We come now to one of the major and most-used types of methods for initial-value problems: Runge–Kutta (RK) methods.[1] (6.2.7) f ( t , u ) f(t,u) f ( t , u )
6.4.1 A second-order method ¶ Consider a series expansion of the exact solution to u ′ = f ( t , u ) u'=f(t,u) u ′ = f ( t , u )
u ^ ( t i + 1 ) = u ^ ( t i ) + h u ^ ′ ( t i ) + 1 2 h 2 u ^ ′ ′ ( t i ) + O ( h 3 ) . \hat{u}(t_{i+1}) = \hat{u}(t_i) + h \hat{u}'(t_i) + \frac{1}{2}h^2 \hat{u}''(t_i) + O(h^3) . u ^ ( t i + 1 ) = u ^ ( t i ) + h u ^ ′ ( t i ) + 2 1 h 2 u ^ ′′ ( t i ) + O ( h 3 ) . If we replace u ^ ′ \hat{u}' u ^ ′ f f f
u ^ ′ ′ = f ′ = d f d t = ∂ f ∂ t + ∂ f ∂ u d u d t = f t + f u f , \hat{u}'' = f' = \frac{d f}{d t} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} \frac{d u}{d t} = f_t + f_u f, u ^ ′′ = f ′ = d t df = ∂ t ∂ f + ∂ u ∂ f d t d u = f t + f u f , where we have applied the multidimensional chain rule to the derivative, because both of the arguments of f f f t t t (6.4.1)
u ^ ( t i + 1 ) = u ^ ( t i ) + h [ f ( t i , u ^ ( t i ) ) + h 2 f t ( t i , u ^ ( t i ) ) + h 2 f ( t i , u ^ ( t i ) ) f u ( t i , u ^ ( t i ) ) ] + O ( h 3 ) . \hat{u}(t_{i+1}) = \hat{u}(t_i) + h\left[f\bigl(t_i,\hat{u}(t_i)\bigr) +
\frac{h}{2}f_t\bigl(t_i,\hat{u}(t_i)\bigr) +
\frac{h}{2}f\bigl(t_i,\hat{u}(t_i)\bigr)\,f_u\bigl(t_i,\hat{u}(t_i)\bigr)\right] \\
+ O(h^3). u ^ ( t i + 1 ) = u ^ ( t i ) + h [ f ( t i , u ^ ( t i ) ) + 2 h f t ( t i , u ^ ( t i ) ) + 2 h f ( t i , u ^ ( t i ) ) f u ( t i , u ^ ( t i ) ) ] + O ( h 3 ) . We have no desire to calculate and then code those partial derivatives of f f f
f ( t i + α , u ^ ( t i ) + β ) = f ( t i , u ^ ( t i ) ) + α f t ( t i , u ^ ( t i ) ) + β f u ( t i , u ^ ( t i ) ) + O ( α 2 + ∣ α β ∣ + β 2 ) . f\bigl(t_i+\alpha,\hat{u}(t_i)+\beta\bigr) = f\bigl(t_i,\hat{u}(t_i)\bigr) +
\alpha f_t\bigl(t_i,\hat{u}(t_i)\bigr) + \beta f_u\bigl(t_i,\hat{u}(t_i)\bigr) +
O\bigl(\alpha^2 + |\alpha\beta| + \beta^2\bigr). f ( t i + α , u ^ ( t i ) + β ) = f ( t i , u ^ ( t i ) ) + α f t ( t i , u ^ ( t i ) ) + β f u ( t i , u ^ ( t i ) ) + O ( α 2 + ∣ α β ∣ + β 2 ) . Matching this expression to the term in brackets in (6.4.3) α = h / 2 \alpha = h/2 α = h /2 β = 1 2 h f ( t i , u ^ ( t i ) ) \beta = \frac{1}{2}h f\bigl(t_i,\hat{u}(t_i)\bigr) β = 2 1 h f ( t i , u ^ ( t i ) )
u ^ ( t i + 1 ) = u ^ ( t i ) + h [ f ( t i + α , u ^ ( t i ) + β ) ] + O ( h α 2 + h ∣ α β ∣ + h β 2 + h 3 ) . \hat{u}(t_{i+1}) = \hat{u}(t_i) + h\left[f\bigl(t_i+\alpha,\hat{u}(t_i)+\beta\bigr)\right] +
O(h\alpha^2 + h|\alpha \beta| + h\beta^2 + h^3). u ^ ( t i + 1 ) = u ^ ( t i ) + h [ f ( t i + α , u ^ ( t i ) + β ) ] + O ( h α 2 + h ∣ α β ∣ + h β 2 + h 3 ) . Truncation of the series here results in a new one-step method.
Thanks to the definitions above of α and β, the omitted terms are of size
O ( h α 2 + h ∣ α β ∣ + h β 2 + h 3 ) = O ( h 3 ) . O(h\alpha^2 + h|\alpha \beta| + h\beta^2 + h^3) = O(h^3). O ( h α 2 + h ∣ α β ∣ + h β 2 + h 3 ) = O ( h 3 ) . Therefore h τ i + 1 = O ( h 3 ) h\tau_{i+1}=O(h^3) h τ i + 1 = O ( h 3 )
6.4.2 Implementation ¶ Runge–Kutta methods are called multistage methods. We can see why if we interpret (6.4.6) t i + 1 2 h t_i+\frac{1}{2}h t i + 2 1 h
k 1 = h f ( t i , u i ) , v = u i + 1 2 k 1 . \begin{split}
k_1 &= h f(t_i,u_i), \\
v &= u_i + \tfrac{1}{2}k_1.
\end{split} k 1 v = h f ( t i , u i ) , = u i + 2 1 k 1 . The second stage employs an Euler-style strategy over the whole time step, but using the value from the first stage to get the slope:
k 2 = h f ( t i + 1 2 h , v ) , u i + 1 = u i + k 2 . \begin{split}
k_2 &= h f\left(t_i+\tfrac{1}{2}h,v\right),\\
{u}_{i+1} &= u_i + k_2.
\end{split} k 2 u i + 1 = h f ( t i + 2 1 h , v ) , = u i + k 2 . Our implementation of IE2 is shown in Function 6.4.1
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function [t, u] = ie2(ivp, a, b, n)
% IE2 Improved Euler method for an IVP.
% Input:
% ivp structure defining the IVP
% a, b endpoints of time interval (scalars)
% n number of time steps (integer)
% Output:
% t selected nodes (vector, length n+1)
% u solution values (array, m by (n+1))
du_dt = ivp.ODEFcn;
u0 = ivp.InitialValue;
p = ivp.Parameters;
% Define time discretization.
h = (b - a) / n;
t = a + h * (0:n)';
% Initialize solution array.
u = zeros(length(u0), n+1);
u(:, 1) = u0(:);
% Time stepping.
for i = 1:n
uhalf = u(:, i) + h/2 * du_dt(t(i), u(:, i), p);
u(:, i+1) = u(:, i) + h * du_dt(t(i) + h/2, uhalf, p);
end1
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def ie2(dudt, tspan, u0, n):
"""
ie2(dudt,tspan,u0,n)
Apply the Improved Euler method to solve the vector-valued IVP u'=`dudt`(u,p,t) over the
interval `tspan` with u(`tspan[1]`)=`u0`, using `n` subintervals/steps. Returns a vector
of times and a vector of solution values/vectors.
"""
# Time discretization.
a, b = tspan
h = (b - a) / n
t = np.linspace(a, b, n + 1)
# Initialize output.
u = np.tile(np.array(u0), (n+1, 1))
# Time stepping.
for i in range(n):
uhalf = u[i] + h / 2 * dudt(t[i], u[i])
u[i+1] = u[i] + h * dudt(t[i] + h / 2, uhalf)
return t, u.T6.4.3 More Runge–Kutta methods ¶ While it is interesting to interpret IE2 as a pair of Euler-like steps, the Taylor series derivation is the only way to see that it will be more accurate than Euler, and it is also the path to deriving other methods. Moving to higher orders of accuracy requires introducing additional stages, each having free parameters so that more terms in the series may be matched. The amount of algebra grows rapidly in size and complexity, though there is a sophisticated theory for keeping track of it. We do not give the derivation details.
A generic s s s
k 1 = h f ( t i , u i ) , k 2 = h f ( t i + c 1 h , u i + a 11 k 1 ) , k 3 = h f ( t i + c 2 h , u i + a 21 k 1 + a 22 k 2 ) , ⋮ k s = h f ( t i + c s − 1 h , u i + a s − 1 , 1 k 1 + ⋯ + a s − 1 , s − 1 k s − 1 ) , u i + 1 = u i + b 1 k 1 + ⋯ + b s k s . \begin{split}
k_1 &= h f(t_i,u_i),\\
k_2 &= h f(t_i+c_1h,u_i + a_{11}k_1),\\
k_3 &= h f(t_i+c_2h, u_i + a_{21}k_1 + a_{22}k_2),\\
&\vdots\\
k_s &= h f(t_i + c_{s-1}h, u_i + a_{s-1,1}k_1 + \cdots +
a_{s-1,s-1}k_{s-1}),\\
\mathbf{u}_{i+1} &= u_i + b_1k_1 + \cdots + b_s k_s.
\end{split} k 1 k 2 k 3 k s u i + 1 = h f ( t i , u i ) , = h f ( t i + c 1 h , u i + a 11 k 1 ) , = h f ( t i + c 2 h , u i + a 21 k 1 + a 22 k 2 ) , ⋮ = h f ( t i + c s − 1 h , u i + a s − 1 , 1 k 1 + ⋯ + a s − 1 , s − 1 k s − 1 ) , = u i + b 1 k 1 + ⋯ + b s k s . This recipe is completely determined by the number of stages s s s a i j a_{ij} a ij b j b_j b j c i c_i c i
0 c 1 a 11 c 2 a 21 a 22 ⋮ ⋮ ⋱ c s − 1 a s − 1 , 1 ⋯ a s − 1 , s − 1 b 1 b 2 ⋯ b s − 1 b s \begin{array}{r|ccccc}
0 & & & & & \\
c_1 & a_{11} & & &\\
c_2 & a_{21} & a_{22} & & &\\
\vdots & \vdots & & \ddots & &\\
c_{s-1} & a_{s-1,1} & \cdots & & a_{s-1,s-1}&\\[1mm] \hline
\rule{0pt}{2.25ex} & b_1 & b_2 & \cdots & b_{s-1} & b_s
\end{array} 0 c 1 c 2 ⋮ c s − 1 a 11 a 21 ⋮ a s − 1 , 1 b 1 a 22 ⋯ b 2 ⋱ ⋯ a s − 1 , s − 1 b s − 1 b s For example, IE2 is given by
0 1 2 1 2 0 1 \begin{array}{r|cc}
\rule{0pt}{2.75ex}0 & & \\
\rule{0pt}{2.75ex}\frac{1}{2} & \frac{1}{2} &\\[1mm] \hline
\rule{0pt}{2.75ex}& 0 & 1
\end{array} 0 2 1 2 1 0 1 Here are two more two-stage, second-order methods, modified Euler and Heun’s method , respectively:
0 1 1 1 2 1 2 0 2 3 2 3 1 4 3 4 \begin{array}{r|cc}
\rule{0pt}{2.75ex}0 & & \\
\rule{0pt}{2.75ex}1 & 1 &\\[1mm] \hline
\rule{0pt}{2.75ex}& \frac{1}{2} & \frac{1}{2}
\end{array}
\qquad \qquad
\begin{array}{r|cc}
\rule{0pt}{2.75ex} 0 & & \\
\rule{0pt}{2.75ex} \frac{2}{3} & \frac{2}{3} &\\[1mm] \hline
\rule{0pt}{2.75ex} & \frac{1}{4} & \frac{3}{4}
\end{array} 0 1 1 2 1 2 1 0 3 2 3 2 4 1 4 3 The most commonly used RK method, and perhaps the most popular IVP method of all, is the fourth-order one given by
0 1 2 1 2 1 2 0 1 2 1 0 0 1 1 6 1 3 1 3 1 6 \begin{array}{r|cccc}
\rule{0pt}{2.75ex}0 & & & & \\
\rule{0pt}{2.75ex}\frac{1}{2} & \frac{1}{2} & & &\\
\rule{0pt}{2.75ex}\frac{1}{2} & 0 & \frac{1}{2} & &\\
\rule{0pt}{2.75ex}1 & 0 & 0 & 1\\[1mm] \hline
\rule{0pt}{2.75ex}& \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6}
\end{array} 0 2 1 2 1 1 2 1 0 0 6 1 2 1 0 3 1 1 3 1 6 1 This formula is often referred to as the fourth-order RK method, even though there are many others, and we refer to it as RK4 . Written out, the recipe is as follows.
Our implementation is given in Function 6.4.2
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function [t, u] = rk4(ivp, a, b, n)
% RK4 Fourth-order Runge-Kutta for an IVP.
% Input:
% ivp structure defining the IVP
% a, b endpoints of time interval (scalars)
% n number of time steps (integer)
% Output:
% t selected nodes (vector, length n+1)
% u solution values (array, m by (n+1))
du_dt = ivp.ODEFcn;
u0 = ivp.InitialValue;
p = ivp.Parameters;
% Define time discretization.
h = (b - a) / n;
t = a + h * (0:n)';
% Initialize solution array.
u = zeros(length(u0), n+1);
u(:, 1) = u0(:);
% Time stepping.
for i = 1:n
k1 = h * du_dt( t(i), u(:, i) , p);
k2 = h * du_dt( t(i) + h/2, u(:, i) + k1/2, p );
k3 = h * du_dt( t(i) + h/2, u(:, i) + k2/2, p );
k4 = h * du_dt( t(i) + h, u(:, i) + k3 , p);
u(:, i+1) = u(:, i) + (k1 + 2*(k2 + k3) + k4) / 6;
end1
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def rk4(dudt, tspan, u0, n):
"""
rk4(dudt,tspan,u0,n)
Apply "the" Runge-Kutta 4th order method to solve the vector-valued IVP u'=`dudt`(u,p,t)
over the interval `tspan` with u(`tspan[1]`)=`u0`, using `n` subintervals/steps.
Return a vector of times and a vector of solution values/vectors.
"""
# Time discretization.
a, b = tspan
h = (b - a) / n
t = np.linspace(a, b, n + 1)
# Initialize output.
u = np.tile(np.array(u0), (n+1, 1))
# Time stepping.
for i in range(n):
k1 = h * dudt(t[i], u[i])
k2 = h * dudt(t[i] + h / 2, u[i] + k1 / 2)
k3 = h * dudt(t[i] + h / 2, u[i] + k2 / 2)
k4 = h * dudt(t[i] + h, u[i] + k3)
u[i+1] = u[i] + (k1 + 2 * (k2 + k3) + k4) / 6
return t, u.TWe solve the IVP u ′ = sin [ ( u + t ) 2 ] u'=\sin[(u+t)^2] u ′ = sin [( u + t ) 2 ] 0 ≤ t ≤ 4 0\le t \le 4 0 ≤ t ≤ 4 u ( 0 ) = − 1 u(0)=-1 u ( 0 ) = − 1
f(u, p, t) = sin((t + u)^2)
tspan = (0.0, 4.0)
u₀ = -1.0
ivp = ODEProblem(f, u₀, tspan)
We use a DifferentialEquations solver to construct an accurate approximation to the exact solution.
u_ref = solve(ivp, Tsit5(), reltol=1e-14, abstol=1e-14);
Now we perform a convergence study of our two Runge–Kutta implementations.
n = [round(Int, 2 * 10^k) for k in 0:0.5:3]
err_IE2, err_RK4 = [], []
for n in n
t, u = FNC.ie2(ivp, n)
push!(err_IE2, maximum(@.abs(u_ref(t) - u)))
t, u = FNC.rk4(ivp, n)
push!(err_RK4, maximum(@.abs(u_ref(t) - u)))
end
pretty_table([n err_IE2 err_RK4], header=["n", "IE2 error", "RK4 error"])
The amount of computational work at each time step is assumed to be proportional to the number of stages. Let’s compare on an apples-to-apples basis by using the number of f f f
plot([2n 4n], [err_IE2 err_RK4], m=3, label=["IE2" "RK4"],
xaxis=(:log10, "f-evaluations"), yaxis=(:log10, "inf-norm error"),
title="Convergence of RK methods", leg=:bottomleft)
plot!(2n, 1e-5 * (n / n[end]) .^ (-2), l=:dash, label=L"O(n^{-2})")
plot!(4n, 1e-10 * (n / n[end]) .^ (-4), l=:dash, label=L"O(n^{-4})")
The fourth-order variant is more efficient in this problem over a wide range of accuracy.
We solve the IVP u ′ = sin [ ( u + t ) 2 ] u'=\sin[(u+t)^2] u ′ = sin [( u + t ) 2 ] 0 ≤ t ≤ 4 0\le t \le 4 0 ≤ t ≤ 4 u ( 0 ) = − 1 u(0)=-1 u ( 0 ) = − 1
ivp = ode;
ivp.ODEFcn = @(t, u, p) sin((t + u)^2);
ivp.InitialValue = -1;
a = 0; b = 4;
We use a built-in solver to construct an accurate approximation to the exact solution.
ivp.AbsoluteTolerance = 1e-13;
ivp.RelativeTolerance = 1e-13;
u_ref = solutionFcn(ivp, a, b);
Now we perform a convergence study of our two Runge–Kutta implementations.
n = round(2 * 10.^(0:0.5:3)');
err = zeros(length(n), 2);
for i = 1:length(n)
[t, u] = ie2(ivp, a, b, n(i));
err(i, 1) = norm(u_ref(t) - u, Inf);
[t, u] = rk4(ivp, a, b, n(i));
err(i, 2) = norm(u_ref(t) - u, Inf);
end
disp(table(n, err(:, 1), err(:, 2), variableNames=["n", "IE2 error", "RK4 error"]))
The amount of computational work at each time step is assumed to be proportional to the number of stages. Let’s compare on an apples-to-apples basis by using the number of f f f
clf, loglog([2*n 4*n], err, '-o')
hold on
loglog(2*n, 1e-5 * (n / n(end)) .^ (-2), '--')
loglog(4*n, 1e-10 * (n / n(end)) .^ (-4), '--')
xlabel("f-evaluations"); ylabel("inf-norm error")
title("Convergence of RK methods")
legend("IE2", "RK4", "O(n^{-2})", "O(n^{-4})", "location", "southwest")
The fourth-order variant is more efficient in this problem over a wide range of accuracy.
We solve the IVP u ′ = sin [ ( u + t ) 2 ] u'=\sin[(u+t)^2] u ′ = sin [( u + t ) 2 ] 0 ≤ t ≤ 4 0\le t \le 4 0 ≤ t ≤ 4 u ( 0 ) = − 1 u(0)=-1 u ( 0 ) = − 1
from scipy.integrate import solve_ivp
du_dt = lambda t, u: sin((t + u)**2)
tspan = (0.0, 4.0)
u0 = -1.0
sol = solve_ivp(du_dt, tspan, [u0], dense_output=True, atol=1e-13, rtol=1e-13)
u_ref = sol.sol
Now we perform a convergence study of our two Runge–Kutta implementations.
n = array([int(2 * 10**k) for k in linspace(0, 3, 7)])
err = {"IE2" : [], "RK4" : []}
results = PrettyTable(["n", "IE2 error", "RK4 error"])
for i in range(len(n)):
t, u = FNC.ie2(du_dt, tspan, u0, n[i])
err["IE2"].append( abs(u_ref(4)[0] - u[0][-1]) )
t, u = FNC.rk4(du_dt, tspan, u0, n[i])
err["RK4"].append( abs(u_ref(4)[0] - u[0][-1]) )
results.add_row([n[i], err["IE2"][-1], err["RK4"][-1]])
print(results)
The amount of computational work at each time step is assumed to be proportional to the number of stages. Let’s compare on an apples-to-apples basis by using the number of f f f
loglog(2 * n, err["IE2"], "-o", label="IE2")
loglog(4 * n, err["RK4"], "-o", label="RK4")
plot(2 * n, 0.5 * err["IE2"][-1] * (n / n[-1])**(-2), "--", label="2nd order")
plot(4 * n, 0.5 * err["RK4"][-1] * (n / n[-1])**(-4), "--", label="4th order")
xlabel("f-evaluations"), ylabel("inf-norm error")
legend()
title("Convergence of RK methods");
The fourth-order variant is more efficient in this problem over a wide range of accuracy.
6.4.4 Efficiency ¶ As with rootfinding and integration, the usual point of view is that evaluations of f f f f f f s s s s s s f f f
The error decreases geometrically as s s s s = 5 s=5 s = 5 s − 1 s-1 s − 1 s − 2 s-2 s − 2 s = 8 s=8 s = 8 s = 9 s=9 s = 9
6.4.5 Exercises ¶ ✍ For each IVP, write out (possibly using a calculator) the first time step of the improved Euler method with h = 0.2 h=0.2 h = 0.2
(a) u ′ = − 2 t u , 0 ≤ t ≤ 2 , u ( 0 ) = 2 ; u ^ ( t ) = 2 e − t 2 u' = -2t u, \ 0 \le t \le 2, \ u(0) = 2;\ \hat{u}(t) = 2e^{-t^2} u ′ = − 2 t u , 0 ≤ t ≤ 2 , u ( 0 ) = 2 ; u ^ ( t ) = 2 e − t 2
(b) u ′ = u + t , 0 ≤ t ≤ 1 , u ( 0 ) = 2 ; u ^ ( t ) = − 1 − t + 3 e t u' = u + t, \ 0 \le t \le 1, \ u(0) = 2;\ \hat{u}(t) = -1-t+3e^t u ′ = u + t , 0 ≤ t ≤ 1 , u ( 0 ) = 2 ; u ^ ( t ) = − 1 − t + 3 e t
(c) ( 1 + x 3 ) u u ′ = x 2 , 0 ≤ x ≤ 3 , u ( 0 ) = 1 ; u ^ ( x ) = [ 1 + ( 2 / 3 ) ln ( 1 + x 3 ) ] 1 / 2 (1+x^3)uu' = x^2,\ 0 \le x \le 3, \ u(0) = 1;\ \hat{u}(x) = [1+(2/3)\ln (1+x^3)]^{1/2} ( 1 + x 3 ) u u ′ = x 2 , 0 ≤ x ≤ 3 , u ( 0 ) = 1 ; u ^ ( x ) = [ 1 + ( 2/3 ) ln ( 1 + x 3 ) ] 1/2
✍ Use the modified Euler method to solve the problems in the preceding exercise.
⌨ Modify Function 6.4.2 n = 30 , 60 , 90 , … , 300 n=30,60,90,\ldots,300 n = 30 , 60 , 90 , … , 300 O ( n − 2 ) O(n^{-2}) O ( n − 2 )
✍ Use Heun’s method to solve the problems in Exercise 1 above
⌨ Modify Function 6.4.2 n = 30 , 60 , 90 , … , 300 n=30,60,90,\ldots,300 n = 30 , 60 , 90 , … , 300 O ( n − 2 ) O(n^{-2}) O ( n − 2 )
✍ Use RK4 to solve the problems in Exercise 1 above
✍ Using (6.4.3) (6.4.4)
✍ Using (6.4.3) (6.4.4)
⌨ For each IVP, compute the solution using Function 6.4.2 n = 300 n=300 n = 300 n = 100 , 200 , 300 , … , 1000 n=100,200,300,\ldots,1000 n = 100 , 200 , 300 , … , 1000
(a) u ′ ′ + 9 u = 9 t , 0 < t < 2 π , u ( 0 ) = 1 , u ′ ( 0 ) = 1 ; u ^ ( t ) = t + cos ( 3 t ) u''+ 9u = 9t, \: 0< t< 2\pi, \: u(0) = 1,\: u'(0) = 1; \: \hat{u}(t) = t+\cos (3t) u ′′ + 9 u = 9 t , 0 < t < 2 π , u ( 0 ) = 1 , u ′ ( 0 ) = 1 ; u ^ ( t ) = t + cos ( 3 t )
(b) u ′ ′ + 9 u = sin ( 2 t ) , 0 < t < 2 π , u ( 0 ) = 2 , u ′ ( 0 ) = 1 u''+ 9u = \sin(2t), \: 0< t< 2\pi, \: u(0) = 2,\: u'(0) = 1 u ′′ + 9 u = sin ( 2 t ) , 0 < t < 2 π , u ( 0 ) = 2 , u ′ ( 0 ) = 1
u ^ ( t ) = ( 1 / 5 ) sin ( 3 t ) + 2 cos ( 3 t ) + ( 1 / 5 ) sin ( 2 t ) \quad \hat{u}(t) = (1/5) \sin(3t) + 2 \cos (3t)+ (1/5) \sin (2t) u ^ ( t ) = ( 1/5 ) sin ( 3 t ) + 2 cos ( 3 t ) + ( 1/5 ) sin ( 2 t )
(c) u ′ ′ − 9 u = 9 t , 0 < t < 1 , u ( 0 ) = 2 , u ′ ( 0 ) = − 1 ; u ^ ( t ) = e 3 t + e − 3 t − t u''- 9u = 9t, \: 0< t< 1, \: u(0) = 2,\: u'(0) = -1; \: \hat{u}(t) = e^{3t} + e^{-3t}-t u ′′ − 9 u = 9 t , 0 < t < 1 , u ( 0 ) = 2 , u ′ ( 0 ) = − 1 ; u ^ ( t ) = e 3 t + e − 3 t − t
(d) u ′ ′ + 4 u ′ + 4 u = t , 0 < t < 4 , u ( 0 ) = 1 , u ′ ( 0 ) = 3 / 4 ; u ^ ( t ) = ( 3 t + 5 / 4 ) e − 2 t + ( t − 1 ) / 4 u''+ 4u'+ 4u = t, \: 0< t< 4, \: u(0) = 1,\: u'(0) = 3/4; \: \hat{u}(t) = (3t+5/4)e^{-2t} + (t-1)/4 u ′′ + 4 u ′ + 4 u = t , 0 < t < 4 , u ( 0 ) = 1 , u ′ ( 0 ) = 3/4 ; u ^ ( t ) = ( 3 t + 5/4 ) e − 2 t + ( t − 1 ) /4
(e) x 2 y ′ ′ + 5 x y ′ + 4 y = 0 , 1 < x < e 2 , y ( 1 ) = 1 , y ′ ( 1 ) = − 1 , y ^ ( x ) = x − 2 ( 1 + ln x ) x^2 y'' +5xy' + 4y = 0,\: 1<x<e^2, \: y(1) = 1, \: y'(1) = -1, \: \hat{y}(x) = x^{-2}( 1 + \ln x) x 2 y ′′ + 5 x y ′ + 4 y = 0 , 1 < x < e 2 , y ( 1 ) = 1 , y ′ ( 1 ) = − 1 , y ^ ( x ) = x − 2 ( 1 + ln x )
(f) 2 x 2 y ′ ′ + 3 x y ′ − y = 0 , 1 < x < 16 , y ( 1 ) = 4 , y ′ ( 1 ) = − 1 , y ^ ( x ) = 2 ( x 1 / 2 + x − 1 ) 2 x^2 y'' +3xy' - y = 0,\: 1<x<16, \: y(1) = 4, \: y'(1) = -1, \: \hat{y}(x) = 2(x^{1/2} + x^{-1}) 2 x 2 y ′′ + 3 x y ′ − y = 0 , 1 < x < 16 , y ( 1 ) = 4 , y ′ ( 1 ) = − 1 , y ^ ( x ) = 2 ( x 1/2 + x − 1 )
(g) x 2 y ′ ′ − x y ′ + 2 y = 0 , 1 < x < e π , y ( 1 ) = 3 , y ′ ( 1 ) = 4 x^2 y'' -xy' + 2y = 0,\: 1<x<e^{\pi}, \: y(1) = 3, \: y'(1) = 4 x 2 y ′′ − x y ′ + 2 y = 0 , 1 < x < e π , y ( 1 ) = 3 , y ′ ( 1 ) = 4
y ^ ( x ) = x [ 3 cos ( ln x ) + sin ( ln x ) ] \quad \hat{y}(x) = x \left[ 3 \cos \left( \ln x \right)+\sin \left( \ln x \right) \right] y ^ ( x ) = x [ 3 cos ( ln x ) + sin ( ln x ) ]
(h) x 2 y ′ ′ + 3 x y ′ + 4 y = 0 , e π / 12 < x < e π , y ( e π / 12 ) = 0 , y ′ ( e π / 12 ) = − 6 x^2 y'' + 3xy' + 4y = 0,\: e^{\pi/12} < x < e^{\pi}, \: y(e^{\pi/12}) = 0, \: y'(e^{\pi/12}) = -6 x 2 y ′′ + 3 x y ′ + 4 y = 0 , e π /12 < x < e π , y ( e π /12 ) = 0 , y ′ ( e π /12 ) = − 6
y ^ ( x ) = x − 1 [ 3 cos ( 3 ln x ) + sin ( 3 ln x ) ] \quad \hat{y}(x) = x^{-1} \left[ 3 \cos \left( 3 \ln x \right)+\sin \left( 3 \ln x \right) \right] y ^ ( x ) = x − 1 [ 3 cos ( 3 ln x ) + sin ( 3 ln x ) ]
⌨ Do Exercise 6.3.4 , but using Function 6.4.2 solve.
✍ Consider the problem u ′ = c u u'=c u u ′ = c u u ( 0 ) = 1 u(0) = 1 u ( 0 ) = 1 c c c t > 0 t>0 t > 0
(a) Find an explicit formula in terms of h h h c c c u i + 1 / u i u_{i+1}/u_i u i + 1 / u i
(b) Show that if c h = − 3 ch=-3 c h = − 3 ∣ u i ∣ → ∞ |u_i|\to\infty ∣ u i ∣ → ∞ i → ∞ i\to\infty i → ∞ u ^ ( t ) \hat{u}(t) u ^ ( t ) t → ∞ t\to\infty t → ∞