Differential Equations

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 2019/04/04   Share

Key concepts

Linear operator
Homogeneous:
- equation: no forcing, or, a multiple of a solution $y$ , $\alpha y$ is also a solution.
- problem: in addition, $\alpha y$ also satisfies B.C.
Boundary conditions
- Dirichlet: the function value on the boundary is defined
- Neumann: the normal derivative value on the boundary is defined

ODE

Homogeneous 2nd order equations

$y''+py'+qy=0$

Independent solutions

Two solutions $y_1$ and $y_2$ are independent iff $Ay_1+By_2=0 \iff A=B=0$ .

Then the general solutions is $Ay_1+By_2$ .

Wronskian

If two solutions are independent, then the Wronskian $W\neq 0$ .
It is intrinsic to an ODE - depends only on $p$ . $W=\exp(-\int p\,dx)$

Proof by considering $W'$ .

The Wronskian can be used to find a second solution when we know one already.

$\frac{W}{y_1^2}=\frac{y_2'}{y_1}-\frac{y_2 y_1'}{y_1^2}=\frac{d}{dx}\frac{y_2}{y_1}\\ \implies y_2=y_1\int\frac{W}{y_1^2}\,dx$

Series solutions

The point about which to expand

Ordinary point if both $p(x)$ and $q(x)$ are analytical; singular point otherwise.
If $(x-x_0)p(x)$ and $(x-x_0)^2q(x)$ are analytical, then regular singular point.

About an ordinary point

Use a Taylor series

$y=\sum_{n=0}^\infty a_n(x-x_0)^n\\ p(x)=\sum_{n=0}^\infty p_n(x-x_0)^n\\ q(x)=\sum_{n=0}^\infty q_n(x-x_0)^n$

and then compare coefficients - we get

$(n+2)(n+1)a_{n+2}+\sum_{r=0}^n p_{n-r}(r+1)a_{r+1}+\sum_{r=0}^n q_{n-r}a_{r}=0$

About a regular singular point

Fuchs’s theorem: at least one solution of the form

$\begin{align} y=\sum_{n=0}^\infty a_n(x-x_0)^{n+\sigma}&&a_0 \neq 0 \end{align}$

Match coefficients with this trail solution, when $n=0$ , we have:

Indicial equation: $\sigma(\sigma-1)+P_0\sigma+Q_0=0$ , where $P=(x-x_0)p;\, Q=(x-x_0)^2q$ .

All is good when $\sigma_1-\sigma_2\notin \mathbb{Z}$ , we have two independent solutions
If $\sigma_1-\sigma_2\in \mathbb{Z}$ , and $\Re(\sigma_1)\geq \Re(\sigma_2)$ , then the solution with $\sigma_2$ will fail (e.g. infinite coefficient).
The second solution may be found by $y_2=y=\sum_{n=0}^\infty b_n(x-x_0)^{n+\sigma_2}+ky_1\ln(x)$ .

Inhomogeneous Problems: Green’s function

Green’s Functions: idea

Idea: (denote the linear PDE by $\mathcal{L}u(\mathbf{r})=f(\mathbf{r})$ ,) We can break up the forcing $f$ into infinite point source forcing $\delta(\mathbf{r}-\mathbf{r_0})$ . If we can find the Green’s function that satisfies

$\mathcal{L}G(\mathbf{r},\mathbf{r_0} )=\delta(\mathbf{r}-\mathbf{r_0}),$

which corresponds to the response of the system due to each of the point source, then we can integrate the effects of these point sources,

$u(\mathbf{r})=\int G(\mathbf{r},\mathbf{r_0} )f(\mathbf{r_0})d^V(\mathbf{r_0})$

to obtain the solution.

We can also understand it as the inverse operator of $\mathcal{L}$ .

The methods depends highly on the boundary conditions given.

Aside: General functions

Functions that has an effect on other functions. Their own specific definition (eg values at discontinuities) is not important.

It is meaningless to multiply together two general functions with the same variable.

Heaviside step function

$H(x)= \begin{cases} 0 &&x<0\\ 1 &&x>0 \end{cases}$

Top hat function

$\delta_\epsilon (x)= \begin{cases} \frac{1}{\epsilon} && 0<x<\epsilon\\ 0 && \mathrm{otherwise} \end{cases}$

Connection: $\lim_{\epsilon \to 0}\delta_\epsilon(x)=H'(x)$

Dirac-Delta function

Has a sampling effect - when integrated, extract the value of a function at a certain point.

$\int_{-\infty}^\infty f(x)\delta(x-\xi)\,dx=f(\xi)$

Note: the delta function itself could be defined smoothly, e.g. a Gaussian

$\delta(x)=\lim_{\epsilon\to 0}\frac{1}{\sqrt{2\pi\epsilon^2}}\exp(-\frac{x^2}{2\epsilon^2})$

D.E.s containing $\delta$ function - matching conditions

Example: $y''+y=\delta(x)$

For $x<0$ and $x>0$ , the equation is homogeneous, hence we have the usual solution:

$y=\begin{cases} A\sin(x)+B\cos(x)&& x<0\\ C\sin(x)+D\cos(x)&& x>0 \end{cases}$

But as a 2nd order DE, we can only have 2 arbitrary constants. Integrate the orginal equation about the origin:

$\int_{-\epsilon}^\epsilon y''\,dx+\int_{-\epsilon}^\epsilon y\,dx=\int_{-\epsilon}^\epsilon \delta(x)\,dx\\ \implies y'(\epsilon)-y'(-\epsilon)+0=1$

where we assumed $y$ is bounded. Hence the relations between $A,B$ and $C,D$ can be found.

Procedure of solving inhomogeneous PDE with Green’s Functions

Find 2 independent homogeneous solutions, $y_1$ and $y_2$ .
Work out the Wronskian

$W= \begin{vmatrix} y_1&y_2\\y_1'&y_2' \end{vmatrix}.$

Solve for $G$ by applying B.C. (same as for $y$ ) and jump condition.
Integrate to find $y$ .

B.C. of initial value problems

The boundary conditions is of the form $y(0)=m; \ y'(0)=n$ .

Suppose the problem is $y''(x)+py'(x)+qy(x)=f(x)$ , subject to $y(0)=y'(0)=0$ .

Then Greens function solves:

$G_{xx}+pG_x+qG=\delta(x-\xi), \tag{*}$

subject to $G(0,\xi)=\partial_xG(0,\xi)=0$ .

Again, by integrating, $G$ must be continuous and the first derivative has a jump. At $x=\xi$ ,

$\begin{align} [G]=0 && [\partial_x G]=1 \end{align}$

Again, the equation is homogeneous when $x\neq\xi$ , so

$G=\begin{cases} A(\xi)y_1(x)+B(\xi)y_2(x)&& 0<x<\xi\\ C(\xi)y_1(x)+D(\xi)y_2(x)&& x>\xi \end{cases}$

Apply B.C.s:

$\begin{align} x=0 && \begin{bmatrix}y_1(0)&y_2(0)\\y_1'(0)&y_2'(0)\end{bmatrix}\begin{bmatrix}A(\xi)\\B(\xi)\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\implies A=B=0\\ x=\xi && \begin{bmatrix}y_1(\xi)&y_2(\xi)\\y_1'(\xi)&y_2'(\xi)\end{bmatrix}\begin{bmatrix}C(\xi)\\D(\xi)\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}\implies \begin{bmatrix}C(\xi)\\D(\xi)\end{bmatrix}=\begin{bmatrix}-y_2(\xi)/W(\xi)\\y_1(\xi)/W(\xi)\end{bmatrix} \end{align}$

Therefore,

$G(x,\xi)=\begin{cases}0 && 0\leq x<\xi \\ \frac{y_1(\xi)y_2(x)-y_1(x)y_2(\xi)}{W(\xi)}&&x\geq \xi\end{cases}$

Boundary value problem

Find complementary functions $y_a(x)$ and $y_b(x)$ satisfying the boundary conditions at $a$ and $b$ respectively. The Green’s function must be

$G=\begin{cases} A(\xi)y_a(x)&& a\leq x\leq\xi\\ B(\xi)y_b(x)&& \xi<x\leq b \end{cases}$

with jump conditions $[G]=0$ and $[\partial_x G]=1$ at $x=\xi$ .

By solving these, the Green’s function is

$G=\frac{1}{W(\xi)}\begin{cases} y_b(\xi)y_a(x)&& a\leq x\leq\xi\\ y_a(\xi)y_b(x)&& \xi<x\leq b \end{cases}$

Sturm-Liouville theory

Inner product of functions

The inner product of two continuous functions $u$ and $v$ , on a range $x \in (a, b)$ is

$<\!\!u|v\!\!>=\int_a^b u^*(x)v(x)\, dx$

Generalise a little by introducing a weight function $w(x)>0$ on $(a,b)$ .

$<\!\!u|v\!\!>_w=\int_a^b u^*(x)v(x)w(x)\, dx$

The subscript $w$ can be omitted.

Self-adjoint and SL operator

The adjoint of a differential operator $\mathcal{L}$ such that (after integration by parts,)

$<\!\!u|\mathcal{L}v\!\!>=<\!\!\mathcal{L}^\dagger u|v\!\!>+ \mathrm{b.t.s}$

An operator is self-adjoint if $\mathcal{L}=\mathcal{L}^\dagger$ , and the boudary terms are 0.

A Sturm-Liouville operator, defined on a range $$\alpha \leq x \leq \beta$ is of the form:

$\mathcal{L}=-\frac{d}{dx}(\rho(x)\frac{d}{dx})+\sigma(x)$

where $\sigma$ and $\rho$ are real functions, and $\rho(x)>0$ for $x \in (\alpha, \beta)$ .

Prove an SL operator is self-adjoint by integrating by part twice.

SL operators resembles Hermitian matrices, and they are useful because they have orthogonal eigenfunctions.

Into SL form

Any 2nd order ordinary differential operator can be put into a SL form.

Consider a general 2nd order operator

$\begin{align} \tilde{\mathcal{L}}y(x)=\lambda y(x)&=a(x)\frac{d^2}{dx^2}+b(x)\frac{d}{dx}+c(x)\\ \mathcal{L}=w(x)\tilde{\mathcal{L}}&=w(x)a(x)\frac{d^2}{dx^2}+w(x)b(x)\frac{d}{dx}+w(x)c(x)\\ \end{align}$

If this is a SL form, then

$\begin{align} \mathcal{L}&=-\frac{d}{dx}(\rho(x)\frac{d}{dx})+\sigma(x)\\ &=-\rho(x)\frac{d^2}{dx^2}-\frac{d\rho}{dx}\frac{d}{dx}+\sigma(x) \end{align}$

Compare and we have

$\begin{align} a(x)w(x)&=-\rho(x)\\ b(x)w(x)&=-\frac{d\rho}{dx}\\ w(x)c(x)&=\sigma(x) \end{align}$

Hence we can solve

$\begin{align} \rho(x)&=-\exp(\int \frac{b}{a})\\ w(x)&=\frac{-\rho(x)}{a(x)}\\ c(x)&=\frac{\sigma(x)}{w(x)} \end{align}$

Eigenvalues and eigenfunctions

Now if a solution $y(x)$ satisfies the eigenvalue equation

$\tilde{\mathcal{L}}y(x)=\lambda y(x)$

then

$\mathcal{L}y(x)=\lambda w(x)y(x)$

We can prove the reality of eigenvalues and orthogonality of eigenfunctions by the same method as in matrices.

Eigenfunction expansion

Let ${y_n}$ be a set of orthonormal eigenfunctions, satisfying the b.c.s, of a self-adjoint operator, w.r.t. a weight function $w$ .

$<\!\!y_n | y_m\!\!>_w = \delta_{nm}$

Any function $f$ satisfying b.c.s may be expressed as

$f(x) = \sum_{n=1}^{\infty}a_n y_n(x) \tag{*}$

and

$a_n=<\!\!y_n | f\!\!>_w$

Completeness relation

Using the above two statements

$\begin{align} f(x) &= \sum_{n=1}^{\infty}a_n y_n(x)\\ &=\sum_{n=1}^{\infty}<\!\!y_n | f\!\!>_w y_n(x)\\ &=\sum_{n=1}^{\infty} y_n(x) \int_\alpha^\beta y_n^*(\xi) f(\xi) w(\xi) \, d\xi\\ &=\int_\alpha^\beta f(\xi) [w(\xi)\sum_{n=1}^{\infty}y_n(x)y_n^*(\xi)] \, d\xi \\ &=\int_\alpha^\beta f(\xi) \delta (x-\xi) \, d\xi \end{align}$

Hence we have the completeness relation, which defines the meaning of “having a complete set of funtions”.

$\sum_{n=1}^{\infty}y_n(x)y_n^*(\xi)=\frac{1}{w(\xi)}\delta (x-\xi)$

Solutions of DEs

How can eigenfunction expansion be used to solve differential equations?

We have the problem

$\mathcal{L}y(x)=f(x)+ \mathrm{b.c.s}$

and we have the eigenfunctions

$\mathcal{L}y(x)=\lambda w(x)y(x)\\ <\!\!y_n | y_m\!\!>_w = \delta_{nm}$

The Green’s function solutions is

$y(x)=\int_\alpha^\beta G(x ; x') f(x') \, dx'$

If we make

$G(x ; x')=\sum_{n=1}^{\infty} \frac{1}{\lambda_n}y_n(x)y_n^*(x') \tag{**}$

then

B.C.s are satisfied
The point-source response if recoverd.

$\begin{align} \mathcal{L}_x G(x ; x')&=w(x) \sum_{n=1}^{\infty} \lambda_n \frac{1}{\lambda_n}y_n(x)y_n^*(x')\\ &=\frac{w(x)}{w(x')}\delta (x-x')\\ &=\delta (x-x') \end{align}$

Note that this will fail if any of the eigenvalues is zero. But if $\lambda_1$ is very small compared to others,

$y(x)=\int_\alpha^\beta \sum_{n=1}^{\infty} \frac{1}{\lambda_n}y_n(x)y_n^*(x') f(x') \, dx' \approx \frac{y_1(x)}{\lambda_1} <\!\!y_1 | f\!\!>$

as long as $<\!\!y_1 | f\!\!>$ is not too small.

Any forcing with non-zero $y_1$ component will cause a large “resonant” response.

Error and Paresval

If the expansion is truncated,

$f(x) \approx \sum_{n=1}^N a_n y_n(x)$

and define the error as

$\begin{align} E_N&=||f(x)-\sum_{n=1}^N a_n y_n(x)||_w^2\\ &=||f||_w^2-\sum_{n=1}^N[<\!\!f |y_n\!\!>_w a_n- a_n^*<\!\!y_n | f\!\!>_w]+\sum_{n=1}^N\sum_{n=1}^N a_n^* a_m <\!\!y_n | y_m!\!\!>_w\\ &=||f||_w^2-\sum_{n=1}^N[<\!\!f |y_n\!\!>_w a_n- a_n^*<\!\!y_n | f\!\!>_w]+\sum_{n=1}^N a_n^* a_n \end{align}$

The partial derivatives

$\frac{\partial E_N}{\partial a_n^*}=a_n-<\!\!y_n | f\!\!>_w\\ \frac{\partial E_N}{\partial a_n}=a_n^*-<\!\!f | y_n\!\!>_w\\$

Of course the error is minimised when $a_n=<\!\!y_n | f\!\!>_w$ , i.e. untracated,

$E_N=||f||_w^2-\sum_{n=1}^N |a_n|^2 \geq 0\\ \implies ||f||_w^2 \geq \sum_{n=1}^N |a_n|^2 \tag{Bessel's inequality}$

with the equality taking at the limit of $N \to \infty$ ,

$||f||_w^2 = \sum_{n=1}^N |a_n|^2 \tag{Parseval's theorem}$

PDE

Laplace and Poisson’s equations

Physical origins

Poisson’s equation is

$\nabla^2 \Psi = \rho(\mathbf{x})$

Examples

$\begin{align} &\mathrm{Diffusion \,equation} &\kappa \nabla^2 u&=\frac{\partial u}{\partial t}\\ &\mathrm{Diffusion \,equation \,with \,source} &\kappa \nabla^2 u&=\frac{\partial u}{\partial t}-S(\mathbf{x})\\ &\mathrm{Electrostatics} &\nabla^2 \Phi&=-\rho(\mathbf{x})/\epsilon_0\\ &\mathrm{Gravitation} &\nabla^2 \Phi&=4\pi G\rho(\mathbf{x})\\ &\mathrm{Schrodinger} &-\frac{\hbar^2}{2m}\nabla^2 \Psi + V(\mathbf{x})\Psi&=E\Psi\\ &\mathrm{Ideal \,fluid \, flow} &\nabla^2 \Phi&=0\\ \end{align}$

Keyword in questions is ‘steady state’.

Separation of variables

Laplace equation is linear in $\Psi$ , so the superposition of any solutions is another solution. Separation of variables in orthogonal coordinates is a method to find basis solutions.

Laplace in Plane polar

The equation becomes

$\nabla^2 \Psi = \frac{1}{r}\frac{\partial}{\partial r} (r \frac{\partial \Psi}{\partial r}) + \frac{1}{r^2}\frac{\partial^2 \Psi}{\partial \phi^2}=0$

Still consider separable solutions $\Psi=R(r)\Phi(\phi)$ , giving

$\begin{align} &\frac{\Phi}{r}\frac{d}{dr}(rR')+\frac{R}{r^2}\Phi''=0\\ \implies & \frac{r}{R}\frac{d}{dr}(r R')=-\frac{\Phi''}{\Phi}=\lambda\\ \end{align}$

Solve the two independent ODEs, with the B.C. that $\Phi(\phi)=\Phi(\phi + 2 \pi)$ ,

$\Phi= \begin{cases} A+B\phi & n=0 \\ A\cos(n\phi)+B\sin(n\phi) & n \neq 0 \end{cases}\\ R= \begin{cases} C+D \ln r& n=0 \\ Cr^n + Dr^{-n} & n \neq 0 \end{cases}\\$

The equation for $R$ can be solved by $u=\ln r$ .

Therefore, the general solution is:

$\Psi = A_0 + B_0 \phi + C_o \ln r + \sum_{n=1}^\infty (A_n r^n+ C_n r^{-n})\cos n\phi + \sum_{n=1}^\infty (B_n r^n+ D_n r^{-n})\sin n\phi \tag{*}$

Eliminate some constants by requiring finite values at $r=0$ or $r=\infty$ .

If there is a barrier, there is no radial component of the flow (1st derivative).

Laplace in spherical polar (axisymmetric)

Axisymmetric means no $\phi$ dependence.

The Laplace equation is

$\nabla^2 \Psi = \frac{1}{r^2}\frac{\partial}{\partial r} (r^2 \frac{\partial \Psi}{\partial r}) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial \Psi}{\partial \theta})=0$

Again, try separable solutions $\Psi=R(r)T(\theta)$ , then

$\frac{1}{R}(r^2R')'=-\frac{1}{T \sin\theta}(T'\sin\theta)'=\lambda$

Rearranging $T$ , with subsitution of $u=\cos \theta$ ,

$\begin{align} \frac{d}{du}((1-u^2)\frac{d T}{du})+\lambda T =0 && \mathrm{Legendre \, equation} \end{align}$

Therefore, $\lambda=l(l+1)$ , and $T=P_l(u)$ .

Then the equation for $R$ is

$(r^2 R')'=l(l+1)R\\ \implies r^2 R''+2rR'-l(l+1)R=0\\ \implies R=Ar^l+Br^{-l-1}$

Solve by $v=\ln r$ .

Therefore, the general solution is:

$\Psi = \sum_{l=0}^{\infty}(A_l r^l+ B_l r^{-l-1})P_l(\cos \theta) \tag{*}$

Remember when $P_l$ is normalised as $P_l(1)=1$ , the orthogonal relation is $\int_{-1}^{1} P_m(u)P_n(u)\, du=2/(2m+1)\delta_{mn}$ .

Uniqueness of solutions to Poisson’s equation

The key identity to use is the Divergence theorem and

$\nabla \cdot (\Psi\nabla\Psi)=|\nabla\Psi|^2+\Psi\nabla^2\Psi$

Green’s function for Poisson’s equation

It is defined that the Green’s function satisfies:

$\begin{align} \nabla_\mathbf{r}^2G(\mathbf{r},\mathbf{r}')&=\delta^3(\mathbf{r}-\mathbf{r}') && \mathbf{r}\, \mathrm{in \, V, }\\ G&=0 && \mathbf{r} \,\mathrm{on \, S \, for \, Dirichlet, \, or}\\ \frac{\partial G}{\partial n}&=\frac{1}{A}&& \mathrm{for \, Neumann} \end{align}$

Fundamental solution in 3D for Dirichlet

Proof required

The fundamental solution is when $V$ is all space. Wlog consider $\mathbf{r}'=\mathbf{0}$ . We know $G$ is a function of $\mathbf{r}$ only.

$\begin{align} & & (r^2G')'&=0\\ &\implies & r^2G'&=C\\ &\implies & G&=\frac{C}{r}+A\\ \end{align}$

We know $A=0$ from the B.C at infinity; to determine C,

$\int_{r<\epsilon}\nabla^2G\, dV = \oint_{r=\epsilon} \nabla G\cdot\hat{\mathbf{n}}\, dS = \oint_{r=\epsilon} \frac{\partial G}{\partial r} \, dS=\oint_{r=\epsilon} \frac{-c}{r^2}\, dS=-4\pi C$

Because we can take $\epsilon$ as small as we like,

$\nabla^2G=-\pi C \delta^3(\mathbf{r})\\ \implies C=-\frac{1}{4\pi}$

Hence, the fundamental solution is

$G(\mathbf{r},\mathbf{r}')=-\frac{1}{4\pi|\mathbf{r}-\mathbf{r}'|}$

Fundamental solution in 2D

Similar to above, we get

$G(\mathbf{r},\mathbf{r}')=\frac{1}{2\pi}\ln|\mathbf{r}-\mathbf{r}'|+ \mathrm{const.}$

Note: divergence theorem in 2D is used.

Method of images

Construct solutions for different geometries from fundamental solutions.

Planes

Remove the plane and replace with a source at the image point

With the opposite sign for Dirichlet (0 potential at the boundary)
With the same sign for Neumann (0 normal derivative at the boundary; insulator)

Sphere (radius a) (Dirichlet)

Strength = $-a/r'$

Position $\mathbf{r}''=\frac{a^2}{r'^2}\mathbf{r}'$

Circle (radius a) (Dirichlet)

Strength = -1

Position $\mathbf{r}''=\frac{a^2}{r'^2}\mathbf{r}'$

Remember $r'r''=a^2$ .

Integral solution of Poisson’s equation

Consider this. Using $\mathbf{F}=\Phi\mathbf{\nabla}\Psi-\Psi\mathbf{\nabla}\Phi$ (divergence theorem) and $\mathbf{\nabla}\cdot(\Phi \mathbf{\nabla}\Psi)=\mathbf{\nabla}\Phi\cdot\mathbf{\nabla}\Psi + \Phi\nabla^2\Psi$ , we can obtain Green’s identity

$\int_V (\Phi\nabla^2\Psi -\Psi\nabla^2\Phi)\, dV = \oint_S (\Phi\mathbf{\nabla}\Psi-\Psi\mathbf{\nabla}\Phi) \cdot \mathbf{n}\, dS= \oint_S(\Phi\frac{\partial \Psi}{\partial n}-\Psi\frac{\partial \Phi}{\partial n})\, dS$

Then, put in $\Psi=G$ (with Dirichlet B.C.),

$\begin{align} \int_V (\Phi\nabla^2G -G\nabla^2\Phi)\, dV &= \oint_S(\Phi\frac{\partial G}{\partial n}-G\frac{\partial \Phi}{\partial n})\, dS && \mathrm{copy \ from \ above}\\ \int_V (\Phi\,\delta(\mathbf{r}-\mathbf{r}') -G\,\rho(\mathbf{r}))\, dV &= \oint_S(f(\mathbf{r})\frac{\partial G}{\partial n}-0\cdot\frac{\partial \Phi}{\partial n})\, dS && \mathrm{put \ in \ knowns}\\ \int_V \Phi\,\delta(\mathbf{r}-\mathbf{r}') &=\int_VG\,\rho(\mathbf{r})\, dV +\oint_Sf(\mathbf{r})\frac{\partial G}{\partial n}\, dS && \mathrm{move \ a \ term}\\ \Phi(\mathbf{r}') &=\int_VG(\mathbf{r} ; \mathbf{r}')\,\rho(\mathbf{r})\, dV +\oint_Sf(\mathbf{r})\frac{\partial G}{\partial n}\, dS && \mathrm{integrate \ the \ \delta} \end{align}$

Notes:

Can use this to solve Laplace - set $\rho$ to 0
If $V$ is all space, then need to make sure the surface integral does go to 0

For Neumann B.C.,

$\begin{align} \int_V \Phi\nabla^2G\, dV &= \int_V G\nabla^2\Phi\, dV+\oint_S(\Phi\frac{\partial G}{\partial n}-G\frac{\partial \Phi}{\partial n})\, dS && \mathrm{copy \ from \ above}\\ \int_V \Phi\delta(\mathbf{r}-\mathbf{r}')\, dV &= \int_V G\rho(\mathbf{r})\, dV+\oint_S(\frac{1}{A}\Phi-Gf(\mathbf{r}))\, dS && \mathrm{sub \ in \ knowns}\\ \Phi(\mathbf{r}') &= \int_V G\rho(\mathbf{r})\, dV+\oint_S(\frac{1}{A}\Phi-Gf(\mathbf{r}))\, dS && \mathrm{integrate \ the \ \delta}\\ \end{align}$

Still seems undoable. But if $V$ is all space, and the surface integral of $\Phi$ is finite, then

$\Phi(\mathbf{r}') = \int_V G(\mathbf{r} ; \mathbf{r}')\rho(\mathbf{r})\, dV-\oint_SG(\mathbf{r} ; \mathbf{r}')f(\mathbf{r})\, dS$

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Author：2LalA猪

Permalink：https://butteraddict.fun/2019/04/Differential-Equations/

Published on：4th April 2019, 5:33 pm

Updated on：9th June 2020, 3:23 pm

License：CC BY 4.0

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CATALOG

1. Key concepts
2. ODE
3. PDE
1. 3.1. Laplace and Poisson’s equations

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