Differential Equations 2 Homework 5 Solutions to the Assigned Exercises

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1 Differentil Equtions Homework Solutions to the Assigned Exercises, # 3 Consider the dmped string prolem u tt + 3u t = u xx, < x <, t >, u, t = u, t =, t >, ux, = fx, u t x, = gx. In the exm you were supposed to write out the solution in the form of terms u n x, t = T n tx n t. A solution to the exm question cn e found t the end of these pges. Use the solution to the prolem to complete the following tsks; n, n refer to the coefficients in my solution.. Find formul for the coefficients n, n in terms of the initil conditions f, g.. Write out the solution for the cse of fx = x + 3x, gx =. 3. For the concrete initil conditions of, sketch u, t s function of t in the intervl t. Solution. It gets pretty messy, which is why I did not wnt you to do it in the exm. The solution is copied from elow ux, t = e 3 + e 3+ x n 9 t n 9 t + n cos nx. n= n= The derivtive with respect to time of the solution is 3 u t x, t = e e 3+ x n 9 n 9 t n 9 t + n + n cos nx n= 3 n 9 t n 9 t n cos nx. Setting t = nd equting to fx, gx, respectively, we get fx = + x + n nx, gx = 3 n= x + Ug the formuls for the coefficients of Fourier e series, we get + = n = = 3 n 9 n + n = n 9 n 3 n nx. n= fx x dx, fx nx dx, n =, 3,,..., gx x dx, gx nx dx, n =, 3,,....

2 Solving we get = = n = n = fx x dx + fx x dx fx nx dx, n =, 3,,..., 6 n 9 Se non è vero, è en trovto fx nx dx + n 9 gx x dx, gx x dx,, #. Consider the het eqution for rectngulr region < x <, < y < u t = κ u x + u y, < x <, < y <, t >, gx nx dx, n =, 3,,.... suject to the initil condition ux, y, = fx, y. Find n expression for the solution if the oundry conditions re s given elow. In every cse, find lim t ux, y, t. { ux, y, t =, u x, y, t =, < y <, t >, Solution. u y x,, t =, u y x,, t =, < x <, t > I will e rief. The solution is given y ux, y, t = with the coefficients determined y m,n= mn e m fx, y cos mx + n κt mx cos ny cos, ny cos dy dx, m >, n >, mn = fx, y cos ny fx, y cos mx dy dx, m =, n >, dy dx, m >, n =, fx, y dy dx, m = n =. One hs lim ux, y, t = fx, y dy dx. t d { u, y, t =, ux, y, t =, < y <, t >, u y x,, t =, u y x,, t =, < x <, t > Solution. By seprtion of vriles, of course. Becuse t every prt of the oundry either u or the norml derivtive of u vnishes, n nlysis ug Ryleigh s quotient tells us there re no non-positive eigenvlues. Tht is looking for solutions u = XY T, plugging this into the eqution, the spce nd time vriles seprte s T κt = X X + Y Y = λ

3 3 nd the only vlues of λ for which we will hve non-zero solutions of the Sturm Liouville prolem will turn out to e positive, We now split the spce vriles s X X = Y Y λ = µ. The spce Sturm Liouville prolem unfolds into the two prolems X + µx =, X =, X =, Y + λ µy =, Y = Y =. The X prolem is seen to hve eigenvlues µ m = m, with corresponding eigenfunctions X m = m x, m =,, 3,.... We consider the Y prolem with µ = λ m. The vlues of λ for which there re non-zero solutions re then λ mn = µ m + n with eigenfunction Y n y = cos ny, n =,,,.... The solution is this given y ux, y, t = m= n= The coefficients re found y ug tht m x The coefficients work out to: cos nx c mn = c mn e k x = = m + κt n m x cos nx. cos lx m x dy dx, k m orl n,, k = m, l = n =,, k = m, l = n >. fx, y fx, y m x m x In this cse, lim t ux, y, t =., #. Consider the wve eqution for rectngulr memrne k x cos ny dx cos nx cos lx dy dx dy dx, m, n >, dy dx, m >, n =. u tt = c u xx + u yy, < x <, < y <, t >, suject to the oundry conditions { u, y, t =, u, y, t =, < y <, t >, ux,, t =, ux,, t =, < x <, t > Find n expression for the solution if the initil conditions re:. ux, y, = xy x y, u t x, y =.. ux, y, =, u t x, y, = x 7y.

4 3. Solution. ux, y, = xy x y, In ll cses, the solution of the PDE hs the form m ux, t = mn cos + n m ct + mn m,n= u t x, y = x 7y. + n ct mx ny. Turning to the three prticulr cses, we get:. In this cse nm = for ll n, m nd nm = xyx y mx ny dy dx = 6 m n 6 n 3 m 3. Notice tht if either n or m is even, nm =, so tht ux, t = 6 j 6 j 3 k 3 cos + j,k= k ct j x k y.. The solution is ux, t = c c x 7y. 3. The solution is the sum of the solutions of Prolem nd Prolem. It works out to: ux, t = 6 j 6 j 3 k 3 cos k j x + ct j,k= + c c x 7y., #. Solve suject to the oundry condition u tt = c u, r <, θ, t >, u, θ, t =, θ, t >, with the following initil conditions. Simplify the solution s much s possile.. ur, θ, =, u t r, θ, = r.. ur, θ, = r θ, u t r, θ =. Solution. k y From my notes on the circulr plte nd memrne, the solution to oth prolems hs the form ur, θ, t = J j kr A,k cos cj k t + C,k cj k t k= + J n j nkr A nk cos nθ + B nk nθ cos cj nk t + n,k= n,k= where j nk is the k-th positive zero of the Bessel function J n, J n j nkr C nk cos nθ + D nk nθ cj nk t, γ nk = rj n j nkr dr

5 SOLUTION TO EXERCISE OF EXAM for n =,,,... ; k =,, 3,... nd the coefficients A nk, B nk, C nk, D nk re given y A nk = B nk = C nk = D nk = γ nk γ nk γ nk cj n,k γ nk cj n,k fr, θrj n j nkr cos nθ dr dθ, n =,,,..., fr, θrj n j nkr nθ dr dθ, n =,,..., gr, θrj n j nkr cos nθ dr dθ, n =,,,..., gr, θrj n j nkr nθ dr dθ, n =,,.... Now to solve the prolems.. The fct tht f = gives A n,k = B n,k = for ll vlues of n, k. For the other coefficients it will e etter to differentite the expression for the solution with respect to t, set t = nd equte to r insted of ug the formuls. We get r = k= J j kr cj k t C,k + n,k= J n j nkr cj nk C nk cos nθ + D nk nθ. Since there is no dependence on θ, we get C nk = D nk = for n, ll k. So ll tht remins is C k. The solution in its simplest form is where ur, θ, t = k= C k = γ k cj,k C,k J j kr cj k t, r 3 J j kr dr.. Since g = ll coefficients C nk, D nk re. The only other non-zero coefficient is the one involving term with θ. The solution is thus ur, θ, t = B k J n j kr θ cos cj k t, where B k = γ k k= r 3 θj n j nkr dr dθ = γ k Solution to Exercise of Exm r 3 J n j nkr dr. If ux, t = XxT t solves the eqution nd the oundry conditions, then seprtion of vriles leds to the eqution/sl prolem T + 3T + λt =, X + λx =, X = X =. The eigenvlues/eigenfunctions of the SL prolem re λ n = n, X n = nx, respectively; n =,,.... For ech n, the T eqution is T + 3T + n T = ; to find the generl solution we need to solve the chrcteristic eqution r + 3r + n =. The solutions of this qudrtic eqution re r = 3 ± 9 n. If n =, then 9 n = > nd the generl solution is T t = e 3 + e 3+,

6 SOLUTION TO EXERCISE OF EXAM 6 for constnts, ; if n then 9 n < nd the generl solution is n 9 t n 9 t T n t = n cos, where n, n re constnts. It follows tht the solution of the PDE nd oundry conditions is given y ux, t = e 3 + e 3+ x n 9 t n 9 t + n cos nx. n=

u(x, y, t) = T(t)Φ(x, y) 0. (THE EQUATIONS FOR PRODUCT SOLUTIONS) Plugging u = T(t)Φ(x, y) in (PDE)-(BC) we see: There is a constant λ such that

u(x, y, t) = T(t)Φ(x, y) 0. (THE EQUATIONS FOR PRODUCT SOLUTIONS) Plugging u = T(t)Φ(x, y) in (PDE)-(BC) we see: There is a constant λ such that Seprtion of Vriles for Higher Dimensionl Wve Eqution 1. Virting Memrne: 2-D Wve Eqution nd Eigenfunctions of the Lplcin Ojective: Let Ω e plnr region with oundry curve Γ. Consider the wve eqution in Ω

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