LINEAR BVP Solve the following BVP using the shooting method. [latex]\overset{..}{u}= 0.02u+ 1, u(0)=10, u(10)=100, \overset{..}{u}=\frac{d^{2}u }{dt^{2} } [/latex]

Since the ODE is second order, there are two state variables: [latex]x_1 = u, x_2 = \overset{.}{u}[/latex] . The state-variable equations are then formed as [latex] \left\{ \begin{matrix} \overset{.}{x_1}=x_2 \\ \overset{.}{x_2}= 0.02x_1+ 1 \end{matrix}, \begin{matrix} x_1(0)=10 \\ x_2(0)= ? \end{matrix}\right . [/latex] (8.1) This system could be solved numerically via RK4 but that would require initial conditions [latex]x_1(0)[/latex] and [latex]x_2(0)[/latex]. Of these two, however, only [latex]x_1(0) = 10[/latex] is available. Note that there is a unique value of [latex]x_2(0)[/latex] for which the above system yields the solution of the original BVP. Strategy to Find [latex]x_2(0)[/latex] We first guess a value for [latex]x_2(0)[/latex] and solve the system in Equation 8.1 using RK4. From the solution, we extract the first state variable [latex]x_1 [/latex] which happens to be u. The value of this variable at the right end (namely, u(10)) is then compared with the boundary condition at that end, that is, u(10) = 100, which will serve as the target here. Next, we guess another value for [latex]x_2(0)[/latex] and go through the process a second time. This way, for each guess of [latex]x_2(0)[/latex], we find a value for u(10). Because the original ODE is linear, these values are linearly related as well. As a result, a linear interpolation of this data will provide the unique value for [latex]x_2(0)[/latex] that will result in the correct solution. As a first guess, we arbitrarily choose [latex]x_2(0)[/latex] = 10 so that Equation 8.1 becomes [latex] \left\{ \begin{matrix} \overset{.}{x_1}=x_2 \\ \overset{.}{x_2}= 0.02x_1 + 1 \end{matrix}, \begin{matrix} x_1(0)=10 \\ \boxed{x_2(0)=10} \end{matrix} \right . [/latex]

Question 8.1: LINEAR BVP Solve the following BVP using the shooting method...

Numerical Methods for Engineers and Scientists Using MATLAB®

LINEAR BVP

Solve the following BVP using the shooting method.

\overset{..}{u}= 0.02u+ 1,   u(0)=10,   u(10)=100,  \overset{..}{u}=\frac{d^{2}u }{dt^{2} }

The 'Blue Check Mark' means that either the MATLAB code/script/answer provided in the answer section has been tested by our team of experts; or the answer in general has be fact checked.

Learn more on how do we answer questions.

Since the ODE is second order, there are two state variables: $x_1 = u, x_2 = \overset{.}{u}$ . The state-variable equations are then formed as

$\left\{ \begin{matrix} \overset{.}{x_1}=x_2 \\ \overset{.}{x_2}= 0.02x_1+ 1 \end{matrix}, \begin{matrix} x_1(0)=10 \\ x_2(0)= ? \end{matrix}\right .$ (8.1)

This system could be solved numerically via RK4 but that would require initial conditions $x_1(0)$ and $x_2(0)$ . Of these two, however, only $x_1(0) = 10$ is available. Note that there is a unique value of $x_2(0)$ for which the above system yields the solution of the original BVP.

Strategy to Find $x_2(0)$
We first guess a value for $x_2(0)$ and solve the system in Equation 8.1 using RK4. From the solution, we extract the first state variable $x_1$ which happens to be u. The value of this variable at the right end (namely, u(10)) is then compared with the boundary condition at that end, that is, u(10) = 100, which will serve as the target here. Next, we guess another value for $x_2(0)$ and go through the process a second time. This way, for each guess of $x_2(0)$ , we find a value for u(10). Because the original ODE is linear, these values are linearly related as well. As a result, a linear interpolation of this data will provide the unique value for $x_2(0)$ that will result in the correct solution. As a first guess, we arbitrarily choose $x_2(0)$ = 10 so that Equation 8.1 becomes

\left\{ \begin{matrix} \overset{.}{x_1}=x_2 \\ \overset{.}{x_2}= 0.02x_1 +  1 \end{matrix},    \begin{matrix} x_1(0)=10 \\ \boxed{x_2(0)=10} \end{matrix} \right .

> f = @(t,x)([x(2);0.02*x(1)+1]);
>> x0_first = [10;10];
>> t = linspace(0,10,20);
>> x_first = RK4System(f,t,x0_first);
>> u_10_first = x_first(1,end)
u_10_first =
217.5208

The result overshoots the target u(10) = 100. As a second guess, we pick $x_2(0) = 0$ so that Equation 8.1 becomes

$\left\{ \begin{matrix} \overset{.}{x_1}=x_2 \\ \overset{.}{x_2}= 0.02x_1 + 1 \end{matrix}, \begin{matrix} x_1(0)=10 \\ \boxed{x_2(0)=10} \end{matrix} \right .$

Note that f and t in our MATLAB code remain unchanged; only the initial state vector is modified.

>> x0_second = [10;0];
>> x_second = RK4System(f,t,x0_second);
>> u_10_second = x_second(1,end)
u_10_second =
80.6910

This time the result is below the target. In summary,

$\begin{matrix} x_2(0)=10 & u(10)=217.5208 \\ x_2(0)=0 & u(10)=80.6910 \end{matrix}$ (8.2)

Linear interpolation of this data will lead to the correct value for $x_2(0)$ ; see Figure 8.1. We will handle this by using linear Lagrange interpolation (Chapter 5), more specifically, the user-defined function LagrangeInterp:

>> yi = LagrangeInterp([u_10_first, u_10_second],[10, 0],100)
yi =
1.4112

Therefore, $x_2(0) = 1.4112$ . Using this information in Equation 8.1, we have

$\left\{ \begin{matrix} \overset{.}{x_1}=x_2 \\ \overset{.}{x_2}= 0.02x_1 + 1 \end{matrix}, \begin{matrix} x_1(0)=10 \\ x_2(0)=1.4112 \end{matrix} \right .$

>> x0_final = [10;yi];
>> x_final = RK4System(f,t,x0_final);
>> x_final(1,end)
ans =
100.0000

The numerical solution u to the original BVP is obtained by extracting the first row of x_final, which contains 20 values of the first state variable $x_1 = u$ . In order to generate a smooth solution trajectory, however, we should use at least 100 points, as opposed to the current 20 points. Similarly, the first row of x_first gives the solution corresponding to the first guess of $x_2(0)$ , and the first row of x_second gives the solution corresponding to the second guess of $x_2(0)$ .
This is summarized in the script below.

>> t = linspace(0,10); % 100 points
>> x_first = RK4System(f,t,x0_first);
>> u_first = x_first(1,:); % Solution based on the first guess
>> x_second = RK4System(f,t,x0_second);
>> u_second = x_second(1,:); % Solution based on the second guess
>> x_final = RK4System(f,t,x0_final);
>> u_final = x_final(1,:); % True solution
>> plot(t,u_first,t,u_second,t,u_final) % Figure 8.2

A linear BVP such as in Example 8.1 is relatively easy to solve using the shooting method because the data generated by two initial-value estimates can be interpolated linearly leading to the correct estimate. This, however, will not be enough in the case of a nonlinear BVP. One remedy would be to apply the shooting method three times and then use a quadratic interpolating polynomial to estimate the initial value. But it is not very likely that this approach would generate the correct solution, and further iterations would probably be required. A more viable option is the following: guess a value for the missing initial condition, solve the ensuing system, and find the value of the solution at the right end. This is either above the unused boundary condition (target) or below it. If it is above the target, we must pick a second guess that leads to a value below the target. If it is below the target, we must pick a second guess that leads to a value above the target. Subsequently, the bisection method (Chapter 3) is employed to find the unique value of the missing initial condition that will result in the true solution. Of course, the bisection method uses iterations that terminate when a tolerance is met. Therefore, the precision of the final outcome is directly dependent on the magnitude of the selected tolerance. The example that follows demonstrates the details of this approach.