pdes - problems (4)
TRANSCRIPT
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8/3/2019 PDEs - Problems (4)
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MATH20401(PDEs) Tony Shardlow Problems Part IV
1. Let u(x, t) for 0 x 1 and t 0 be the solution of
ut = uxx u, ux(0, t) = ux(1, t) = 0
with initial condition u(x, 0) = u0(x) for 0 x 1. Show that E(t) is decreasing, whereE(t) =
1
0u(x, t)2 dx. Hence prove that the PDE problem has at most one solution.
2. For a time step k , define the explicit Euler approximations Xn to the solution X(t) of
dX
dt= f(X), X(0) = X0
at t = nk , where X0 is given.
For the case f(X) = X2 and X0 = 1, determine the exact solution X(t) and the explicitEuler approximation at t = 0.2 with time step k = 0.1.
Compute the error in approximating X(0.2) by the explicit Euler method for k = 0.1.
3. The solution X(t) to the ODE
dX
dt= f(X), X(0) = X0,
where f: R R , can be approximated by X(tn) Xn , where tn = nk and
Xn+1 = Xn + kf((Xn + Xn+1)/2).
This is known as the trapezoidal method.
(a) Let f(X) = X and X0 = 2. Compute the approximation by the trapezoidal methodto X(1) with time step k = 0.2 and k = 0.1. Compare with the exact solution.
(b) The linear test equation is the case f(X) = X, for parameter . Find all values ofk such that |Xn| 0 as n . Sketch the region k in the complex plane.
(c) Is the trapezoidal method implicit or explicit?
4. Consider the following PDE on x [0, 1]
ut = uxx u, ux(0, t) = ux(1, t) = 0
with initial condition u(x, 0) = u0(x). Assume u(x, t) can be written as the series
u(x, t) =
j=0
uj(t)cos(jx),
for some uj(t).
(a) Derive a system of ODEs for uj(t).
(b) Find the initial condition for uj(0) as an integral involving u0(x).
(c) Describe the methods of lines approximation for the above PDE. In particular, de-scribe the explicit Euler method and why the implicit Euler method is preferred inpractical computations.
MATH20401(PDEs): Problem Sheet IV: Page 1
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8/3/2019 PDEs - Problems (4)
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5. Let u(x, t) satisfy the heat equation:
ut = uxx, 0 < x < 1, t > 0,
subject to u(0, t) = 0 = u(1, t) for t > 0, and the initial condition u(x, 0) = u0(x).
(a) Consider the initial condition
u0(x) = sin(x) + sin (3x).
Take six intervals in space and the timestep k = 0.002 and apply the explicit fi-nite difference scheme to approximate to u(xj , 0.002). Compare the finite differencesolution with the exact solution at the grid points given by separation of variables:
u(x, t) = e2t sin(x) + e9
2t sin(3x) .
(b) If the ratio = k/h2 is fixed, show that the truncation error of the explicit finitedifference scheme satisfies |Tmj | = O(k), so that the consistency is first order.
6. Consider the explicit finite difference method for the heat equation
um+1j umj =
umj1 2u
mj + u
mj+1
um0 =umn = 0,
where = k/h2 and h = 1/n and j = 1, . . . , n 1 and m = 0, 1, . . . .
We know from class that the method is stable if 1 2 0. We show this condition isnecessary.
(a) For N = n 1, letumj = am sin(jNh).
By substituting umj into the finite difference method, show that umj is a solution ifam satisfies
am+1 am = 2(cos Nh 1)am.
(b) By considering the limit of cos Nh as h 0, show that am if am = 0 and > 1/2.
7. Let u(x, t) satisfy the heat equation:
ut = uxx, 0 < x < 1, t > 0,
subject to u(0, t) = 0 = u(1, t) for t > 0, and the initial condition u(x, 0) = u0(x).
If the ratio = k/h2
is fixed, show that the truncation error of the implicit finite differencescheme satisfies |Tmj | = O(k).
8. Determine whether or not the finite difference scheme
1
2k(um+1j u
m1j ) =
1
h2
umj+1 (u
m+1j + u
m1j ) + u
mj1
is consistent with the heat equation in the two cases of being the fixed ratio satisfying
(a) = k/h .
(b) = k/h2 .
MATH20401(PDEs): Problem Sheet IV: Page 2