Math 311, Partial Differential Equations, Winter 2015, Midterm

Transcription

1 Score: Name: Math 3, Partial Differential Equations, Winter 205, Midterm Instructions. Write all solutions in the space provided, and use the back pages if you have to. 2. The test is out of 60. There are six questions each worth 0 marks. The marks for each part are written in the margin. 3. For full credit you must show all work. Providing only an answer will result in very few marks. 4. If you don t understand the wording of any of the questions please ask. 5. One double-sided page of notes is allowed.

2 (0). For each of the following equations write down its order, and in the subsequent columns check off whether the equation is linear or non-linear, and whether it is homogeneous or inhomogeneous. Note that x, y, z, t are the independent variables and u is the dependent one. Equation Order Linearity Homogeneity Linear Non-linear Homogeneous Inhomogeneous u t = u xt + (sin x)u t + u 2 u tx = uu xxyy + e t 4 e x u xx + 6u xy + u yy = cos x 2 u = 2 (u2 ) x uu x 0 u yz + (z 3 cos y)u zz = cos 2 y 2

3 2. Consider the following PDE for u = u(x, y), u x + yu y = 0 (4) (a) Find and sketch the characteristic curves. (These can be very rough sketches). Solution: The characteristic curves satisfy y (x) = y. The general solution is y = C exp(x). The characteristic curves are ye x = C for C R 40 y = Cexp(x) 20 y 0 C C = 2 C = 0.5 C = 0.5 C = x (4) (b) Solve the initial value problem with u(0, y) = ( + y 2 ). Solution: The general solution is u(x, y) = f(ye x ). Put in the initial condition = u(0, y) = f(y) + y2 = u(x, y) = + y 2 e 2x (2) (c) What is lim x u(x, y)? Solution: lim u(x, y) = x + y 2 =. lim x e 2x

4 3. Consider the following initial value problem for the wave equation on R, ρu tt = T u xx u(x, 0) = 0 and u t (x, 0) = { x a 0 otherwise, where a > 0 is some fixed constant. This models the behaviour of an infinite piano string that is hit by a hammer with head diameter is 2a. (2) (a) A flea is sitting at a distance l away from zero. Assume that l > a. How long does it take for the disturbance caused by the hammer hit to reach the flea? Solution: Information travels at speed c = T/ρ. If a > l, it takes ( a l)/c amount of time. If a < l, it takes (l a)/c amount of time. (4) (b) How does the string at position 0 evolve in time? Ie., find u(0, t) as a function of t. Solution: The solution of the wave equation is Check cases: u(x, t) = 2 (φ(x + ct) + φ(x ct)) + = = u(0, t) = x+ct x ct ct ct (since φ 0) x+ct x ct ct a, ct a, u(0, t) = u(0, t) = ct ct a a ds = t ds = a/c (4) (c) Describe what the string looks like at time t = a/c. Ie., find u(x, a/c) as a function of x. Solution: u(x, a/c) =

5 Split into cases. 0 x 2a, 2a x 0, x 2a, x 2a, = a = 2a x = = 0 = 0 a = x + 2a

6 4. Recall that we call a function f : R R odd if f( x) = f(x) for all x R. We call f even if f( x) = f(x) for all x R. (3) (a) Determine whether the following functions are odd, even, or neither. i. f(x) = x 2 + x ii. f(x) = exp(sin 2 (x)) iii. f(x) = sin x + sinh x (Recall: sinh x = 2 (ex e x )) Solution: (i) even (ii) even (iii) odd (7) (b) Suppose u satisfies the one dimensional, homogeneous wave equation on R with initial condition u(x, 0) = φ(x), u t (x, 0) = ψ(x). i. Show that if φ and ψ are odd then for any t, u(x, t) is an odd function of x. ii. Similarly if φ and ψ are even then for any t, u(x, t) is an even function of x. Solution: Using D Alemberts formula, u( x, t) = 2 (φ( x + ct) + φ( x ct)) + = 2 (φ( (x + ct)) + φ( (x ct))) + x+ct x ct x+ct x ct ψ( r)dr. We used the change of variable r = s in the integral and just factored the minus sign in φ. Using this equality and plugging in the definition of odd and even for φ and ψ gives the result.

7 5. Consider the following heat equation with a constant sink (also known as a negative source): { ut = ku xx b, x R, t 0 u(x, 0) = φ(x), x R with k > 0 and b > 0. (2) (a) Find a function f(x) such that kf (x) = b. Solution: f(x) = b 2κ x2 (4) (b) Let w(x, t) = u(x, t) f(x). What PDE does w satisfy? What are the initial conditions for w? Solution: Notice w t = u t, so κw xx = κu xx κf (x) = κu xx b = w t. So w satisfies the diffusion equation with initial condition, w(x, 0) = u(x, 0) f(x) = φ(x) b 2κ x2. (4) (c) Write down a solution formula for u in terms of φ. You may use the following standard integral results: 4πkt y 2 e (x y)2 /(4kt) dy = 2kt + x 2, e (x y)2 /(4kt) dy =. 4πkt 4πkt ye (x y)2 /(4kt) dy = x Solution: Use the solution to the diffusion equation w(x, t) = = S(x y, t)w(y, 0)dy S(x y, t)φ(y) b S(x y, t)y 2. 2κ Using the given formula, S(x y, t)y2 = 2kt + x 2. So w(x, t) = = u(x, t) = w(x, t) + f(x) = S(x y, t)φ(y) bt b 2κ x2 S(x y, t)φ(y) bt

8 6. Consider the following boundary value problem for the backwards heat equation in the interval [0, ] (note carefully the minus sign) in one spatial dimension: u t = u xx 0 < x <, t > 0 () u(0, t) = 0, u(, t) = 0 (2) (5) (a) Use separation of variables to find the general solution to () and (2) of the form u(x, t) = X(x)T (t). (You should should get an eigenvalue equation for X. What can you say about the eigenvalues?) Solution: u t = XT, u xx = X T and so using the PDE, This gives the ODE s XT = X T = T T = X X = λ. T = λt (3) X = λx with X(0) = 0 = X(). (4) In class we showed that solutions of (4) must have λ > 0. We take λ = β 2. The solutions to the eigenvalue equation (??) are then λ n = (nπ) 2, X n (x) = sin(nπx). For each λ n = (nπ) 2, we can solve (3) to get T n (t) = a n exp(n 2 π 2 t). The general solution is then u(x, t) = a n exp(n 2 π 2 t) sin(nπx). n= (2) (b) Fix N N. For the initial condition u(x, 0) = ɛ sin(nπx) give the solution to () and (2) using part (a). Solution: ɛsin(nπx) = u(x, 0) = a n sin(nπx) means that n = 0 for n N and a N = ɛ. Ie., the solution is ɛ exp(n 2 π 2 t) sin(nπx) () (c) What is the solution to () and (2) with initial condition u(x, 0) = 0? Solution: u(x, t) 0 n= (2) (d) Use part (a) and (b) to discuss the stability of the backwards heat equation. Solution: For any ɛ > 0 we can find a solution u(x, t) such that u(x, 0) ɛ. Namely u(x, t) = ɛ exp(n 2 π 2 t) sin(nπx) for any N a positive integer. Stability of solutions would mean that if ɛ is small such solutions should stay close to the solution with initial condition u(x, 0) = 0 which is u(x, t) 0. But u(x, t) = ɛ exp(n 2 π 2 t) sin(nπx) which gets arbitrarily big for any t > 0 by picking N large. In other words, this PDE is not stable.