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# 48 Class Note for MATH 251 at PSU

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Date Created: 02/06/15

Differential Equations LECTURE 40 Separation of Variables and Heat Equation IVPS 1 Initial Value Problems Partial differential equations generally have lots of solutions To specify a unique one we ll need some additional conditions These conditions are usually motivated by the physics and come in two varieties initial conditions and boundary conditions An initial value problem for a PDE consists of the equation initial conditions and boundary conditions An initial condition speci es the physical state at a given time to For example an initial condition for the heat equation would be the starting temperature distribution ux 0 This is the only initial condition required because the heat equation is rst order with respect to time The wave equation which we will look at later is second order with respect to time and so needs two initial conditions PDEs are also generally only valid on a certain domain Boundary conditions specify how the solution is to behave on the boundary of this domain These need to be speci ed because the solution isn t on the one side ofthe boundary meaning we might have problems with differentiabiltiy there Our heat equation was derived for a one dimensional bar of length i so the relevant domain in question can be taken to be the interval 0 lt z lt l and the boundary consists of the two points z 0 and z Z We could have derived a two dimensional heat equation for example in which case the domain would be some region in the zy plane with boundary some closed curve It will usually be clear from a physical description of the problem what the appropriate boundary conditions are We might know that at the endpoints z 0 and z L the temperatures u0t and ult are xed Boundary conditions that give the value of the solution are called Dirichlet conditions Or we might insulate the ends meaning there should be no heat ow out of the boundary this would yield the boundary conditions um0t umlt 0 If the boundary conditions specify the derivative of the solution they re called Neumann conditions We could also specify that we have one insulated end and at the other we control the temperature this is an example of a mixed boundary condition As we ve seen changing boundary conditions can signi cantly change the character of a prob lem lnitially to get a feel for our solution method we ll work with the homogeneous Dirichlet conditions u0 t ul t 0 giving us the following initial value problem DE u 1w 401a BC u0t 14125 0 40110 10 limo my 401c After we ve seen the general method we ll see what happens with homogeneous Neumann conditions leaving nonhomogeneous conditions for a later discussion Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs 2 Separation of Variables In Section 1 we derived the heat equation 401a for our bar of length L Suppose we ve got an initial value problem such as 401 How do we proceed We ll want to try to build up a general solution out of smaller solutions that are easier to nd A rather straightforward approach is to start by assuming these mini solutions have some nice form For example we might suppose we have a separated solution where uzt XmTt That is our solution is the product of a function that depends only on x and a function that depends only on t We can then try to write down an equation depending only on x and another depending only on 25 before using our knowledge of ODEs to try to solve them It should be noted that this is very special and should not be expected in general In fact this method cannot always be used and even when it can be used often it s hard to move past the rst step However it works for all of the equations we will be considering in this class and is a valuable starting point How does the method work We begin by plugging our separated solution into the heat equation 401a 2 8 8 a lXTtl kg XTtl XzT t lacam Now notice that we can move everything depending on x to one side and everything depending on t to the other T t X 95 kTt Xm This equation should look a little funny to you On the left we have an expression which depends only on 25 while on the right we have an expression that depends only on x Yet these two sides have to be equal for any choice of z and t we make The only way this is possible is if both sides of the equation are the same constant In other words T t X kTt T X 7 We ve written the minus sign explictly for convenience it will turn out that gt 0 but these expressions should be negative The equation above really contains a pair of separate ordinary differential equations X X 0 402a T AkT 0 40210 Notice that with these separated equations our boundary conditions 394b become X0 0 and Xl 0 Now 402b is easy enough to solve we have T ikkT so that Tt A54 What about 402a This just gives us the boundary value problem X X 0 X0 0 Xl 0 This should look familiar it s very similar to Example 386 The only difference is instead of our second condition occuring at z 27139 it s at some z 1 As in that example it will turn out that Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs our eigenvalues have to be positive Letting 32 for B gt 0 our general solution is Xz Bcos z Csin m The rst boundary condition says B 0 The second condition says that Xl Csin l 0 To avoid only having the trivial solution we must have Bl 717139 In other words 2 An and Xnz sin for n 123 So we end up having found an in nite number of solutions to our boundary value problem 401a and 401b one for each positive integer n and no other values of n since our eigenvalues are all positive They are WK 2 unm t Ane T kt sin The heat equation is linear and homogeneous As such the Principle of Superposition still holds a linear combination of solutions is again a solution So any function of the form ux t iAne 2kt sin 403 710 is also a solution to 401a and 401b Notice that we haven t used our initial condition 401c yet which is why we referred to 396 as a solutions to just the boundary value problem How does our initial data come into play We have N 7171391 m m 0 ZAnsm n0 So if our initial condition has this form 403 works perfectly for us with the coef cients An just being the associated coef cients from EXAMPLE 401 Find the solution to the following heat equation problem on a 7011 of length 2 Ut Umm u0 t u2 t 0 uz0 sin 7 5sin37rz In this problem we have k 1 Now we know that our solution will have the form of something like 403 since our initial condition is just the difference of two sine functions We just need to gure out which terms are represented and what the coef cients An are This isn t too hard to do our initial condition is 35 sin 7 5sin37rx Looking at 403 with l 2 we can see that the rst term corresponds to n 3 and the second n 6 and there are no other terms Thus we have A3 1 A6 75 and all other An 0 Our solution is then Wat l sin 7 559W2t sin 37m Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs This isn t very satisfying there s no reason to suppose that our initial distribution is a nite sum of sine functions Physically7 such situations are clearly very special What do we do if we have a more general initial temperature distribution Let s consider what happens if we take an in nite sum of our separated solutions Then our solution to 401a and 401b is 742525 WWW Sm y Now the initial condition speci es that the coef cients must satisfy 00 35 Z A sin 404 n This idea is due to the French mathematician Joseph Fourierl7 and 404 is called the Fourier sine series for There are several very important questions raised by this7 however Why should we believe that our initial condition x ought to be able to be written as an in nite sum of sines Why should we believe that such a sum would converge to anything We ll come to these in good time7 but keep them in mind 21 Neumann boundary conditions Now let s consider a heat equation problem with homogeneous Neumann conditions7 DE ut kn 405a BC um07 t nzl7 t 405b lC uz0 405c We ll start by again supposing that our solution to 405a is separable7 so we have umt XzTt and7 as we re using the same differential equation as before7 we obtain the pair of ODEs 402a and 402b The solution to 402b is still Tt Ae wt Now we need to determine the boundary conditions for 402a Our boundary conditions 405b are on iim07 t and umlt thus they are conditions on X 0 and X l7 as the t derivative isn t controlled at all So we have the boundary value problem X X 0 X 0 0 X l 0 Along the lines of the analogous computation last lecture7 this has eigenvalues and eigenfunctions An er cos for n O7 17 27 So individual separable solutions to 405a and 405b have the form 77471713 unmt Anemiykt cos 1Fourier 17681830 was a big promoter of the French Revolution and traveled with Napoleon to Egypt where he was made governor of Lower Egypt Fourier has been credited with being one of the rst to understand the greenhouse e ect and more generally planetary energy balance There s a great though almost certain apocryphal story about what led Fourier to study the heat equation Reportedly what deeply concerned Fourier was being able to nd the ideal depth to build his wine cellar so that the wine would be stored at the perfect temperature yearround and so he proceeded to try to understand the way heat propagated through the ground Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs Taking nite linear combinations of these work similarly to the Dirichlet case and is the solution to 405 when x is a nite linear combination of constants and cosines in direct analogy to 403 but in general we re interested in knowing when we can take in nite sums ie 1 m 2 umt 5A0 ZAWAT kt cos ri1 Notice how we wrote the n 0 case as A0 the reason for this will be made clear in the future The initial condition 405c means that we need 406 7mm x 140 2A cos ri1 An expression of the form 406 is called the Fourier cosine series of 22 Other boundary conditions It s also possible for certain boundary conditions to re quire the full Fourier series of the initial data this is an expression of the form 7mm 7mm fm 140 0 An cos Bn sin 407 n1 but for most of our purposes we ll want to solve problems with Dirichlet or Neumann conditions However in the process of learning about Fourier sine and cosine series we ll also learn how to compute the full Fourier series of a function

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