Chapter 27, Problem 28P

A heated rod with a uniform heat source can be modeled with the Poisson equation,

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

Given a heat source $f\left(x\right)=25$ and the boundary conditions, $T\left(x=0\right)=40$ $\text{and }T\left(x=10\right)=200$, solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method $\left(\Delta x=2\right)$.

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 $f\left(x\right)=25$ and the boundary conditions, $T\left(x=0\right)=40$ and $T\left(x=10\right)=200$

, also $\Delta x=2$.

Answer to Problem 28P

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

$\begin{array}{ccc}x& T& z\\ 0& 40& 141\\ 2& 272& 91\\ 4& 404& 41\\ 6& 436& -9\\ 8& 368& -59\\ 10& 200& -109\end{array}$

Explanation of Solution

Given:

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

, where heat source $f\left(x\right)=25$ 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, $f\left(x\right)=25$

. Then,

$\frac{d^{2}T}{d{x}^2}=-25$

Also, $T\left(0\right)=40,T\left(10\right)=200$.

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

$\frac{dT}{dx}=z\text{and }\frac{d^{2}T}{d{x}^2}=\frac{dz}{dx}$

But $\frac{d^{2}T}{d{x}^2}=-25$

. Therefore,

$\frac{dT}{dx}=z\text{and }\frac{dz}{dx}=-25$

Now, 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

$\begin{array}{l}\frac{dT}{dx}=z\text{and }\frac{dz}{dx}=-25\\ T\left(0\right)=40,z\left(0\right)=-1\end{array}$

Now, solve the above system of differential equation

Then,

$\begin{array}{c}\frac{dz}{dx}=-25\\ dz=-25dx\end{array}$

Integrate on both the sides of the above differential equation

Then, $z\left(x\right)=-25x+C_1$

Where $C_1$ is constant of integration

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

. Then,

$\begin{array}{c}z\left(0\right)=-25\times 0+C_1\\ -1=-25\times 0+C_1\\ C_1=-1\end{array}$

Therefore, $z\left(x\right)=-25x-1$.

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

. But $z\left(x\right)=-25x-1$

. Thus,

$\begin{array}{c}\frac{dT}{dx}=-25x-1\\ dT=\left(-25x-1\right)dx\end{array}$

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

$T\left(x\right)=\left(-25\frac{x^2}{2}-1x\right)+C_2$

Where $C_2$

is constant of integration.

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

. Then,

$\begin{array}{c}T\left(0\right)=\left(-25\frac{{\left(0\right)}^2}{2}-1\times \left(0\right)\right)+C_2\\ 40=0+C_2\\ C_2=40\end{array}$

Therefore,

$T\left(x\right)=\left(-25\frac{x^2}{2}-x\right)+40$

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

. Thus,

$\begin{array}{c}T\left(10\right)=\left(-25\frac{{10}^2}{2}-10\right)+40\\ =-1250-10+40\\ =-1220\end{array}$

Hence, $T\left(10\right)=-1220$.

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,

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{dz}{dx}=-25\\ T\left(0\right)=40,z\left(0\right)=-0.5\end{array}$

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

$\begin{array}{c}\frac{dz}{dx}=-25\\ dz=-25dx\end{array}$

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

$z\left(x\right)=-25x+C_1$

Where $C_1$ is constant of integration

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

Then,

$\begin{array}{c}z\left(0\right)=-25\times 0+C_1\\ -0.5=-25\times 0+C_1\\ C_1=-0.5\end{array}$

Therefore, $z\left(x\right)=-25x-0.5$.

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

But $z\left(x\right)=-25x-0.5$

. Therefore,

$\begin{array}{c}\frac{dT}{dx}=-25x-0.5\\ dT=\left(-25x-0.5\right)dx\end{array}$

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

$T\left(x\right)=\left(-25\frac{x^2}{2}-0.5x\right)+C_2$

Where $C_2$

is constant of integration.

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

. Thus,

$\begin{array}{c}T\left(0\right)=\left(-25\frac{0^2}{2}-0.5\times 0\right)+C_2\\ 40=0+C_2\\ C_2=40\end{array}$

Therefore, $T\left(x\right)=\left(-25\frac{x^2}{2}-0.5x\right)+40$.

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

Then,

$\begin{array}{l}T\left(10\right)=\left(-25\frac{{10}^2}{2}-0.5\times 10\right)+40\\ =-1250-5+40\\ =-1215\end{array}$

Hence, $T\left(10\right)=-1215$.

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

corresponds to $T\left(10\right)=-1220$ and the second-guess value $z\left(0\right)=-0.5$

corresponds to $T\left(10\right)=-1215$

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

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

Then, by linear interpolation formula

$\begin{array}{c}z(0)=-1+\frac{-0.5+1}{-1215-\left(-1220\right)}\left(200-\left(-1220\right)\right)\\ =-1+\frac{0.5\times 1420}{5}\\ =-1+142\\ =141\end{array}$

Therefore, the right value of $z\left(0\right)$ which yields $T\left(10\right)=200$ is $z\left(0\right)=141$.

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

$\begin{array}{c}\frac{dT}{dx}=z\\ \frac{dz}{dx}=-25\\ T\left(0\right)=40,z\left(0\right)=141\end{array}$

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 $f\left(x_n,T_n,z_n\right)=z_n\text{ and }g\left(x_n,T_n,z_n\right)=-25$

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

And compute the values as shown below,

$\begin{array}{c}{k}_0=hf\left(x_0,T_0,z_0\right)=2\times z_0=2\times 141=282\\ l_0=hg\left(x_0,T_0,z_0\right)=2\times \left(-25\right)=-50\\ k_1=hf\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_0,z_0+\frac{1}{2}{l}_0\right)=2\left(z_0+\frac{1}{2}{l}_0\right)=2\left(141+\frac{1}{2}\left(-50\right)\right)=232\\ l_1=hg\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_0,z_0+\frac{1}{2}{l}_0\right)=2\left(-25\right)=-50\end{array}$

Also,

$\begin{array}{c}{k}_2=hf\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_1,z_0+\frac{1}{2}{l}_1\right)=2\left(z_0+\frac{1}{2}{l}_1\right)=2\left(141+\frac{1}{2}\left(-50\right)\right)=232\\ l_2=hg\left(x_0+\frac{1}{2}h,T_0+\frac{1}{2}{k}_1,z_0+\frac{1}{2}{l}_1\right)=2\times \left(-25\right)=-50\\ k_3=hf\left(x_0+h,T_0+k_2,z_0+l_2\right)=2\times \left(z_0+l_2\right)=2\left(141-50\right)=182\\ l_3=hg\left(x_0+h,T_0+k_2,z_0+l_2\right)=2\times \left(-25\right)=-50\end{array}$

Then,

$\begin{array}{c}{T}_1=T_0+\frac{1}{6}\left(k_0+2k_1+2k_2+k_3\right)\\ =40+\frac{1}{6}\left(282+464+464+182\right)\\ =272\end{array}$

And

$\begin{array}{c}{z}_1=z_0+\frac{1}{6}\left(l_0+2l_1+2l_2+l_3\right)\\ =141+\frac{1}{6}\left(-50-2\times 50-2\times 50-50\right)\\ =141-50\\ =91\end{array}$

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

. Thus,

$T_2=404\text{ and }{z}_2=41$

$T_3=436\text{ and }{z}_3=-9$

$T_4=368\text{ and }{z}_4=-59$

And

$T_5=200\text{ and }{z}_5=-109$

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

$\begin{array}{ccc}x& T& z\\ 0& 40& 141\\ 2& 272& 91\\ 4& 404& 41\\ 6& 436& -9\\ 8& 368& -59\\ 10& 200& -109\end{array}$

Hence, the required graph of the temperature distribution of the boundary value problem 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 $f\left(x\right)=25$ and the boundary conditions, $T\left(x=0\right)=40$ and $T\left(x=10\right)=200$

, also $\Delta x=2$.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

$\begin{array}{cc}x& T\\ 0& 40\\ 2& 272\\ 4& 404\\ 6& 436\\ 8& 368\\ 10& 200\end{array}$

Explanation of Solution

Given:

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

, where heat source $f\left(x\right)=25$ 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, $f\left(x\right)=25$

. Thus,

$\frac{d^{2}T}{d{x}^2}=-25$

Also, $T\left(0\right)=40,T\left(10\right)=200$.

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,

$\begin{array}{c}\frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}=-25\\ \frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}+25=0\\ T_{i+1}-2T_i+T_{i-1}+25\Delta x^2=0\end{array}$

Since, $\Delta x=2$

. Then,

$\begin{array}{c}{T}_{i+1}-2T_i+T_{i-1}+25{\left(2\right)}^2=0\\ T_{i+1}-2T_i+T_{i-1}+100=0\\ -T_{i-1}+2T_i-T_{i+1}=100\end{array}$

For the first node, $i=1$.

$\begin{array}{c}-T_{1-1}+2T_1-T_{1+1}=100\\ -T_0+2T_1-T_2=100\end{array}$

But, $T_0=40$

. Thus,

$\begin{array}{c}-40+2T_1-T_2=100\\ 2T_1-T_2=140\end{array}$

For the second node, $i=2$.

$\begin{array}{c}-T_{2-1}+2T_2-T_{3+1}=100\\ -T_1+2T_2-T_3=100\end{array}$

For the third node, $i=3$.

$\begin{array}{c}-T_{3-1}+2T_3-T_{3+1}=100\\ -T_2+2T_3-T_4=100\end{array}$

For the fourth node, $i=4$.

$\begin{array}{c}-T_{4-1}+2T_4-T_{4+1}=100\\ -T_3+2T_4-T_5=100\\ -T_3+2T_4-200=100\\ -T_3+2T_4=300\end{array}$

Then, write the above system in matrix form

$\left[\begin{array}{cccc}2& -1& 0& 0\\ -1& 2& -1& 0\\ 0& -1& 2& -1\\ 0& 0& -1& 2\end{array}\right]\left\{\begin{array}{c}{T}_1\\ T_2\\ T_3\\ T_4\end{array}\right\}=\left\{\begin{array}{c}140\\ 100\\ 100\\ 300\end{array}\right\}$

Now, use the Gauss-Seidel iterative technique. The Gauss-Seidel iterative method is given by,

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

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

$\begin{array}{cc}x& T\\ 0& 40\\ 2& 272\\ 4& 404\\ 6& 436\\ 8& 368\\ 10& 200\end{array}$

Therefore, the graph of temperature distribution is,