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Unformatted text preview: Week 3 Lectures, Math 716, Tanveer 1 Wave Equation as 1st order homogeneous system of PDEs Another approach to linear wave equation u tt c 2 u xx = 0 (1) is to notice that for C 2 functions, it is equivalent to a system of 1st order equation u t = cv x , v t = cu x (2) It is clear that if (2) is satisfied by ( u, v ), then elimination of v between the two equations clearly leads to (1). On the otherhand, if u ( x, t ) solves (1), then clearly ( u, v ), with v ( x, t ) = 1 c integraltext x u t ( y, t ) dy c integraltext t u x (0 , ) d satisfies (2). If ( u, v ) satisfies (2), then we observe by taking a linear combination of the two equations that u 1 = u + v , u 2 = u v satisfies t u 1 c x u 1 = 0 ; t u 2 + c x u 2 = 0 , (3) which are a decoupled system of 1st order nonlinear equation. Using method of characteristics, u 1 = u + v is a constant on characteristics x + ct = constant. Further, u 2 = u v is a constant on characteristics x ct = constant. Initial data on each of u 1 , u 2 can be specified on a non-characteristic curve that intersects the corresponding characteristics transverally. We can directly conclude from this that u 1 = f ( x + ct ) and u 2 = g ( x ct ) and hence recover the prior result u ( x, t ) = f ( x + ct ) + g ( x ct ), for f and g related to f and g . Remark 1 In general for hyperbolic equations, we specify as many conditions on a curve , adjoining a domain U as the number of different sets of characteristics that enter U through . We illustrate this in Fig. 1 for wave equation, which as two sets of characteristics x + ct = const and x ct = const . . is the boundary of the domain in the x t plane. Note, in Fig. 1, we are seeking solutions for increasing t . The characteristic arrows will be reversed if we go backwards in time t . 2 Heat equation and Maximum Principle We now consider the solution u ( x , t ) to heat equation The text discusses this for 1 space dimen- sion in section 2.3, but the generalization to arbitrary dimension is not very difficult. Consider u t = u for x R n , t > (4) We will assume to be a bounded set. We will prove that both the maximum and minimum of u in [0 , T ] is attained either at t = 0 or on . Physically, this makes sense. As you know, heat flows from hot to colder part of the domain. If you have a domain with no internal heat source, the hottest or coldest spot can occur only 1 x-ct=const. x+ct=const. 1 condition 1 condition 2 conditions No conditions x-axis x=l x=0 t-axis. Domain U t=T Figure 1: Characteristics entering the region U for linear wave equation initially or on on the boundary . A hot spot at time zero will cool off, unless heat is fed into the domain through the boundary. But heat can be fed from a boundary point to the interior only if the boundary temperature is higher that its adjoining interior point. The same argument can be made for a cold spot....
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This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.
- Spring '08