Boundary value problems (BVPs)

• Initial value problems (IVP) have initial conditions a starting location (one side of the domain).
  • Solution methods begin at this initial condition and march in time (or space) to evolve the solution.
  • IVP: BC at $t=0$ $\rightarrow$ march in time (or space if we have $x$ instead of $t$, or some other independent coordinate…)
• Boundary value problems involve higher order derivatives with boundary conditions at one or more locations.
  • These are often problems on a spatial domain rather than a temporal domain.
  • BVP: Closed domain (space) with B.C. at both ends.
    • Second order ODE $y^{\prime\prime}$ or higher.

Linear: $$ \frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = F(x).$$

• Example (heat flux with $k = k(x)$) $$\frac{d}{dx}\left(k\frac{dT}{dx}\right) = q(x),,,\rightarrow$$ $$(k)\frac{d^2T}{dx^2} + \left(\frac{dk}{dx}\right)\frac{dT}{dx} = q.$$
  • Can divide through by $k(x)$ to show the linear form above with $P(x)$, $Q(x)$, $F(x)$.

Nonlinear: * Terms have products of $y$, $y^{\prime}$, $y^{\prime\prime}$. * Nonlinear means nonlinear in $y$ or derivatives of $y$, not nonlinear in $x$. * The above is nonlinear if $k=k(T)$.

Boundary conditions: BCs

1. Dirichlet Specify $y$ at boundary
2. Neumann Specify $y^{\prime}$ at boundary
3. Robin Specify additive combination of $y$ and $y^{\prime}$ at boundary.
   • this is sometimes called a mixed boundary condition.

Examples for heat transfer: * Dirichlet? * Neumann? * Robin?

Examples: heat conduction.

1. Dirichlet: Specify $T$
2. Neumann: Specify $q=k\frac{dT}{dx}$
3. Robin: Specify $h(T-T_{\infty}) = -k\frac{dT}{dx}$ * This is conjugate heat transfer, where we have a heat flux conservation boundary condition (equating convection to a wall to conduction through the wall).

Solution approaches include:

• Shooting methods
• Relaxation methods.

Shooting Method

• Second order ODE $\rightarrow$ system of first order ODEs
• Solve by marching, like an IVP.
• Then need initial conditions for each variable.
• But the boundary conditions may not be located at the initial grid position.
• So,
  • guess B.C. as needed at one end of the domain.
  • solve the IVP to the other end of the domain.
  • with the solution, check to see if the boundary condition is satisfied.
  • if not, update the guess and repeat.
• This has the form of a nonlinear problem.
  • We are solving for the unknown boundary condition that gives the known boundary condition.
  • $G(BC_{unknown})=BC_{known}$ or $G(BC_{unknown})-BC_{known}=0$, or
  • $F(BC_{unknown}) = 0$.
  • But our function is not a simple equation. Our function is

put in a value, solve an ODE IVP, take the difference between the known BC, and the BC you got with the guess. Return that difference.

Example, gas stripper

• A counter-current gas stripper flows liquid down a column, and gas up the column.
• We want the concentration of gas and liquid species throughout the column.
• We are given the inlet concentration of the liquid and the inlet concentration of the gas.
• But these are at opposite ends of the column.
• To use an IVP marching method, we would have to guess liquid concentrations at the bottom, then march from the bottom to the top, then compare the liquid concentrations at the top to the known values, then adjust our guesses, and repeat. (Or vice-versa.)

Example: heat conduction

• Conduction in a 1-D bar of length $H$, with some heat source distribution.

• Dirichlet boundary conditions at each end of the bar. $$\frac{d^2T}{dx^2} = f(x),$$ $$T(0) = T_L,$$ $$T(H) = T_R.$$

• Convert to a system of ODEs, for solution via IVP marching method.

  • Let $z=T^{\prime}$ $$\frac{dT}{dx} = z,$$ $$\frac{dz}{dx} = f(x).$$

• So, we don’t know $z$ at $x=0$.
• Guess $z_{0g1}$ at $x=0$.
• March to $x=H$ by solving the ODE system.
• This gives $T_{R1}$, which is probably not $T_R$.
• Guess $z_{0g2}$ at $x=0$ and march to get $T_{R2}$.
• Now, we can interpolate to get the next guess: $$\frac{z_{Og3}-z_{Og2}}{T_R-T_{R2}} = \frac{z_{Og2}-z_{Og1}}{T_{R2}-T_{R1}},,,\rightarrow$$

$$z_{Og3} = z_{Og2} + (z_{Og2}-z_{Og1})\frac{T_R-T_{R2}}{T_{R2}-T_{R1}}.$$

• This solution approach with interpolation amounts to the Secant method.

• Repeat, using this formula to update guesses.

• (Note, if the problem is linear, you’ll hit the target on the first interpolation.)

Shooting is just root finding!

But with a function that involves solving an ODE and then checking the resulting boundary condition against the specified boundary condition.

BC types

• D,D
  • $T_0 = T_L$ $\rightarrow$ $T_0 = T_L$
  • $T_H = T_R$ $\rightarrow$ $\mathbf{z_0 = z_{0,g}}$
    • Guess $z(x=0)$, and shoot for $T(x=H)=T_R$.
• D,N
  • $T_0 = T_L$ $\rightarrow$ $T_0 = T_L$
  • $T^\prime_H = T^\prime_R,,\rightarrow,,\mathbf{z_0 = z_{0,g}}$
    • Guess $z(x=0)$, and shoot for $z(x=H)=T^{\prime}_R$.
• N,D
  • $T_H = T_R$ $\rightarrow$ $\mathbf{T_0 = T_{0,g}}$
  • $T^\prime_0 = T^\prime_L$ $\rightarrow$ $z_0 = T^\prime_L$
    • Guess $T(x=0)$, and shoot for $T(x=H)=T_R$.
• N,N
  • $T^\prime_H = T^\prime_R$ $\rightarrow$ $\mathbf{T_0 = T_{0,g}}$
  • $T^\prime_0 = T^\prime_L$ $\rightarrow$ $z_0 = T^\prime_L$
    • Guess $T(x=0)$, and shoot for $z(x=H) = T^\prime_R$.
• Consider combinations with Robin conditions similarly.