Partial Differential Equations/The transport equation

Introduction and first examples Test functions In the first chapter, we had already seen the one-dimensional transport equation. In this chapter we will see that we can quite easily generalise the solution method and the uniqueness proof we used there to multiple dimensions. Let . The inhomogenous -dimensional transport equation looks like this: , where is a function and is a vector. Solution The following definition will become a useful shorthand notation in many occasions. Since we can use it right from the beginning of this chapter, we start with it. 0 100% #FFFFFF Definition 2.1: Let be a function and . We say that is times continuously differentiable iff all the partial derivatives exist and are continuous. We write . Before we prove a solution formula for the transport equation, we need a theorem from analysis which will play a crucial role in the proof of the solution formula. 0 100% #FFFFFF Theorem 2.2: (Leibniz' integral rule) Let be open and , where is arbitrary, and let . If the conditions for all , for all and , exists there is a function such that hold, then We will omit the proof. 0 100% #FFFFFF Theorem 2.3: If , and , then the function solves the inhomogenous -dimensional transport equation Note that, as in chapter 1, that there are many solutions, one for each continuously differentiable in existence. Proof: 1. We show that is sufficiently often differentiable. From the chain rule follows that is continuously differentiable in all the directions . The existence of follows from the Leibniz integral rule (see exercise 1). The expression we will later in this proof show to be equal to , which exists because just consists of the derivatives 2. We show that in three substeps. 2.1 We show that This is left to the reader as an exercise in the application of the multi-dimensional chain rule (see exercise 2). 2.2 We show that We choose so that we have By the multi-dimensional chain rule, we obtain But on the one hand, we have by the fundamental theorem of calculus, that and therefore and on the other hand , seeing that the differential quotient of the definition of is equal for both sides. And since on the third hand , the second part of the second part of the proof is finished. 2.3 We add and together, use the linearity of derivatives and see that the equation is satisfied. Initial value problem 0 100% #FFFFFF Theorem and definition 2.4: If and , then the function is the unique solution of the initial value problem of the transport equation Proof: Quite easily, . Therefore, and due to theorem 2.3, is a solution to the initial value problem of the transport equation. So we proceed to show uniqueness. Assume that is an arbitrary other solution. We show that , thereby excluding the possibility of a different solution. We define . Then Analogous to the proof of uniqueness of solutions for the one-dimensional homogenous initial value problem of the transport equation in the first chapter, we define for arbitrary , Using the multi-dimensional chain rule, we calculate : Therefore, for all is constant, and thus , which shows that and thus . Exercises Let and . Using Leibniz' integral rule, show that for all the derivativeis equal toand therefore exists. Let and . Calculate . Find the unique solution to the initial value problem. Sources Martin Brokate Partielle Differentialgleichungen, Vorlesungsskript german http://www-m6.ma.tum.de/~brokate/pde_ws11.pdf 2011/2012 Introduction and first examples Test functions