Homework #3 Solutions

Transcription

1 Homework #3 Solutions Math 82, Spring 204 Instructor: Dr. Doreen De eon Exercises 2.2: 2, 3 2. Write down the heat equation (homogeneous) which corresponds to the given data. (Throughout, heat is measured in calories, temperature is measured in C and the other basic units are measured in centimeters, grams and seconds.) (a) Thermal diffusivity =.72 cm 2 /sec Solution: Since α 2 is the thermal diffusivity, we obtain u t = 0.72u xx. (b) Specific heat =.25 cal/g- C Density = 2.7 g/cm 3 Thermal conductivity =.63 cal/cm-sec- C Solution: The thermal diffusivity is defined as α 2 = k, where k is the thermal conductivity, σρ σ is the specific heat, and ρ is the mass density. So, we obtain α 2.63 = (.25)(2.7).085, giving u t =.085u xx. (c) Specific heat =.09 cal/g- C Density = 8.9 g/cm 3 Thermal conductivity =.92 cal/cm-sec- C Solution: The thermal diffusivity is defined as α 2 = k, where k is the thermal conductivity, σρ σ is the specific heat, and ρ is the mass density. So, we obtain α 2.92 =.49, giving (.09)(8.9) u t =.49u xx. 3. Use Taylor series to show that we re justified in writing f(x + x) f(x) = f (x) x in these physical derivations. Thus, in (2.5), we can immediately write u x (x + x, t) u x (x, t) = u xx (x, t) x.

2 Solution: f(x + x) f(x) = ) (f(x) + xf (x) + ( x)2 f (x) + f(x) 2! = xf (x) + ( x)2 f (x) +. 2! Then, since x is much less than, ( x) 2 is much smaller than x, and when we take the limit as x 0, the right-hand side of the equation behaves like xf (x). 2 Exercises 2.3:, 2. Write down the homogeneous wave equation for the following data: tension = 6 dynes density = 2 g/cm Solution: The wave equation takes the form u tt = c 2 u xx, where c 2 = T, with T being the tension ρ and ρ being the density. So, we have c 2 = 6 2 = 3, and the requisite wave equation is u tt = 3u xx. 2. Show that, if u is a solution of the wave equation u tt = c 2 u xx + F (x, t), then so is u + c x + c 2 t + c 3 for any choice of the constants c, c 2, and c 3. Solution: et u = u + c x + c 2 t + c 3. Then, 2 u x 2 = u xx, and 2 u t 2 = u tt, and since u satisfies the wave equation, so does u for any choice of c, c 2, and c 3. 3 Exercises 2.4: 6, 7 6. Apply the change of variable τ = t t 0 to the problem u t = α 2 u xx u(x, t 0 ) = f(x) u(0, t) = u(, t) = 0. Thus, without loss of generality (mathematicians say WOG), we need only consider such heat/diffusion problems with initial time t 0 = 0. (You should convince yourself that this is true for any of the initial-boundary-value problems we have discussed.) Solution: We first need to determine u τ, and the changes to the initial and/or boundary values. u τ = u t dt dτ 2

3 Since τ = t t 0, t = τ + t 0. Therefore, u τ = u t = u t. In terms of τ, t 0 becomes 0, so the initial condition becomes u(x, 0) = u(x, t 0 ) = f(x) the boundary conditions have the following change: u(0, t) = u(0, τ + t 0 ), u(, t) = u(, τ + t 0 ). Therefore, we obtain the following problem. u τ = α 2 u xx u(0, τ + t 0 ) = u(, τ + t 0 ) = (a) Given the heat problem with nonhomogeneous boundary conditions u t = u xx u(0, t) = 0, u(5, t) = 30, find a function v(x) = c x + c 2 so that the new unknown w(x, t) = u(x, t) v(x) satisfies the heat problem with homogeneous boundary conditions. What is the new initial condition for w? Solution: To determine the constants c and c 2, we simply plug w into the boundary conditions and set them equal to 0. 0 = w(0, t) = u(0, t) v(0) = 0 (c (0) + c 2 ) = 0 c 2 = c 2 = 0. 0 = w(5, t) = c = 4. = u(5, t) v(5) = 30 (c (5) + 0) = 20 5c 3

4 Therefore, v(x) = 4x + 0, and The initial condition that w satisfies is (b) Do the same for the general problem w(x, t) = u(x, t) 4x 0. w(x, 0) = u(x, 0) 4x 0 u t = α 2 u xx = f(x) 4x 0. u(0, t) = T, u(, t) = T 2, where T and T 2 are constants. Therefore, if we can solve the heat problem with homogeneous Dirichlet conditions, then we know how to solve it when it has nonhomogeneous, constant Dirichlet conditions. Solution: Again, to determine the constants c and c 2 so that v(x) = c x + c 2 and w(x, t) = u(x, t) v(x), we simply plug w into the boundary conditions and set them equal to 0. 0 = w(0, t) = u(0, t) v(0) = T (c (0) + c 2 ) = T c 2 = c 2 = T. 0 = w(, t) = u(, t) v() = T 2 (c () + T ) = (T 2 T ) c Therefore, and = c = T 2 T. v(x) = T 2 T x + T 2, The initial condition that w satisfies is w(x, t) = u(x, t) T 2 T x T 2. w(x, 0) = u(x, 0) T 2 T x T 2 = f(x) T 2 T x T 2. 4

5 4 Exercises 2.5: 2, 3 2. Find all polynomials of the form ax 2 + bxy + cy 2 + dx + fy + g which satisfy aplace s equation. Solution: We let u(x, y) = ax 2 + bxy + cy 2 + dx + fy + g. Then, u x = 2ax + by + d = u xx = 2a and u y = bx + 2cy + f = u yy = 2c. Therefore, 0 = u xx + u yy = 2a + 2c = c = a. Therefore, all polynomials of the form u(x, y) = a(x 2 y 2 ) + bxy + dx + fy + g solves aplace s equation for all constants a, b, d, f, and g. 3. Show that the function u = ln x 2 + y 2 satisfies aplace s equation (except at the origin, of course). This function is called the logarithmic potential. Solution: u x = x 2 + y 2 x(x2 + y 2 ) 2 = x x 2 + y 2 = u xx = x2 y 2 (x 2 + y 2 ) 2. u y = x 2 + y 2 y(x2 + y 2 ) 2 = y x 2 + y 2 = u yy = y2 x 2 (x 2 + y 2 ) 2. Therefore, u xx + u yy = x2 y 2 (x 2 + y 2 ) 2 + y2 x 2 (x 2 + y 2 ) 2 = 0. 5