Practice_Problems_Weeks_2-13_2024

17. Suppose the interpolation shown in Q8 was obtained using a series of second-order Lagrange polynomial fits. For how many of the points would you expect there to be an undefined gradient? 18. Which of the following statements is true (select any that apply): A. Least squares fitting is very useful for fitting a line or low-order polynomial to noisy data in order to estimate the underlying signal. B. Least squares fitting is typically not reliable for high-order fits to noisy data. C. A low-order least squares fit is always a better option than a low-order (e.g. cubic) spline since it assumes data are noisy. D. Piecewise constant, piecewise linear or second-order Lagrange interpolation would be perfectly fine if we only want to integrate a set of data points and are not interested in computing gradients. x U

FURTHER EXAMPLES WITH SIMULATIONS Now that you’ve done a few examples to practice using the Euler formulae to obtain Fourier series coefficients for odd and even functions, we’ll demonstrate a few more examples; these examples are used in the FOURIER DEMOS which you can download from our Canvas site. You’ll get the most out of these examples by running the FOURIER DEMOS alongside so that you can visualise what is happening. EXAMPLE 1: Calculate the Fourier sine series representing the odd square wave shown in Figure 1 . Figure 1 . Odd square wave, amplitude 1. SOLUTION Since 𝑓𝑓 ( 𝑥𝑥 ) is an odd function which is periodic over the interval [ −𝐿𝐿 , 𝐿𝐿 ] , it can be represented as a Fourier sine series. The Fourier coefficients for this series are given by the Euler formula on the data sheet: 𝐵𝐵 𝑛𝑛 = 1 𝐿𝐿 ∫ 𝑓𝑓 ( 𝑥𝑥 ). sin � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 −𝐿𝐿 (1) where 𝑓𝑓 ( 𝑥𝑥 ) is our square wave function shown in Figure 1. Since both 𝑓𝑓 ( 𝑥𝑥 ) and sin � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � are odd functions, the integrand is even (i.e. odd x odd = even). Whenever you integrate an even function over symmetric limits you can simplify the integral by integrating from 0 to L and doubling the result, viz: 𝐵𝐵 𝑛𝑛 = 2 𝐿𝐿 ∫ 𝑓𝑓 ( 𝑥𝑥 ). sin � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 0 (2) Now we can go ahead and substitute 𝑓𝑓 ( 𝑥𝑥 ) and perform the integration: 𝐵𝐵 𝑛𝑛 = 2 𝐿𝐿 ∫ (1). sin � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 / 2 0 + 2 𝐿𝐿 ∫ ( − 1). sin � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 𝐿𝐿 / 2 (3) = − 2 𝑛𝑛𝑛𝑛 � cos � 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 �� 0 𝐿𝐿 / 2 + 2 𝑛𝑛𝑛𝑛 � cos � 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 �� 𝐿𝐿 / 2 𝐿𝐿 = − 2 𝑛𝑛𝑛𝑛 � cos � 𝑛𝑛𝑛𝑛 2 � − 1 � + 2 𝑛𝑛𝑛𝑛 � cos( 𝑛𝑛𝑛𝑛 ) − cos � 𝑛𝑛𝑛𝑛 2 ��

Figure 5. Considering our initial condition on [0, 𝐿𝐿 ] as part of a periodic odd function on [ −𝐿𝐿 , 𝐿𝐿 ] . Because both 𝑓𝑓 ( 𝑥𝑥 ) and cos � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � are even functions, these formulae simplify to: 𝐴𝐴 0 = 1 𝐿𝐿 ∫ 𝑓𝑓 ( 𝑥𝑥 ). 𝑑𝑑𝑥𝑥 𝐿𝐿 0 (8) 𝐴𝐴 𝑛𝑛 = 2 𝐿𝐿 ∫ 𝑓𝑓 ( 𝑥𝑥 ) cos � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 0 , 𝑛𝑛 = 1, 2, … (9) The initial condition in Figure 3 is given by: 𝑓𝑓 ( 𝑥𝑥 ) = � 2 𝐿𝐿 𝑥𝑥 0 ≤ 𝑥𝑥 < 𝐿𝐿 2 −2𝑛𝑛 𝐿𝐿 + 2 𝐿𝐿 2 ≤ 𝑥𝑥 ≤ 𝐿𝐿 (10) Applying our initial condition to (8) and (9) gives: 𝐴𝐴 0 = 1 𝐿𝐿 ∫ 2𝑛𝑛 𝐿𝐿 𝑑𝑑𝑥𝑥 𝐿𝐿 / 2 0 + 1 𝐿𝐿 ∫ � −2𝑛𝑛 𝐿𝐿 + 2 � 𝑑𝑑𝑥𝑥 𝐿𝐿 𝐿𝐿 / 2 = 1 𝐿𝐿 2 [ 𝑥𝑥 2 ] 0 𝐿𝐿 / 2 − 1 𝐿𝐿 2 [ 𝑥𝑥 2 ] 𝐿𝐿 / 2 𝐿𝐿 + 2 𝐿𝐿 [ 𝑥𝑥 ] 𝐿𝐿 / 2 𝐿𝐿 = 1 4 − 3 4 + 1 = 1 2 𝐴𝐴 𝑛𝑛 = 2 𝐿𝐿 ∫ 2𝑛𝑛 𝐿𝐿 cos � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 / 2 0 + 2 𝐿𝐿 ∫ � −2𝑛𝑛 𝐿𝐿 + 2 � . cos � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � 𝑑𝑑𝑥𝑥 𝐿𝐿 𝐿𝐿 / 2 (11) = 4 𝐿𝐿 2 � 𝑥𝑥 cos � 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 �𝑑𝑑𝑥𝑥 𝐿𝐿 / 2 0 − 4 𝐿𝐿 2 � 𝑥𝑥 cos � 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 �𝑑𝑑𝑥𝑥 𝐿𝐿 𝐿𝐿 2 + 4 𝐿𝐿 � cos � 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 �𝑑𝑑𝑥𝑥 𝐿𝐿 𝐿𝐿 2 The first two integrals here contain 𝑥𝑥 cos � 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿 � and thus to solve them we need to use integration by parts. This is left as an exercise for you to try. Solving these integrals you should obtain:

PRACTICE EXERCISES: INITIAL CONDITION DEMO Note: When you click on the Initial Condition Demo Application, you’ll most likely be prompted with a security warning: • If using Windows, select ‘Run anyway’ • If using Mac, follow the instructions at the end of this document to enable the demo to run The Initial Condition Demo is designed to demonstrate how various initial conditions on the domain [0, L ] can be represented as either a Fourier sine series or a Fourier cosine series. The following general instructions and practice exercises will help you to explore the functionality of the demo. GENERAL INSTRUCTIONS TO USE THE DEMO (i) SELECT AN INITIAL CONDITION from the list of examples This will initialise a plot showing the chosen initial condition on the domain [0, L ], with the functional form of the initial condition displayed at the top. (ii) SELECT “ODD” OR “EVEN” This will represent your chosen initial condition as a Fourier sine or cosine series, with the first few terms of the Fourier series shown below the plot. (iii) ADJUST THE NUMBER OF TERMS in the Fourier series using the slider and “+1”/”-1” This will demonstrate what happens to the Fourier series representation of the periodic initial condition as you allow more or less terms. PRACTICE EXERCISES 1. Select each initial condition example in turn and verify that the functional form of the initial condition (shown at the top of the demo) corresponds to the plotted red curve. 2. Choose one of the initial condition examples and select “Odd”. Notice how the initial condition on the domain [0, L ] is being considered as part of a periodic odd function on the domain [- L , L ]. Now select “Even” and notice how the initial condition on the domain [0, L ] is being considered as part of a periodic even function on the domain [- L , L ]. 3. When you select “Odd”, which terms appear in the Fourier series representation? Why? Which terms appear in the Fourier series representation when you select “Even”? 4. Select the different initial conditions and choose “Odd” or “Even”. Adjust the number of Fourier terms, starting from n =1, and increase one term at a time. How does the Fourier approximation to the initial condition change as you add more terms? 5. Select the “Sinusoid” initial condition and choose “Even”. Starting from n=1, increase the number of Fourier terms one at a time. Which parts of the initial condition show a good fit even with a small number of terms? Which part requires more terms to fit it exactly? Can you explain this behaviour? 6. Select the different initial conditions and choose “Odd” or “Even”. Adjust the number of Fourier terms, starting from n =1, and increase one term at a time. Which modes appear in the Fourier series? Can you explain why? 7. Using the Euler formulae on the data sheet, see if you can derive the Fourier series for some of the odd/even representations of these initial conditions. (Refer to the pre-lecture video for Week 2 to see some examples of using the Euler formulae.)

Week 6 Practice Problems 1. Using a Taylor Series determine if the expression ( f i +1 - 3 * f i + f i - 1 + f i - 2 ) / Δ x 2 a consistent approximation to d 2 f/dx 2 ( x i ). 2. Calculate the leading truncation error of the following approximation: ( - f i - 3 + 3 f i - 2 - 3 f i - 1 + f i ) / Δ x 3 = d 3 f/dx 3 + where is the error . 3. The linear advection equation u t + au x = 0, where a is a positive constant is solved using the following scheme: u n +1 i = u n i - Δ ta 2Δ x ( u n i +1 - u n i - 1 ) (4) Apply von Neumann stability analysis to show that the scheme is unconditionally unstable. Note that if G is composed of a real and imaginary part then | G | 2 = Re [ G ] 2 + Im [ G ] 2 , e.g. if G = cos β m + I sin β m then | G | 2 = cos 2 β m + sin 2 β m = 1 . . 4. Using von Neumann stability analysis, show that the Backwards in Time Central in Space scheme for the heat diffusion equation is stable for all σ = D Δ t/ Δ x 2 5. The instability in the Forward-Time Centred-Space scheme found in question 3. was corrected for by Peter Lax, in a scheme known as the Lax Method . This approach modifies the FTCS scheme by taking the spacial average of the u n i term on the RHS of (4), i.e.; u n +1 i = 1 2 ( u n i - 1 + u n i +1 ) - Δ ta 2Δ x ( u n i +1 - u n i - 1 ) (5) Using the von Neumann stability analysis, determine the conditions for which the Lax Method is stable ( Note: the importance of this condition will appear in future lectures ). .

4. Consider a basilar membrane of length 50 µ m undergoing torsional motion with a node halfway along the length (Fourier mode n =2) and period 4 s. The governing equation for the deflection angle θ ( x , t ) in radians is: 𝜃𝜃 𝑡𝑡𝑡𝑡 = 𝑐𝑐 2 𝜃𝜃 𝑥𝑥𝑥𝑥 where c 2 = 1.61045 x 10 -10 m 2 /s 2 and is computed from the ratio of the torsional stiffness and the mass polar moment of inertia. Derive the general solution for the arbitrary initial conditions and boundary conditions. Simplify the resulting equation by noting that θ (0, t ) = θ ( L , t ) = 0, and calculate the frequency and period of the second mode. How does this compare to the actual measured frequency and period? 5. Sketch the following functions and work out the Fourier Sine Series coefficients B n for each of them: (a) L x L x x f ≤ ≤ − = for ) ( (b)      ≤ ≤ + ≤ ≤ − − − = L x x x L x x f 0 if 1 0 if 1 ) ( 2 2 (c)            ≤ ≤ − ≤ ≤ − ≤ ≤ − + − = L x L L x L k L x L L kx L x L L x L k x f 2 / if ) ( 2 2 / 2 / if 2 2 / if ) ( 2 ) (

5. Solving for the steady-state concentration of a chemical species inside a biological compartment (Note: This is the BMET tutorial problem for Week 8) A biological ‘compartment’ is any distinct space within a biological system such as a sub-cellular structure, a single cell or group of cells, or an entire organ or system such as the arterial vasculature. Consider Figure 2 where we model a rectangular 2D biological compartment of width a in the x - direction and height b in the y -direction. Due to certain metabolic and cellular transport processes, the concentration of a particular chemical species is fixed at the four boundaries of the compartment as shown in Figure 2. Note that these are Dirichlet boundary conditions and together they completely determine the steady-state behaviour of the concentration of the chemical inside the compartment. The concentration of the chemical as a function of time and space can be modelled using Fick’s second law as: 𝐶𝐶 𝑡𝑡 = 𝐷𝐷 ( 𝐶𝐶 𝑥𝑥𝑥𝑥 + 𝐶𝐶 𝑦𝑦𝑦𝑦 ) (13) where D is the mass diffusivity (analogous to the thermal diffusivity ‘ c 2 ’ in the heat equation). Note that (1) is simply the 2D version of the heat equation, which we are now very familiar with. By definition, steady-state means no change with respect to time – that is, C t = 0. Substituting this condition into (1) gives the steady-state equation for the concentration of the chemical: 𝐶𝐶 𝑥𝑥𝑥𝑥 + 𝐶𝐶 𝑦𝑦𝑦𝑦 = 0 (14) which is the Laplace equation. Derive the analytical solution to (2) for the boundary conditions shown in Figure 2.

Week 9 Practice Problems 1. Suppose a bomb containing a toxic gas was released in a narrow corridor in the Melbourne CBD during a terrorist attack. If the initial condition describing the gas is: 𝑌𝑌 ( 𝑥𝑥 , 0) = 𝑓𝑓 ( 𝑥𝑥 ) = 1 for − 10 𝑚𝑚 < 𝑥𝑥 < 10 𝑚𝑚 and zero elsewhere (1) and there are no boundary conditions, derive the solution to the diffusion equation using the Fourier Integral approach. 2. A very long uniform bar initially at 0 ° C is heated resistively at t =0 to 25 ° C in a patch - 0.25 ≤ x ≤ 0.25. Derive the temperature at x = 0 and t = 200 s. Assume the bar is copper and c 2 = 1.14 x 10 -4 m 2 /s. 3. Given a function Q ( x ) defined on the interval 0 ≤ x ≤ L : 𝑄𝑄 ( 𝑥𝑥 ) = ⎩ ⎪ ⎨ ⎪ ⎧ − 4 𝑥𝑥 2 𝐿𝐿 2 + 1 for 0 ≤ 𝑥𝑥 < 𝐿𝐿 2 0 for 𝐿𝐿 2 ≤ 𝑥𝑥 ≤ 𝐿𝐿 (2) Derive coefficients so that Q ( x ) may be represented as the Fourier series: 𝑄𝑄 ( 𝑥𝑥 ) = 𝐴𝐴 0 + � 𝐴𝐴 𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 + 𝐵𝐵 𝑛𝑛 ∞ 𝑛𝑛=1 𝑐𝑐𝑠𝑠𝑛𝑛 𝑛𝑛𝑛𝑛𝑥𝑥 𝐿𝐿 (3) 4. A wire of length L =1 m and density ρ =0.5 kg/m 3 is under tension T =80 N. Initially it has a velocity of: 𝑌𝑌 𝑡𝑡 ( 𝑥𝑥 , 0) = 1 4 − 1 2 𝐿𝐿 �𝑥𝑥 − 𝐿𝐿 2 � (4) and no initial displacement. The ends of the wire are fixed at Y (0, t ) = Y ( L , t ) = 0. Derive the Fourier series solution to the wave equation which governs the behaviour of the wire. Implement your solution in Matlab and animate one period of oscillation. 5. Perform a Fourier transform of the equation f ( x ) = 1- x 2 over the domain 0 ≤ x ≤ 3 – i.e. compute the integral: 𝑓𝑓 ̑ ( 𝑤𝑤 ) = 1 √ 2 𝑛𝑛 � 𝑓𝑓 ( 𝑥𝑥 ) 𝑒𝑒𝑥𝑥𝑒𝑒 ( −𝑠𝑠𝑤𝑤𝑥𝑥 ) 𝑑𝑑𝑥𝑥 3 0 (5)

Week 10 Practice Problems 1. Find the Laplace transform of (i) t + 4, (ii) at + b , (iii) 4 t 3 + t 2 . 2. Find the inverse Laplace transform of (i) 1 s 2 +1 , (ii) 1 s 5 + 1 s 2 and (iii) 1 ( s +1)( s +2) 3. Using Fourier Integrals, obtain a solution to the heat equation on a domain -∞ < x < ∞ with initial conditions T ( x, 0) = 1 for x > 0 and T ( x, 0) = 0 for x ≤ 0. Hint: Start with the ‘box car’ problem in the lecture notes, derive the solution for arbitrarily locations of the two discontinuities (say x 1 and x 2 ), and then consider what happens as the location of one of them tends to infinity. 4. [ Note: You would not be asked to solve a heat equation question using the Fourier transform, but we include this example to demonstrate how it can be used instead of the Fourier integral] Solve using the Fourier Transform: T t = c 2 T xx (4) for 0 ≤ x ≤ ∞ and t > 0. The initial condition is T ( x, 0) = f ( x ) and the boundary condition is T (0 , t ) = 0. This boundary condition which passes through zero means that the resultant function is odd. If the gradient was zero then it would be an even function. Thus, using the Fourier sine transform obtain: T ( x, t ) = 2 π Z ∞ 0 f ( v ) Z ∞ 0 sin ( wv ) sin ( wx ) e - c 2 w 2 t dw dv (5) Then use: sin 1 2 ( a + b ) sin 1 2 ( a - b ) = 1 2 cos a + 1 2 cos b (6) and Z ∞ 0 cos (2 bs ) e - s 2 ds = √ π 2 e - b 2 (7) and a change in the integration basis using m = ( v + x ) /τ and n = ( v - x ) /τ to obtain T ( x, t ) = 1 √ π " Z ∞ - x/τ f ( x + τn ) e - n 2 dn - Z ∞ x/τ f ( - x + τm ) e - m 2 dm # (8) where τ = 2 √ c 2 t . Setting f ( x ) = 1, obtain the final solution in terms of error functions (Ans: T ( x, t ) = erf ( x/τ )). Roughly sketch the solution for a range of times.

Figure 1 . Two types of vibrations in an earthquake, P-Waves (top) and S-Waves (bottom In an earthquake there are two basic forms of wave, a primary P-wave and a secondary S-wave. The P-wave is a stretching/compression of the surface in the longitudinal direction, whereas the secondary waves, called S-waves, travel more slowly, but are more destructive as they move up and down. We need a quick crude estimate of how fast a stress wave generated by an earthquake travels. To do this, we decide to model it as an isolated travelling longitudinal wave, with the governing equation: u tt = λ + 2 μ ρ u xx , (1) where u is the displacement in the longitudinal direction (e.g. squashing or extending of the crust), λ + 2 μ = 90GPa is the modulus of elasticity of the Earth’s crust, and ρ = 2660kg/m 3 is the density assuming it is granite. Given an initial disturbance at the epicentre of u (0 , t ) = f ( t ) = 0 . 1 sin (2 πt/ 10) e - 0 . 01 t , calculate: • the solution to the governing equation for this specific forcing term following the example in class • the time it takes for the disturbance to reach 100km from the epicentre • the speed the wave travels. Implement your solution in Matlab and plot the variation in time of the longitudinal vibration. An alternative name for the unit step function is the ‘Heaviside’ function. In Matlab, it is computed using the command ‘y = heaviside(t-a)’ for example. The equation governing displacement of the Earth’s surface due to S-waves is: y tt = μ ρ y xx (2) where μ ≈ 30GPa. What is the velocity of the S-waves compared to the P-waves? Week 11 Practice Problems 1. Wave Equation With Time-Varying BC: Earthquake Tremors (AMME2000 example)

2. Wave Equation With Time-Varying BC: Pressure Pulse in Blood (Biomed example) Cardiac pumping can cause pressure waves to propagate down blood vessels. Here the heart causes a time-varying disturbance that initiates the pressure wave along the vessel. We’ll assume the vessel is long (“semi-infinite”) and model this situation as an isolated travelling longitudinal wave with the governing equation: 𝑝𝑝 𝑡𝑡𝑡𝑡 = 𝐴𝐴 𝜌𝜌𝜌𝜌 𝑝𝑝 𝑥𝑥𝑥𝑥 (1) where p ( x , t ) is the pressure distribution as a function of space ( x ) and time ( t ), A is the cross-sectional area of the vessel (m 2 ), ρ is the density of blood (kg/m 3 ), and C is the compliance of the vessel (m 2 /Pa). The disturbance caused by the heart is equivalent to a time-varying boundary condition at x =0. Assuming the pressure distribution in the vessel is initially 0 and at rest, and that the disturbance generated by the heart is given by: 𝑝𝑝 (0, 𝑡𝑡 ) = 𝑓𝑓 ( 𝑡𝑡 ) = 𝑠𝑠𝑠𝑠𝑠𝑠 � 2 𝜋𝜋𝑡𝑡 0.004 �𝑒𝑒 −250𝑡𝑡 (2) Derive the analytical solution for this problem using the Laplace transform, following the approach covered in the lectures.

3. Heat Equation With Time-Varying BC: Thermal Heating of Soil Figure 2 shows a series of heat pipes used to ensure that the frozen ground around an Alaskan pipeline does not melt. A new pipeline is planned for installation and you need to decide whether heat pipes should also be installed. Assuming an average Summer daytime temperature of 20◦C in the pipeline location, that the pipes are buried at a depth of 1.5m, and that the soil is mostly composed of sand with a 0.5 volume fraction of water, how long it will take before the soil at a depth of 1.5 m will defrost, in days? The thermal diffusivity of soil types with varying volume fractions of water is plotted in Figure 2. Assume the soil is initially at 0 ◦ C and is fully melted when it reaches 1 ◦ C. As an engineer, the first assumption should be to take the worst case scenario and assume that there is sun for 18 hours a day and that in the six hours of darkness there is no cooling of the soil. Make a recommendation as to whether heat pipes need to be installed. Implement the solution to the governing equation in Matlab, and plot the variation of temperature with depth into the soil for day 1, 10 and 100. Figure 2. Typical use of heat pipes to cool the soil around a buried pipeline (left-top), and a diagram of the heat pipes (left-bottom), and the thermal diffusivity of different soil types (right) 4. Inhomogeneous Heat Equation With Time-Varying BC: Thermal Heating of Soil In our soil melting example above, how would the solution to the governing equation change if the answer was required in Kelvin?

5. Heat Equation With Time-Varying BC: Hypothermic Elephant You’ve managed to get a job as an engineer at Sydney’s Taronga Zoo where you consult on a range of animal-related issues. Today there is a situation where one of the elephants is suffering from hypothermia. The zoo staff are wondering how effective an active thermal blanket will be to restore the core temperature of this huge animal. Using the tools from AMME2000/BMET2960/BMET9960, you decide to get a quick estimate for them using the 1D heat equation: 𝑇𝑇 𝑡𝑡 = 𝑘𝑘 𝜌𝜌𝜌𝜌 𝑇𝑇 𝑥𝑥𝑥𝑥 (1) where T ( x , t ) is the temperature distribution as a function of space ( x ) and time ( t ), and k , ρ and σ are the thermal conductivity (W/m-K), tissue density (kg/m 3 ) and tissue heat capacity (J/kg-K), respectively, all of which we’ll assume are constant. The elephant is lying on its side with the active thermal blanket draped over the main part of its body. You can assume that the elephant’s body temperature everywhere is initially steady at 34 ° C and that the blanket delivers heat to the body surface (at x =0) according to: 𝑇𝑇 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑡𝑡 = 𝑓𝑓 ( 𝑡𝑡 ) (2) Using the Laplace Transform approach, determine the analytical solution T ( x , t ) if f ( t ) = 60 ° C.

We want to calculate the response of a semi-infinite wire subjected to a forcing f ( t ). It is clamped at x=0, but is free at the other end so the appropriate boundary condition as x → ∞ is y x ( ∞ , t ) = 0. Initially the wire is at rest and has zero displacement. Thus, the governing equation is: y tt = c 2 y xx + f ( t ) ( 1 ) subject to boundary conditions: y (0 , t ) = 0 , y x ( ∞ , t ) = 0 ( 2 ) and the initial conditions: y ( x, 0) = 0 , y t ( x, 0) = 0 ( 3 ) Show that the Laplace transform of the governing equation gives the following inhomogeneous ODE: Y xx - s 2 c 2 Y = - F ( s ) c 2 ( 4 ) where L { y } = Y . This is an inhomogeneous equation, so the solution must be gained from the homogeneous solution Y H (where F(s)=0) plus the ‘Particular Integral’ Y P I , which is the solution of the above equation. The full solution is then Y = Y H + Y P I . The particular integral is gained in this case by assuming that it takes the form Y P I = K ( s ). Substitute this into Eqn ( 4 ), and rearrange to gain K ( s ). Show thus that the general solution is: Y ( x, s ) = A ( s ) e sx/c + B ( s ) e - sx/c + F ( s ) s 2 ( 5 ) Apply the boundary conditions and initial conditions to show that: Y ( x, s ) = F ( s ) s 2 - F ( s ) s 2 e - sx/c ( 6 ) Given a constant forcing term f ( t ) = f 0 , take the inverse Laplace transform to gain the solution: y ( x, t ) = f 0 2 t 2 - t - x c 2 u t - x c ( 7 ) Plot the variation of y as a function of x for several instants in time. 6. Wave Equation With Forcing in the Governing Equation

Week 12 Practice Problems 1. Solve u xx + f = 0 for 0 ≤ x ≤ 1, with u (0) = 1 and u x (1) = 2 using two elements and given f = 10. What is the deflection u (0.25)? 2. Solve the above problem but for f = x 2 . 3. Derive the weak form of the governing equation T t + t = 0 for 0 ≤ x ≤ 1 .

Week 13 Practice Problems 1. Determine the analytical solution to the 1D Poisson equation u xx = - q ( x ) for q ( x ) = - x 2 , where u (0) = 0 and u x ( L ) = 0. (ans: u ( x ) = x 4 / 12 - L 3 x/ 3) 2 Steady state temperature i n a one dimensional bar of l ength L heated by a source q ( x ) satisfies the equation kT xx = - q ( x ), where k i s the the thermal conductivity of the material. Given Dirichlet boundary conditions at the first node at x = 0 of T 0 = 273K and assuming that the bar is insulated at x = L show that the finite element solution is given by solving the following matrix problem: k l e          2 - 1 0 · · · 0 - 1 2 - 1 . . . . . . 0 . . . . . . . . . 0 . . . . . . - 1 2 - 1 0 · · · 0 - 1 1                 T ˆ 1 T ˆ 2 . . . T ˆ n - 1 T ˆ n        = l e 6        q ˆ 0 + 4 q ˆ 1 + q ˆ 2 q ˆ 1 + 4 q ˆ 2 + q ˆ 3 . . . q ˆ ne - 2 + 4 q ˆ n - 1 + q ˆ n q ˆ ne - 1 + 2 q ˆ n               + T 0 k/l e 0 . . . 0        3. Solve the above problem for L = 0 . 5m, q = 10W/m 3 and k = 401W/m-K using 2 elements to estimate the temperature at the end of the bar ( x = L ). How does this compare to the analytical? (ans: analytical: T ( x ) = - qx ( x/ 2 - L ) + 273, numerical: T 3 = 273 . 0031K . ) 4. Modify your Matlab code to solve this with an arbitrary number of elements.