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)
ut+Фx-f=0 conservation law states this must equal zero
Change of Coordinates->ξ=x-ct, τ=t ; u(x,t)->U(ξ, τ)
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:
, 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)
r=
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
Heat Kernel/Fundamental Solution
G(x,t)=
Temperature surface from x=0 and t=0
u(x,t)=
erf(z)=
erfc=
=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)=
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)=
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)=
where w(x,t;τ) solves
wt=kwxx
w(x,0;τ)=f(x, τ)
u(x,t)=
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)=
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
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 |