Understanding Partial Differential Equations: Key Concepts and Applications

Key Partial Differential Equations (PDEs)

Wave Equation:                       utt-c2uxx=0

Laplace’s Equation:                 uxx+uyy=0

Heat Equation:                         ut=kuxx

Diffusion Equation:                  ut-Duxx=0

Transport/Advection Equation: ut+cux=0

Advection-Diffusion Equation: ut+cux-Duxx=0

Advection-Diffusion-Decay Equation:

                                                    ut+cux-Duxx=-ʎu

Completing the Square

ax2+bx+c=0 to a(x+d)2+e=0

d=b/2a

e=c-b2/2a

Section 1.1: PDE Models

PDE integration yields arbitrary functions, whereas ODE integration yields constants.

Cauchy Problems: ϵ x, no boundary conditions, initial condition(s) are needed.

Operator Functions

Linear when: L(u+w)=L(u)+L(w) & L(cu)=cL(u)

Section 1.2: Conservation Laws

Conservation law: The rate at which a quantity changes is equal to the rate of flow (flux) across the boundary plus the rate at which it is created or destroyed.

  • u=density (amount/volume)
  • AФ=flux, amount crossing cross section at location x at time t (positive if to the right, negative if to the left)
  • f(x,t)=sink (negative) or source (positive)

utx-f=0 conservation law states this must equal zero

Change of Coordinates->ξ=x-ct, τ=t   ;   u(x,t)->U(ξ, τ)

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 ut=Uξ ξt+Uττt, ux=Uξ if ξ=x-ct, τ=t

For example:

ut+cux+au=f -> Uτ+aU=F(ξ, τ)

 if ut+ux-3u=t -> Uτ-U=τ -> integrate and then return to normal coordinates

When c=f(x,t), dx/dt=c, separate and integrate and solve for constant.

For example:

if ut+xtux=0, c=xt

dx/dt=xt

dx/x=tdt

lnx=t2/2+c

c=ξ=lnx-t2/2                            

Section 1.3: Diffusion

Fick’s Law: Ф=-Dux

Fourier’s Law: Ф=-KƟx (K=thermal conductivity, Ɵ(x,t)=temperature assuming energy is proportional to temperature)

If uxx>0, or concentration profile is concave up, ut>0

Dirichlet Condition: u(0,t)=g(t), t>0

Neumann Condition: =Kux(0,t)=h(t), t>0

            If h(t)=0 it is insulated and homogeneous.

Newton’s Law of Cooling: -Kux(0,t)=-ɳ(u(0,t)-Ѱ(t)), t>0

            Ѱ(t)=environment temperature

            ɳ=constant of proportionality (heat loss factor)

Steady state-> time derivatives vanish, integrations are constants not functions.

Section 1.4: Diffusion and Randomness

Fischer’s Equation: ut=Duxx+ru(1-u/K)

  • Advection graph– shifts
  • Diffusion graph- “smears”
  • Advection diffusion– shifts and “smears”

Point source=fundamental solutions-> those that satisfy boundary conditions=Green’s functions

Section 1.5: Vibrations and Acoustics

Often only assume movement in one direction (string being plucked, assume no horizontal, just vertical movement).

General Solution to the Wave Equation (utt=c2uxx): u(x,t)=F(x-ct)+G(x+ct)

Section 1.7: Heat Conduction in Higher Dimensions

The Laplacian: Δu= ∇∇u= ∇2u=uxx+uyy+uzz

Gradient acts on a scalar to make a vector:

grad(u)= ∇u=(ux,uy,uz) ; u is scalar, ∇u is vector.

Divergence acts on a vector to make it scalar:

div(Ф)= ∇Ф

div(grad(u))= ∇·(ux,uy,uz)= uxx+uyy+uzz

            dot product

Divergence of a scalar and a vector yields a scalar:

div(u*ṽ)= ṽ·grad(u) + udiv(ṽ)

Can’t take divergence of a scalar, take gradient.

dV=dxdydz

Divergence Theorem: nPFPgNuYBZGMOB7tkAAAAASUVORK5CYII= , ɳ=normal vector

Section 1.8: Laplace’s Equation

u(x,y)=(1/4)*[u(x-h,y)+u(x+h,y)+u(x,y-h)+u(x,y+h)] -temperature>

The Maximum Principle: If u is not a constant function, then the maximum and minimum values of u are attained on the boundary of D.

u(x,y)->u(r)

qnibY7jcNd9unOsXvJf96Xr6P9X95i1uijvKoFaF

r=hn6z4B7MsWHVHJcL8AAAAASUVORK5CYII=

ux=(d/dx) u(r)=ur (dr/dx)

uy=(d/dy) u(r)=ur(dr/dy)

Solve for all u derivatives and plug in to original equation; solve for u(r) using boundary conditions.

Section 1.9: Classification of PDEs

Auxx+Buxt+Cutt+F(x,t,u,ux,ut)=0

D=B2-4AC, the discriminant

ξ=ax+bt, τ=cdt

D>0, hyperbolic (wave)

If a=c=1

b=(-B+sqrt(D))/2C

d=(-B-sqrt(D))/2C

if C=0, b=d=1

a=(-B+sqrt(D))/2A

c=(-B-sqrt(D))/2A

D<0, elliptic (Laplace’s)

D=0, parabolic (diffusion)

If a=c=1

d=-B/2C

b=0                                                  

ux= Uξξx+Uττx =aUξ+cUτ

ut = Uξξt+Uττt =bUξ+dUτ

uxx= a2Uξξ+2acUξτ+c2Uττ

utt= b2Uξξ+2bdUξτ+d2Uττ

uxt= abUξξ+(ad+cb)Uξτ+cdUττ

Auxx+Buxt+Cutt= Uξξ(Aa2+Bab+Cb2)+Uξτ(2Aac+B(ad+cb)+C(2bd))+Uττ(Ac2+Bcd+Cd2)

Section 2.1: Cauchy Problem for the Heat Equation

Cauchy problem=pure initial value problem

ABeMNSoY8r35AAAAAElFTkSuQmCC     Heat Kernel/Fundamental Solution

G(x,t)=MKfANAujPEHIGrDQAAAABJRU5ErkJggg==      Temperature surface from x=0 and t=0

u(x,t)=wKkku8HwyxiPAAAAABJRU5ErkJggg==

erf(z)=PS8UH2H7+vwBcrRClLwGzRVQAAAABJRU5ErkJggg

erfc=5HoA5usHt6KwWpgAAAAASUVORK5CYII= =1-erf

*Don’t forget to change limits and derivatives when replacing r value.

Section 2.2: Cauchy Problem for the Wave Equation

u(x,t)=F(x-ct)+G(x+ct) general solution

d’Alembert’s Formula

u(x,t)=AO6T4AO3Cu2gwAAAABJRU5ErkJggg==

u(x,0)=f(x) initial displacement

ut(x,0)=g(x) initial velocity

Section 2.3: Well-Posed Problems

A problem is well-posed if:

1. (existence) it has a solution

2. (uniqueness) the solution is unique

3. (stability) the solution depends continuously on the initial and/or boundary data

If a problem does not meet these requirements, it is ill-posed.

Steps

1.Plug in 0 for t in u(x,0)

2.Set f(x) equal to 1 or another arbitrary number to solve for x.

3.Change n to a large number, like 100, and plug in to f(x) (where t=0).

4.Compare value calculated to value expected (should be close).

5.Now pick arbitrary numbers for t in u(x,t) and set n at the same large number—should be a large number, and incorrect, thus ill-posed.

Section 2.4: Semi-Infinite Domains

Even Functions: f(-x)=f(x)

Odd Functions: -f(x)=f(-x)

Break apart integrals in sections of different initial conditions, use dummy variables for -y when necessary.

Section 2.5: Duhamel’s Principle

PDE contains a source term, f(x,t)

Duhamel’s Principle

The solution to the problem,

y’(t)+ay=F(t), t>0, y(0)=0

is given by

y(t)=eNJjmh8vIXAAAAAElFTkSuQmCC

where w(t, τ) solves the homogeneous problem,

w’(t;τ)+aw(t;τ)=0, t>0; w(0;τ)=F(τ)

u(x,t)=REdkQGtlbDBjeJiMnsKiK8JwFvIzw3Tt8gezb8k3

where w(x,t;τ) solves

wt=kwxx

w(x,0;τ)=f(x, τ)

u(x,t)=3WWZeiNMtveAAAAAElFTkSuQmCC

Section 2.6: Laplace Transforms

Linear

L(u+v)=L(u)+L(v)

L(cu)=cL(u)

L-1(U)=u

Convolution theorem

L(u*v)(s)=U(s)V(s)

Where,

(u*v)(t)=YGB86ISsg41VtNip1X19twB0aW2K77bmemL8a1fw

Replace given equations with Laplace transforms, solve, and revert back.

Must be bounded with exponentials (set function to 0 in front of exponential).

Solve for remaining function with boundary condition(s).

t changes, x does not.

Section 2.7: Fourier Transforms

x changes, t does not.

Schwartz Class of Functions

S=the set of rapidly decreasing functions of all real numbers that have continuous derivatives of all orders.

Definition of S: Rapidly decreasing functions are functions that along with all their derivatives decay to zero as x->infinity faster than any power function.

Integration by Parts

6rL2dUw9U1AAAAAElFTkSuQmCC

u=f(x) given -> take derivative for du

du=f’(x)

v=g(x)

dv=g’(x) given -> integrate for v

1st Order ODE

Separate and Integrate

Integrating Factor

y’+P(x)y=Q(x)

µ(x)=e∫P(x)dx

µ(x)*[y’+P(x)y=Q(x)]

Bernoulli

y’+p(x)y=q(x)yn

v=y1-n

-solve for y in terms of v

-plug in to solve for v

-convert back to y

2nd Order Homogeneous ODE

Auxiliary Equation

am2+bm+c=0

      1. M1≠M2

y=C1*em1*x+C2*em2*x

      2. M1=α+iβ, M2= α-iβ

y=C1*eαx*cos(βx)+ C2*eαx*sin(βx)

3. M1=M2

y=C1*em*x+C2*em*x

Memorize

y’’+a2y=0

y=C1cos(ax)+C2sin(ax)

y’’-a2y=0

y=C1cosh(ax)+C2sinh(ax)

Knowing

cosh(x)= [ex+e-x]/2

sinh(x)= [ex-e-x]/2

2nd Order Non-Homogeneous ODE

y’’+y=f(x)

yc=homogeneous

Method of Undetermined Coefficients (polynomials, cos(ax), sin(ax), eax)

yp=U1y1+U2y2

U1=-∫y2*f(x)/W dx

U2=∫y1*f(x)/W

W=y1*y’2-y2*y’1

Variation of Parameters

yp=f(x)

y’p=f’(x)

y’’p=f’’(x)

Plug in and find yp.

y=yc+yp