Chapter 27, Problem 29P

Repeat Prob. 27.28, but for the following heat source: $f\left(x\right)=0.12x^3-2.4x^2+12x$.

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, $\frac{d^{2}T}{d{x}^2}=-f\left(x\right)$

, where heat source $f\left(x\right)=0.12x^3-2.4x^2+12x$ and the boundary conditions, $T\left(x=0\right)=40$ and $T\left(x=10\right)=200$

, also $\Delta x=2$.

Answer to Problem 29P

Solution: The table of the solution of the boundary value problem is,

$\begin{array}{ccc}x& T& z\\ 0& 40& 76\\ 2& 179.04& 57.92\\ 4& 261.056& 23.52\\ 6& 276.544& -6.08\\ 8& 246.592& -21.28\\ 10& 200& -24\end{array}$

Explanation of Solution

Given:

A differential equation, $\frac{d^{2}T}{d{x}^2}=-f\left(x\right)$

, where heat source $f\left(x\right)=0.12x^3-2.4x^2+12x$ and the boundary conditions, $T\left(x=0\right)=40$ and $T\left(x=10\right)=200$

, also $\Delta x=2$.

Formula used:

Linear-interpolation formula:

$f_1\left(x\right)=f\left(x_0\right)+\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\left(x-x_0\right)$

Calculation:

Consider the following Poisson equation, $\frac{d^{2}T}{d{x}^2}=-f\left(x\right)$.

Since, $f\left(x\right)=0.12x^3-2.4x^2+12x$

. Then,

$\begin{array}{c}\frac{d^{2}T}{d{x}^2}=-\left(0.12x^3-2.4x^2+12x\right)\\ T\left(0\right)=40,T\left(10\right)=200\end{array}$

Now, change the above boundary value problem into equivalent initial-value problem. Then,

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{d^{2}T}{d{x}^2}=\frac{dz}{dx}\end{array}$

But $\frac{d^{2}T}{d{x}^2}=-\left(0.12x^3-2.4x^2+12x\right)$.

Thus,

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{dz}{dx}=-\left(0.12x^3-2.4x^2+12x\right)\end{array}$

Use the shooting method in the above system of first order linear differential equation

Suppose, $z\left(0\right)=-1$ and given that $T\left(0\right)=40$.

Then, the system of system of first order linear differential equation with initial condition is,

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{dz}{dx}=-\left(0.12x^3-2.4x^2+12x\right)\\ T\left(0\right)=40,z\left(0\right)=-1\end{array}$

Now, solve the above system of differential equation.

Thus,

$\begin{array}{c}\frac{dz}{dx}=-\left(0.12x^3-2.4x^2+12x\right)\\ dz=-\left(0.12x^3-2.4x^2+12x\right)dx\end{array}$

Integrate on both the sides of the above differential equation, to get

$\begin{array}{c}z\left(x\right)=-\left(0.12\frac{x^4}{4}-2.4\frac{x^3}{3}+12\frac{x^2}{2}\right)+C_1\\ =-\left(0.03x^4-0.8x^3+6x^2\right)+C_1\end{array}$

Where $C_1$ is constant of integration.

Now, use the initial condition, $z\left(0\right)=-1$

. Thus,

$\begin{array}{c}z\left(0\right)=-\left(0.03\times 0^4-0.8\times 0^3+6\times 0^2\right)+C_1\\ -1=0+C_1\\ C_1=-1\end{array}$

Put the value of $C_1$ in $z\left(x\right)=-\left(0.03x^4-0.8x^3+6x^2\right)+C_1$.

Thus,

$z\left(x\right)=-\left(0.03x^4-0.8x^3+6x^2\right)-1$

Since, $\frac{dT}{dx}=z$

. But $z\left(x\right)=-\left(0.03x^4-0.8x^3+6x^2\right)-1$.

Therefore,

$\begin{array}{c}\frac{dT}{dx}=-\left(0.03x^4-0.8x^3+6x^2\right)-1\\ dT=\left[-\left(0.03x^4-0.8x^3+6x^2\right)-1\right]dx\end{array}$

Integrate on both the sides of the above differential equation, to get

$T\left(x\right)=\left[-\left(0.03\frac{x^5}{5}-0.8\frac{x^4}{4}+6\frac{x^3}{3}\right)-x\right]+C_2$

Where $C_2$

is constant of integration.

Use the initial condition, $T\left(0\right)=40$

. Then,

$\begin{array}{c}T\left(0\right)=\left[-\left(0.03\frac{0^5}{5}-0.8\frac{0^4}{4}+6\frac{0^3}{3}\right)-0\right]+C_2\\ 40=0+C_2\\ C_2=40\end{array}$

Put the value of $C_2$ in $T\left(x\right)$

, then

$T\left(x\right)=\left[-\left(0.03\frac{x^5}{5}-0.8\frac{x^4}{4}+6\frac{x^3}{3}\right)-x\right]+40$

Now, evaluate the above for $T\left(10\right)$

. Thus,

$\begin{array}{c}T\left(10\right)=\left[-\left(0.03\frac{{10}^5}{5}-0.8\frac{{10}^4}{4}+6\frac{{10}^3}{3}\right)-10\right]+40\\ =\left(-600+2000-2000-10\right)+40\\ =-570\end{array}$

But the above of $T\left(10\right)$ is much different from the boundary condition of $T\left(10\right)=200$.

Then, put another guess. Suppose $z\left(0\right)=-0.5$.

Then, the system of system of first order linear differential equation with initial condition is,

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{dz}{dx}=-\left(0.12x^3-2.4x^2+12x\right)\\ T\left(0\right)=40,z\left(0\right)=-0.5\end{array}$

Now, solve the above system of differential equation. Then,

$\begin{array}{c}\frac{dz}{dx}=-\left(0.12x^3-2.4x^2+12x\right)\\ dz=-\left(0.12x^3-2.4x^2+12x\right)dx\end{array}$

Integrate on both the sides of the above differential equation

Thus,

$\begin{array}{c}z\left(x\right)=-\left(0.12\frac{x^4}{4}-2.4\frac{x^3}{3}+12\frac{x^2}{2}\right)+C_3\\ =-\left(0.03x^4-0.8x^3+6x^2\right)+C_3\end{array}$

Where $C_3$ is constant of integration.

Now, use the initial condition, $z\left(0\right)=-0.5$.

Therefore,

$\begin{array}{c}z\left(0\right)=-\left(0.03\times 0^4-0.8\times 0^3+6\times 0^2\right)+C_3\\ -0.5=0+C_3\\ C_3=-0.5\end{array}$

Put the value of $C_3$ in $z\left(x\right)$, then

$z\left(x\right)=-\left(0.03x^4-0.8x^3+6x^2\right)-0.5$

Since, $\frac{dT}{dx}=z$

. But $z\left(x\right)=-\left(0.03x^4-0.8x^3+6x^2\right)-0.5$.

Thus,

$\begin{array}{c}\frac{dT}{dx}=-\left(0.03x^4-0.8x^3+6x^2\right)-0.5\\ dT=\left[-\left(0.03x^4-0.8x^3+6x^2\right)-0.5\right]dx\end{array}$

Integrate on both the sides of the above differential equation. Then,

$T\left(x\right)=\left[-\left(0.03\frac{x^5}{5}-0.8\frac{x^4}{4}+6\frac{x^3}{3}\right)-0.5x\right]+C_4$

Where $C_4$ is constant of integration.

Use the initial condition, $T\left(0\right)=40$

. Thus,

$\begin{array}{c}T\left(0\right)=\left[-\left(0.03\frac{0^5}{5}-0.8\frac{0^4}{4}+6\frac{0^3}{3}\right)-0.5\times 0\right]+C_4\\ 40=0+C_4\\ C_4=40\end{array}$

Put the value of $C_4$ in $T\left(x\right)$, then

$T\left(x\right)=\left[-\left(0.03\frac{x^5}{5}-0.8\frac{x^4}{4}+6\frac{x^3}{3}\right)-0.5x\right]+40$

Now, evaluate the above for $T\left(10\right)$

. Thus,

$\begin{array}{c}T\left(10\right)=\left[-\left(0.03\frac{{10}^5}{5}-0.8\frac{{10}^4}{4}+6\frac{{10}^3}{3}\right)-0.5\times 10\right]+40\\ =\left(-600+2000-2000-5\right)+40\\ =-565\end{array}$

Since, the first guess value $z\left(0\right)=-1$

corresponds to $T\left(10\right)=-570$

and the second-guess value $z\left(0\right)=-0.5$

corresponds to $T\left(10\right)=-565$.

Now, use these values to compute the value of $z\left(0\right)$

that yields $T\left(10\right)=200$.

Then, by linear interpolation formula,

$\begin{array}{c}z(0)=-1+\frac{-0.5+1}{-565-\left(-570\right)}\left(200-\left(-570\right)\right)\\ =-1+\frac{0.5\times 770}{5}\\ =-1+77\\ =76\end{array}$

Therefore, the right value of $z\left(0\right)$

which yields $T\left(10\right)=200$ is $z\left(0\right)=76$.

Then, the equivalent initial value problem corresponding to the boundary value problem is,

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{dz}{dx}=-\left(0.12x^3-2.4x^2+12x\right)\\ T\left(0\right)=40,z\left(0\right)=76\end{array}$

Now, use the fourth order RK method with step size $h=2$.

The RK method for above system of first order linear differential equation with initial condition is,

$\begin{array}{c}{T}_{n+1}=T_n+\frac{1}{6}\left(k_0+2k_1+2k_2+k_3\right)\\ z_{n+1}=z_n+\frac{1}{6}\left(l_0+2l_1+2l_2+l_3\right)\end{array}$

Where $x_{n+1}=x_n+nh$

And

$\begin{array}{c}{k}_0=hf\left(x_n,T_n,z_n\right)\\ k_1=hf\left(x_n+\frac{1}{2}h,T_n+\frac{1}{2}{k}_0,z_n+\frac{1}{2}{l}_0\right)\\ k_2=hf\left(x_n+\frac{1}{2}h,T_n+\frac{1}{2}{k}_1,z_n+\frac{1}{2}{l}_1\right)\\ k_3=hf\left(x_n+h,T_n+k_2,z_n+l_2\right)\end{array}$

And

$\begin{array}{c}{l}_0=hg\left(x_n,T_n,z_n\right)\\ l_1=hg\left(x_n+\frac{1}{2}h,T_n+\frac{1}{2}{k}_0,z_n+\frac{1}{2}{l}_0\right)\\ l_2=hg\left(x_n+\frac{1}{2}h,T_n+\frac{1}{2}{k}_1,z_n+\frac{1}{2}{l}_1\right)\\ l_3=hg\left(x_n+h,T_n+k_2,z_n+l_2\right)\end{array}$

Where $f\left(x_n,T_n,z_n\right)=z_n\text{ and }g\left(x_n,T_n,z_n\right)=-\left(0.12x^3-2.4x^2+12x\right)$.

Then, for $n=0$

$\begin{array}{c}{x}_1=x_0+0\times 2\\ =0\end{array}$

And

$\begin{array}{c}{k}_0=hf\left(x_0,T_0,z_0\right)=2\times z_0=2\times 76=152\\ l_0=hg\left(x_0,T_0,z_0\right)=2\times \left[-\left(0.12x_0{}^3-2.4x_0{}^2+12x_0\right)\right]=0\\ k_1=hf\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_0,z_0+\frac{1}{2}{l}_0\right)=2\left(z_0+\frac{1}{2}{l}_0\right)=2\left(76+\frac{1}{2}\left(0\right)\right)=152\\ l_1=hg\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_0,z_0+\frac{1}{2}{l}_0\right)=-2\left[\begin{array}{l}0.12{\left(x_0+\frac{1}{2}\times 2\right)}^3-2.4{\left(x_0+\frac{1}{2}\times 2\right)}^2\\ +12\left(x_0+\frac{1}{2}\times 2\right)\end{array}\right]=-19.44\end{array}$

Also,

$\begin{array}{c}{k}_2=hf\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_1,z_0+\frac{1}{2}{l}_1\right)=2\left(z_0+\frac{1}{2}{l}_1\right)=2\left(76+\frac{1}{2}\left(-19.44\right)\right)=132.56\\ l_2=hg\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_1,z_0+\frac{1}{2}{l}_1\right)=-2\left[\begin{array}{l}0.12{\left(x_0+\frac{1}{2}\times 2\right)}^3-2.4{\left(x_0+\frac{1}{2}\times 2\right)}^2\\ +12\left(x_0+\frac{1}{2}\times 2\right)\end{array}\right]=-19.44\\ k_3=hf\left(x_0+h,T_0+k_2,z_0+l_2\right)=2\times \left(z_0+l_2\right)=2\left(76-19.44\right)=113.12\\ l_3=hg\left(x_0+h,T_0+k_2,z_0+l_2\right)=-2\times \left[0.12{\left(x_0+2\right)}^3-2.4{\left(x_0+h\right)}^2+12\left(x_0+h\right)\right]=-30.72\end{array}$

Thus,

$\begin{array}{c}{T}_1=T_0+\frac{1}{6}\left(k_0+2k_1+2k_2+k_3\right)\\ =40+\frac{1}{6}\left(152+304+265.12+113.12\right)\\ =40+139.04\\ =179.04\end{array}$

And

$\begin{array}{c}{z}_1=z_0+\frac{1}{6}\left(l_0+2l_1+2l_2+l_3\right)\\ =76+\frac{1}{6}\left(0-38.88-38.88-30.72\right)\\ =57.92\end{array}$

In the similar way, find the remaining $T_n\text{ and }{z}_n$

. Then,

$T_2=261.056\text{ and }{z}_2=23.52$

$T_3=276.544\text{ and }{z}_3=-6.08$

$T_4=246.592\text{ and }{z}_4=-21.28$

And

$T_5=200\text{ and }{z}_5=-24$

Therefore, the table of the solution of the boundary value problem is

$\begin{array}{ccc}x& T& z\\ 0& 40& 76\\ 2& 179.04& 57.92\\ 4& 261.056& 23.52\\ 6& 276.544& -6.08\\ 8& 246.592& -21.28\\ 10& 200& -24\end{array}$

Hence, the graph of the temperature distribution is

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, $\frac{d^{2}T}{d{x}^2}=-f\left(x\right)$

, where heat source $f\left(x\right)=0.12x^3-2.4x^2+12x$ and the boundary conditions, $T\left(x=0\right)=40$ and $T\left(x=10\right)=200$

, also $\Delta x=2$.

Answer to Problem 29P

Solution:

The table of the solution of boundary value problem is

$\begin{array}{cc}x& T\\ 0& 40\\ 2& 184.128\\ 4& 266.816\\ 6& 280.384\\ 8& 247.872\\ 10& 200\end{array}$

Explanation of Solution

Given:

A differential equation, $\frac{d^{2}T}{d{x}^2}=-f\left(x\right)$

, where heat source $f\left(x\right)=0.12x^3-2.4x^2+12x$ and the boundary conditions, $T\left(x=0\right)=40$ and $T\left(x=10\right)=200$

, also $\Delta x=2$.

Formula used:

(1) The finite difference method is:

$\frac{d^{2}T}{d{x}^2}=\frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}$

(2) The Gauss-Seidel iterative method is:

$T_i^{\left(k\right)}=\frac{1}{a_{ii}}\left[-{\displaystyle \sum _{j=1}^{i-1}\left(a_{ij}{T}_j^{\left(k\right)}\right)}-{\displaystyle \sum _{j=i+1}^n\left(a_{ij}{T}_j^{\left(k-1\right)}\right)}+b_i\right]$

Calculation:

Consider the following Poisson equation, $\frac{d^{2}T}{d{x}^2}=-f\left(x\right)$.

Since, $f\left(x\right)=0.12x^3-2.4x^2+12x$.

Thus,

$\begin{array}{c}\frac{d^{2}T}{d{x}^2}=-\left(0.12x^3-2.4x^2+12x\right)\\ T\left(0\right)=40,T\left(10\right)=200\end{array}$

The finite difference method is given by,

$\frac{d^{2}T}{d{x}^2}=\frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}$

Now, substitute the value of second order derivative in the boundary value problem.

Then, the boundary value problem becomes,

$\begin{array}{c}\frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}=-\left(0.12x_i{}^3-2.4x_i^2+12x_i\right)\\ T_{i+1}-2T_i+T_{i-1}=-\Delta x^2\left(0.12x_i{}^3-2.4x_i^2+12x_i\right)\end{array}$

Or

$-T_{i-1}+2T_i-T_{i+1}=\Delta x^2\left(0.12x_i{}^3-2.4x_i^2+12x_i\right)$

Since, $\Delta x=2$

. Then,

$\begin{array}{l}-T_{i-1}+2T_i-T_{i+1}={\left(2\right)}^2\left(0.12x_i{}^3-2.4x_i^2+12x_i\right)\\ -T_{i-1}+2T_i-T_{i+1}=0.48x_i{}^3-9.6x_i^2+48x_i\end{array}$

For the first node, $i=1$.

$\begin{array}{c}-T_{1-1}+2T_1-T_{1+1}=0.48x_1{}^3-9.6x_1^2+48x_1\\ -T_0+2T_1-T_{1+1}=0.48x_1{}^3-9.6x_1^2+48x_1\\ -40+2T_1-T_{1+1}=0.48x_1{}^3-9.6x_1^2+48x_1\\ 2T_1-T_2=101.44\end{array}$

For the second node, $i=2$.

$\begin{array}{c}-T_{2-1}+2T_2-T_{2+1}=0.48x_2{}^3-9.6x_2^2+48x_2\\ -T_1+2T_2-T_3=69.12\end{array}$

For the third node, $i=3$.

$\begin{array}{c}-T_{3-1}+2T_3-T_{3+1}=0.48x_3{}^3-9.6x_3^2+48x_3\\ -T_2+2T_3-T_4=46.08\end{array}$

For the fourth node, $i=4$.

$\begin{array}{c}-T_{4-1}+2T_4-T_{4+1}=0.48x_4{}^3-9.6x_4^2+48x_4\\ -T_3+2T_4-200=0.48x_4{}^3-9.6x_4^2+48x_4\\ -T_3+2T_4=215.36\end{array}$

Then, write the system of equations in matrix form

$\left[\begin{array}{cccc}2& -1& 0& 0\\ -1& 2& -1& 0\\ 0& -1& 2& -1\\ 0& 0& -1& 2\end{array}\right]\left\{\begin{array}{c}{T}_1\\ T_2\\ T_3\\ T_4\end{array}\right\}=\left\{\begin{array}{c}101.44\\ 69.12\\ 46.08\\ 215.36\end{array}\right\}$

Since, the coefficient matrix is tridiagonal matrix, then use Gauss-Seidel iterative technique

The Gauss-Seidel iterative method is,

$T_i^{\left(k\right)}=\frac{1}{a_{ii}}\left[-{\displaystyle \sum _{j=1}^{i-1}\left(a_{ij}{T}_j^{\left(k\right)}\right)}-{\displaystyle \sum _{j=i+1}^n\left(a_{ij}{T}_j^{\left(k-1\right)}\right)}+b_i\right]$

Now, evaluate $T_i$

by above Gauss-Seidel method

Then,

$T_1=184.128$

$T_2=266.816$

$T_3=280.384$

And

$T_4=247.872$

Then, the table of the solution of boundary value problem is

$\begin{array}{cc}x& T\\ 0& 40\\ 2& 184.128\\ 4& 266.816\\ 6& 280.384\\ 8& 247.872\\ 10& 200\end{array}$

Therefore, the graph of temperature distribution is