Sunday, September 16, 2007

In Diffeqs

An nth-order diffeq has the general form F(x, y, dy/dx, ... d^ny/dx^n)=0. Often it is restated with the highest order derivative term isolated on one side. Add an initial condition in the form of y(m) = k, and you've got an initial value problem.

We started with Initial Value problems for n-th order diffeq's. Generally, these consist of a stated differential equation dy/dx=f(x,y), and a specified value for y(0)= some number. We use he Theorem which states that if f and its partial derivative df/dy are continuous over a rectangle, then the problem has a unique solution function F(x) in some interval within that rectangle.

The solution to a diffeq is often an implicit solution, a relation G(x,y)=0 which when differentiated implicitly can be shown to be equivalent to the diffeq. The relation then implicitly defines one or more explicit solutions on an interval I. An explicit solution is a function F(x) that when substituted for y in the diffeq, satisfies the equation for all x in the interval. You verify this simply by substituting.

Newtons Law of Heating and Cooling
dT/dt = k(A(t)-T(t)) + H(t) + U(t),
where H(t) is a composite of heat sources and U is the HVAC system, each having similar formulas with constants of proportionality. A(t) is the temp of the exterior environment, T(t) is the temp of an object such as a cup, bottle, auto, or interior air of a building.
In the case of U(t), the formula involves a different kf (Td(t)-T(t)) where Td(t) is the setpoint for the thermostat and T(t) .

The constants of proportionality in both cases are positive, so if the interior temperature is higher than the setpoint, the HVAC system should be working to get it down and the differential - the slope corresponding to the change in temperature - should be negative. Similarly when the external air temp is high and the internal temp is lower, a building or cup should be experiencing an increasing temperature, an upward slope, and a positive differential. Both should level out as the differences get smaller and smaller. Note that sometimes an HVAC system may work with the outside temperatures, but most often against it.

We often solve these problems with the Integration Factor method, because they are first order linear equations. Sometimes they are separable too.

We also re-visited Separation of Variables as a method, most often to be applied against non-linear first order diffeqs. If the variables can be separated to different sides of the equation, do so, then integrate both sides.

The other method we looked at was that of Exact Equations. If an equation is exact, you can put it into the form of a Total Differential. First you check for exactness - it will be exact if the second-mixed-partial derivatives of each component are equivalent. That done, you read off the "M(x,y)dx" first-order-mixed-partial derivative, and integrate wrt x to get F(x,y)= something +g(y) ; solve for g(y) by taking the partial derivative wrt y, and using the fact that "N(x,y)" is also the partial dF/dy and solve for g'(y), then integrate back again to get g(y). Substitute g(y) back into the earlier integral of M to get F(x,y)=C. Specific solutions can be found by solving for C using initial conditions.

Other things we covered: direction fields; basically sketching in the dy/dx as marks representing the slope over a given rectangle on an xy grid; the method of isoclines (level curves); take the diffeq y'=f(x,y) and fix the equation to equal constants f(x,y)=(c) at fixed intervals. Then solve it for y to get a line you can sketch in along which the dy/dx is always the same (c). Draw tick marks representing the dy/dx=f(x,y) at points along those lines, then erase the lines. I sucked at this. The book doesn't explain it well, and the constant "c" is not the slope of the isocline, so it can get confusing. It is also not general: where the diffeq is not linear it can get convoluted.

We also studied Euler's method of approximation, using the first couple of terms of a Taylor's series to estimate the solution value of a diffeq at a point, when given another point a fixed distance away. y=y0+(x-x0)*f(x0,y0)... the next x value is x plus a bit of increment, the next y value is the first y value plus the product of the bit of increment along x times the evaluted diffeq at the first point. Work a little bit of slope into it. To extend it requires taking higher order n-th derivatives, and dividing by the n-th factorial, so it is analytically more complicated to get better estimates this way.

Runge-Kutta goes further, defining four composite functions k1...k4, where the next y value is the first y value plus 1/6th of the sum (k1+ 1/2(k2+k3) + k4), and the next x value is simply x plus the bit of increment. k1 is like Euler's, the increment times f(xo,y0). k2 is the increment times the f() evaluated at (xo+increment/2, y0+k1/2). k3 is the incrment times the f() evaluated at (x0+increment/2, y0+k2/2). k4 is the increment times the f() evaluated at x0+increment, y+k3). More convoluted numerically, but less so analytically than Euler's.
Post a Comment