Math 5440 Problem Set 2 Solutions

Transcription

1 Math Math Problem Set Solutions Aaron Fogelson Fall, 1: (Logan, 1. # 1) How would the derivation of the basic conservation law u t + φ = f change if the tube had variable cross-sectional area A = A() rather than a constant cross-sectional area? Rewriting Equation (1.7) with A() instead of A, we have d b b u(, t)a() d = A(a)φ(a, t) A(b)φ(b, t) + f (, t)a() d. dt a a By the Fundamental Theorem of Calculus, we can turn the first two terms of the righthand side into an integral. Assuming then that u has continuous first partials, we can move the d dt inside the left-hand integral to yield b a b b (u(, t)a()) d = (A()φ(, t)) d + f (, t)a() d. t a a Rewriting this as one large integral, we have b a Au t + (φa) A f d = for any interval [a, b]. Therefore, since this interval is arbitrary, the integrand must be identically, so Au t + (φa) = A f. If we suppose that A() > everywhere, then our PDE becomes u t + φ + φ A () A() = f. 1

2 : (Logan, 1. # ) Solve the initial boundary value problem u t + cu = λu,, t > u(, ) =, >, u(, t) = g(t), t >. In this problem you will have to treat the domains > ct and < ct differently; the boundary condition affects the region < ct and the initial condition affects the region > ct. Let ξ = ct and τ = t. Then the first partials of u(, t) = u(ξ + cτ, τ) = U(ξ, τ) are u = U ξ ξ + U τ τ = U ξ and u t = U ξ ξ t + U τ τ t = cu ξ + U τ. Substituting these quantities into the orginial PDE yields cu ξ + U τ + cu ξ = U τ = λu. Therefore, U τ + λu =. Solving for U(ξ, τ) yields U(ξ, τ) = h(ξ)e λτ where h is some unknown function of ξ. Then, converting back to the original (, t) variable system, our solution to the advection-decay equation is u(, t) = U(ξ, τ) = h(ξ)e λτ = h( ct)e λt. (1) We now match this solution to the initial and boundary conditions. Suppose < ct. Then at =, ct > and t >, so we can apply the boundary condition u(, t) = g(t). Thus, letting z = ct, we see u(, t) = h( ct)e λt = g(t), h(z) = g( z/c)e λz/c for z <. Now suppose > ct. Then at t =, >, so we can apply our initial condition u(, ) =. Then u(, ) = h()e = h() = for >. We ve now determined h(z) for positive and negative z; substituting back into (1) yields the solution to this IBVP as { g(t /c)e λ/c, < ct, u(, t) =, > ct. We leave the reader to ponder what happens when = ct.

3 : (Logan, 1. # ) To study the absorption of nutrients by a grasshopper we model its digestive tract by a tube of length l and cross-sectional area A. Nutrients of concentration n = n(, t) flow through the tract at speed c, and they are adsorbed locally at a rate proportional to n. What is the PDE model? If the tract is empty at t = and then nutrients are introduced at the concentration n at the mouth ( = ) for t >, formulate an initial boundary value problem for n = n(, t). Solve this PDE model and sketch a graph of the nutrient concentration eiting the tract (at = l) for t >. Physically, why is the solution n(, t) = for > ct? Here, nutrients flow through the tract at speed c and are removed from the tract at a rate proportional to n, so f (, t, n) = k n. The PDE model for this system is then n t + cn = k n, where k > is some constant. The digestive tract is empty at t =, so n(, ) = for l. Nutrients are introduced at the mouth ( = ) at concentration n for t >, so n(, t) = n for t >. This is our IBVP for n. Switching to characteristic coordinates ξ = ct and τ = t and defining n(, t) = n(ξ + cτ, τ) = N(ξ, τ) we find via the chain rule that n = N ξ and n t = cn ξ + N τ, so that our PDE is now N τ = k N. We must be careful solving this ODE; applying some physical insight will keep us from going astray. First note that N = is one solution to this ODE. Second, note for N >, that N τ <, so that N is decreasing. We can solve the ODE by separating variables and integrating with respect to τ, we have N = dn N = k dτ = kτ + g(ξ), where g(ξ) is some unknown function of ξ. Therefore, solving for n we have n(, t) = N(ξ, τ) = 1 (g(ξ) kτ) However, when we check this solution, we find that for kτ > g(ξ), that N is increasing with τ in fact the solution is spurious in this range, corresponding to taking N <. The correct solution is { 1 (g(ξ) kτ) kτ g(ξ) N(ξ, τ) kτ g(ξ) which, as the reader should check, is both continuous and differentiable at kτ = g(ξ). We now apply our initial and boundary conditions. Suppose > ct; at t =, > and we apply the initial condition n(, ) =. Since = n(, ) = N(, ) = 1 (g()) for >

4 we conclude that g() = for >. Suppose < ct; at =, t > and we apply the boundary condition n(, t) = n. Since n = n(, t) = N( ct, t) = 1 (g( ct) kt) for t > we conclude that g( ct) = kt + n for t >. Combining the two we find { n kt/c z < g(z) = z > or ( n ) k n(, t) = N(ξ, τ) c c n /k and < ct, > c n /k or > ct. Note that if the parameters n, k, c, and the length of the digestive tract l satisfy l c n /k, then < c n /k for all l, and the solution formula simplifies to ( n ) k n(, t) = N(ξ, τ) c < ct, > ct. but if the length of the digestive track is longer than c n /k, we have to use the first formula () for the solution. The solution for the first case in which l > c n /k is graphed here: () Figure.1: Plots of n(, t) at t = 1,,,,, for parameters n = 1, c = 1, k = 1, and l = 1. and the solution for the other case l c n /k is graphed here: Figure.: Plots of n(, t) at t = 1,,,,,, 7 for parameters n = 1, c = 1, k = 1, and l =.

5 We also see that n(l, t) is piecewise constant. When l > c n /k all the nutrients have been digested before it reaches the end of the digestive track and n(l, t) =. If l < c n /k the solution at = l is ( n ) kl n(l, t) c l/c < t, l/c > t, which we graph here: t Figure.: Plot of n(l, t) for parameters n = 1, c = 1, k = 1, and l =. Basically, the solution becomes non-zero at time t = l/c when the nutrients ingested at (, t) = (, ) first reach the end of the track. In fact the solution vanishes for < ct for the same reason the nutrients which enter the digestive track at = for t > haven t had time to reach the point yet.

6 : (Logan, 1. # 1) Heat flows longitudinally through a metal bar of length 1 centimeters, and the temperature u = u(, t) satisfies the diffusion equation u t = ku, where k =. square centimeters per second. Suppose the temperatures at some fied time T at =,, cm are, and 7 degrees, respectively. Estimate u (, T) using a difference approimation. Will the temperature at = increase or decrease in the net instant of time? Estimate the temperature at = at T +. seconds. (Recall from calculus that f f (a h) f (a) + f (a + h) (a) h where h is a small increment.) Using the difference approimation suggested we have u(, T) u(, T) + u(, T) u t (, T) k =.1deg/sec The temperature at = will increase. An estimate of the temperature at = at time t = T +. is obtained using Taylor series in t, and the above approimation for u t (, T): u(, T +.) u(, T) + u t (, T)(.) = +.1. =..

7 : (Logan, 1. # ) Let u = u(, t) satisfy the PDE model Show that l u t = ku, < < l, t >, u(, t) = u(l, t) =, t >, u(, ) = u (), l. u(, t) d l u () d, t. Hint: Let E(t) = l u(, t) d and show that E (t). What can be said about u(, t) if u () =? From the definition of E(t), we see that E (t) = l l l t u(, t) d = uu t d = k uu d. The last step used the PDE u t = ku. Integrating the last epression by parts we obtain E (t) = k {uu l } l (u ) d. Using the boundary conditions, the first term in braces is zero, so E (t) = k l (u ) d. Since the integrand is non-negative, so is the integral. Hence the right hand side is nonpositive, E (t), and E(t) E() which is the statement to be proved. If u (), then E() =, and E(t) =. This implies that u(, t) = for all t. 7

8 : (Logan, 1. # ) Show that the problem u t = ku, < < l, t >, u(, t) = g(t), u(l, t) = h(t), t >, u(, ) = u (), l. with nonhomogeneous boundary conditions can be transformed into a problem with homogeneous boundary conditions. Hint: Introduce a new dependent variable w by subtracting from u a linear function of that satisfies the boundary conditions at any fied t. In the transformed problem for w, notice that the PDE picks up a source term, so you are really trading boundary conditions for source terms. Let s(, t) = h(t) l + (1 l )g(t). Then, s(, t) = g(t) and s(l, t) = h(t). Note that s t = h (t) l + (1 l )g (t) and s =. Let w(, t) = u(, t) s(, t). Then w t = u t s t = u t h (t) l + (1 l )g (t) and w = u. So u t = ku if and only if w t = kw s t = kw h (t) l + (1 l )g (t). Also w(, t) = u(, t) s(, t) = and w(l, t) = u(l, t) s(l, t) =. So w(, t) satisfies an nonhomogeneous diffusion equation, with homogeneous boundary conditions.

9 7: (Logan, 1. # ) Considering all cases, find the form of the steady-state solutions to the advection-diffusion equation u t = Du cu and the advection-diffusion-growth equation u t = Du cu + ru. A steady-state solution of the advection-diffusion-growth equation satisfies = Du () cu () + ru This is a linear second order ODE with constant coefficients; we seek solutions of the form u() = ep(α), and substituting into the ODE we find that α must satisfy Dα cα + r =, so α ± = c± c Dr D. There are three cases, depending on whether α ± are real and distinct, real and equal, or comple. A ep(α + ) + B ep(α ) if α ± R and α + = α u() = A ep(α) + B ep(α) if α ± R and α + = α A ep(a) cos(b) + B ep(a) sin(b) if α ± = a ± ib. The advection-diffusion equation is a special case of this with r = from which we see that α + = c/d and α =. So in this case the solution has the form u() = A ep(c/d) + B. 9

10 : Let erf() be the error function defined by erf() = π e s ds. a) Graph erf() for 1 < < 1. What are erf(), erf( ) and erf( )? b) Show u(, t) = 1 [ ( )] 1 + erf t solves the heat equation u t = u. You may use MAPLE to do the algebra. c) Plot u(, t) at time t =.1,.,., 1, on the same aes. What is the initial temperature distribution? d) Plot the solution surface u(, t) for < < and t <. e) Show u (, t) also solves the heat equation. Find and plot this solution at some sample times. See the Maple worksheet hw.mws. 1

11 9: Consider the problem: U t + U = e t >, < < U(, ) = a) Use the method of characteristics to find the solution for the problem. Remember that along a characteristic you can write as a function of t. Your answer will contain an integral or an error function. b) Use MAPLE to plot U(, t) at t =, 1,,, for 1 < < 1. c) Plot the solution surface U(, t) for < t < 1 and 1 < < 1. d) Give a physical interpertation of the PDE and the solution. What happens as t? a) Let ξ = t and τ = t. Then the first partials of u(, t) = u(ξ + τ, τ) = U(ξ, τ) are u = U ξ ξ + U τ τ = U ξ and u t = U ξ ξ t + U τ τ t = U ξ + U τ. Substituting these quantities into the orginial PDE yields Solving for U(ξ, τ) yields U(ξ, τ) = U τ = e = e (ξ+τ). τ e (ξ+τ ) dτ + A(ξ) where A is some unknown function of ξ. Then, converting back to the original (, t) variable system, our solution is u(, t) = U(ξ, τ) = t e ( t+τ ) dτ + A( t) = t e s ds + A( t) where we have made the change of variables in the integral s = t + τ. We now apply the initial condition, u(, ) = A() = which leads to the solution u(, t) = t e s ds b)-d) The solution is graphed in the accompanying MAPLE worksheet hw.mws. 11

12 1: Let u(, t) be defined by u(, t) = 1 +t t e s ds. a) Show u(, t) solves the wave equation u tt = u. What is the initial displacement (u(, )) and velocity (u t (, ))? You may use MAPLE to do the algebra. b) Plot u(, t) at time t =, 1,, on the same aes. c) Plot the solution surface u(, t) for < < and t <. See the Maple worksheet hw.mws. 1