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Lecture 13 2 nd order partial differential equations Remember Phils Problems and your notes = everything Only.

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Presentation on theme: "Lecture 13 2 nd order partial differential equations Remember Phils Problems and your notes = everything Only."— Presentation transcript:

1 Lecture 13 2 nd order partial differential equations http://www.hep.shef.ac.uk/Phil/PHY226.htm Remember Phils Problems and your notes = everything Only 8 lectures left !!!!!!!!!!!!!! Read through your notes in conjunction with lecture presentations Try some examples from tutorial questions or Phils Problems Email me or come to E47 if anything is unclear

2 x Introduction to PDEs A wave equation is an example of a partial differential equation Think of a photo of waves (i.e. time is fixed) If you make either x or t constant, then you return to the expected SHM case You all know from last year that a possible solution to the above is and that this can be demonstrated by substituting this into the wave equation Or a movie of the sea in which you focus on a specific spot (i.e. x is fixed) y t y

3 Poisson’s equation Introduction to PDEs In many physical situations we encounter quantities which depend on two or more variables, for example the displacement of a string varies with space and time: y(x, t). Handing such functions mathematically involves partial differentiation and partial differential equations (PDEs). As Laplace but in regions containing mass, charge, sources of heat, etc. Electromagnetism, gravitation, hydrodynamics, heat flow. Laplace’s equation Heat flow, chemical diffusion, etc. Diffusion equation Quantum mechanics Schrödinger’s equation Elastic waves, sound waves, electromagnetic waves, etc. Wave equation

4 Introduction to PDEs Before we start solving this type of equation let’s recap the skills we’ve learned so far during this course that we’ll have to draw on….. If find all 1 st and 2 nd order differentials of y with respect to x and t. Example in your notes…. So 1 st order differential equations 2 nd order differential equations Fourier series and be reminded of a useful principle …….

5 Introduction to PDEs gives 1 st order differential equations 2 nd order differential equations Harmonic oscillator Unstable equilibrium

6 Step 4: Boundary conditions could then be applied to find A and B Step 2: The auxiliary is then and so roots are Step 1: Let the trial solution be So and Unstable equilibrium Step 3: General solution for real roots is Introduction to PDEs Thing to notice is that x(t) only tends towards x=0 in one direction of t, increasing exponentially in the other

7 Step 1: Let the trial solution be So and Introduction to PDEs Harmonic oscillator Step 2: The auxiliary is then and so roots are Step 3: General solution for complex is where  = 0 and   =  so Thing to notice is that x(t) passes through the equilibrium position (x=0) more than once !!!! Step 4: Boundary conditions could then be applied to find C and D

8 Introduction to PDEs Half-range Fourier sine series where A guitarist plucks a string of length d such that it is displaced from the equilibrium position as shown. This shape can then be represented by the half range sine (or cosine) series.

9 For such equations there is a fundamental theorem called the superposition principle, which states that if and are solutions of the equation then is also a solution, for any constants c 1, c 2. The principle of superposition The wave equation (and all PDEs which we will consider) is a linear equation, meaning that the dependent variable only appears to the 1st power. i.e. In x never appears as x 2 or x 3 etc. Waves and Quanta: The net amplitude caused by two or more waves traversing the same space (constructive or destructive interference), is the sum of the amplitudes which would have been produced by the individual waves separately. All are solutions to the wave equation. Can you think when you used this principle last year?? Electricity and Magnetism: Net voltage within a circuit is the sum of all smaller voltages, and both independently and combined they obey V=IR.

10 The One-Dimensional Wave Equation Today we’ll learn how to solve this equation and using the boundary conditions predict the displacement at any x or t Waves on strings are governed by the equation, where y(x, t) is the displacement of the string at position x and time t. You met this equation last year and learned that c is the wave velocity.

11 Find the solution to the wave equation to predict the displacement of the guitar string at any later time t The One-Dimensional Wave Equation A guitarist plucks a string of length L such that it is displaced from the equilibrium position as shown at t = 0 and then released. Let’s go thorugh the steps to solve the PDE for our specific case …..

12 Step 1: Separation of the Variables Since Y(x,t) is a function of both x and t, and x and t are independent of each other then the solutions will be of the form where the big X and T are functions of x and t respectively. Substituting this into the wave equation gives … Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t Step 2: Rearrange equation Rearrange the equation so all the terms in x are on one side and all the terms in t are on the other: The One-Dimensional Wave Equation

13 (i) (ii) Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 3: Equate to a constant Since we know that X(x) and T(t) are independent of each other, the only way this can be satisfied for all x and t is if both sides are equal to a constant: Suppose we call the constant N. Then we have which rearrange to … and (i) (ii)

14 Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 4: Decide based on situation if N is positive or negative We have ordinary differential equations for X(x) and T(t) which we can solve but the polarity of N affects the solution ….. Linear harmonic oscillator Unstable equilibrium Which case we have depends on whether our constant N is positive or negative. We need to make an appropriate choice for N by considering the physical situation, particularly the boundary conditions. If N is positive If N is negative Decide now whether we expect solutions of X(x) and T(t) to be exponential or trigonometric …..

15 Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 4: Continued ….. Remember that if N is negative, solutions will pass through zero displacement many times, whilst if N is positive solutions only tend to zero once. From this we deduce N must be negative. Let’s write So (i) becomes From lecture 3, this has general solution and in the same way and From before

16 Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 5: Solve for the boundary conditions for X(x) In our case the boundary conditions are Y(0, t) = Y(L, t) = 0. This means X(0) = X(L) = 0, i.e. X(x) is equal to zero at two different points. (This was crucial in determining the sign of N.) Now we apply the boundary conditions: X(0) = 0 gives A = 0. Saying B ≠ 0 then X(L) = 0 requires sin kL = 0, i.e. kL = n . So k can only take certain values where n is an integer for n = 1, 2, 3, …. So we have

17 From previous page Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 6: Solve for the boundary conditions for T(t) By standard trigonometric manipulation we can rewrite this as

18 Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 7: Write down the special solution for Y(x,t) Hence we have special solutions: We see that each Y n represents harmonic motion with a different wavelength (different frequency). In the diagram below of course time is fixed constant (as it’s a photo not a movie!!): (Mistake in notes – please correct harmonic numbers in diagram below)

19 This is the most general answer to the problem. For example, if a skipping rope was oscillated at both its fundamental frequency and its 3 rd harmonic, then the rope would look like the dashed line at some specific point in time and its displacement could be described just by the equation :- (mistake in notes at top of page 4, 3 rd not 2 nd harmonic) Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 8: Constructing the general solution for Y(x,t) We have special solutions: Bearing in mind the superposition principle, the general solution of our equation is the sum of all special solutions: NB. The Fourier series is a further example of the superposition principle.

20 Since Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation At t = 0 the string is at rest, i.e., if we differentiate we find For this to be true for all n and x, and this is only true if So the general solution becomes Step 8 continued: Constructing the general solution for Y(x,t)

21 The guitarist plucked the string of length L such that it was displaced from the equilibrium position as shown and then released at t = 0. This shape can therefore be represented by the half range sine (or cosine) series. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 9: Use Fourier analysis to find values of B n Half-range sine series where It can be shown (see Phils Problems 5.10) that this shape can be represented by

22 and we see above that the coefficients B n are the coefficients of the Fourier series for the given initial configuration at t = 0. Therefore we can write the general solution at t = 0 as ….. Since Fourier series at t = 0 is Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 9 continued: Use Fourier series to find values of B n for and the general solution is then at t = 0 the general solution is

23 Hence, by trusting the superposition principle treating each harmonic as a separate oscillating sinusoidal waveform which is then summed together like a Fourier series to get the resulting shape, we deduce that at later times the configuration of the string will be:- The solution at t = 0 is Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 10: Finding the full solution for all times But we also know that the general solution at all times is

24 Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation What does this all mean ???? This means that if you know the initial conditions and the PDE that defines the relationships between all variables, the full solution can be found which describes the shape at any later time. Want to know how heat passes down a rod, how light waves attenuate and interfere through a prism, how to define time dependent Schrodinger eigenfunctions, or how anything else that is a linear function with multiple variables interacts ???? Then this is what you should use.

25 SUMMARY of the procedure used to solve PDEs 9. The Fourier series can be used to find the full solution at all times. 1. We have an equation with supplied boundary conditions 2. We look for a solution of the form 3. We find that the variables ‘separate’ 4. We use the boundary conditions to deduce the polarity of N. e.g. 5. We use the boundary conditions further to find allowed values of k and hence X(x). 6. We find the corresponding solution of the equation for T(t). 7. We hence write down the special solutions. 8. By the principle of superposition, the general solution is the sum of all special solutions.. so


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