PDE 83215 Solutions to exercises 5 1. (a) The general solution of utt ...

We just need to implement the intial conditions. The u condition gives A n = 0,the u t conditions gives that all the B n should be zero except B 3 and B 3 = 118u(x, t) = 1 6( 13 sin 3t − t cos 3t )sin 3xOn reflection it is clear that only the sin 3x mode is excited. and we could havesaved some time by using this. See the next question.(d) This time we will be smarter! We guess a solution in the formu(x, t) = T 0 (t) + T 1 (t) cos x + T 2 (t) cos 2xIt satisfies the boundary conditions. To make it satisfy the differential equationwe need to imposeT ′′0 (t) + kT ′ 0(t) = 1T ′′1 (t) + kT ′ 1(t) + T 1 (t) = qT ′′2 (t) + kT ′ 2(t) + 4T 2 (t) = 0The initial conditions (don’t forget cos 2 x = 1 (1 + cos 2x)) give2T 0 (0) = 1 2, T ′ 0(0) = 0T 1 (0) = 0 , T ′ 1(0) = 0T 2 (0) = 1 2, T ′ 2(0) = 1Solving the 3 differential equations for T 0 , T 1 , T 2 is a straightforward exercise inODEs. I used Maple to get the following:T 0 (t) = 1 2 − 1 k + t 2 k + 1 k 2 e−kt⎛ ⎛√T 1 (t) = q − qe −kt/2 ⎝cos ⎝1 − k22 t ⎞⎠ +⎛√k√ sin ⎝ 1 − k24 − k2⎛ ⎛T 2 (t) = e −kt/2 ⎝ 1 √ ⎞⎛2 cos ⎝ 4 − k24 t ⎠ + 2 + k √2sin ⎝16 − k2⎞⎞2 t ⎠⎠√ ⎞⎞4 − k24 t ⎠⎠We could have made our lives even easier here had we noticed that q only excites thecos x mode.6

3. Let us write u(x, t) in the formThenand∞∑( ) nπxu(x, t) = f n (t) sinn=1L∞∑( ) nπxu t (x, t) = f n(t) ′ sinn=1L∞∑ ∞∑( ) ( )nπx mπxu t (x, t) 2 = f n(t)f ′ m(t) ′ sin sinn=1 m=1L Lwhen we integrate this expression over [0, L] the only terms that will contribute arethose with m = n. Using the principle that the average of sin 2 is 1 2we have∫ L0u t (x, t) 2 dx = L 2∞∑f n(t) ′ 2n=1Similarlyand∞∑( )nπ nπxu x (x, t) =n=1L f n(t) cosL∞∑ ∞∑u x (x, t) 2 nmπ 2( ) ( )nπx mπx=fn=1 m=1L 2 n (t)f m (t) cos cosL L∫ LPutting everything together we haveE = ρ 2∫ LNow we use the form of f n (t):Differentiating gives00u x (x, t) 2 dx = L 2(u2t + c 2 u 2 x)dx =ρL4∞∑n=1n 2 π 2L 2 f n(t) 2(∞∑f n(t) ′ 2 + c2 n 2 π 2 )fn=1L 2 n (t) 2( ) ( )nπctnπctf n (t) = α n cos + β n sinLLf ′ n(t) = nπcL( ( nπct−α n sinL) ( )) nπct+ β n cosLand thusf n(t) ′ 2 + c2 n 2 π 2fL 2 n (t) 2 = c2 n 2 π 2 ( )α2L 2 n + βn2The energy is thus writtenE = ρc2 π 24L∞∑n ( )2 αn 2 + βn2n=17

It is not dependent on time!The case of Neumann boundary conditions is almost identical. The solution can bewritten∞∑( ) ( ( ) ( ))nπx nπctnπctu(x, t) = α 0 + β 0 t + cos α n cos + β n sinn=1LLLThe “zero mode” does not contribute to u x , but adds β 0 to u t . The energy turns outthe same as in the Dirichlet case with the addition of a further contribution from thezero mode of size.12 ρLβ2 0Note that α 0 does not contribute — it represents a translation of the string and doesnot contribute to the energy. β 0 (which has the units of speed) represents an overallmotion of the string, which has mass ρL. So the above contribution is just the usual“ 1 2 mv2 ” expression for kinetic energy.8

4. (a) To make the boundary conditions homogeneous we writeThe function v must satisfyWe look for a solution in the formu(x, t) = xt + v(x, t)v tt − v xx = cos 2t cos 3xv x (0, t) = v x (π, t) = 0v(x, 0) = cos 2 xv t (x, 0) = 0v(x, t) = T 0 (t) + T 2 (t) cos 2x + T 3 (t) cos 3xThe boundary conditions are satisfied. The differential equation givesand the initial conditions giveSolving these equations givesT ′′0 = 0T ′′2 + 4T 2 = 0T ′′3 + 9T 3 = cos 2tT 0 (0) = 1 2, T ′ 0(0) = 0T 2 (0) = 1 2, T ′ 2(0) = 0T 3 (0) = 0 , T ′ 3(0) = 0The final solution isT 0 (t) = 1 2T 2 (t) = 1 cos 2t2T 3 (t) = 1 (cos 2t − cos 3t)5u(x, t) = xt + 1 2 + 1 2 cos 2t cos 2x + 1 (cos 2t − cos 3t) cos 3x5(b) To make the boundary conditions homogeneous we write(u(x, t) = v(x, t) + x sin t + x 1 − x )2π9

The function v must satisfyv tt − 4v xx = − 4 π + sin tv x (0, t) = v x (π, t) = 0(v(x, 0) = x2 xπ π − 3 2)v t (x, 0) = 0The boundary conditions and the initial condition for v t suggest we should lookfor a solution of this in the form∞∑v(x, t) = T 0 (t) + v n cos nx cos 2ntn=1To satisfy the differential equation T 0 must satisfyT ′′0 = − 4 π + sin tTo satisfy the initial condition for v t , T 0 must satisfyThus we must haveT ′ 0(0) = 0T 0 = − 2t2π + t + T 0(0) − sin tImposing the intial condition for v givesx 2π( xπ − 3 ∞∑= T 0 (0) + v n cos nx2)n=1Integrating givesT 0 (0) = 1 ∫ π x 2 ( xπ 0 π π − 3 dx = −2)π 4Multiplying by cos nx and integrating givesThe answer:v n = 2 π∫ π0x 2π⎧( xπ − 3 ⎨cos nx dx =2)⎩(u(x, t) = (x − 1) sin t + x 1 − x )− 2t22π π + t − π 4 +0 n even24π 3 n 4∑n=1,3,5,...n odd24cos nx cos 2ntπ 3 n4 10