Partial Differential Equations, Winter 2015

Transcription

1 Partial Differential Equations, Winter 15 Homework #1 Due: Wednesday, January 8th, 15 in tutorial. 1. (Chapter 1.1, 3) For each of the following, state the order of the equation and whether it is nonlinear, linear homogeneous, or linear inhomogeneous. (a) u t u xx + 1 = (b) u t u xx + xu = (c) u t u xxt + uu x = (d) u tt u xx + x = order, linear inhomogeneous order, linear homogeneous order 3, nonlinear (e) u x (1 + u x) 1/ + u y (1 + u y) 1/ = (f) u x + e y u y = (g) u t + u xxxx u = order, linear inhomogeneous order 1, linear homogeneous order, nonlinear order 1, nonlinear. (Chapter 1.1, ) Let u 1 and u be two solutions of the inhomogeneous linear equation Lu = g. Show that u 1 u is a solution of the homogeneous equation Lu =. If Lu 1 = g = Lu, then by linearity 3. (Chapter 1.1, 1) Verify by direct substitution that L(u 1 u ) = Lu 1 Lu = g g u n (x, y) = sin(nx) sinh(ny) is a solution of u xx + u yy = for every n >. d dy sinh(ny) = n sinh(ny) d dx sin(nx) = n sin(nx) 1

2 And so (u n ) yy (x, y) = n u(x, y) and (u n ) xx (x, y) = n u(x, y).. (Chapter 1., ) Solve the equation 3u y + u xy =. (Hint: let v = u y. What is the PDE for v?) The PDE for v = v y is 3v + v x = = (exp(3x) v) = x = exp(3x) v = f(y) = v(x, y) = f(y) exp( 3x), where we have used the integrating factor exp(3x) and f is a function of one variable. Since v = u y, we integrate with respect to y and get u(x, y) = F (y) exp( 3x) + g(x), where F = f and g is a function of one variable. 5. (Chapter 1., 1) Solve u x + u y + u = e x+y with u(x, ) =. Hint: use the change of coordinates: ζ = x + y, η = x y. Make the change of variable ζ = x + y and η = x y, then u x = u ζ + u η u y = u ζ u η So u x + u y = u ζ and x + y = (3ζ η)/. The PDE, u x + u y + u = e x+y becomes u ζ + u = exp(3ζ/ η/) u ζ + 1 u = exp( η/) exp(3ζ/) Note that exp( η/) is a constant with respect to ζ so this is really an ODE in ζ. We use the integrating factor exp(ζ/) and get d ( ) ue ζ/ = exp( η/) exp(ζ) dζ = ue ζ/ = exp( η/) exp(ζ) + f(η) = u = exp( η/) exp(3ζ/) + f(η) exp( ζ/)

3 where f is an arbitrary function of one variable. Changing back to x and y variables and skipping some algebra, u(x, y) = exp(x + y) + f(x y) exp( (x + y)/). Finally we plug in the initial condition, u(x, ) = which gives, Summing up, u(x, y) = = = exp(x)/ + f(x) exp( x/) ( ) 3x = f(x) = exp /. exp(x + y) 1 ( 3x exp exp(x + y) exp ((x y)/). ) exp( (x + y)/) 6. (Chapter 1., ) Heat conduction is also described by the diffusion equation u t = κu xx where u(x, t) is the temperature at x at time t and κ is some constant. A homogeneous thin pipe occupying the region < x < l is completely insulated. Its initial temperature is f(x). After a sufficiently long time, the temperature of the object eventually reaches a steady (or equilibrium) state. We call this the steady state temperature. Find a formula for the steady-state temperature in terms of f. Hints: The steady state temperature does not change in time. What does this mean for u t? The ammount of heat is constant. None is lost or gained. We are trying to solve for u(x, t) which solves the diffusion equation with the above conditions but which does not change in time. This means that u t = and so u(x, t) = g(x) for some arbitrary function of g. Ie., u does not depend on t. From the diffusion equation = u t = κu xx and so since κ, g (x) = u xx (x, t) = = g(x) = ax + b where a and b are constants. Also, since the pipe is insulated at the ends, this means there is no heat flow in or out so by Fick s law, u x (, t) = = u x (l, t) for all t,

4 (Neumann condition) which gives that a = and u(x, t) = b. Finally, we have that heat is conserved. The total heat in the bar at time t is H(t) = We differentiate this with respect to t, H (t) = = u(x, t)dx. u t (x, t)dx κu xx (x, t)dx = κ(u x (l, t) u x (, t)) =. So H(t) is constant. H() = f(x)dx while H(t) = lb, so b = 1 l 7. (Chapter 1.5, ) Consider the problem with f(x) a given function. (a) Is the solution unique? Explain. f(x)dx u (x) + u (x) = f(x) u () = u() = 1 [u (l) + u(l)], (b) Does a solution necessarily exist, or is there a condition that f(x) must satisfy for existence? Explain. Hint: Consider w = u + u. (a) The solution is not unique. Suppose there is a solution u to the above ODE. Let s find a solution to the homogeneous ODE w (x) + w (x) = w () = w() = 1 [w (l) + w(l)]. We sub in w(x) = A exp(rx) to the ODE to get r + r = = r = 1,

5 So the general solution is w(x) = a + b exp( x). Plugging in the boundary conditions gives that w b (x) = b + b exp( x) where b is an arbitrary constant. Now notice that u + w solves the inhomogeneous ODE with the given boundary conditions. So the solution (if it exists) is not unique. In fact there are infinitely many solutions because there are infinitely many w that satisfy the homogeneous ODE (just choose a different b). (b) Let w = u + u. Then the ODE for w is w (x) = f(x) w() = w(l) Solving the ODE gives that w(x) = x f(s) ds+c. From the boundary conditions, we must have that f(s) ds =. In other words we only have a solution for w and hence a solution for u when this integral is zero. 8. (Chapter 1.5, 5) Consider the equation with the initial condition u(x, ) = φ(x). u x + yu y = (a) For φ(x) = x, show that no solution exists. (b) For φ(x) 1, show that there are many solutions. The ODe for the characteristic curves are given by y (x) = y, whose solution is y(x) = C exp(x) for C a constant. The characteristic curves are therefore ye x = C and so the general solution to the PDE is u(x, y) = f(e x y). Notice that φ(x) = u(x, ) = f(). If φ(x) = x we cannot find any f that will satisfy this equation but if φ(x) 1 any f with the property that f() = 1 will work. There are infinitely many such functions.