Past Paper 2 Flashcards
explain how equations of the form a(x,y,u)du/dx + b(x,y,u)du/dy = c(x,y,u) can be solved by integrating along characteristic lines
- the equation can be written as (a,b) . ∇u = c
- (a,b) . ∇u is the derivative in the direction (a,b)
- this equation is one for the rate of change in a characteristic direction parallel to (a,b)
- if ξ is introduced as a variable that varies along the line derived by following the (a,b) direction, the equation becomes an ODE for u
what is a well-posed problem in the context of second order differential equations
- a problem is well posed if:
- the solution exists
- the solution is unique
- and it depends continuously on the data
- in other words, its robust and stable to changes in the boundary condition data
what is the maximum principle in the context of ∇^2(u) = 0 in a region V, with u = U(x) for x on the boundary S surrounding V
- the maximum principle for ∇^2(u) = 0 states that the max and min of u will occur on the boundary
- this means that any changes to u in the interior are automatically smaller than changes in U
for ∇^2(u) = 0, what does it equal if u1 and u2 are solutions and why
- if u1 and u2 are solutions then ∇^2(u1 - u2) = 0 in V
- and u1 - u2 = U - U = 0 on S
- the max and min values of u1 - u2 are thus 0
- that means that if the solution exists, its unique
if a function φ satisfies ∇^2(φ) = f in a region V, and dφ/dn = Ψ(x) on the boundary S surrounding V, what conditions must be applied to make this problem well-posed
- int[∇^2(φ) dV] = int[dφ/dn dS]
- it needs φ to be specified at some point on S to make the solution unique
the vector r satisfies the system of equations dr/dt + A*dr/dx = 0 where A is a 3x3 matrix. how can you show that the system of equations is hyperbolic
- by proving all the eigenvalues of A are real
- using det(A - Iλ) = 0
how would you find the characteristic speeds if the matrix A had diagonals of u(x,t) and some c(x,t)
- let Wdw = dr
- sub into the original expression
- get dw/dt on its own to give W^-1AW on dw/dx
- if W^-1AW can be chosen to be diagonal, then the system equation takes the form dw/dt + λ_i*dw/dx = 0
- then you find det(A - Iλ) = 0 and the λs are your characteristic speeds
what are the two possible expressions for int[u∇^2G - G∇^2u] dV
- int[u∇^2G - G∇^2u] dV = int[udG/dn - Gdu/dn] dS
- int[u∇^2G - G∇^2u] dV = int[∇ . (u∇G - G∇u)] dS = int[ (u∇G - G∇u).n] dS
- these are the same thing just different steps
if youre told to prove that the greens function for a certain case is given by an expression, what are you trying to do
- do some algebra to prove that G = 0 all things considered
when doing green function questions and working on r = a, what is a relationship you can abuse
- if r = a, then the radius and normal to the circles surface are pointing in the same direction
- so d/dn = d/dr
- this is useful in case you have dG/dn and need dG/dr
if u(x0) is written as a function of G(x,x0) and u(x) with integrals and derivatives of both functions, how can that be used to derive solutions for u which satisfy certain boundary conditions
- if G is chosen to satisfy the boundary condition G = 0 for x, then the expression for u(x0) with G = 0 is a solution for u(x)
how do you go about deriving the general representation theorem for greens function questions
- start with int[u∇^2G - G∇^2u] dV = int[udG/dn - Gdu/dn] dS
- you know ∇^2G = δ(x - x0)
- you’ll be given ∇^2u = something
- insert and rearrange
- remember the evaluation of δ(x - x0) = 1 in an integral is at x0 for any other thing thats a function of x
for second order PDEs, what are the steps for showing that ξ = … and n = … represent characteristic variables for a given equation
- write x and y as functions of ξ and n alone
- find dx/dξ, dx/dn, dy/dξ and dy/dn as functions of x and y
- find d/dξ using chain rule (a pair for dx diagonals and a pair for dy diagonals)
- do the same for d/dn
- find d^2u/dξdn = d/dξ * d/dn * u
- relate it to the original expression to see what it equals
- youre showing that with these variables, the equation becomes the standard form and so ξ and n are characteristics
for second order PDEs, after you have found d^2u/dξdn, how would you find the general solution u
- you should have d^2u/dξdn = … c where c is a constant
- integrate wrt ξ to get du/dn = cξ + A(n) where A(n) is a constant thats a function of n
- integrate again wrt n to get u = cξn + F(ξ) + G(n) where F(ξ) is a constant thats a function of ξ
- input the expressions for ξ and n in this to get u as a function of x and y
after having found the general solution, how would you solve for u subject to boundary conditions, eg u(1,y) = 0 and du(1,y)/dx = -y-1
- use b.c. u(1,y) = 0 for inputs into the general solution to get an expression
- use b.c. du(1,y)/dx = -y-1 for inputs separately to get another expression
- usually one of these will give you the answer for what the function F(ξ) or G(n) is
- then you can use it in the other expression to find the other function
- input found functions into the general solution to get u(x,y)
