will base our considerations on the physical properties of the model. The temperature u x, t develops at every point x0 in terms of the following two phenomena: 1.
Making use of this fact, we try to express the solution of the initial boundary value problem 7. If we substitute 7. Equation 7. The reader is asked to verify it. The exponential term in this case reflects the motion of the medium, w x, t corresponds only to the diffusion process. Notice the wave reflection on the boundary. Sketch the graph of the solution on several time levels.
What will be the temperature of the rod in the steady state that will be achieved after a sufficiently long time? Realize that in the steady state u depends only on x. Does the absence of an initial condition cause any trouble? Consider the case that the heat leaks from the rod over its lateral surface at a speed proportional to its temperature u.
Draw the temperature distribution in the steady state and discuss how the heat flows in the rod and across its boundary. What will be the density distribution if we wait long enough? Consider a bar of length l which is insulated in such a way that there is no exchange of heat with the surrounding medium. Illustrate the string motion by a graphic represen- tation of a partial sum of the resulting series for various values t.
Plot the graph of the solution on several time levels. Find the string vibrations provided the initial displacement was zero. Assume that the string is vibrating in the medium that resists the vibrations. The resistance is proportional to the velocity with the constant of proportionality 0. Consider heat flow in a thin circular ring of unit radius that is insulated along its lateral surface. The temperature distribution in the ring can be described by the standard one-dimensional diffusion equation, where x rep- resents the arc length along the ring.
Determine if the following PDEs are separable. If so, separate them into appropriate ODEs. If not, explain why. Chapter 8 Solutions of Boundary Value Problems for Stationary Equations In this chapter we consider two-dimensional boundary value problems for the Laplace or Poisson equation.
In particular cases, some of the boundary segments can be empty. On a rectangle or on a strip, on a half-plane , the solution of the Laplace equation can be found using the separation of variables the Fourier method. The general scheme is the same as in the case of evolution equations. It is in this step that the geometry of the rectangle is very important. There are several special domains which can be transformed to a rectangle. For example, this is the case with the disc or its suitable parts if we use the transformation into polar coordinates.
Section 8. The rectangle R and boundary conditions of 8. Then expression 8. In that case, problem 8. This means that 8. Remark 8. Let us consider again the Laplace equation on a rectangle, but this time let all four boundary conditions be nonhomogeneous. It does not matter which types of boundary conditions Dirichlet, Neumann, or Robin are given on particular sides. Using the linearity of the problem, we can decompose the totally nonhomogeneous problem into four partially homogeneous problems which are easy to solve.
Schematically, we illustrate the decomposition in Figure 8. Decomposition of the nonhomogeneous boundary value problem for the Laplace equation on a rectangle. We solve the equation on the disc D with the center at the origin and radius a. If we use the transformation formula 6. Both these ODEs are easily solvable.
We only have to add the appropriate boundary conditions. If we put 8. The expression 8. Now, we go back to the Cartesian coordinates. The reader is asked to draw a picture. The Poisson formula in polar coordinates 8. The above calculations are summarized in the following assertion. Theorem 8. Then the Poisson formula 8. Some important consequences of the Poisson Formula are summarized in Section Search for the solution in the form of a quadratic polynomial in x and y.
Write the solution using the Fourier coefficients of the function f. Derive the Poisson formula for the exterior of the disc in R2. Search for a function independent of r. These are the so called methods of integral transforms. The funda- mental ones are the Laplace and the Fourier transforms. This idea can be easily extended to PDEs, where the transformation decreases of the number of independent variables.
The function U is called the Laplace image of the function u, which is then called the original. Some of Laplace images and their originals can be found in tables, or the transformation can be done using various software packages.
However, without loss of generality, we can assume that the function u and its derivatives are continuous from the right at 0.
Relations 9. Applying the Laplace transform to a linear ODE with constant coefficients, we obtain a linear algebraic equation for the unknown function U s. Example 9. Using the Laplace transform, we solve the following initial boundary value problem for the diffusion equation.
This is an ODE with respect to the variable x and with real positive parame- ter s. Since we require the solution u to be bounded in both variables x and t, the image U must be bounded in x as well.
However, it has a theoretical character, and from the practical point of view, it is used very rarely. In most cases, it is more or less useless. In some cases, instead of above mentioned inverse formula, we can exploit another useful tool, which is stated in the Convolution Theorem given below.
Theorem 9. Remark 9. In particular, it follows from Theorem 9. Notice that the Laplace transform is additive, however, it is not multiplicative! Here we have used the boundedness assumption. The Convolution Theorem and Remark 9. The string is set in motion by acting of a force f t. This is an ODE in the x-variable with constant coefficients and a non-zero right-hand side. If the only acting external force in Example 9.
A string falling due to the gravitation. Contrary to the Laplace transform, which usually uses the time variable, the Fourier transform is applied to the spatial variable on the whole real line.
First, we start with functions of one spatial variable. However, the theory of the Fourier transform usually works with a smaller set of functions. Section 9. We say that the Schwartz space S is closed with respect to the Fourier transform.
The reader should keep this fact in mind while working with various sources or using the transformation tables. Nevertheless, it is convenient to use the transformation tables or some software packages when solving particular problems. When using the Fourier transform, we have obtained the solu- tion 9. It can be proved, for instance, that the function u in 9. We apply again the Fourier transform with respect to the spatial variable to the equation and both initial conditions.
Indeed, substituting the complex represen- tation of the sine and cosine functions into 9. In some cases, the methods of integral transforms are applicable also to equations with non-constant coefficients. Since the varying coefficient is — in this case — the time variable t, we use the Fourier transform with time playing the role of a parameter. By the inverse Fourier transform e. The Laplace and Pois- son equations can also be solved, in some cases, by the method of integral transforms.
In the following exercises we suppose that all the solutions we search for are bounded. Using the Laplace transform method, solve the following initial boundary value problems.
Simplify the results as much as possible. Check your answer by direct differentiation. Using the Fourier transform method, solve the following Cauchy problems.
However, these properties follow directly from the wave equation itself and the knowledge of the formula for the solution is not needed. To prove it, we proceed in the following way.
Domain of influence of the point x0 , 0 and domain of dependence of the point x, t. Section Trapezoid of characteristic triangle.
Now, we integrate Thus, the integral in Moreover, since this result holds true for a trapezoid of any height, we obtain that ut and ux are zero and thus u is constant in the whole characteristic triangle. The principle of causality can be obtained also by other and probably sim- pler methods. However, the advantage of this approach is its applicability in any dimension see Section We suppose that the Cauchy problem has a classical solution. Relations Remark Formula The potential energy represents the product of the force and the extension caused by this force cf.
In our case, the acting force is represented by the tension T. Example The potential energy is expressed by relation The initial conditions in problems So, problem This phenomenon has also its physical explanation: Diffusion, heat flow, the so called Brownian motion, etc.
The length of side on the x-axis is l, the length of side on the t-axis is T. The reason why we use these terms is that the maximum principle can be derived in the same way also for the diffusion equation in more spatial variables where the idea of the cylinder is more realistic.
We will prove the following assertion. Theorem Actually, a stronger assertion holds true see, e. Proof of Theorem The proof will be done for the maximum value. The idea of the proof uses the fact that the first partial derivatives of the function must be zero and the second derivatives must be non-positive at the inner maximum point.
This contradiction with the diffusion equation would imply that the maximum point must lie somewhere on the boundary. We have seen a variant of Maximum Principle also in the case of the diffusion equation on the whole real line. Corollary Let u1 x, t and u2 x, t be two classical solutions of problem Let u1 , u2 be two classical solutions of the initial boundary value problem In particular, the classical solution of The technique of the proof is called the energy method.
Let us consider again two solutions u1 x, t , u2 x, t of problem In other words, we obtain again uniqueness of the solution of the initial boundary value problem We will state it later and use the Poisson formula for its proof, see Section For now, however, we put up with the weaker version formulated in Theo- rem The Maximum Principle Theorem Let u be a harmonic function on a disc D, continuous in its closure D.
We shift the coordinate system so that the origin 0 is placed at the center of the disc this can be done because the Laplace operator is invariant with respect to translations. Another important consequence of the Poisson formula is the strong version of the Maximum Principle. According to the Mean Value Theorem, u xM equals the mean value of u over the circle. Moreover, the same holds for an arbitrary circle with the center xM and smaller radius.
The last consequence of the Poisson formula which we state here is the fol- lowing differentiability assertion. This property of harmonic functions is — in a certain sense — similar to the property that we have seen when studying the diffusion equation see Chap- ter 5. First, let us consider an open disc D with the center at the origin.
In the Poisson formula 8. Since we can change the order of integration and differentiation, the function u has partial derivatives of all orders in D as well. It then follows from above that u is differentiable in D. This arithmetic average can be neither greater nor less than all the surrounding values. Thus, even here, we meet a certain numerical analogue of the Maximum Principle and the Mean Value Theorem. For a solution u x, t of the wave equation Draw the characteristic parallelogram with vertices formed by the arguments in the previous relation.
What does it mean with regard to stability? Using the Maximum Prin- ciple, show that an odd even initial condition leads to an odd even solution. Is M T decreasing or as a increasing function of T? Is m T decreasing or increasing as a function of T? Send-to-Kindle or Email Please login to your account first Need help?
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