## my mathematics cheat sheet

April 2, 2017 compiled on — Sunday April 02, 2017 at 08:33 PM

### Contents

A place to keep quick notes about Math that I keep forgetting. This is meant to be a scratch notes to write math notes. Can contain errors and/or not complete. Use at your own risk.

### 1 general notes

1. Methods to find Green function are

1. Fredholm theory
2. methods of images
3. separation of variables
4. Laplace transform

reference wikipedia I need to make one example and apply each of the above methods on it.

2. we can't just say is Analytic and stop. Have to say is analytic at a point or in a region. is analytic at a point if the power series for expanded around converges to evaluated at . (i.e. has Taylor series that converges to at the point).
For a region, the above test will be true for every point in that region. For example, is not analytic at . This is because when evaluated at is a singularity (a fancy name for saying it blows up at ). But , so we see that is not analytic at since its power series does not converge to the same value as the function at
But this is only for this point . The function is analytic at another point, say .
3. In solving an ODE with constant coefficient just use the characteristic equation to solve the solution.
4. In solving an ODE with coefficients that are functions that depends on the independent variable, as in , first classify the point type. This means to check how and behaves at . We are talking about the ODE here, not the solution yet. There are 3 kinds of points. can be normal, or regular singular point, or irregular singular point. Normal means and have Taylor series expansion that converges to at .
Regular singular point means that the above test fails, but has a convergent Taylor series, and also that now has a convergent Taylor series at . All this just means we can get rid of the singularity. i.e. is a removable singularity. If this is the case, then the solution at can be assumed to have a Frobenius series where and must be integer values.
The third type of point, is the hard one. Called irregular singular point. We can't get rid of it using the above trick. So we also say the ODE has an essential singularity at (another fancy name for irregular singular point, I think). What this means is that we can't approximate the solution at using either Taylor nor Frobenius series.
If the point is an irregular singular point, then use the methods of asymptotic. See advanced mathematical methods for scientists and engineers chapter 3. For normal point, use , for regular singular point use . Remember, to solve for first. This should give two values. If you get one root, then use reduction of order to find second solution.
5. For Taylor series, leading behavior is no controlling factor? For Frobenius series, leading behavior term is and controlling factor is . For asymptotic series, controlling factor is assumed to be always. proposed by Carlini (1817)
6. Method to find the leading behavior of the solution near irregular singular point using asymptotic is called the dominant balance method.
7. When solving for very small then use WKB method, if there is no boundary layer between the boundary conditions. If the ODE non-linear, can't use WKB, has to use boundary layer (B.L.).  Example with then use BL.
8. good exercise is to solve say with using both B.L. and WKB and compare the solutions, they should come out the same. with BL had to do the matching between the outer and the inner solutions. WKB is easier. But can't use it for non-linear ODE.
9. When there is rapid oscillation over the entire domain, WKB is better. Use WKB to solve Schrodinger equation where becomes function of (Planck's constant, mkg/s)
10. In second order ODE with non constant coefficient, , if we know one solution , then a method called the reduction of order can be used to find the second solution . Write , plug this in the ODE, and solve for . The final solution will be . Now apply I.C.'s to find .
11. To find particular solution to , we can use a method called undetermined coefficients.  But a better method is called variation of parameters, In this method, assume where are the two linearly independent solutions of the homogeneous ODE and are to be determined. This ends up with and . Remember to put the ODE in standard form first, so , i.e. . In here, is the Wronskian
12. Two solutions of are linearly independent if , where is the Wronskian.
13. Green function takes the homogeneous solution and the forcing function and constructs a particular solution. For PDE's, we always want a symmetric Green's function.
14. To get a symmetric Green's function given an ODE, start by converting the ODE to a Sturm-Liouville form first. This way the Green's function comes out symmetric.
15. For numerical solutions of field problems, there are basically two different problems: Those with closed boundaries and those with open boundaries but with initial conditions. Closed boundaries are elliptical problems which can be cast in the form , and the other are either hyperbolic or parabolic.
16. For numerical solution of elliptical problems, the basic layout is something like this: Always start with trial solution such that where the are the unknowns to be determined and the are set of linearly independent functions (polynomials) in . How to determine those comes next. Use either residual method (Galerkin) or variational methods (Ritz). For residual, we make a function based on the error . It all comes down to solving over the domain. This is a picture
|
+---------------+-------------------------------------+
|                                                     |
residual                       Variational (sub u_trial in I(u)
|                          where I(u) is functional to minimize.
|
+----------------+-------------+----------+
|                |             |          |
Absolute error   collocation   subdomain orthogonality
....
+----------------------+------------+
|                      |            |
method of moments   Galerkin     least squares
17. Geometric probability distribution. Use when you want an answer to the question: What is the probability you have to do the experiment times to finally get the output you are looking for, given that a probability of showing up from doing one experiment.

For example: What is the probability one has to flip a fair coin times to get a head? The answer is . So for a fair coin, that a head will show up from one flip. So the probability we have to flip a coin times to get a head is which is very low as expected.

18. To generate random variable drawn from some distribution different from uniform distribution, by only using uniform distribution do this: Lets say we want to generate random number from exponential distribution with mean . This distribution has , the first step is to find the cdf of exponential distribution, which is known to be . Now find the inverse of this, which is . Then generate a random number from the uniform distribution . Let this value be called . Now plug this value into , this gives a random number from exponential distribution, which will be (take the natural log of both side of ).

This method can be used to generate random variables from any other distribution by knowing on . But it requires knowing the CDF and the inverse of the CDF for the other distribution. This is called the inverse CDF method. Another method is called the rejection method

19. Given , a r.v. from uniform distribution over [0,1], then to obtain , a r.v. from uniform distribution over [A,B], then the relation is .
20. When solving using F.E.M. is best to do everything using isoparametric element (natural coordinates), then find the Jacobian of transformation between the natural and physical coordinates to evaluate the integrals needed. For the force function, using Gaussian quadrature method.
21. A solution to differential equation is a function that can be expressed as a convergent series. (Cauchy. Briot and Bouquet, Picard)
22. To solve a first order ODE using integrating factor.

then as long as it is linear and are integrable functions in , then follow these steps

1. multiply the ODE by function , this is called the integrating factor.

2. We solve for such that the left side satisfies

3. Solving the above for gives

Integrating both sides gives

4. Now eq(1) can be written as

We now integrate the above to give

Where is given by (2). Hence

23. a polynomial is called ill-conditions if we make small change to one of its coefficients and this causes large change to one of the roots.
24. to find rank of matrix by hand, find the row echelon form, then count how many zero rows there are. subtract that from number of rows, i.e. .
25. to find the basis of the column space of , find the row echelon form and pick the columns with the pivots, there are the basis (the linearly independent columns of ).
26. For symmetric matrix , its second norm is its spectral radius which is the largest eigenvalue of (in absolute terms).
27. The eigenvalues of the inverse of matrix is the inverse of the eigenvalues of .
28. If matrix of order , and it has distinct eigenvalues, then it can be diagonalized  , where

and is matrix that has the eigenvectors as its columns.

29. only if converges uniformly over .
30. , has infinite number of solutions. Think of as 3 rotations, each of , going back to where we started. Each rotation around a straight line. Hence infinite number of solutions.
31. How to integrate .

Let , then and the above becomes

Now let or , hence and the above becomes

But hence

Substituting back

Substituting back

32. (added Nov. 4, 2015) Made small diagram to help me remember long division terms used.
33. If a linear ODE is equidimensional, as in for example then use ansatz this will give equation in only. Solve for and obtain and the solution will be

For example, for the above ode, the solution is . This ansatz works only if ODE is equidimensional. So can't use it on for example.

If is multiple root, use as solutions.

34. for , where , write it as hence
35. For Sturm Liouville, write it as . Now it is easy to compare to. For example, , then we see that . And for then . Do not divide by first to re-write as and then compare, as this makes it hard to compare to standard form and will make error.  Books write Sturm Liouville as but I find this confusing to compare to this way. Better to expand it to first so now it is in a more familiar form and can read it out more directly.
36. Some integral tricks: use . For use and for use .
37. is called Emden-Fowler form.
38. For second order ODE, boundary value problem, with eigenvalue (Sturm-Liouville), remember that having two boundary conditions is not enough to fully solve it. One boundary condition is used to find first constant of integration, second boundary condition is used to find the eigenvalues. We still need another input to find the second constant of integration. This is normally done by giving the initial value. This problem happens as part of initial value, boundary value problem. The point is, with boundary value and eigenvalue also present, we need 3 inputs to fully solve it. Two boundary conditions is not enough.
39. If given ODE  and we are asked to classify singular at , then let and check what happens at . The operator becomes and operator becomes . And write the ode now where is the independent variable, and follow standard operating procedures. i.e. look at and and see if these are finite or not. To see how the operator are mapped, always start with then write and write . For example, and

Then the new ODE becomes

The above is how the ODE will always become after the transformation. Remember to change to using and same for . Now the new is and the new is . Then do and as before.

40. If the ODE ,  and say , is either non-linear or linear, and there is essential singularity at either end, then use boundary layer or WKB. But Boundary layer method works on non-linear ODE's (and also on linear ODE) and only if the boundary layer is at end of the domain, ie. at or . WKB method on the other hand, works only on linear ODE, but the singularity can be any where (i.e. inside the domain). As rule of thumb, if the ODE is linear, use WKB. If the ODE is non-linear, we must use boundary layer. Another difference, is that with boundary layer, we need to do matching phase at the interface between the boundary layer and the outer layer in order to find the constants of integrations. This can be tricky and is the hardest part of solving using boundary layer. Using WKB, no matching phase is needed. We apply the boundary conditions to the whole solution obtained. See my HWs for NE 548 for problems solved from Bender and Orszag text book.

### 2 things to find out

1. find out why I get different answers here. branch cut handling?

Mathematica 9.01

Clear[x, y, k1, k2, k3];
sol = Solve[Cos[x - y] == k1 + k2*Cos[y] + k3*Cos[x], y]
sol /. {x -> 1, k1 -> 2, k2 -> 3, k3 -> 4}
N[%]

gives

{{y -> -2.81197 - 1.04751 I},
{y -> 2.81197 + 1.04751 I},
{y -> -2.81197 + 1.04751 I},
{y -> 2.81197 - 1.04751 I}}

Maple 18

restart;
eq:= cos(x-y)=k1+k2*cos(y)+k3*cos(x):
sol:=solve(eq,y):
evalf(subs({x=1,k1=2,k2=3,k3=4},{sol}));

gives

{2.811969896 - 1.047512037 I,
2.811969896 + 1.047512037 I}

### 3 Solving exact first order ODE

(Added Sept. 30, 2016). When solving an exact ODE setup the following two ODE's are setup

Next, the first ODE is integrated w.r.t. , leading to

 (3)

Where replaces the ”integration constant”. Now the above is differentiated w.r.t. and the resulting equation is compared to (2) to solve for . Next is found by integrating. Then now that is found, then is found from (3). And since is some constant, then an implicit solution for is thus obtained.

### 4 Direct solving of some simple PDE's

Some simple PDE's can be solved by direct integration, here are few examples.

Example 1

Integrate w.r.t. ., and remember that now constant of integration will be function of , hence

Example 2

Integrating once w.r.t. gives

Integrating again gives

Example 3

Integrating once w.r.t. gives

Integrating again gives

Example 4

Integrating once w.r.t gives

Integrating again w.r.t. gives

### 6 Linear combination of two solution is solution to ODE

If are two solutions to then to show that is also solution:

Multiply the first ODE by and second ODE by

Add the above two equations, using linearity of differentials

Therefore satisfies the original ODE. Hence solution.

### 7 To find the Wronskian ODE

Since

Where are two solutions to Write

Multiply the first ODE above by and the second by

Subtract the second from the first

 (1)

But

 (2)

And

Substituting (2,3) into (1) gives the Wronskian differential equation

Whose solution is

Where is constant of integration.

Remember: does not mean the two functions are linearly dependent. The functions can still be Linearly independent on other interval, It just means can't be in the domain of the solution for two functions to be solutions. However, if the two functions are linearly dependent, then this implies everywhere.  So to check if two functions are L.D., need to show that everywhere.

### 8 Sturm-Liouville Notes

#### 8.1 Definitions

Regular Sturm-Liouville ODE is an eigenvalue boundary value ODE. This means, it the ODE has an eigenvalue in it , where solution exists only for specific values of eigenvalues. The ODE is

Or

With the restrictions that are real functions that are continuous everywhere over and also we need . The is called the weight function. But this is not all. The boundary conditions must be linear homogeneous, of this form

Where are just real constants. Some of them can be zero but not all. For example, is OK. So boundary conditions do not have to mixed, but they can be in general. But they must be homogeneous. Notice that periodic boundary conditions are not allowed. Well, they are allowed, but then the problem is no longer called Sturm-Liouville. The above is just the definition of the equation and its boundary conditions. Below is list of the important properties of this ODE. Each one of these properties have a proof.

1. All eigenvalues are real. No complex .
2. Each eigenvalue , will have one and only one real eigenfunction associated with it. (in higher dimensions, the eigenvalue problem can have more than one eigenfunction associated with one eigenvalue. For example, heat PDE in rectangle). But we can use Gram-schmidt to make these eigenfunctions linearly independent if we need to). But for exam, just worry about 1D for now.
3. There is smallest , called and there are infinite number of eigenvalues. Hence eigenvalues are ordered
4. Each eigenfunction will have zeros in . Note that this does not include the end points. This means, will have two zeros inside the domain. i.e. there are two locations where in .
5. Eigenfunctions make a complete set of basis. This means any piecewise continuous function can be represented as . This is called the generalized Fourier series.
6. This follows from (5). Each eigenfunction is orthogonal to another eigenfunction, but with the weight in there. This means if .
7. Rayleigh quotient relates an eigenvalue to its eigenfunction. Starting with the SL ODE , then multiplying by and integrating, we obtain and solving for , gives

Carrying integration by parts on first part of the integral in numerator, it becomes

But this becomes much simpler when we plug-in the boundary conditions that we must use, making the above

And if and , then it becomes Rayleigh quotient is useful to show that can be positive without solving for and also used to estimate the value of the minimum by replacing by a trial function and actually solving for using to obtain a numerical estimate of .

8. There is symmetry relation. In operator form, let , then we have where are any two different eigenfunctions.
9. There is also what is called Lagrange identity, which says . This comes from simply expanding and simplifying things. Just calculus.
10. Green formula, follows from Lagrange identity, which just gives the integral form . But form Sturm-Liouville, we know that (from 8). So this really just says that . But we know this already from boundary conditions. So I am not sure why this is useful for Sturm-Liouville now. Since it is just saying the same thing again.
11. If at the left or right end (boundaries), then the problem is now called singular Sturm-Liouville. This is actually the important case. In regular S.L., must be positive everywhere. We only consider at the ends, not in the middle or any other place. When this happens, then solutions to the ODE generate special functions, such as Bessel, Legendre, Chebyshev. These special functions are solution to the Singular S.L. ODE, not the regular S.L. ODE. In addition, at the end where , the boundary conditions must be bounded. Not the regular boundary conditions above. For example, if where is say the left end, the the boundary conditions at the left end must be . Notice that singular can be at left end, right end or at both at same time. At the end where , the boundary conditions must be bounded type.
12. Symmetry relation

where are are two eigenfunction. But remember, this does NOT mean the integrand is identically zero, or . Only when happened to have same eigenvalue, only then we can say that . But for unique eigenfunctions only the integral form is zero This is important difference, used in the proof below, that different eigenfunctions have different eigenvalues (for the scalar SL case). For higher dimensions, we can have more than one eigenfunction for same eigenvalue.

Proofs for these properties follows

#### 8.2 Proof symmetry of operator

Given regular Sturm Liouville (RSL) ODE

In operator form

And (1) becomes

When solving RSL ode, since it is an eigenvalue ODE with associated boundary conditions, we will get infinite number of non-negative eigenvalues. For each eigenvalue, there is associated with it one eigenfunction (for 1D case). Looking at any two different eigenfunctions, say , then the symmetry relation says the following

Now we will show the above is true. This requires integration by parts two times. We start from the LHS expression and at the end we should end up with the integral on the RHS. Let , then

Where we used in the above. Hence

 (2)

Now we will do integration by parts on the first integral. Let . Using , and if we let , then . Hence

We now apply integration by parts again to the second integral above. But now let and , hence and , therefore the above becomes

Substituting the above into (2) gives

 (3)

Now comes the part where the boundary conditions are important. In RSL, the boundary conditions are such that all the terms vanish. This is because the boundary conditions are

And

So now (3) becomes

Therefore, we showed that . The only thing to watch for here, is which term to make and which to make when making integration by parts. To remember, always start with and make the term with in them during integration parts.

#### 8.3 Proof eigenvalues are real

Assume is complex. The corresponding eigenfunctions are complex also. Let be corresponding eigenfunction

 (1)

Taking complex conjugate of both sides, and since is real, we obtain

But since all the coefficients of the ODE are real. The above becomes

 (2)

But by symmetry, we know that

Substituting  (1),(2) into the above gives

But which is positive. Also the weight is positive by definition. Hence for the above to zero, it must be that . Which means is real. QED.

So the main tools to use in this proof: Definition of and symmetry relation and that . This might come up in the exam.

#### 8.4 Proof eigenfunctions are unique

Now will show that there is one eigenfunction associated with each eigenvalue (Again, this is for 1D, it is possible to get more than one eigenfunction for same eigenvalue for 2D, as mentioned earlier).  By contradiction, assume that has two eigenfunctions associated with it. Hence

From the first equation, , substituting this into the second equation gives

By Lagrange identity, , hence this means that

Where is some constant. This is the main difference between the above argument, and between Largrange identity. This can be confusing. So let me talk more about this. In Lagrange identity, we write

And when also satisfy the SL boundary condition, only then we say that

But the above is not the same as saying . This is important to keep in mind. Only the integral form is zero for any two functions with the SL B.C..Now we continue. We showed that . In SL, this constant is zero due to B.C. Hence

But by definition. Hence or

or . So the eigenfunctions are linearly dependent. One is just scaled version of the other. But eigenfunction must be linearly independent. Hence assumption is not valid, and there can not be two linearly independent eigenfunctions for same eigenvalue. Notice also that is the just the Wronskian. When it is zero, we know the functions are linearly dependent. The important part in the above proof, is that only when happened to have same eigenvalue.

#### 8.5 Proof eigenfunctions are real

The idea of this proof is to assume the eigenfunction is complex, then show that its real part and its complex part both satisfy the ODE and the boundary conditions. But since they are both use the same eigenvalue, then the real part and the complex part must be linearly dependent. This implies the eigenfunction must be real. (think of Argand diagram)

Assume that is complex eigenfunction with real part and complex part . Then since

The above is just writing the Sturm-Liouville ODE in operator form, where is the operator as above.  Now we have

By linearity of operator

Which implies

So we showed that the real and complex part satisfy S.L. Now we need to show they are satisfy S.L. boundary conditions. Since

Where are the left and right ends of the domain. Then

Hence

So the above means both and satisfy the boundary conditions of S.L. But since both have the same eigenvalue, then they must be linearly dependent, since we know with S.L. each eigenfunction (or one linearly dependent to it) have only one eigenvalue. This means

Where is some constant. In other words,

Where is new constant. (OK it happens to be complex constant, but it is OK to do so, we always do this trick in other places, if it will make me feel better, I could take the magnitude of the constant). So all what the above says, is that we assumed to be complex, and found that it is real. So it can't be complex.

#### 8.6 Proof eigenfunctions are orthogonal with weight

Given two different eigenfunctions . Hence

From symmetry integral relation, since these eigenfunctions also satisfy S.L. boundary conditions, we can write

Replacing (1,2) into the above

But since there are different eigenvalues for different eigenfunctions. Hence

Which means are orthogonal to each others with weight

#### 8.7 Special functions generated from solving singular Sturm-Liouville

When S.L. is singular, meaning at one or both ends, we end with important class of ODE's, whose solutions are special functions (not or ) as the case with the regular S.L. Recall that S.L. is

With the regular S.L., we say that over the whole domain, including end points. But with singular, this is not the case.  Here are three important S.L. ODE's that are singular.

Bessel equation:

Or in standard form

So we see that . We see that at , then , which what makes it singular. (also the weight happens to be zero also), but we only care about being zero or not, at one of the ends. So to check if S.L. is singular or not, just need to check if is zero or not at one of the ends. As mentioned before, when at one of the ends, we can't use the standard B.C. for the regular S.L. instead, at the end where , we must use what is called bounded boundary conditions, which is in this case . The solution to this ODE will be in terms of Bessel functions. Notice that this happened to be singular, due to the domain starting at . If the domain happened to be from to say , then it is no longer singular S.L. but regular one.

Legendre equation

Or it can be written as

We see that . And now it happens to be that at . So it is singular at the other end compared to Bessel. In this case . Again, the boundary conditions at must now be bounded. i.e. . On the other end, where is not zero, we still use the standard boundary conditions  .

Chebyshev equation

Or

So we see that . So where is here? At then and at then also. So this is singular at both ends. So we need to use bounded boundary conditions at both ends now to solve this.

The solution to this singular S.L. is given in terms of special function Chebyshev.

#### 8.8 Some notes

1. Solution to . Set in S-L form, we get , then . For with . This is singular S-L. Using Frobenius series, the solution comes out to be , where is Bessel function of first kind. are roots of . (check if this should be . is constant, which still needs to be found.
2. We can find approximate solution to S-L ODE for large eigenvalue, without solving the ODE using series method, but by using WKB. Given which is standard form S-L. Using physical optics correction, we obtain the solution as

Where are found from boundary conditions now. The above is valid for ”large” , and is found by first letting and then assuming . And working though the WKB method. Remember, WKB only works for linear homogeneous ODE and is used to estimate solution for large (or small ).

3. in 1D, S-L is with , with B.C. given by and and in higher dimensions, this problem becomes with boundary conditions now written as on boundary . In higher dimensions, .
4. Lagrange's identity in 1D is

where above is the S-L operator, as in so that we write .

5. When it says eigenfunctions are normalized, it should mean that   where is the weight function (comes from SL equation) and is the domain. This is for 1D SL. Remember for example, that . where . Here, it is not normalized since the result is not one. The weight here is just one and here
6. Heat PDE in cylinder is For steady state, it becomes, say cylinder has length

And for axis symmetric (no dependency), it becomes

The solution is found by separation of variables. For the above is in terms of Bessel function order zero

which is found using series method . The eigenvalues are zeros of where is disk radius.

7. Heat PDE in 2D polar is and for steady state it becomes . After separating, the ODE becomes Euler ODE. Use guess for solution. The solution will be

8. To change to S.L. form, multiply the ODE by
9. In using method of eigenfunction expansion to solve nonhomogeneous B.C. PDE, the eigenfunction used are the ones from solving the PDE with homogeneous B.C. For example, if given with nonhomogeneous B.C. , say , and we want to use on the nonhomogeneous B.C. PDE, then those are from the corresponding homogeneous B.C. PDE. They should be something like and so on. See the table at the top. That is why we write above and not and remember not to do term-by-term differentiation in when this is the case. We can do term-by-term differentiation only of the PDE itself also has homogeneous B.C. , even if it had a source term there also. The point is, always come from solving the nonhomogeneous B.C. PDE. (This normally means solving a Sturm-Liouville ODE).
10. I found new relation for eigenvalue , but it is probably not useful as the one the book has. Here how to derive it
Given S-L
 (1)

Let be the eigenfunction associated with eigenvalue . Since eigenfunctions satisfy the ODE itself, then we can write, for any arbitrary eigenfunction (subscript removed for clarity in what follows)

Integrating both sides

 (2)

Looking at . Integrating by part. Let , hence

Substituting (3) into (2) gives

Hence

compare to the one in the book .

#### 8.9 Methods of solutions to some problems

Problem 1 If the problem gives S-L equations, and asks to find estimate on the smallest eigenvalue, then use Rayleigh quotient for . And write . Now we do not need to solve the SL to find solution , this is the whole point. Everything in RHS is given, except for, of course the solution . Here comes the main idea: Come up with any trial and use it in place of . This trial function just needs to satisfy the boundary conditions, which is also given. Then all what we need to do is just evaluate the integral. Pick the simplest function of which satisfies boundary conditions. All other terms we can read from the given problem. At the end, we should get a numerical value for the integral. This is the upper limit of the lowest eigenvalue

Problem 2 We are given SL problem with boundary conditions, and asked to show that without solving the ODE to find : Use Rayleigh quotient and argue that the denominator can't be zero (else eigenfunction is zero, which is not possible) and it also can't be negative. Argue about the numerator not being negative. Do not solve the ODE! this is the whole point of using Rayleigh quotient.

Problem We are given an SL problem with boundary conditions and asked to estimate large and corresponding eigenfunction. This is different from being asked to estimate the smallest eigenvalue, where we use Rayleigh quotient and trial function. In this one, we instead use WKB. For 1D, just use what is called the physical optics correction method, given by . Where in this are all known functions (from the problem itself). Notice that is not there. Now use the boundary conditions and solve for one of the constants, should come to be zero. Use the second boundary condition and (remember, the boundary conditions are homogenous), and we get one equation in and for non-trivial solution, solve for allowed values of . This gives the large eigenvalue estimate. i.e. for when is very large. Depending on problem, does not have be too large to get good accurate estimate compared with exact solution. See HW7, problem 5.9.2 for example.

Problem We are given 1D heat ODE . If B.C. are homogenous, we are done. We know the solution. Separation of variables. If the B.C. is not homogenous, then we have a small problem. We can't do separation of variables. If you can find equilibrium solution, , where this solution only needs to satisfy the non-homogenous B.C. then write solution as , and plug this in the PDE. This will give but this now satisfies the homogenous B.C. This is the case if the original non-homogenous B.C were Dirichlet. If they were Neumann, then we will get where is extra term that do not vanish because do not vanish in this case after double differentiating. We solve for now, since it has homogenous B.C., but it has extra term there, which we treat as new source. We apply initial conditions now also to final all the eigenfunction expansion coefficients. We are done. Now we know

But if we can find the reference function or we do not want to use this method, then we can use another method, called eigenfunction expansion. We assume and plug this in the PDE.  But now we can't do term by term differentiation, since are the eigenfunctions that satisfies the homogenous B.C., while has non-homogenous B.C. So the trick is to use Green formula, to re-writes as plus the contribution from the boundary terms. (this is like doing integration by parts twice, but much easier). Now rewrite . See page 355-356, Haberman.

#### 8.10 Examples converting second order linear ODE to Sturm-Liouville

Any linear second order ODE can be converted to S-L form. The S-L form is the following

 (1)

Sometimes there is a minus sign there. I really never understood why some book put a minus sign and some do not. May be I'll find out one day. But for now, (1) is used as the S-L form. The goal now, is given any general form eigenvalue ODE, we want to convert it (rewrite it) in the above form. The second order linear ODE will have this form

 (2)

The parameter in the S-L form, is the eigenvalue. We really only use S-L form for eigenvalue problems. First, we will show how to convert (2) to (1), and then show few examples. We are given (2), and want to convert it to (1). The first step is to convert (2) to standard form

Then multiply (2) by some unknown which is assumed positive.

Now, rewrite as . These are the same thing. Now we replace this in the above and obtain

Here comes the main trick in all of this. We want to force to be zero. This implies . (this comes from just solving the ODE . Therefore, if , then (3) becomes

We are done.  Hence

And

Let see how this works on some examples

##### 8.10.1 Example 1

Convert to S-L.  We see the general form here is , with . Hence , therefore the SL form is

 (1)

Where

Hence (1) becomes

This was easy, since the ODE given was already in SL form.

##### 8.10.2 Example 2

Convert

Rewrite as

We see the general form here is , with . Hence

Therefore the SL form is

 (1)

Where

Hence (1) becomes

##### 8.10.3 Example 3

Convert

Rewrite as

We see the general form here is , with . Hence

Therefore the SL form is

 (1)

Where

Hence (1) becomes

##### 8.10.4 Example 4

Convert

We see the general form here is , with . Hence

Therefore the SL form is

 (1)

Where

Hence (1) becomes

I seem to have a sign mistake there. I just do not see it now. This should come out to be

### 9 Green functions notes

1. Green function is what is called impulse response in control. But it is more general, and can be used for solving PDE also. Given a differential equation with some forcing function on the right side. To solve this, we replace the forcing function with an impulse. The solution of the DE now is called the impulse response, which is the Green's function of the differential equation. Now to find the solution to the original problem with the original forcing function, we just convolve the Green function with the original forcing function. Here is an example. Suppose we want to solve    with zero initial conditions. Then we solve . The solution is . Now . This is for initial value problem.  For example. , with . Then we solve . The solution is , this is for causal system. Hence . The nice thing here, is that once we find , we can solve for any by just convolving the Green function (impulse response) with the new .
2. We can think of Green function as an inverse operator. Given , we want to find solution . So in a sense, is like .
3. Need to add notes for Green function for Sturm-Liouville boundary value ODE. Need to be clear on what boundary conditions to use. What is B.C. is not homogeneous?
4. Green function properties:

1. is continuous at . This is where the impulse is located.
2. The derivative just before is not the same as just after . i.e. . This means there is discontinuity in derivative.
3. should satisfy same boundary conditions as original PDE or ODE (this is for Sturm-Liouville or boundary value problems).
4. for
5. is symmetric. i.e. .
5. When solving for , in context of 1D, hence two boundary conditions, one at each end, and second order ODE (Sturm-Liouville), we now get two solutions, one for and one for . So we have constants of integrations to find (this is for second order ODE) not just two constants as normally one would get , since now we have 2 different solutions. Two of these constants from the two boundary conditions, and two more come from property of Green function as mentioned above.

### 10 Laplace transform notes

1. Remember that and . For example, if we are given , then . Do not do ! That will be a big error. We use this allot when asked to write a piecewise function using Heaviside functions.

### 11 References

Too many references used, but will try to remember to start recording books used from now on. Here is current list

1. Applied partial differential equation, by Haberman
2. Advanced Mathematical Methods for Scientists and Engineers, Bender and Orszag, Springer.
3. Boundary value problems in physics and engineering, Frank Chorlton, Van Norstrand, 1969
4. Class notes. Math 322. University Wisconsin, Madison. Fall 2016. By Professor Smith. Math dept.
5. various pages Wikipedia.
6. good note on Sturm-Liouville http://ramanujan.math.trinity.edu/rdaileda/teach/s12/m3357/lectures/lecture_4_10_short.pdf