Math 447/547 Partial Differential Equations Prof. Carlson Homework 7 Text section Solve the diffusion problem. u(t,0) = 0 = u x (t,l).

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1 Math 447/547 Partia Differentia Equations Prof. Carson Homework 7 Text section Sove the diffusion probem with the mixed boundary conditions u t = ku xx, < x <, u(t,) = = u x (t,). Soution The anaysis proceeds initiay as on p. 85. The probem for X(x) is X = λx, X() = = X (). For positive eigenvaues λ = β 2 soutions must have the form X n (x) = A n sin(βx) because of the boundary condition at x =. The condition at x = becomes cos(β) = so that β = (n + 1/2)π/, n =,1,2,... It is easy to see that is not an eigenvaue. For genera compex eigenvaues λ we have X(x) = Ce i λx + De i λx. C + D =, i λce λ i λde λ =. Together we find that e i λ + e i λ =, 1

The probem for X(x) is X = λx, X() = = X ().

2 or This gives e i λ [e 2i λ + 1] =. 2 λ = (π + 2nπ)/, or λ = (π/2 + nπ)/ n =,1,2,... Thus the positive eigenvaues are the ony ones. The genera soution then has the form u(t,x) = n X n (x)t n (t) = A n exp( (n + 1/2) 2 π 2 kt/ 2 )sin((n + 1/2)πx/). n= 2. Consider the equation u tt = c 2 u xx for < x < with the boundary conditions u x (t,) = = u(t,). a) Show that the eigenfunctions are cos((n + 1/2)πx/). b) Write the series expansion for a soution u(t,x). Soution a) The anaysis proceeds pretty much as in the previous probem. One checks easiy that cannot be an eigenvaue. For other compex λ we have X(x) = Ce i λx + De i λx. i λc i λd =, Ce i λ + De i λ =. Together we find that or e i λ + e i λ =, e i λ [e 2i λ + 1] =. This gives λ = (π/2 + nπ)/, n =,1,2,... 2

Consider the equation u tt = c 2 u xx for < x < with the boundary conditions u x (t,) = = u(t,). a) Show that the eigenfunctions are cos((n + 1/2)πx/).

3 as before. We now have X n (x) = C n cos( (n + 1/2)π (n + 1/2)π x) + D n sin( x). To satisfy the boundary condition at we must have D n =, so eigenfunctions are nonzero mutipes of cos( (n+1/2)π x). b) As in p. 83 (5) the equation for T n is This gives and u(t,x) = T n + λc 2 T n =. T n (t) = A n cos((n + 1/2)πct/) + B n sin((n + 1/2)πct/), [A n cos((n+ 1 2 )πct/)+b n sin((n+ 1 2 )πct/)]cos((n+1/2)πx). n= 3. Sove the Schrödinger equation u t = iku xx, u x (t,) =, u(t,) =, x. As in probem 2, The equation for T n is X n (x) = cos( (n + 1/2)π x). so T n + 1/2)π (t) + ik[(n ] 2 T n (t) =, T n (t) = C n exp( ik(n + 1/2)2 π 2 t), 2 3

83 (5) the equation for T n is This gives and u(t,x) = T n + λc 2 T n =.

4 and u(t,x) = n= C n exp( ik(n + 1/2)2 π 2 2 t)cos((n + 1/2)πx). 4. (NOT ASSIGNED) Consider diffusion inside an encosed circuar tube. Let its ength (circumference) be 2. Let x denote the arc ength parameter where x. Then the concentration of the diffusing substance satisfies u t = ku xx, x, u(t, ) = u(t,), u x (t, ) = u x (t,). a) Show that the eigenvaues are λ = (nπ/) 2 for n =,1,2,... Soution There are constant eigenfunctions for eigenvaue λ =. For other λ we have X(x) = Ce λx + De λx. Ce λ + De λ = Ce λ + De λ, and λ[ce λ De λ ] = λ[ce λ De λ ]. Some arithmetic eads to or so that e λ = e λ 1 = e 2 λ, λ = nπ/. b) Show that the concentration is u(t,x) = a 2 + (a n cos( nπx ) + b n sin( nπx ))exp( n 2 π 2 kt/ 2 ). n=1 Soution So far we have identified the eigenvaues, but not the eigenfunctions. Notice that for each n = 1,2,3,... both functions cos( nπx ) and 4

a) Show that the eigenvaues are λ = (nπ/) 2 for n =,1,2,... Soution There are constant eigenfunctions for eigenvaue λ =. For other λ we have X(x) = Ce λx + De λx.

5 sin( nπx ) satisfy the periodic boundary conditions. The ony exceptiona case is n = when the function 1 satisfies the boundary conditions, but x does not. This resuts in the given form for u(t,x). 9. On the interva x 1 of ength 1, consider the eigenvaue probem X = λx, X () + X() =, X(1) =. a) Find an eigenfunction with eigenvaue. Ca in X. Soution Any mutipe of the function 1 x satisfies the boundary conditions. b) Find an equation for the positive eigenvaues λ = β 2. Soution We may write X n (x) = C n cos(βx) + D n sin(βx). βd n + C n =, C n cos(β) + D n sin(β) =. Together these give or β cos(β) + sin(β) =, tan(β) = β. c) Show graphicay that there are an infinite number of positive eigenvaues. Soution Pot the functions tan(β) and β and see where they intersect. d) Is there a negative eigenvaue? Soution Suppose that λ = γ 2, γ >. Then we woud have X n (x) = C n cosh(βx) + D n sinh(βx) and the boundary conditions woud give γd n + C n =, C n cosh(γ) + D n sinh(γ) =. 5

Soution Any mutipe of the function 1 x satisfies the boundary conditions. b) Find an equation for the positive eigenvaues λ = β 2. Soution We may write X n (x) = C n cos(βx) + D n sin(βx).

6 The equation for γ is γ = tanh(γ). Since we see that tanh() = and tanh(γ) = eγ e γ e γ + e γ tanh(γ) < 1, γ <. Compute tanh (γ) = 4 (e γ + e γ ) 2. Notice that e γ + e γ has a minimum vaue of which is achieved ony at γ =. Thus < tanh (γ) < 1, < γ <. If there were a point γ 1 > such that γ = tanh(γ 1 ), then the function γ tanh(γ 1 ) woud vanish at and at γ 1. By Roe s theorem the function woud have a positive root for the derivative. But the above anaysis shows that 1 tanh (γ) >, < γ <. Thus there are no negative eigenvaues. 11. a) Prove that the tota energy is conserved for the wave equation with Dirichet boundary conditions, where the energy is defined to be E = 1 2 (c 2 u 2 t + u2 x ) dx. Soution We have de dt = 1 2 (c 2 2u t u tt + 2u x u xt ) dx = 1 2 (2u t u xx + 2u x u xt ) dx. 6

If there were a point γ 1 > such that γ = tanh(γ 1 ), then the function γ tanh(γ 1 ) woud vanish at and at γ 1. By Roe s theorem the function woud have a positive root for the derivative.

7 Integrate the second term by parts to get de dt = 1 2 The boundary conditions are Differentiation gives (2u t u xx + 2u xx u t ) dx + u x u t. u(t,) = = u(t,). u t (t,) = = u t (t,), so de dt =. b) Do the same for Neumann boundary conditions. Soution The ony difference is that the vanishing of the boundary terms comes from u x (t,) = = u x (t,). c) For the Robin boundary conditions, show that E R = 1 2 (c 2 u 2 t + u 2 x) dx a u 2 (t,) a u 2 (t,). Soution Differentiation and integration by parts gives de R dt = u x u t + a u(t,)u t (t,) + a u(t,)u t (t,) = u t (t,)[u x (t,) + a u(t,)] u t (t,)[u x (t,) a u(t,)] =. 7

Soution The ony difference is that the vanishing of the boundary terms comes from u x (t,) = = u x (t,).

Math 2280 - Assignment 6

Math 2280 - Assignment 6 Dylan Zwick Spring 2014 Section 3.8-1, 3, 5, 8, 13 Section 4.1-1, 2, 13, 15, 22 Section 4.2-1, 10, 19, 28 1 Section 3.8 - Endpoint Problems and Eigenvalues 3.8.1 For the eigenvalue