A heated rod with a uniform heat source can be modeled with the Poisson equation, d 2 T d x 2 = − f ( x ) Given a heat source f ( x ) = 25 and the boundary conditions, T ( x = 0 ) = 40 and T ( x = 10 ) = 200 , solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method ( Δ x = 2 ) .

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Answer 1

Textbook Question

Chapter 27, Problem 28P

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

d 2 T d x 2 = − f ( x )

Given a heat source f ( x ) = 25 and the boundary conditions, T ( x = 0 ) = 40 and T ( x = 10 ) = 200 , solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method ( Δ x = 2 ) .

(a)

Expert Solution

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

(b)

Expert Solution

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

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Answer 2

Textbook Question

Chapter 27, Problem 28P

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

d 2 T d x 2 = − f ( x )

Given a heat source f ( x ) = 25 and the boundary conditions, T ( x = 0 ) = 40 and T ( x = 10 ) = 200 , solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method ( Δ x = 2 ) .

(a)

Expert Solution

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

(b)

Expert Solution

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

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Answer 3

Textbook Question

Chapter 27, Problem 28P

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

d 2 T d x 2 = − f ( x )

Given a heat source f ( x ) = 25 and the boundary conditions, T ( x = 0 ) = 40 and T ( x = 10 ) = 200 , solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method ( Δ x = 2 ) .

(a)

Expert Solution

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

(b)

Expert Solution

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

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Answer 4

Textbook Question

Chapter 27, Problem 28P

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

d 2 T d x 2 = − f ( x )

Given a heat source f ( x ) = 25 and the boundary conditions, T ( x = 0 ) = 40 and T ( x = 10 ) = 200 , solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method ( Δ x = 2 ) .

(a)

Expert Solution

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

(b)

Expert Solution

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

(b)

Expert Solution

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

Answer 8

(a)

Expert Solution

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer 13

Answer to Problem 28P

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

xTz0401412272914404416436−98368−5910200−109

Answer 14

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Then,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

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

dTdx=z and d2Tdx2=dzdx

But d2Tdx2=−25

. Therefore,

dTdx=z and dzdx=−25

Now, use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=−1 and given that T(0)=40.

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

dTdx=z and dzdx=−25T(0)=40, z(0)=−1

Now, solve the above system of differential equation

Then,

dzdx=−25dz=−25dx

Integrate on both the sides of the above differential equation

Then, z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−1

. Then,

z(0)=−25×0+C1−1=−25×0+C1C1=−1

Therefore, z(x)=−25x−1.

Since, dTdx=z

. But z(x)=−25x−1

. Thus,

dTdx=−25x−1dT=(−25x−1)dx

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

T(x)=(−25x22−1x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=(−25(0)22−1×(0))+C240=0+C2C2=40

Therefore,

T(x)=(−25x22−x)+40

Now, evaluate the above for T(10)

. Thus,

T(10)=(−251022−10)+40=−1250−10+40=−1220

Hence, T(10)=−1220.

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

Then, put another guess. Suppose z(0)=−0.5.

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

dTdx=zdzdx=−25T(0)=40,z(0)=−0.5

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

dzdx=−25dz=−25dx

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

z(x)=−25x+C1

Where C1 is constant of integration

Now, use the initial condition, z(0)=−0.5.

Then,

z(0)=−25×0+C1−0.5=−25×0+C1C1=−0.5

Therefore, z(x)=−25x−0.5.

Since, dTdx=z.

But z(x)=−25x−0.5

. Therefore,

dTdx=−25x−0.5dT=(−25x−0.5)dx

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

T(x)=(−25x22−0.5x)+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=(−25022−0.5×0)+C240=0+C2C2=40

Therefore, T(x)=(−25x22−0.5x)+40.

Now, evaluate the above for T(10).

Then,

T(10)=(−251022−0.5×10)+40=−1250−5+40=−1215

Hence, T(10)=−1215.

Since, the first guess value z(0)=−1

corresponds to T(10)=−1220 and the second-guess value z(0)=−0.5

corresponds to T(10)=−1215

, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula

z(0)=−1+−0.5+1−1215−(−1220)(200−(−1220))=−1+0.5×14205=−1+142=141

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

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

dTdx=zdzdx=−25T(0)=40,z(0)=141

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,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=−25

Then, for n=0, x1=x0+0×2=0

And compute the values as shown below,

k0=hf(x0,T0,z0)=2×z0=2×141=282l0=hg(x0,T0,z0)=2×(−25)=−50k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(141+12(−50))=232l1=hg(x0+12h,T0+12k0,z0+12l0)=2(−25)=−50

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(141+12(−50))=232l2=hg(x0+12h,T0+12k1,z0+12l1)=2×(−25)=−50k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(141−50)=182l3=hg(x0+h,T0+k2,z0+l2)=2×(−25)=−50

Then,

T1=T0+16(k0+2k1+2k2+k3)=40+16(282+464+464+182)=272

And

z1=z0+16(l0+2l1+2l2+l3)=141+16(−50−2×50−2×50−50)=141−50=91

In the similar way, find the remaining Tn and zn

. Thus,

T2=404 and z2=41

T3=436 and z3=−9

T4=368 and z4=−59

And

T5=200 and z5=−109

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

xTz0401412272914404416436−98368−5910200−109

Hence, the required graph of the temperature distribution of the boundary value problem is,

Answer 15

(b)

Expert Solution

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer 20

Answer to Problem 28P

Solution:

The table of the solution of boundary value problem is

xT040227244046436836810200

Answer 21

Explanation of Solution

Given:

A differential equation, d2Tdx2=−f(x)

, where heat source f(x)=25 and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=−f(x).

Since, f(x)=25

. Thus,

d2Tdx2=−25

Also, T(0)=40,T(10)=200.

The finite difference method is given by,

d2Tdx2=Ti+1−2Ti+Ti−1Δx2

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

Then, the boundary value problem becomes,

Ti+1−2Ti+Ti−1Δx2=−25Ti+1−2Ti+Ti−1Δx2+25=0Ti+1−2Ti+Ti−1+25Δx2=0

Since, Δx=2

. Then,

Ti+1−2Ti+Ti−1+25(2)2=0Ti+1−2Ti+Ti−1+100=0−Ti−1+2Ti−Ti+1=100

For the first node, i=1.

−T1−1+2T1−T1+1=100−T0+2T1−T2=100

But, T0=40

. Thus,

−40+2T1−T2=1002T1−T2=140

For the second node, i=2.

−T2−1+2T2−T3+1=100−T1+2T2−T3=100

For the third node, i=3.

−T3−1+2T3−T3+1=100−T2+2T3−T4=100

For the fourth node, i=4.

−T4−1+2T4−T4+1=100−T3+2T4−T5=100−T3+2T4−200=100−T3+2T4=300

Then, write the above system in matrix form

[ 2−100−12−100−12−100−12 ]{ T1T2T3T4 }={ 140100100300 }

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

Ti(k)=1aii[ −∑j=1i−1(aijTj(k))−∑j=i+1n(aijTj(k−1))+bi ]

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

xT040227244046436836810200

Therefore, the graph of temperature distribution is,

Answer 22

Ch. 27 - A steady-state heat balance for a rod can be...Ch. 27 - 27.2 Use the shooting method to solve Prob. 27.1....Ch. 27 - 27.3 Use the finite-difference approach with to...Ch. 27 - 27.4 Use the shooting method to solve Ch. 27 - Solve Prob. 27.4 with the finite-difference...Ch. 27 - 27.7 Differential equations like the one solved...Ch. 27 - 27.8 Repeat Example 27.4 but for three masses....Ch. 27 - 27.9 Repeat Example 27.6, but for five interior...Ch. 27 - Use minors to expand the determinant of...Ch. 27 - 27.11 Use the power method to determine the...

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