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Course: MATH 442, Spring 2008
School: Texas A&M
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for Notes TUT course MAT-51316 Robert Pich e 2.9.2007 1 Transport Equation Initial Value Problem how to derive the one dimensional transport equation how to solve initial value problems for this equation using the method of characteristics how to compute and plot solutions using Maple function PDEplot 1.1 Transport Equation Let u(x, t) denote the density (units [quantity] [volume]-1 ) of a substance (e.g....

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for Notes TUT course MAT-51316 Robert Pich e 2.9.2007 1 Transport Equation Initial Value Problem how to derive the one dimensional transport equation how to solve initial value problems for this equation using the method of characteristics how to compute and plot solutions using Maple function PDEplot 1.1 Transport Equation Let u(x, t) denote the density (units [quantity] [volume]-1 ) of a substance (e.g. mass, energy, . . . ) as a function of position x and time t. (This is a one dimensional model, it is assumed that there are no density variations in the y and z directions.) The amount of substance in an interval a x b of a tube-shaped region of b constant cross section A is a u(x, t)A dx. A a b x Let (x, t) be the flux (units [quantity] [area]-1 ). The net flux into the tubular interval is (a, t)A-(b, t)A. Let f (x, t, u) be the rate (units [quantity] [time] -1 [volume] -1 ) at which substance density increases by processes other than flux, for example chemical reaction. This is called a source term. The rate of increase of the total amount of substance in the interval is d dt b b [time]-1 u(x, t)A dx = (a, t)A - (b, t)A + a a f (x, t, u)A dx, which can be rearranged to give b (ut + x - f ) dx = 0. a Because [a, b] is arbitrary, this implies the conservation equation ut + x = f. The constitutive law = cu models a flow with given velocity c(x, t) (units [length] [time]-1 ). Substituting this into the conservation equation gives the transport equation ut + (cu)x = f. (1) In an initial value problem for the transport equation, one seeks the function u(x, t) that satisfies (1) and that satisfies u(x, 0) = u0 (x) for some given initial density profile u0 . 1.2 Method of Characteristics The transport equation (1) can be written as cux + ut = f - cx u, that is, as [c, 1] u = g where g = f - cx u. The geometric interpretation of this equation is that the directional derivative of u in the direction of vector [c, 1] is specified by g(x, t, u). This fact can be exploited to decouple the transport equation (a PDE) into a family of ordinary differential equations (ODEs). 1 Curves x = X(t) in the (x, t) plane that are tangential to the vector field [c(x, t), 1] are called characteristic curves, and are defined by the ODE dX = c(X, t). dt (2) The ODE describes a family of curves. The initial condition X(0) = k specifies the particular characteristic curve that goes through the point (x, t) = (k, 0) on the x-axis. t [ c,1] k x Denoting the value of u along a characteristic curve by U (t) = u(X(t), t), we have d u dX u U= + = cux + ut = g, dt x dt t that is, the value of u along the characteristic curve is determined by the ODE U = g(U (t), X(t), t). Thus, to find the solution u(x, t) of an initial value problem for given x and t: follow the characteristic curve that goes through (x, t) (i.e. solve the ODE (2)) and find the point (k, 0) where it intersects the x-axis; solve then the ODE (3) with initial condition U (0) = u0 (k) and "follow the characteristic curve back" to find U (t); this is the value of u(x, t). The Maple code PDEplot uses a slightly different approach. For given k, it computes the space curve [X(t), t, U (t)] that contains the point [k, 0, u0 (k)], using numerical algorithms to solve the ODEs (23). The family of such curves corresponding to a range of k defines the solution surface. (3) 1.3 Example: ut + 2ux = 0 This equation models transport with constant velocity c(x, t) = 2 and no source term. The characteristic ODE is X = 2. The solution of this ODE satisfying the initial condition X(0) = k is the straight line X = 2t + k. The characteristic curve (line) through a given point (x, t) crosses the x axis at (k, 0) with k = x - 2t. t x 2t = k k x The ODE describing the value of u along a characteristic line is U (t) = 0, i.e. the value is constant along the line. The solution of this ODE satisfying the initial condition U (0) = u0 (k) is U (t) = u0 (k). The solution of the PDE initial value problem is therefore u(x, t) = u0 (x - 2t). 2 The solution u(x, y) = e(x-2t) corresponding to the initial profile u0 (x) = 2 e-x can be plotted in Maple by the commands > PDE:=diff(u(x,t),t)+2*diff(u(x,t),x)=0; > with(PDEtools): PDEplot(PDE,[x,0,exp(-x^2)],x=-3..3,t=0..2); 2 0.8 2 0.6 u(x,t) 0.4 0.2 1 0.5 -2 0 2 x 4 0 t 1.5 6 Notice how the initial profile translates to the right at constant speed without changing shape. 1.4 Example: ut + xux = 0 This equation models transport in a velocity field whose magnitude increases with distance from the origin (c(x, t) = x). The equation can also be written as ut + (xu)x = u, which is of the form of equation (1) with source term f (x, t, u) = u. The source term models the generation of substance at a rate proportional to the amount of substance. The characteristic ODE is X = X. The solution of this ODE satisfying the initial condition X(0) = k is X = ket . The characteristic curve through a given point (x, t) crosses the x axis at (k, 0) with k = xe-t . t x = ket k x The ODE describing the value of u along a characteristic curve is U = 0, i.e. the value is constant along the curve. The solution of this ODE satisfying the initial condition U (0) = u0 (k) is U (t) = u0 (k). The solution of the PDE initial value problem is therefore u(x, t) = u0 (xe-t ). -t 2 The solution u(x, y) = e-(xe -3) corresponding to the initial profile u0 (x) = 2 e-(x-3) can be plotted in Maple by the commands > PDE:=diff(u(x,t),t)+x*diff(u(x,t),x)=0; > PDEplot(PDE,[x,0,exp(-(x-3)^2)],x=0..6,t=0..2); 2 0.8 0.6 u(x,t) 0.4 0.2 0 10 20 30 x 40 1.5 1 t 0.5 0 The PDE solution spreads out as time advances, but the surface height remains constant along the characteristic curves. Notice that the total amount of substance increases as time advances; this is because of the source term. 3
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f (t ) = L 1. 3. 5. 7. 9. 11. 1-1{F ( s )}Table of Laplace Transforms F ( s ) = L { f ( t )} f ( t ) = L -1 { F ( s )} 1 s n! s n +1 2. 4. 6. 8.2F ( s ) = L { f ( t )} 1 s-a G ( p + 1) s p +1 1 3 5L ( 2n - 1) p 2n s 2 s 2 s + a2 s2 - a22
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