Use the shooting method to solve the mixed boundary value problem u″ = u – 4x e^x, 0

Question

Use the shooting method to solve the mixed boundary value problem

[latex] \begin{gathered} u^{\prime \prime}=u-4 x e^{x}, 0

Use the Taylor series method

[latex] \begin{aligned} & u_{j+1}=u_{j}+h u_{j}^{\prime}+\frac{h^{2}}{2} u_{j}^{\prime \prime}+\frac{h^{3}}{6} u_{j}^{\prime \prime \prime} . \ & u_{j+1}^{\prime}=u_{j}^{\prime}+h u_{j}^{\prime \prime}+\frac{h^{2}}{2} u_{j}^{\prime \prime \prime} \end{aligned} [/latex]

to solve the initial value problems. Assume [latex]h=0.25[/latex]. Compare with the exact solution [latex]u(x)=x(1-x) e^{x}[/latex].

Accepted Answer

We assume the solution in the form

[latex]u(x)=u_{0}(x)+\mu_{1} u_{1}(x)+\mu_{2} u_{2}(x)[/latex]

where [latex]u_{0}(x), u_{1}(x)[/latex] and [latex]u_{2}(x)[/latex] satisfy the differential equations

[latex] \begin{aligned} & u_{0}^{\prime \prime}-u_{0}=-4 x e^{x}, u_{1}^{\prime \prime}-u_{1}=0 ,\ & u_{2}^{\prime \prime}-u_{2}=0. \end{aligned} [/latex]

The initial conditions may be assumed as

[latex] \begin{array}{ll} u_{0}(0)=0, & u_{0}^{\prime}(0)=0, \ u_{1}(0)=1, & u_{1}^{\prime}(0)=0, \ u_{2}(0)=0, & u_{2}^{\prime}(0)=1. \end{array} [/latex]

We solve the three, second order initial value problems

[latex] \begin{array}{ll} u_{0}^{\prime \prime}=u_{0}-4 x e^{x}, & u_{0}(0)=0, u_{0}^{\prime}(0)=0, \ u_{1}^{\prime \prime}=u_{1} . & u_{1}(0)=1, u_{1}^{\prime}(0)=0, \ u_{2}^{\prime \prime}=u_{2}, & u_{2}(0)=0, u_{2}^{\prime}(0)=1 \end{array} [/latex]

by using the given Taylor series method with [latex]h=0.25[/latex]. We have the following results. (i) [latex]i=0, u_{0,0}=0, u_{0,0}^{\prime}=0[/latex].

[latex] \begin{aligned} ext{Hence,}\qquad u_{0, j}^{\prime \prime} & =u_{0, j}-4 x_{j} e^{x_{j}}, u_{0, j}^{\prime \prime \prime}=u_{0, j}^{\prime}-4\left(x_{j}+1 ight) e^{x_{j}}, j=0,1,2,3 . \ u_{0, j+1} & =u_{0, j}+h u_{0, j}^{\prime}+\frac{h^{2}}{2}\left(u_{0, j}-4 x_{j} e^{x_{j}} ight)+\frac{h^{3}}{6}\left[u_{0, j}^{\prime}-4\left(x_{j}+1 ight) e^{x_{j}} ight] \ & =\left(1+\frac{h^{2}}{2} ight) u_{0, j}+\left(h+\frac{h^{3}}{6} ight) u_{0, j}^{\prime}-\left[\frac{2}{3} h^{3}\left(1+x_{j} ight)+2 h^{2} x_{j} ight] e^{x_{j}} \ & =1.03125 u_{0, j}+0.25260 u_{0, j}^{\prime}-\left(0.13542 x_{j}+0.01042 ight) e^{x_{j}} \ u_{0, j+1}^{\prime} & =u_{0, j}^{\prime}+h\left[u_{0, j}-4 x_{j} e^{x_{j}} ight]+\frac{h^{2}}{2}\left[u_{0, j}^{\prime}-4\left(x_{j}+1 ight) e^{x_{j}} ight] \ & =h u_{0, j}+\left(1+\frac{h^{2}}{2} ight) u_{0, j}^{\prime}-2\left[2 h x_{j}+h^{2}\left(1+x_{j} ight) ight] e^{x_{j}} \ & =0.25 u_{0, j}+1.03125 u_{0, j}^{\prime}-2\left(0.5625 x_{j}+0.0625 ight) e^{x_{j}} . \end{aligned} [/latex]

Therefore, we obtain

[latex] \begin{aligned} & u_{0}(0.25) \approx u_{0,1}=-0.01042, \quad u_{0}^{\prime}(0.25) \approx u_{0,1}^{\prime}=-0.12500 ,\ & u_{0}(0.50) \approx u_{0,2}=-0.09917, \quad u_{0}^{\prime}(0.50) \approx u_{0,2}^{\prime}=-0.65315, \ & u_{0}(0.75) \approx u_{0,3}=-0.39606, \quad u_{0}^{\prime}(0.75) \approx u_{0,3}^{\prime}=-1.83185 ,\ & u_{0}(1.00) \approx u_{0,4}=-1.10823, \quad u_{0}^{\prime}(1.00) \approx u_{0,4}^{\prime}=-4.03895. \end{aligned} [/latex]

(ii) [latex]i=1, u_{1,0}=1, u_{1,0}^{\prime}=0[/latex].

[latex] \begin{aligned} u_{1, j}^{\prime \prime} & =u_{1, j}, u_{1, j}^{\prime \prime \prime}=u_{1, j}^{\prime}, j=0,1,2,3 . \ u_{1, j+1} & =u_{1, j}+h u_{1, j}^{\prime}+\frac{h^{2}}{2} u_{1, j}+\frac{h^{3}}{6} u_{1, j}^{\prime} \ & =\left(1+\frac{h^{2}}{2} ight) u_{1, j}+\left(h+\frac{h^{3}}{6} ight) u_{1, j}^{\prime}=1.03125 u_{1, j}+0.25260 u_{1, j}^{\prime} \ u_{1, j+1}^{\prime} & =u_{1, j}^{\prime}+h u_{1, j}+\frac{h^{2}}{2} u_{1, j}^{\prime} \ & =h u_{1, j}+\left(1+\frac{h^{2}}{2} ight) u_{1, j}^{\prime}=0.25 u_{1, j}+1.03125 u_{1, j}^{\prime} \end{aligned} [/latex]

Hence,

[latex] \begin{array}{ll} u_{1}(0.25) \approx u_{1,1}=1.03125, & u_{1}^{\prime}(0.25) \approx u_{1,1}^{\prime}=0.25, \ u_{1}(0.50) \approx u_{1,2}=1.12663, & u_{1}^{\prime}(0.50) \approx u_{1,2}^{\prime}=0.51563, \end{array} [/latex]

[latex] \begin{array}{ll} u_{1}(0.75) \approx u_{1,3}=1.29209, & u_{1}^{\prime}(0.75) \approx u_{1,3}^{\prime}=0.81340, \ u_{1}(1.00) \approx u_{1,4}=1.53794, & u_{1}^{\prime}(1.00) \approx u_{1,4}^{\prime}=1.16184. \end{array} [/latex]

(iii) [latex]i=2, u_{2,0}=0, u_{2,0}^{\prime}=1[/latex].

[latex]u_{2, j}^{\prime \prime}=u_{2, j}, u_{2, j}^{\prime \prime \prime}=u_{2, j}^{\prime}, j=0,1,2,3 .[/latex]

Since the differential equaton is same as for [latex]u_{1}[/latex], we get

[latex] \begin{aligned} u_{2, j+1} & =1.03125 u_{2, j}+0.25260 u_{2, j}^{\prime} \ u_{2, j+1}^{\prime} & =0.25 u_{2, j}+1.03125 u_{2, j}^{\prime} \end{aligned} [/latex]

[latex] \begin{array}{ll} ext{Hence,}\qquad u_{2}(0.25) \approx u_{2,1}=0.25260, & u_{2}^{\prime}(0.25) \approx u_{2,1}^{\prime}=1.03125, \ u_{2}(0.50) \approx u_{2,2}=0.52099, & u_{2}^{\prime}(0.50) \approx u_{2,2}^{\prime}=1.12663, \ u_{2}(0.75) \approx u_{2,3}=0.82186, & u_{2}^{\prime}(0.75) \approx u_{2,3}^{\prime}=1.29208, \ u_{2}(1.00) \approx u_{2,4}=1.17393, & u_{2}^{\prime}(1.00) \approx u_{2,4}^{\prime}=1.53792 . \end{array} [/latex]

From the given boundary conditions, we have

[latex] \begin{gathered} a_{0}=a_{1}=1, b_{0}=b_{1}=1, \gamma_{1}=-1, \gamma_{2}=-e. \ \mu_{1}-\mu_{2}=-1 \ {\left[u_{1}(1)+u_{1}^{\prime}(1) ight] \mu_{1}+\left[u_{2}(1)+u_{2}^{\prime}(1) ight] \mu_{2}=-e-\left[u_{0}(1)+u_{0}^{\prime}(1) ight]} \end{gathered} [/latex]

or [latex]2.69978 \mu_{1}+2.71185 \mu_{2}=2.42890[/latex].

Solving these equations, we obtain [latex]\mu_{1}=-0.05229, \mu_{2}=0.94771[/latex].

We obtain the solution of the boundary value problem from

[latex]u(x)=u_{0}(x)-0.05229 u_{1}(x)+0.94771 u_{2}(x) .[/latex]

The solutions at the nodal points are given in the Table 5.1. The maximum absolute error which occurs at [latex]x=0.75[/latex], is given by

max. abs. error = 0.08168.

Alternative

Here, we solve the initial value problems

[latex] \begin{aligned} & u_{1}^{\prime \prime}-u_{1}=-4 x e^{x}, u_{1}(0)=0, u_{1}^{\prime}(0)=-\left(\gamma_{1} / a_{1} ight)=1, \ & \left.u_{2}^{\prime \prime}-u_{2}=-4 x e^{x}, u_{2}(0)=1, u_{2}^{\prime}(0)=\left[\left(a_{0}-\gamma_{1} ight) ight) / a_{1} ight]=2. \end{aligned} [/latex]

Using the given Taylor's series method with [latex]h=0.25[/latex], we obtain (as done earlier)

[latex] \begin{aligned} u_{i, j+1} & =1.03125 u_{i, j}+0.25260 u_{i, j}^{\prime}-\left(0.13542 x_{j}+0.01042 ight) e^{x_{j}} \ u_{i, j+1}^{\prime} & =0.25 u_{i, j}+1.03125 u_{i, j}^{\prime}-2\left(0.5625 x_{j}+0.0625 ight) e^{x_{j}} \ i & =1,2 ext { and } j=0,1,2,3 . \end{aligned} [/latex]

Using the initial conditions, we obtain

[latex] \begin{array}{ll} u_{1}(0.25) \approx u_{1,1}=0.24218, & u_{1}^{\prime}(0.25) \approx u_{1,1}^{\prime}=0.90625, \ u_{1}(0.50) \approx u_{1,2}=0.42182, & u_{1}^{\prime}(0.50) \approx u_{1,2}^{\prime}=0.47348 ,\ u_{1}(0.75) \approx u_{1,3}=0.42579, & u_{1}^{\prime}(0.75) \approx u_{1,3}^{\prime}=-0.53976 ,\ u_{1}(1.00) \approx u_{1,4}=0.06568, & u_{1}^{\prime}(1.00) \approx u_{1,4}^{\prime}=-2.50102 ,\ u_{2}(0.25) \approx u_{2,1}=1.52603, & u_{2}^{\prime}(0.25) \approx u_{2,1}^{\prime}=2.18750, \ u_{2}(0.50) \approx u_{2,2}=2.06943, & u_{2}^{\prime}(0.50) \approx u_{2,2}^{\prime}=2.11573, \ u_{2}(0.75) \approx u_{2,3}=2.53972, & u_{2}^{\prime}(0.75) \approx u_{2,3}^{\prime}=1.56571, \ u_{2}(1.00) \approx u_{2,4}=2.77751, & u_{2}^{\prime}(1.00) \approx u_{2,4}^{\prime}=0.19872. \end{array} [/latex]

Using the boundary condition at [latex]x=1[/latex], we obtain, on using (5.98),

[latex] \lambda=\frac{\gamma_2-\left[b_0 \phi_2(b)+b_1 \phi_2^{\prime}(b) ight]}{\left[b_0 \phi_1(b)+b_1 \phi_1^{\prime}(b) ight]-\left[b_0 \phi_2(b)+b_1 \phi_2^{\prime}(b) ight]} [/latex] (5.98)

[latex] \lambda=\frac{-e-\left[u_{2}(1)+u_{2}^{\prime}(1) ight]}{\left[u_{1}(1)+u_{1}^{\prime}(1) ight]-\left[u_{2}(1)+u_{2}^{\prime}(1) ight]}=\frac{-5.69451}{-2.43534-2.97623}=1.05228. [/latex]

Hence, we have

[latex]u(x)=\lambda u_{1}(x)+(1-\lambda) u_{2}(x)=1.05228 u_{1}(x)-0.05228 u_{2}(x)[/latex]

Substituting [latex]x=0.25,0.5,0.75[/latex] and 1.0 , we get

[latex] \begin{aligned} & u(0.25) \approx 0.17506, u(0.50) \approx 0.33568, \ & u(0.75) \approx 0.31527, u(1.00) \approx-0.07609. \end{aligned} [/latex]

These values are same as given in the Table except for the round off error in the last digit.

Table 5.1. Solution of Problem 5.55.
$x_j$	Exact : $u(x_j)$	$u_j$
0.25	0.24075	0.17505
0.50	0.41218	0.33567
0.75	0.39694	0.31526
1.00	0.0	-0.07610

Question 5.55: Use the shooting method to solve the mixed boundary value pr......

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