MATH 3150: PDE FOR ENGINEERS MIDTERM TEST #2 VERSION B Name: Work out everything as far as you can before making decimal approximations. 1. Consider the general solution of the wave equation for a vibrating string with fixed ends: ∞ πnx X u(x, t) = sin (bn cos (λn t) + b∗n sin (λn t)) L n=1 where πcn . L Suppose that at initial time we have initial position f (x) = 0 (a flat string). At what later time t will the string be completely flat again? λn = Date: July 16, 2001. 1 2 MATH 3150: PDE FOR ENGINEERS MIDTERM TEST #2 VERSION B 2. Use separation of variables to find the general solution of the heat equation 2 ∂u ∂ u ∂2u = c2 + ∂t ∂x2 ∂y 2 in a rectangular plate with edges of lengths a and b, and with top, bottom and right edges insulated, while the left edge is kept at 0o . Hint: recall the solutions of the harmonic oscillator type equation 1 d2 h =λ h dt2 √ √ A cos λt + B sin λt h(t) = A + Bt A cosh p|λ|t + B sinh p|λ|t − are if λ > 0 if λ = 0 if λ < 0 where A and B can be chosen to be any constants. Also recall the identities about cosh and sinh: d cosh x = sinh x dx d sinh x = cosh x dx sinh 0 = 0 cosh 0 = 1 and the fact that cosh x is positive for every value of x. MATH 3150: PDE FOR ENGINEERS (continued) MIDTERM TEST #2 VERSION B 3 4 MATH 3150: PDE FOR ENGINEERS (continued) MIDTERM TEST #2 VERSION B MATH 3150: PDE FOR ENGINEERS MIDTERM TEST #2 VERSION B 5 3. Recall that the steady state of the heat equation in a disk of radius R with edge temperature f (θ) is ∞ n X r u(r, θ) = a0 + (an cos (nθ) + bn sin (nθ)) . R n=1 Find the total heat ZZ Q= ZZ u(x, y) dx dy = u(r, θ) r dr dθ where the integral is carried out over the whole disk. 6 MATH 3150: PDE FOR ENGINEERS MIDTERM TEST #2 VERSION B 1 0.8 0.6 0.4 0.2 –1 –0.5 y0 0.5 1 1 0.5 0 x –0.5 –1 Figure 1. A round peak 4. Looking at the heat equation in polar coordinates 2 ∂u ∂ u 1 ∂u 1 ∂2u = c2 + + ∂t ∂r2 r ∂r r2 ∂θ2 and recalling first and second derivatives from calculus, explain mathematically why a round peak in the graph of u (as in figure 1) representing a blob of heat will be pushed downward (will cool).