MATH 121B Notes



Bessel ODE	\(x^{2}\ y^{\prime \prime }+x\ y^{\prime }+\left (x^{2}-p^{2}\right ) y=0\ \ \ \ \ \ \ \) deﬁned for INTEGER and NON integer \(P\)
ﬁrst solution	\(y_{1}=J_{p}\relax (x) =\sum \frac {-1^{n}}{\Gamma \left (n+1\right ) \Gamma \left (n+p+1\right ) }\left (\frac {x}{2}\right ) ^{2n+p}\) (for \(p\) an integer or not)
second solution	\(y_{2}=N_{p}\relax (x) =Y_{p}\relax (x) =\frac {\cos \left (\pi p\right ) J_{p}\relax (x) -J_{-p}}{\sin \pi p}\) (note: \(p\) here is NOT an integer)
second solution	\(y_{x}\) contains a log function. note: \(p\) here IS an integer.
Orthogonality	\(\int _{0}^{1}x\ J_{p}\left (ax\right ) \ J_{p}\left ( bx\right ) \ dx=\left \{ \begin {array} [c]{ll}0 & if\ a\neq b\\ \frac {1}{2}J_{p+1}^{2}\relax (a) =\frac {1}{2}J_{p-1}^{2}\left ( a\right ) =\frac {1}{2}J_{p}^{^{\prime }2}\relax (a) & if\ a=b \end {array} \right . \ a,b\) are zeros of \(J_{p}\)

recursive formula	\(\frac {d}{dx}\left [ x^{p}J_{p}\right ] =x^{p}J_{p-1}\), \(\frac {d}{dx}\left [ \frac {1}{x^{p}}J_{p}\right ] =-\frac {1}{x^{p}}J_{p+1},\) \(J_{p-1}+J_{p+1}=\frac {2p}{x}J_{p},\) \(J_{p-1}-J_{p+1}=2J_{p}^{\prime }\)

	\(J_{p}^{\prime }=-\frac {p}{x}J_{p}+J_{p-1}=\frac {p}{x}J_{p}-J_{p+1}\) NOTICE: No Rodrigues formula for Bessel func, since not polyn.

notes:	We used a generalized power series method to ﬁnd the solutions.
	IF \(p\) is NOT an integer, then \(J_{p}\) and \(J_{-p}\) (or \(N_{p}\)) are two independent solutions
	IF \(p\) is an integer, then \(J_{p}\) and \(J_{-p}\) are NOT two independent solutions, use \(\log \) for \(y_{2}\)
	\(J_{p}\) is called Bessel function of ﬁrst kind, and \(Y_{p}\) is called second kind. \(p\) is called the ORDER.
	IF \(p=n+\frac {1}{2}\,\), a special case, we get spherical bessel functions \(j_{n}\relax (x) \) and \(y_{n}\relax (x) \)
	\(j_{n}\relax (x) =\sqrt {\frac {\pi }{2x}}J_{n+\frac {1}{2}}\left ( x\right ) =x^{n}\left (-\frac {1}{x}\frac {d}{dx}\right ) ^{n}\left ( \frac {\sin x}{x}\right ) \)



Legendre ODE	\(\left (1-x^{2}\right ) \ y^{\prime \prime }-2x\ y^{\prime }+l\left (l+1\right ) y=0\) deﬁned for INTEGER \(l\) only
ﬁrst solution	\(y_{1}=P_{l}\relax (x) \) examples: \(P_{0}\left ( x\right ) =1,P_{1}\relax (x) =x,P_{2}\relax (x) =\frac {1}{2}\left (3x^{2}-1\right ) ,P_{3}\relax (x) =\frac {1}{2}\left ( 5x^{3}-3x\right ) \)
	\(P_{4}\relax (x) =\frac {1}{8}\left (35x^{4}-30x^{2}+3\right ) ,\ P_{5}\relax (x) =\frac {1}{8}\left (63x^{5}-70x^{3}+15x\right ) \)
second sol.	We do not use this. Called Legendre polynomials of second kind \(Q_{l}\relax (x) \)
Orthogonality	\(\int _{-1}^{1}P_{l}\relax (x) \ \ P_{m}\left ( x\right ) \ dx=0\) if \(m\neq n\) , also \(\int _{-1}^{1}P_{l}\relax (x) \ \ \times \) \(\left (\text {any poly degree}<l\ \right ) \ dx=0\)
Normalization	\(\int _{-1}^{1}\left [ P_{l}\relax (x) \right ] ^{2}dx=\frac {2}{2l+1}\)

Generating function	\(\Phi \left (x,h\right ) =\frac {1}{\sqrt {\left ( 1-2xh+h^{2}\right ) }}\), \(\left \vert h\right \vert <1\), \(\Phi \left ( x,h\right ) =P_{0}\relax (x) +hP_{1}\relax (x) +h^{2}P_{2}\relax (x) +\cdots ={\displaystyle \sum \limits _{l=0}^{\infty }} h^{l}P_{l}\relax (x) \)

recursive formula	later, see book page 491

Rodrigues	\(P_{l}=\frac {1}{2^{l}l!}\frac {d^{l}}{dx^{l}}\left (x^{2}-1\right ) ^{l}\)

notes:	we used series method to ﬁnd the solution (not generalized series method).
	\(x\) must be less than 1, this is needed to have convergence. Hence Legendre solution only deﬁned
	over \(-1,1\). Also, \(l\) is assumed to be a non-negative integer.\(l\) is called the ORDER of legendre poly.



Associated Legendre	\(\left (1-x^{2}\right ) \ y^{\prime \prime }-2x\ y^{\prime }+\left [ l\left (l+1\right ) -\frac {m^{2}}{1-x^{2}}\right ] y=0\)
ﬁrst solution	\(y_{1}=P_{l}^{m}\relax (x) \) \(=\) \(\left ( 1-x^{2}\right ) ^{\frac {m}{2}}\ \frac {d^{m}}{dx^{m}}\left (p_{l}\left ( x\right ) \right ) \)
second solution	do not use
Orthogonality	did not cover, but should be the same as Legendre polynomials \(P_{l}\)
Normalization	\(\int _{-1}^{1}\left [ P_{l}^{m}\relax (x) \right ] ^{2}dx=\frac {2}{2l+1}\left (\frac {l+m!}{l-m!}\right ) \)

example using recursive formula for Legendre: \(\Phi \left (x,h\right ) =\frac {1}{\sqrt {\left (1-2xh+h^{2}\right ) }}\), let \(y=2xh-h^{2}\) then \(\Phi \left (x,h\right ) =\frac {1}{\sqrt {\left (1-y\right ) }}=1+\frac {1}{2}y+\frac {\frac {1}{2}\frac {3}{2}}{2!}y^{2}+\cdots \), then sub back for \(y\), and simplify we get \(\Phi =1+xh+h^{2}\left (\frac {3}{2}x^{2}-\frac {1}{2}\right ) +\cdots =P_{0}+hP_{1}+hP_{2}+\cdots \), hence \(P_{0}=1\,,P_{1}=x,P_{2}=\left (\frac {3}{2}x^{2}-\frac {1}{2}\right ) ,\)etc..

Generalized series solution: \(y=a_{0}x^{s}+a_{1}x^{s+1}+a_{2}x^{s+2}+\cdots \) solve for \(s\), we get indicial eq. for each \(s\) we solve again to ﬁnd the \(a_{0}\) and the \(a_{1}\) solutions. Final solution is the sum of the solutions for both \(s\) values. Will only get 2 solutions in total (for second order ODE).

2.4.1.1 Leibniz Rule for diﬀerentiation of product

\begin {align*} \frac {d^{n}}{dx^{n}}\left (fg\right ) & =\left ( \begin {array} [c]{l}n\\ 0 \end {array} \right ) \frac {d^{0}}{dx^{0}}f\frac {d^{n}}{dx^{n}}g+\left ( \begin {array} [c]{l}n\\ 1 \end {array} \right ) \frac {d}{dx}f\ \frac {d^{n-1}}{dx^{n-1}}g+\left ( \begin {array} [c]{l}n\\ 2 \end {array} \right ) \frac {d^{2}}{dx^{2}}f\ \frac {d^{n-2}}{dx^{n-2}}g+\cdots +\left ( \begin {array} [c]{l}n\\ n \end {array} \right ) \frac {d^{n}}{dx^{n}}f\frac {d^{0}}{dx^{0}}g\\ & =\ \frac {d^{0}}{dx^{0}}f\frac {d^{n}}{dx^{n}}g+n\ \frac {d}{dx}f\ \frac {d^{n-1}}{dx^{n-1}}g+\ \frac {n\times n-1}{2!}\frac {d^{2}}{dx^{2}}f\ \frac {d^{n-2}}{dx^{n-2}}g+\cdots +\frac {d^{n}}{dx^{n}}f\frac {d^{0}}{dx^{0}}g \end {align*}

For example \(\frac {d^{9}}{dx^{9}}\left (x\ \sin x\right ) =x\ \frac {d^{9}}{dx^{9}}\sin x+9\times \frac {d}{dx}x\ \frac {d^{3}}{dx^{3}}\sin x\ +\) rest is ZERO terms, so we get \(\frac {d^{9}}{dx^{9}}\left (x\ \sin x\right ) =x\ \cos x+9\sin x\) this is much faster than actually diﬀerentiating 9 times !

\begin {align*} \left (f+g\right ) ^{n} & =\left ( \begin {array} [c]{l}n\\ 0 \end {array} \right ) f^{0}\ g^{n}+\left ( \begin {array} [c]{l}n\\ 1 \end {array} \right ) f^{1}\ g^{n-1}+\left ( \begin {array} [c]{l}n\\ 2 \end {array} \right ) \ f^{2}\ \ g^{n-1}+\cdots +\left ( \begin {array} [c]{l}n\\ n \end {array} \right ) \ f^{n}\ g^{0}\\ \left (f+g\right ) ^{9} & =\ \ g^{9}+\ 9\ f\ g^{8}+\ \frac {9\times 8}{2!}\ f^{2}\ \ g^{7}+\cdots +\ f^{9}\ \ \end {align*}

2.4.1.2 Finding second solution for ODE

when the indicial equation gives only one value for \(s\) (from the generalized power series method), we can ﬁnd the second solution by assuming \[ y_{2}=y_{1}\ln \relax (x) +\sum _{n=0}^{\infty }b_{n}\ x^{n+s}\] Then ﬁnd \(y^{\prime },y^{\prime \prime }\), from these, and sub back into ODE and set \(n=0\) to solve for the new indicial equation, ﬁnd \(s\) from it (should get one solution), then most likely you’ll ﬁnd \(b_{n}=0\) for all \(n>0\) (for the HW’s we did), and so just need to use \(b_{0}x^{n+s}\) and this gives the complete solution. \(y=A\ y_{1}+B\ y_{2}=A\ y_{1}+\left (y_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}\ x^{n+s}\right ) \)

note: IF when solving the indicial equation, 2 values for \(s\) that diﬀer by an integer from each others (say \(4,6\)), then must use the value \(6\), also when we solve for the second solution, \(s\) there must come out to be the ﬁrst \(s\) which we did not use for the ﬁrst solution, i.e. \(4\) in this example (so I really do not need to solve for \(s\) again!, expect I need to ﬁnd the recursive formula).

2.4.2 Chapter 16. Probability

\(P_{A}\relax (B) \) is the probability that B will happen KNOWING that A has already happened.

\begin {align*} P\left (AB\right ) & =P\relax (A) P_{A}\relax (B) \\ & =P\relax (B) P_{B}\relax (A) \end {align*}

If \(A\) and \(B\) are independent, then \(P_{A}\relax (B) =P\left ( B\right ) \)

\[ P\left (AB\right ) =P\relax (A) \ \relax (B) \ \ \ \ \ \ \ \ \ \ \ \text {If A,B independent}\]

\begin {align*} P\left (A+B\right ) & =P\relax (A) +\ \relax (B) -P\left ( AB\right ) \ \ \ \ \ \ \ \ \ \ \\ P\left (A+B\right ) & =P\relax (A) +\ \relax (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text {IF\ A,B are mutually exclusive } \end {align*}

This means that \(P\left (AB\right ) \ =0\) if they are mutually exclusive (obvious)

\[ P(A+B+C)=P\relax (A) +P\relax (B) +P\relax (C) -\left \{ P\left (AB\right ) +P\left (AC\right ) +P\left (BC\right ) \right \} +P\left (ABC\right ) \]

note:\(\ P_{r}^{n}=\) number of permutations (arrangements) or \(n\) things taken \(r\) at a time.\(\ P_{r}^{n}=\frac {n!}{\left (n-r\right ) !}\) Here order is important. i.e. ABC is DIFFERENT from CAB, hence this number will be larger than the one below.

\[ \left ( \begin {array} [c]{l}n\\ r \end {array} \right ) =C_{r}^{n}=\frac {n!}{\left (n-1\right ) !\ r!}\] Number of combinations OR selections of \(n\) things \(r\) at a time. here order is NOT important. so ABC is counted the same as CAB, hence this number will be smaller.

A)\(\left ( \begin {array} [c]{l}10\\ 4 \end {array} \right ) 4!=\frac {10!}{6!4!}4!=\frac {10!}{6!}=10\times 9\times 8\times 7\)

To understand this: \(\left ( \begin {array} [c]{l}10\\ 4 \end {array} \right ) \) is the number of ways 4 people can be selected out of 10. ONCE those 4 people have been selected, then there are \(4!\) diﬀerent ways they can be arranged on the bench. Hence the answer is we multiply these together.

note: Find number of ways of putting \(r\) particles in \(n\) boxes according to the 3 kinds of statistics.

note: If asked this: there is box A which has 5 red balls and 6 black balls, and box B which has 5 red balls and 8 white balls, what is the prob. of picking a red ball? Answer:

P(pick box A) P(pick red ball from it) + P(pick box B) P(pick red ball from it)\(\ \)

note: If we get a problem such as 2 boxes A,B, and more than more try picking balls, it is easier to draw a tree diagram and pull the chances out the tree than having to calculate them directly in the exam. Tree can be drawn in 2 minutes and will have all the info I need.

note: random variable \(x\) is a function deﬁned on the sample space (for the example, the sum of 2 die throw). The probability density is the probability of each random variable.

average or mean of a random variable \(\mu =\sum x_{i}\ P_{i}\) where \(P_{i}\) is the probability of the random variable.

The Variance Var measures the spread of the random variable around the average, also called dispersion deﬁned as

Standard deviation is another measure of the dispersion, deﬁned as \(\sigma \relax (x) =\sqrt {Var\relax (x) }\)

Distribution function is just a histogram of the probability density. it tells one what the probability of a random variable being less than a certain \(x\) value. see page 711.

2.4.3 Chapter 13. PDE



\(PDE\)	\(solution\)	\(equation\)	\(notes\)


Laplace	\(u\left (x,y,z\right ) \)	\(\nabla ^{2}u=0\)	describes steady state (no time) of region with no source
			for example, gravitional potential with no matter, electrostatic
			potential with no charge, or steady state Temp. distribution

Poisson	\(u\left (x,y,z\right ) \)	\(\nabla ^{2}u=f\left (x,y,z\right ) \)	Same as Laplace, i.e. sescribes steady state, howevere
			here the source of the ﬁeld is present. \(f\left (x,y,z\right ) \) is called
			the source density. i.e. it is a function that describes the
			density distribution of the source of the potential.

Diﬀusion or	\(u\left (t,x,y,z\right ) \)	\(\nabla ^{2}u=\frac {1}{\alpha ^{2}}\ \frac {\partial u}{\partial t}\)	Here \(u\) is usually the temperature \(T\) function. Now time
heat equation			is involved. So this equation is alive.

Wave equation	\(u\left (t,x,y,z\right ) \)	\(\nabla ^{2}u=\frac {1}{v^{2}}\ \frac {\partial ^{2}u}{\partial t^{2}}\)	Here \(u\) is the position of a point on the wave at time \(t\).
			Notice the wave equation has second derivative w.r.t. time
			while the diﬀusion is ﬁrst derivative w.r.t. time

Helmholtz	\(F\left (x,y,z\right ) \)	\(\nabla ^{2}F+k^{2}F=0\)	The diﬀusion and wave equation generate this. This is the
equation			SPACE only solution of the wave and heat equations. i.e.
			\(u=F\left (x,y,z\right ) T\relax (t) \) is the solution for both heat and wave eq.

Each of these equations has a set of candidate solutions, which we start with and try to ﬁt the boundary and initial condition into to eliminate some solution of this set that do not ﬁt until we are left with the one candidate solution. We then use this candidate solution to ﬁnd the general solution, which is a linear combination of it. We use fourier series expansion in this part of the solution.

In table below I show for each equation what the set of candidate solutions are. Use these to start the solution with unless the question asks to start at an earlier stage, which is the separation of variables.



PDE	candidate solutions	notes


\(\nabla ^{2}u=0\)	\(u\left (x,y\right ) =\left \{ \begin {array} [c]{lll}e^{ky}\ \cos kx & & \\ e^{ky}\ \sin kx & & \\ e^{-ky}\ \cos kx & & \\ e^{-ky}\ \sin kx & & \end {array} \right . \)	for 2 dimensions

\(\nabla ^{2}u=f\left (x,y,z\right ) \)	\(u\left (x,y,z\right ) =-\frac {1}{4\pi }\int \int \int \frac {f\left (x^{\prime },y^{\prime },z^{\prime }\right ) }{\sqrt {\left (x-x^{\prime }\right ) ^{2}+\left (y-y^{\prime }\right ) ^{2}+\left (z-z^{\prime }\right ) ^{2}+}}dx^{\prime }\ dy^{\prime }\ dz^{\prime }\)	\(f\left (x^{\prime },y^{\prime },z^{\prime }\right ) \) is a function
		that describes mass density
		distribution evaluated at point
		\(x^{\prime },y^{\prime },z^{\prime }\). The point \(x,y,z\) is
		where we are calculating the

		potential \(u\) itself

\(\nabla ^{2}u=\frac {1}{\alpha ^{2}}\ \frac {\partial u}{\partial t}\)	\(u\left ( t,x\right ) =\left \{ \begin {array} [c]{lll}e^{-k^{2}\alpha ^{2}t}\ \cos kx & & \\ \ & & \\ \ e^{-k^{2}\alpha ^{2}t}\ \sin kx & & \end {array} \right . \)	for one space dimension



\(\nabla ^{2}u=\frac {1}{v^{2}}\ \frac {\partial ^{2}u}{\partial t^{2}}\)	\(Y=XT \), where \(X\relax (x) =\left \{ \begin {array} [c]{l}\cos kx\\ \sin kx \end {array} \right . \ T\relax (t) =\left \{ \begin {array} [c]{l}\cos \omega t\\ \sin \omega t \end {array} \right . \)	\(v\) is the wave velosity, 1D
	\(Y\left (x,t\right ) =\left \{ \begin {array} [c]{lll}\cos kx\ \cos \omega t & & \\ \cos kx\ \sin \omega t & & \\ \sin kx\ \sin \omega t & & \\ \sin kx\ \cos \omega t & & \end {array} \right . \)

	\(Z=XYT\), where \(X\relax (x) =\left \{ \begin {array} [c]{l}\cos k_{x}x\\ \sin k_{x}x \end {array} \right . \ \ Y\relax (x) =\left \{ \begin {array} [c]{l}\cos k_{y}x\\ \sin k_{y}x \end {array} \right . \)	2D case in rectangular coord
	\(T\relax (t) =\left \{ \begin {array} [c]{l}\cos \omega t\\ \sin \omega t \end {array} \right . \)

\(\nabla ^{2}F+k^{2}F=0\)

\begin {align*} X\relax (x) & =\left \{ \begin {array} [c]{l}\cos kx\\ \sin kx \end {array} \right . \\ T\relax (t) & =\left \{ \begin {array} [c]{l}\cos \omega t\\ \sin \omega t \end {array} \right . \end {align*}

2.4.3.1 Laplace equation in cylindrical coordinates

The Laplacian in cylindrical is \(\nabla ^{2}u=\frac {1}{r}\frac {\partial }{\partial r}\left (r\frac {\partial u}{\partial r}\right ) +\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}+\frac {\partial ^{2}u}{\partial z^{2}}=0\), the solution can be written as \(u=R\relax (r) \Theta \left ( \theta \right ) Z\relax (z) \)

\(Z\relax (z) =\left \{ \begin {array} [c]{l}e^{kz}\\ e^{-kz}\end {array} \right . \), we quickly eliminate the \(e^{kz}\) since we do not want the potential to blow up as \(z\) becomes larger. \(\Theta \left (\theta \right ) =\left \{ \begin {array} [c]{l}\sin n\theta \\ \cos n\theta \end {array} \right . \), \(R\relax (r) =J_{n}\left (kr\right ) \) where \(J_{n}\left ( kr\right ) \) is Bessel function of order \(n\), we do not use \(N_{n}\left ( kr\right ) \) solutions since we origin is on base of cylinder. see book for more details, all problems we will get will be like this. We ﬁnd \(k\) from boundary conditions, it will turn out to be the zeros of \(J_{n}\). From above, the set of candidate solutions for Laplace on cylindrical is \[ u\left (r,\theta ,z\right ) =\left \{ \begin {array} [c]{l}J_{n}\left (kr\right ) \ \sin n\theta \ e^{-kz}\\ \\ J_{n}\left (kr\right ) \ \cos n\theta \ e^{-kz}\end {array} \right . \] Now usually we eliminate the \(\theta \) dependency if boundary condition is such that it is not dependent of angle. So we get \(u\left (r,\theta ,z\right ) =J_{0}\left (k_{m}r\right ) \ \ e^{-k_{m}z}\) and from this we need to solve \(u=\sum _{m=1}^{\infty }c_{m}\ J_{0}\left (k_{m}r\right ) \ \ e^{-k_{m}z}\), now we use boundary condition to ﬁnd \(c_{m}\), for example if given that base (\(z=0\)) was at temp (or potential) = 100, then we need to solve \(100=\sum _{m=1}^{\infty }c_{m}\ J_{0}\left (k_{m}r\right ) \ \) and here to use orthogonality of bessel functions to expand RHS.

2.4.3.2 Laplace equation in spherical coordinates

The Laplacian in spherical is \(\nabla ^{2}u=\frac {1}{r^{2}}\frac {\partial }{\partial r}\left (r^{2}\frac {\partial u}{\partial r}\right ) +\frac {1}{r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left (\sin \theta \ \frac {\partial u}{\partial \theta }\right ) +\frac {1}{r^{2}\sin \theta }\frac {\partial ^{2}u}{\partial \phi ^{2}}=0\). Separate using \(u=R\left ( r\right ) \Theta \left (\theta \right ) \Phi \left (\phi \right ) \)

The solutions are \(\Phi =\left \{ \begin {array} [c]{l}\sin m\phi \\ \cos m\phi \end {array} \right . \) and \(\Theta =P_{l}^{m}\left (\cos \theta \right ) \) and \(R\left ( r\right ) =\left \{ \begin {array} [c]{l}r^{l}\\ r^{-l-1}\end {array} \right . \) here \(l\) is an integer (came from separation of constants by setting \(k=l\left (l+1\right ) \)), Here \(P_{l}^{m}\) is the associated Legendre function.

Now we quickly discard solution \(r^{-l-1}\) because we want solution inside the sphere, so our set of candidate solutions are \(u=R\relax (r) \Theta \left (\theta \right ) \Phi \left (\phi \right ) =r^{l}\ \ \ P_{l}^{m}\left (\cos \theta \right ) \ \ \left \{ \begin {array} [c]{l}\sin m\phi \\ \cos m\phi \end {array} \right . \). For symmetry w.r.t. \(\phi \) we set \(m=0\) and solution reduces to \(r^{l}\ \ \ P_{l}\left (\cos \theta \right ) \ \ \) and then the general solution is \(u=\sum c_{l}\ r^{l}\ \ \ P_{l}\left (\cos \theta \right ) \)

2.4 MATH 121B Notes