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Calculus/Partial differential equations

First order
Any partial differential equation of the form
 * $$ h_1 \frac {\partial u}{\partial x_1}

+ h_2 \frac {\partial u}{\partial x_2} \cdots + h_n \frac {\partial u}{\partial x_n} = b$$ where h1, h2 … hn, and b are all functions of both u and Rn can be reduced to a set of ordinary differential equations.

To see how to do this, we will first consider some simpler problems.

Special cases
We will start with the simple PDE
 * $$u_z(x,y,z)=u(x,y,z) \quad (1)$$

Because u is only differentiated with respect to z, for any fixed x and y we can treat this like the ODE, du/dz=u. The solution of that ODE is cez, where c is the value of u when z=0, for the fixed x and y

Therefore, the solution of the PDE is
 * $$u(x, y, z)=u(x,y,0) e^z $$

Instead of just having a constant of integration, we have an arbitrary function. This will be true for any PDE.

Notice the shape of the solution, an arbitrary function of points in the xy, plane, which is normal to the 'z' axis, and the solution of an ODE in the 'z' direction.

Now consider the slightly more complex PDE
 * $$a_x u_x +a_y u_y + a_z u_z=h(u) \quad (2)$$

where h can be any function, and each a is a real constant.

We recognize the left hand side as being a·∇, so this equation says that the differential of u in the a direction is h(u). Comparing this with the first equation suggests that the solution can be written as an arbitrary function on the plane normal to a combined with the solution of an ODE.

Remembering from Calculus/Vectors that any vector r can be split up into components parallel and perpendicular to a,
 * $$\mathbf{r}= \mathbf{r}_\perp + \mathbf{r}_\| = \left( \mathbf{r}-\frac{(\mathbf{r} \cdot \mathbf{a})\mathbf{a}}{|\mathbf{a}|^2} \right) + \frac{(\mathbf{r} \cdot \mathbf{a})\mathbf{a}}{|\mathbf{a}|^2} $$

we will use this to split the components of r in a way suggested by the analogy with (1).

Let's write
 * $$\mathbf{r} = (x,y,z) = \mathbf{r}_\perp + s\mathbf{a} \quad

s = \frac{\mathbf{r} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}}$$ and substitute this into (2), using the chain rule. Because we are only differentiating in the a direction, adding any function of the perpendicular vector to s will make no difference.

First we calculate grad s, for use in the chain rule,
 * $$\nabla s = \frac{\mathbf{a}}{a^2}$$

On making the substitution into (2), we get,
 * $$h(u)= \mathbf{a} \cdot \nabla s \frac{d}{ds} u(s)=

\frac{\mathbf{a} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \frac{d}{ds} u(s)= \frac{du}{ds}$$ which is an ordinary differential equation with the solution
 * $$s=c(\mathbf{r}_\perp) + \int^u \frac{dt}{h(t)}$$

The constant c can depend on the perpendicular components, but not upon the parallel coordinate. Replacing s with a monotonic scalar function of s multiplies the ODE by a function of s, which doesn't affect the solution.

Example:
 * $$u(x,t)_x=u(x,t)_t$$

For this equation, a is (1, -1), s=x-t, and the perpendicular vector is (x+t)(1, 1). The reduced ODE is du/ds=0 so the solution is
 * u=f(x+t)

To find f we need initial conditions on u. Are there any constraints on what initial conditions are suitable?

Consider, if we are given
 * u(x,0), this is exactly f(x),
 * u(3t,t), this is f(4t) and f(t) follows immediately
 * u(t3+2t,t), this is f(t3+3t) and f(t) follows, on solving the cubic.
 * u(-t,t), then this is f(0), so if the given function isn't constant we have a inconsistency, and if it is the solution isn't specified off the initial line.

Similarly, if we are given u on any curve which the lines x+t=c intersect only once, and to which they are not tangent, we can deduce f.

For any first order PDE with constant coefficients, the same will be true. We will have a set of lines, parallel to r=at, along which the solution is gained by integrating an ODE with initial conditions specified on some surface to which the lines aren't tangent.

If we look at how this works, we'll see we haven't actually used the constancy of a, so let's drop that assumption and look for a similar solution.

The important point was that the solution was of the form u=f(x(s),y(s)), where (x(s),y(s)) is the curve we integrated along -- a straight line in the previous case. We can add constant functions of integration to s without changing this form.

Consider a PDE,
 * $$a(x,y)u_x+b(x,y)u_y=c(x,y,u)$$

For the suggested solution, u=f(x(s),y(s)), the chain rule gives
 * $$ \frac{du}{ds} = \frac{dx}{ds}u_x+ \frac{dy}{ds}u_y$$

Comparing coefficients then gives
 * $$ \frac{dx}{ds}=a(x,y) \quad \frac{dy}{ds}=b(x,y) \quad \frac{du}{ds}=c(x,y,u) $$

so we've reduced our original PDE to a set of simultaneous ODE's. This procedure can be reversed.

The curves (x(s),y(s)) are called characteristics of the equation.

Example: Solve $$yu_x=xu_y$$ given u=f(x) for x≥0 The ODE's are
 * $$ \frac{dx}{ds}=y \quad \frac{dy}{ds}=-x \quad \frac{du}{ds}=0 $$

subject to the initial conditions at s=0,
 * $$x(0)=r \quad y(0)=0 \quad u(0)=f(r) \quad r \ge 0$$

This ODE is easily solved, giving
 * $$x(s)=r \cos s \quad y(s)=r \sin s \quad u(s)=f(r)$$

so the characteristics are concentric circles round the origin, and in polar coordinates u(r,θ)=f(r)

Considering the logic of this method, we see that the independence of a and b from u has not been used either, so that assumption too can be dropped, giving the general method for equations of this quasilinear form.

Quasilinear
Summarising the conclusions of the last section, to solve a PDE
 * $$ a_1(u,\mathbf{x}) \frac {\partial u}{\partial x_1}

+ a_2(u,\mathbf{x}) \frac {\partial u}{\partial x_2} \cdots + a_n(u,\mathbf{x}) \frac {\partial u}{\partial x_n} = b(u,\mathbf{x})$$ subject to the initial condition that on the surface, (x1(r1,…,rn-1, …xn(r1,…,rn-1), u=f(r1,…,rn-1) --this being an arbitrary paremetrisation of the initial surface--

 * we transform the equation to the equivalent set of ODEs,
 * $$\frac{dx_1}{ds}=a_1 \ \ldots \ \frac{dx_n}{ds}=a_n \quad \frac{du}{ds}=b$$
 * subject to the initial conditions
 * $$x_i(0)=f(r_1, \ldots ,r_{n-1}) \quad u=f(r_1,r_2, \ldots r_{n-1})$$

 * Solve the ODE's, giving xi as a function of s and the ri.
 * Invert this to get s and the ri as functions of the xi.
 * Substitute these inverse functions into the expression for u as a function of s and the ri obtained in the second step.

Both the second and third steps may be troublesome.

The set of ODEs is generally non-linear and without analytical solution. It may even be easier to work with the PDE than with the ODEs.

In the third step, the ri together with s form a coordinate system adapted for the PDE. We can only make the inversion at all if the Jacobian of the transformation to Cartesian coordinates is not zero,
 * $$\begin{vmatrix}

\frac{\partial x_1}{\partial r_1}&\cdots &\frac{\partial x_1}{\partial r_{n-1}} & a_1 \\ \vdots & \ddots & & \vdots \\ \frac{\partial x_n}{\partial r_1}&\cdots &\frac{\partial x_n}{\partial r_{n-1}} & a_n \end{vmatrix} \ne 0$$ This is equivalent to saying that the vector (a1, &hellip:, an) is never in the tangent plane to a surface of constant s.

If this condition is not false when s=0 it may become so as the equations are integrated. We will soon consider ways of dealing with the problems this can cause.

Even when it is technically possible to invert the algebraic equations it is obviously inconvenient to do so.

Example
To see how this works in practice, we will a/ consider the PDE,
 * $$uu_x+u_y+u_t=0$$

with generic initial condition,
 * $$u=f(x,y) \mbox{ on } t=0$$

Naming variables for future convenience, the corresponding ODE's are
 * $$\frac{dx}{d\tau}=u \quad \frac{dy}{d\tau}=1 \quad \frac{dt}{d\tau}=1 \quad \frac{du}{d\tau}=0 \quad$$

subject to the initial conditions at τ=0
 * $$x=r \quad y=s \quad t=0 \quad u=f(r,s)$$

These ODE's are easily solved to give
 * $$x=r+f(r,s)\tau \quad y=s+\tau \quad t=\tau \quad u=f(r,s)$$

These are the parametric equations of a set of straight lines, the characteristics.

The determinant of the Jacobian of this coordinate transformation is
 * $$\begin{vmatrix}

1+\tau\frac{\partial f}{\partial r} & \tau\frac{\partial f}{\partial s} & f \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{vmatrix}=1+\tau\frac{\partial f}{\partial r}$$

This determinant is 1 when t=0, but if fr is anywhere negative this determinant will eventually be zero, and this solution fails.

In this case, the failure is because the surface $$sf_r=-1$$ is an envelope of the characteristics.

For arbitrary f we can invert the transformation and obtain an implicit expression for u
 * $$u=f(x-tu,y-x)$$

If f is given this can be solved for u.

1/ $$f(x,y)=ax$$, The implicit solution is
 * $$u=a(x-tu) \Rightarrow u=\frac{ax}{1+at}$$

This is a line in the u-x plane, rotating clockwise as t increases. If a is negative, this line eventually become vertical. If a is positive, this line tends towards u=0, and the solution is valid for all t.

For f(x,y)=x2 The implicit solution is:
 * $$u=(x-tu)^2 \Rightarrow u=\frac{1+2tx-\sqrt{1+4tx}}{2t^2}$$

This solution clearly fails when $$1+4tx<0$$, which is just when $$sf_r=-1$$. For any t > 0 this happens somewhere. As t increases this point of failure moves toward the origin.

Notice that the point where u=0 stays fixed. This is true for any solution of this equation, for all values of f

We will see later that we can find a solution after this time, if we consider discontinuous solutions. We can think of this as a shockwave.

3/ $$f(x,y)=\sin (xy)$$ The implicit solution is
 * $$u(x,y,t)=\sin \left( (x-tu)(y-x) \right)$$

and we can not solve this explitely for u. The best we can manage is a numerical solution of this equation.

We can also consider the closely related PDE
 * $$uu_x+u_y+u_t=y$$

The corresponding ODE's are
 * $$\frac{dx}{d\tau}=u \quad \frac{dy}{d\tau}=1 \quad \frac{dz}{d\tau}=1 \quad \frac{du}{d\tau}=y \quad$$

subject to the initial conditions at τ=0
 * $$x=r \quad y=s \quad t=0 \quad u=f(r,s)$$

These ODE's are easily solved to give
 * $$x=r+\tau f+\frac{1}{2}s\tau^2+\frac{1}{6}\tau^3 \quad

y=s+\tau \quad t=\tau \quad u=f+s\tau+\frac{1}{2}\tau^2$$ Writing f in terms of u, s, and τ, then substituting into the equation for x gives an implicit solution
 * $$u(x,y,t)=f(x-ut+\frac{1}{2}yt^2-\frac{1}{6}t^3,y-t)+yt-\frac{1}{2}t^2$$

It is possible to solve this for u in some special cases, but in general we can only solve this equation numerically. However, we can learn much about the global properties of the solution from further analysis

Characteristic initial value problems
What if initial conditions are given on a characteristic, on an envelope of characteristics, on a surface with characteristic tangents at isolated points?

Discontinuous solutions
So far, we've only considered smooth solutions of the PDE, but this is too restrictive. We may encounter initial conditions which aren't smooth, e.g.

 * $$u_t = c u_x \quad u(x,0)=

\left\{ \begin{matrix} 1, & x \ge 0 \\ 0, & x < 0 \end{matrix} \right. $$

If we were to simply use the general solution of this equation for smooth initial conditions,

 * $$ u(x,t) = u(x + ct, 0) \,$$

we would get

 * $$ u(x,t) =

\left\{ \begin{matrix} 1, & x+ct \ge 0 \\ 0, & x+ct < 0 \end{matrix} \right. $$

which appears to be a solution to the original equation. However, since the partial differentials are undefined on the characteristic x+ct=0, so it becomes unclear what it means to say that the equation is true at that point.

We need to investigate further, starting by considering the possible types of discontinuities.

If we look at the derivations above, we see we've never use any second or higher order derivatives so it doesn't matter if they aren't continuous, the results above will still apply.

The next simplest case is when the function is continuous, but the first derivative is not, e.g. |x|. We'll initially restrict ourselves to the two-dimensional case, u(x, t) for the generic equation.

 * $$a(x,t)u_x + b(x,t)u_t = c(u,x,t) \quad (1)$$

Typically, the discontinuity is not confined to a single point, but is shared by all points on some curve, (x0(s), t0(s) )

Then we have
 * $$\begin{matrix}

x>x_0 & \lim_{x \rarr x_0} = u_+ \\ x<x_0 & \lim_{x \rarr x_0} = u_- \end{matrix}$$

We can then compare u and its derivatives on both sides of this curve.

It will prove useful to name the jumps across the discontinuity. We say

 * $$ [u]=u_+-u_- \quad [u_x]=u_{x+}-u_{x-} \quad [u_t]=u_{t+}-u_{t-}$$

Now, since the equation (1) is true on both sides of the discontinuity, we can see that both u+ and u-, being the limits of solutions, must themselves satisfy the equation. That is,

 * $$\begin{matrix}

a(x,t)u_{+x} + b(x,t)u_{+t} & = & c(u_+,x,t) \\ a(x,t)u_{-x} + b(x,t)u_{-t} & = & c(u_-,x,t) \end{matrix} \mbox{ where  } \begin{matrix}x=x_0(s) \\ t=t_0(s)\end{matrix}$$

Subtracting then gives us an equation for the jumps in the differentials

 * $$a(x,t)[u_x] + b(x,t)[u_t] = 0 \, $$

We are considering the case where u itself is continuous so we know that [u]=0. Differentiating this with respect to s will give us a second equation in the differential jumps.

 * $$ \frac{dx_0}{ds}[u_x] + \frac{dt_0}{ds}[u_t] =0 $$

The last two equations can only be both true if one is a multiple of the other, but multiplying s by a constant also multiplies the second equation by that same constant while leaving the curve of discontinuity unchanged, hence we can without loss of generality define s to be such that

 * $$ \frac{dx_0}{ds}=a \quad \frac{dt_0}{ds}=b $$

But these are the equations for a characteristic, i.e. discontinuities propagate along characteristics. We could use this property as an alternative definition of characteristics.

We can deal similarly with discontinuous functions by first writing the equation in conservation form, so called because conservation laws can always be written this way.

 * $$ (au)_x+(bu)_t=a_x u + b_t u +c \quad (1)$$

Notice that the left hand side can be regarded as the divergence of (au, bu). Writing the equation this way allows us to use the theorems of vector calculus.

Consider a narrow strip with sides parallel to the discontinuity and width h

We can integrate both sides of (1) over R, giving

 * $$\int_R (au)_x+(bu)_t \, dx dt = \int_R (a_x + b_t) u +c \, dx dt $$

Next we use Green's theorem to convert the left hand side into a line integral.

 * $$\oint_{\partial R} au dt - bu dx = \int_R (a_x + b_t) u +c \, dx dt $$

Now we let the width of the strip fall to zero. The right hand side also tends to zero but the left hand side reduces to the difference between two integrals along the part of the boundary of R parallel to the curve.

 * $$ \int au_+dt-bu_+dx - \int au_-dt-bu_-dx =  0 $$

The integrals along the opposite sides of R have different signs because they are in opposite directions.

For the last equation to always be true, the integrand must always be zero, i.e.

 * $$ \left( a \frac{dt_0}{ds} - b \frac{dx_0}{ds} \right) [u] = 0 $$

Since, by assumption [u] isn't zero, the other factor must be, which immediately implies the curve of discontinuity is a characteristic.

Once again, discontinuities propagate along characteristics.

Above, we only considered functions of two variables, but it is straightforward to extend this to functions of n variables.

The initial condition is given on an n-1 dimensional surface, which evolves along the characteristics. Typical discontinuities in the initial condition will lie on a n-2 dimensional surface embedded within the initial surface. This surface of discontinuity will propagate along the characteristics that pass through the initial discontinuity.

The jumps themselves obey ordinary differential equations, much as u itself does on a characteristic. In the two dimensional case, for u continuous but not smooth, a little algebra shows that

 * $$\frac{d[u_x]}{ds}=[u_x]\left( \frac{\partial c}{\partial u} + a\frac{b_x}{b}-a_x \right) $$

while u obeys the same equation as before,

 * $$\frac{du}{ds}=c $$

We can integrate these equations to see how the discontinuity evolves as we move along the characteristic.

We may find that, for some future s, [ux] passes through zero. At such points, the discontinuity has vanished, and we can treat the function as smooth at that characteristic from then on.

Conversely, we can expect that smooth functions may, under the right circumstances, become discontinuous.

To see how all this works in practice we'll consider the solutions of the equation

 * $$u_t+uu_x=0 \quad u(x,0)=f(x)$$

for three different initial conditions.

The general solution, using the techniques outlined earlier, is

 * $$ u=f(x-tu) \,\! $$

u is constant on the characteristics, which are straight lines with slope dependent on u.

First consider f such that

 * $$ f(x)= \left\{ \begin{matrix}

1 & x>a \\ \frac{x}{a} & a \ge x > 0 \\ 0 & x \le 0 \end{matrix} \right. \quad a>0 $$

While u is continuous its derivative is discontinuous at x=0, where u=0, and at x=a, where u=1. The characteristics through these points divide the solution into three regions.

All the characteristics to the right of the characteristic through x=a, t=0 intersect the x-axis to the right of x=1, where u=1 so u is 1 on all those characteristics, i.e whenever x-t>a.

Similarly the characteristic through the origin is the line x=0, to the left of which u remains zero.

We could find the value of u at a point in between those two characteristics either by finding which intermediate characteristic it lies on and tracing it back to the initial line, or via the general solution.

Either way, we get

 * $$ f(x)= \left\{ \begin{matrix}

1 & x-t>a \\ \frac{x}{a+t} & a+t \ge x > 0 \\ 0 & x \le 0 \end{matrix} \right. $$

At larger t the solution u is more spread out than at t=0 but still the same shape.

We can also consider what happens when a tends to 0, so that u itself is discontinuous at x=0.

If we write the PDE in conservation form then use Green's theorem, as we did above for the linear case, we get

 * $$[u]\frac{dx_0}{ds}=\frac{1}{2}[u^2]\frac{dt_0}{ds}$$

[u²] is the difference of two squares, so if we take s=t we get

 * $$\frac{dx_0}{dt}=\frac{1}{2}(u_-+u_+)$$

In this case the discontinuity behaves as if the value of u on it were the average of the limiting values on either side.

However, there is a caveat.

Since the limiting value to the left is u- the discontinuity must lie on that characteristic, and similarly for u+; i.e the jump discontinuity must be on an intersection of characteristics, at a point where u would otherwise be multivalued.

For this PDE the characteristic can only intersect on the discontinuity if

 * $$u_- > u_+ \, $$

If this is not true the discontinuity can not propagate. Something else must happen.

The limit a=0 is an example of a jump discontinuity for which this condition is false, so we can see what happens in such cases by studying it.

Taking the limit of the solution derived above gives

 * $$ f(x)= \left\{ \begin{matrix}

1 & x>t \\ \frac{x}{t} & t \ge x > 0 \\ 0 & x \le 0 \end{matrix} \right. $$

If we had taken the limit of any other sequence of initials conditions tending to the same limit we would have obtained a trivially equivalent result.

Looking at the characteristics of this solution, we see that at the jump discontinuity characteristics on which u takes every value between 0 and 1 all intersect.

At later times, there are two slope discontinuities, at x=0 and x=t, but no jump discontinuity.

This behaviour is typical in such cases. The jump discontinuity becomes a pair of slope discontinuities between which the solution takes all appropriate values.

Now, lets consider the same equation with the initial condition

 * $$ f(x)= \left\{ \begin{matrix}

1 & x \le 0 \\ 1-\frac{x}{a} & a \ge x > 0 \\ 0 & x > a \end{matrix} \right. \quad a>0$$

This has slope discontinuities at x=0 and x=a, dividing the solution into three regions.

The boundaries between these regions are given by the characteristics through these initial points, namely the two lines

 * $$ x=t \quad x=a $$

These characteristics intersect at t=a, so the nature of the solution must change then.

In between these two discontinuities, the characteristic through x=b at t=0 is clearly

 * $$ x=\left( 1- \frac{b}{a} \right) t + b \quad 0 \le b \le a $$

All these characteristics intersect at the same point, (x,t)=(a,a).

We can use these characteristics, or the general solution, to write u for t t \\ 0 & x > a \end{matrix} \right. \quad a >t \ge 0$$

As t tends to a, this becomes a step function. Since u is greater to the left than the right of the discontinuity, it meets the condition for propagation deduced above, so for t>a u is a step function moving at the average speed of the two sides.

 * $$ u(x,t)= \left\{ \begin{matrix}

1 & x \le \frac{a+t}{2} \\ 0 & x > \frac{a+t}{2} \end{matrix} \right. \quad t \ge a \ge 0 $$

This is the reverse of what we saw for the initial condition previously considered, two slope discontinuities merging into a step discontinuity rather than vice versa. Which actually happens depends entirely on the initial conditions. Indeed, examples could be given for which both processes happen.

In the two examples above, we started with a discontinuity and investigated how it evolved. It is also possible for solutions which are initially smooth to become discontinuous.

For example, we saw earlier for this particular PDE that the solution with the initial condition u=x² breaks down when 2xt+1=0. At these points the solution becomes discontinuous.

Typically, discontinuities in the solution of any partial differential equation, not merely ones of first order, arise when solutions break down in this way and propagate similarly, merging and splitting in the same fashion.

Fully non-linear PDEs
It is possible to extend the approach of the previous sections to reduce any equation of the form
 * $$F(x_1,x_2, \ldots ,x_n,u,u_{x_1},u_{x_2}, \ldots, u_{x_n})=0$$

to a set of ODE's, for any function, F.

We will not prove this here, but the corresponding ODE's are
 * $$\frac{dx_i}{d\tau}=\frac{\partial F}{\partial u_i} \quad \frac{du_i}{d\tau} = -\left( \frac{\partial F}{\partial x_i} +

u_i \frac{\partial F}{\partial u} \right) \quad \frac{du}{d\tau}=\sum_{i=1}^n u_i \frac{\partial F}{\partial u_i} $$

If u is given on a surface parameterized by r1…rn then we have, as before, n initial conditions on the n, xi
 * $$\tau=0 \quad x_i=f_i(r_1,r_2, \ldots ,r_{n-1})$$

given by the parameterization and one initial condition on u itself,
 * $$\tau=0 \quad u=f(r_1,r_2, \ldots ,r_{n-1})$$

but, because we have an extra n ODEs for the ui's, we need an extra n initial conditions.

These are, n-1 consistency conditions,
 * $$\tau=0 \quad \frac{\partial f}{\partial r_i}=

\sum_{j=1}^{n-1} u_i \frac{\partial f_i}{\partial r_j}$$ which state that the ui's are the partial derivatives of u on the initial surface, and one initial condition
 * $$\tau=0 \quad F(x_1,x_2, \ldots ,x_n,u,u_1,u_2, \ldots, u_n)=0$$

stating that the PDE itself holds on the initial surface.

These n initial conditions for the ui will be a set of algebraic equations, which may have multiple solutions. Each solution will give a different solution of the PDE.

Example
Consider
 * $$u_t=u_x^2+u_y^2, \quad u(x,y,0)=x^2+y^2$$

The initial conditions at τ=0 are
 * $$\begin{matrix} x=r & y=s & t=0 & u=r^2+s^2 \\

u_x=2r & u_y=2s & u_t=4(r^2+s^2) & \end{matrix}$$ and the ODE's are
 * $$\begin{matrix}

\frac{dx}{d\tau}=-2u_x & \frac{dy}{d\tau}=-2u_y & \frac{dt}{d\tau}=1 & \frac{du}{d\tau}=u_t-2(u_x^2+u_y^2)\\ \frac{du_x}{d\tau}=0 & \frac{du_y}{d\tau}=0 & \frac{du_t}{d\tau}=0 & \end{matrix}$$

Note that the partial derivatives are constant on the characteristics. This always happen when the PDE contains only partial derivatives, simplifying the procedure.

These equations are readily solved to give
 * $$x=r(1-4\tau ) \quad y=s(1-4\tau ) \quad t=\tau \quad u=(r^2+s^2)(1-4\tau )$$

On eliminating the parameters we get the solution,
 * $$u=\frac{x^2+y^2}{1-4t}$$

which can easily be checked. abc

Second order
Suppose we are given a second order linear PDE to solve

$$a(x,y)u_{xx}+b(x,y)u_{xy}+c(x,y)u_{yy} = d(x,y)u_x+e(x,y)u_y +p(x,y)u +q(x,y) \quad (1)$$

The natural approach, after our experience with ordinary differential equations and with simple algebraic equations, is attempt a factorisation. Let's see how for this takes us.

We would expect factoring the left hand of (1) to give us an equivalent equation of the form

 * $$ a(x,y)(D_x+\alpha_+(x,y)D_y)(D_x+\alpha_-(x,y)D_y)u \, $$

and we can immediately divide through by a. This suggests that those particular combinations of first order derivatives will play a special role.

Now, when studying first order PDE's we saw that such combinations were equivalent to the derivatives along characteristic curves. Effectively, we changed to a coordinate system defined by the characteristic curve and the initial curve.

Here, we have two combinations of first order derivatives each of which may define a different characteristic curve. If so, the two sets of characteristics will define a natural coordinate system for the problem, much as in the first order case.

In the new coordinates we will have

 * $$ D_x+\alpha_+(x,y)D_y=D_r \quad D_x+\alpha_-(x,y)D_y =D_s $$

with each of the factors having become a differentiation along its respective characteristic curve, and the left hand side will become simply urs giving us an equation of the form

 * $$u_{rs}=A(r,s)u_r+B(r,s)u_s+C(r,s)u+D(r,s) \, $$

If A, B, and C all happen to be zero, the solution is obvious. If not, we can hope that the simpler form of the left hand side will enable us to make progress.

However, before we can do all this, we must see if (1) can actually be factored.

Multiplying out the factors gives

$$u_{xx}+\frac{b(x,y)}{a(x,y)}u_{xy}+\frac{c(x,y)}{a(x,y)}u_{yy}= u_{xx}+(\alpha_++\alpha_-)u_{xy}+\alpha_+\alpha_-u_{yy}$$

On comparing coefficients, and solving for the α's we see that they are the roots of

 * $$a(x,y)\alpha^2+b(x,y)\alpha+c(x,y)=0 \,$$

Since we are discussing real functions, we are only interested in real roots, so the existence of the desired factorization will depend on the discriminant of this quadratic equation.

 * $$\mbox{If } b(x,y)^2 > 4 a(x,y)c(x,y) $$
 * then we have two factors, and can follow the procedure outlined above. Equations like this are called hyperbolic

 * $$\mbox{If } b(x,y)^2 = 4 a(x,y)c(x,y) $$
 * then we have only factor, giving us a single characteristic curve. It will be natural to use distance along these curves as one coordinate, but the second must be determined by other considerations.
 * The same line of argument as before shows that use the characteristic curve this way gives a second order term of the form urr, where we've only taken the second derivative with respect to one of the two coordinates. Equations like this are called parabolic

 * $$\mbox{If } b(x,y)^2 < 4 a(x,y)c(x,y) $$
 * then we have no real factors. In this case the best we can do is reduce the second order terms to the simplest possible form satisfying this inequality, i.e urr+uss
 * It can be shown that this reduction is always possible. Equations like this are called elliptic

It can be shown that, just as for first order PDEs, discontinuities propagate along characteristics. Since elliptic equations have no real characteristics, this implies that any discontinuities they may have will be restricted to isolated points; i.e., that the solution is almost everywhere smooth.

This is not true for hyperbolic equations. Their behavior is largely controlled by the shape of their characteristic curves.

These differences mean different methods are required to study the three types of second equation. Fortunately, changing variables as indicated by the factorisation above lets us reduce any second order PDE to one in which the coefficients of the second order terms are constant, which means it is sufficient to consider only three standard equations.

 * $$u_{xx}+u_{yy}=0 \quad u_{xx}-u_{yy}=0 \quad u_{xx}-u_y=0 $$

We could also consider the cases where the right hand side of these equations is a given function, or proportional to u or to one of its first order derivatives, but all the essential properties of hyperbolic, parabolic, and elliptic equations are demonstrated by these three standard forms.

While we've only demonstrated the reduction in two dimensions, a similar reduction applies in higher dimensions, leading to a similar classification. We get, as the reduced form of the second order terms,

 * $$a_1\frac{\partial^2 u}{\partial x_1^2}+

a_2\frac{\partial^2 u}{\partial x_2^2} + \cdots + a_n\frac{\partial^2 u}{\partial x_n^2} $$

where each of the ais is equal to either 0, +1, or -1.

If all the ais have the same sign the equation is elliptic

If any of the ais are zero the equation is parabolic

If exactly one of the ais has the opposite sign to the rest the equation is hyperbolic

In 2 or 3 dimensions these are the only possibilities, but in 4 or more dimensions there is a fourth possibility: at least two of the ais are positive, and at least two of the ais are negative.

Such equations are called ultrahyperbolic. They are less commonly encountered than the other three types, so will not be studied here.

When the coefficients are not constant, an equation can be hyperbolic in some regions of the xy plane, and elliptic in others. If so, different methods must be used for the solutions in the two regions.

Elliptic
Standard form, Laplace's equation:
 * $$\nabla^2 h=0$$

Quote equation in spherical and cylindrical coordinates, and give full solution for cartesian and cylindrical coordinates. Note averaging property Comment on physical significance, rotation invariance of laplacian.

Hyperbolic
Standard form, wave equation:
 * $$\nabla^2 h=c^2 h_{tt}$$

Solution, any sum of functions of the form
 * $$h=f(\mathbf{k}\cdot \mathbf{x}-\omega t) \quad \omega=|\mathbf{k}|c $$

These are waves. Compare with solution from separating variables. Domain of dependence, etc.

Parabolic
The canonical parabolic equation is the diffusion equation:

 * $$\nabla^2 h=h_t$$

Here, we will consider some simple solutions of the one-dimensional case.

The properties of this equation are in many respects intermediate between those of hyperbolic and elliptic equation.

As with hyperbolic equations but not elliptic, the solution is well behaved if the value is given on the initial surface t=0.

However, the characteristic surfaces of this equation are the surfaces of constant t, thus there is no way for discontinuities to propagate to positive t.

Therefore, as with elliptic equations but not hyberbolic, the solutions are typically smooth, even when the initial conditions aren't.

Furthermore, at a local maximum of h, its Laplacian is negative, so h is decreasing with t, while at local minima, where the Laplacian will be positive, h will increase with t. Thus, initial variations in h will be smoothed out as t increases.

In one dimension, we can learn more by integrating both sides,

 * $$\begin{matrix}

\int_{-a}^b h_t dt & = & \int_{-a}^b h_{xx} dx \\ \frac{d}{dt} \int_{-a}^b h \,dt & = & \left[ h_x \right]_{-a}^b \end{matrix}$$

Provided that hx tends to zero for large x, we can take the limit as a and b tend to infinity, deducing

 * $$\frac{d}{dt} \int_{-\infty}^\infty h \,dt$$

so the integral of h over all space is constant.

This means this PDE can be thought of as describing some conserved quantity, initially concentrated but spreading out, or diffusing, over time.

This last result can be extended to two or more dimensions, using the theorems of vector calculus.

We can also differentiate any solution with respect to any coordinate to obtain another solution. E.g. if h is a solution then

 * $$\nabla^2 h_x = \partial_x \nabla^2 h = \partial_x \partial_t h

=\partial_t h_x $$

so hx also satisfies the diffusion equation.

Similarity solution
Looking at this equation, we might notice that if we make the change of variables

 * $$ r = \alpha x \quad \tau =\alpha^2 t $$

then the equation retains the same form. This suggests that the combination of variables x²/t, which is unaffected by this variable change, may be significant.

We therefore assume this equation to have a solution of the special form

 * $$h(x,t)=f(\eta) \mbox{ where } \eta = \frac{x}{t^{1 \over 2}} $$

then

 * $$h_x = \eta_x f_{\eta} = t^{-{1\over 2}}f_{\eta} \quad

h_t = \eta_t f_{\eta} = -\frac{\eta}{2t}f_{\eta} $$

and substituting into the diffusion equation eventually gives

 * $$ f_{\eta \eta} + \frac{\eta}{2} f_{\eta}=0 $$

which is an ordinary differential equation.

Integrating once gives

 * $$f_{\eta} = A e^{-\frac{\eta^2}{4}}$$

Reverting to h, we find

 * $$\begin{matrix}

h_x & = & \frac{A}{\sqrt{t}} e^{-\frac{\eta^2}{4}} \\ h & = & \frac{A}{\sqrt{t}} \int_{-\infty }^x e^{-s^2/4t} ds +B \\ & = & A \int_{-\infty }^{x/2\sqrt{t}} e^{-z^2} dz +B \end{matrix}$$

This last integral can not be written in terms of elementary functions, but its values are well known.

In particular the limiting values of h at infinity are

 * $$h(-\infty ,t)=B \quad h(\infty ,t)= B + A \sqrt{\pi } ,$$

taking the limit as t tends to zero gives

 * $$h= \left\{ \begin{matrix}

B & x<0 \\ B + A \sqrt{\pi } & x>0 \end{matrix} \right. $$

We see that the initial discontinuity is immediately smoothed out. The solution at later times retains the same shape, but is more stretched out.

The derivative of this solution with respect to x

 * $$h_x =  \frac{A}{\sqrt{t}} e^{-x^2/4t}$$

is itself a solution, with h spreading out from its initial peak, and plays a significant role in the further analysis of this equation.

The same similarity method can also be applied to some non-linear equations.

Separating variables
We can also obtain some solutions of this equation by separating variables.

 * $$h(x,t)=X(x)T(t) \Rightarrow X^{\prime \prime}T=X\dot{T} $$

giving us the two ordinary differential equations

 * $$ \frac{d^2X}{dx^2}+k^2 X=0 \quad \frac{dT}{dt} = -k T $$

and solutions of the general form

 * $$ h(x,t)= A e^{-kt} \sin (kx+\alpha) \,$$

title: Calculus/Partial differential equations


text:
Intelligence Intensification/Information Sifting

Working in the IT field I come across an unimaginable amount of data on a regular basis in just dealing with day-to-day issues. Any task is like that these days - and especially to someone who knows nothing of the task until it's presented.

The main thing to remember here is organization. The method you use to organize the data is not important - only that you stick to it for the most part. If something doesn't fit then you either change the item you're having a problem with...or you change your system of organization.

Ever watch a flock of birds? One thing you'll notice is that if you try to focus on all the birds at once you'll most likely get confused. If you focus on, say, one bird out of the flock, then you've mentally organized that one bird out of the flock. Guess what...if you keep adding birds out of the flock until you've looked at them all then you've organized the flock in your mind. Information in the day-to-day world is no different.

Breaking down information into chunks you can deal with is absolutely vital to understanding and remembering. After all, without training you couldn't recall a whole book from memory, right? But you could recall important parts of it. Eventually, if you keep adding to those parts then you would at some point recall the entire book. In another context, this techique has been known for roughly two millennia. Remember the Latin phrase divide et impera (divide and rule)?

Now, how this relates to sifting is most of the time pretty simple. Organize the data, but focus on the stuff you want. For instance if you want to observe only the red birds in the flock then focus on the color red. Sounds pretty easy right? Well...yes and no.

When dealing with large amounts of data you have to do a little estimation of chance up front.

What's that mean? Let's say your family bakes a pie with a ring in it for the holiday. Whoever gets the ring wins a prize. You really want that prize so you focus on the ring. Thus you ask some questions of yourself:

Does the ring have any special properties I can use to identify it?

Does the ring affect the piece of pie it's in in any way?

Is there anywhere in the pie I know the ring is not located?

Suppose the ring is copper and turns the pie green. Then you know that any piece of pie that is the normal color is less likely to have the ring in it.

That narrows your focus by a certain amount.

Suppose the ring is also pretty big, so the pie bulges where the ring is. Then you know that any flat piece of pie will not have the ring in it.

That narrows your focus again.

You keep going with this method until you've found the ring.

Some things that help this process:

''Learning to speed-read. There are plenty of courses out there for this.''

Good language skills for the materials you're searching.

''Frequent breaks. Spending too much time sifting often blurs your focus.''

Focus on one thing at a time. You get more done that way. Remember, sunlight spread out only warms things up, but sunlight focused on one place with a magnifying glass can burn holes in things.

Know what you want. Know what you are looking for. If you don't know, how will you know if you find it?

Delegate. Get people to help you. Parallel searching is helpful, as long as you make sure they know what they are looking for.

It's horribly cliche, but big mountains are just piles of pebbles stacked on top of each other. Information is the same...

Next
../Speed Reading/

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Talk:Intelligence Intensification/Information Sifting

Obviously this doesn't work as I want :/  Instead of removing the II Speed Reading page and cut-and-paste the text to this page, I would like to import the text from the speed reading page to this page. Why? Alfred Korzybski coined the phrase "The map is not the territory" and anyone familiar with maps know that there are different kinds of maps concentrating on different aspects of the same territory. As this is a modern book I would like to have different maps (Table of contents) to the same territory (text). At the same time, I don't want redundancy. Therefore it would be nice if I could "import" text from other wiki pages. --DanielJanzon 12:45, 9 Apr 2004 (UTC)

Try "move this page" I'll write more about transclusion later. r3m0t (cont) (talk) 17:43, 9 Apr 2004 (UTC)

Moving is not what I want. Transclusion is precisly what I'm looking for. Another idea might be to make the whole book one page. The manual of style says "Try and keep everything in one block because its easier to edit and read". I don't really understand what they mean. Of course, having the whole book in one page isn't compatible with the idea of sveral simulatenous organizations of the book. --DanielJanzon 22:46, 11 Apr 2004 (UTC)

 * Transclusion: You put your text-to-transclude at MediaWiki:Whatever. You insert it with or you can substitute it in with.
 * For example, I'm going to subst the TOC into the bottom of the page - it'll copy the source of Template:IITOC over. r3m0t (cont) (talk) 22:49, 11 Apr 2004 (UTC)

 * Damn how nice it would be to be able to transclude any page rather then only pages beginning with "MediaWiki:".
 * You can always copy and paste to a "MediaWiki:" page.


text:
User talk:Paul~enwikibooks

What maths should we cover, you ask? -- Ultimately, I'd say, we'd like to cover every topic at every level, but that's a task that needs taking in small steps.

I suggest assigning each book to an ability range- under 12, 12-16, 16-1st year undergrad, undergrad, postgrad, postdoc (allowing for differing syllabuses)- the higher levels having more, but more narrowly focused, books. If any of the current books cover too wide a range, split them up.

Then in each book we can say 'you must know this first,' and 'This book stops here.' which I've been doing a bit. Carandol 20:08, 9 Apr 2004 (UTC)

Sounds like a very good suggestion. Also sounds like it would take a lot of guts for a single person to try to do it :). This WikiBook site is fascinating to me, I want to help as much as I can - keep contributing everyone!

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Paul. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Paul~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:33, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:21, 19 April 2015 (UTC)

title: User talk:Paul~enwikibooks


text:
User:Carandol

I'm English, with a maths MA (Camb), and a reasonably comprehensive knowledge of many other matters. Mostly, I suspect, I'll contribute to the books using maths, but not solely.

title: User:Carandol


text:
User:Andrewa

G'day

I'm a newbie and sometime lurker on this Wiki, but you can learn about me on my Wikipedia user page and talk to me at my Wikipedia talk page there. Or, see my talk page here if you don't have an English Wikipedia signon.

Andrew Alder

Some links I use here:

User:Andrewa/monobook.css

title: User:Andrewa


text:
User talk:Andrewa

''Welcome! I'm an occasional lurker on this Wiki, so it's probably best to contact me on my Wikipedia talk page. Or, if you don't feel like signing into the English Wikipedia, leave your message here but consider leaving a brief unsigned message on my Wikipedia talk page just saying that I have mail at Wikibooks. Looking forward to hearing from you! Andrewa 21:02, 9 Apr 2004 (UTC)''

Wikiversity
I saw your name on the Wikiversity participant's list, you may be very experienced in mathematics already, but I'm trying out instructing a Calculus course if you're interested, go to Wikiveristy, check out the mathematics department, pure mathematics, and in the course list is included Calculus. Fephisto 22:43, 9 July 2006 (UTC)

 * Looks interesting. School of Mathematics:Calculus. Andrewa 13:48, 10 July 2006 (UTC)

title: User talk:Andrewa


text:
Cookbook:Beef Gravy Soup

| Soups

This recipe for beef gravy soup is derived from the Household Cyclopedia of 1881.

Ingredients

 * 1 pound lean beef, sliced
 * 2 medium onions, sliced
 * 1 pound sliced root vegetables, such as carrots, turnips, or radishes
 * 2 tablespoons beef broth
 * 2 cups vegetable broth
 * Salt, to taste

Procedure

 * 1) Place the onions and root vegetables in a large stock pot. Add the beef and the beef stock on top.
 * 2) Bring to a simmer over low heat for 30 minutes, stirring well.
 * 3) When the beef and vegetables look cooked, add the vegetable broth and salt.
 * 4) Bring back to a simmer and serve.

title: Cookbook:Beef Gravy Soup


text:
Cookbook talk:Beef Gravy Soup

 * 1) (cur) (last)  . . m 00:21, Oct 25, 2003 . . Big Bob the Finder (Update of "ancient" entries)
 * 2) (cur) (last) . . 10:35, Mar 28, 2002 . . Szopen (see wikipedia cookbook)
 * 3) (cur) (last) . . m 06:48, Mar 28, 2002 . . 62.98.249.56

This module uses text adapted from the 1881 Household Cyclopedia, which is now in the public domain.

title: Cookbook talk:Beef Gravy Soup


text:
User:Calmypal

See w:User:Calmypal.

My current project:

 * Chess for beginners

Completed projects:

 * So far I've been a bit useless here, actually.

title: User:Calmypal


text:
User talk:Calmypal

Hi Cal. BTW we have Chess already, if you want to pool your efforts :) Dysprosia 06:31, 10 Apr 2004 (UTC)

title: User talk:Calmypal


text:
Field Guide/Chuparosa

The Chuparosa (hummingbird in Spanish) is a nectar-rich flowers which attracts hummingbirds, linnets and sparrows. It is said to have been eaten by the Papago tribe.

title: Field Guide/Chuparosa


text:
Talk:Electronics/TTL

As far as I can see, this page describes RTL (Resistor-Transistor Logic) and not TTL. There is also DTL (Diode-transistor logic). DTL superceeded RTL and was in turn superceeded by TTL. TTL if I recall correctly was developed at least partly as a requirement of the Apollo moon program of the 1960s.

TTL as the name implies has 2 (at least) transistors in each gate. Key is the input transistor which often has multiple emitters. The emitters are free and used as the inputs for the gates. The base is connected via a resistor to +Vcc. Pulling down any of the emmiters will result in the collector of the input transistor being pulled down, which further transistors will pass to the output (perhaps with inversion to change functionality).

Errors
There are lots of things wrong here. It is RTL logic not TTL. The explanation of how current 'chooses' the path of least resistance is wrong. The battery in the diagrams is connected backwards. The base of the transistor is left floating instead of being either high or low voltage.

title: Talk:Electronics/TTL


text:
User:Djp~enwikibooks

I'm English, with a Degree in Electronics and a PhD in Laser Physics (sounds good !). Got my first chemistry set at age 8 and bought my first electronics magazine at 12 (1971 before they became popular). After working for quite a few years with Optoelectronics and Optical Fibres, I have spent the last 10 years (since 1993) working in the Internet and Telco business. I spent 5 years working in Munich and speak German.

title: User:Djp~enwikibooks


text: UK Constitution and Government/British Monarchs


text:
Cookbook:Cuisine of New Zealand

| Pacific cuisines

Characteristics & influences
New Zealand cuisine is characterised by its freshness and diversity. Diversity is owed to its relative youth, in world terms, which brings a willingness to experiment with food. Freshness is owed to its surrounding ocean and fertile lands. Its distinctiveness is more in the way New Zealanders eat - generally preferring as relaxed and unaffected as possible.

New Zealand’s cuisine has been described as Pacific Rim, drawing inspiration from Europe, Asia, Polynesia and its indigenous people, the Maori. For dishes that have a distinctly New Zealand style, there’s lamb, pork and cervena (venison), salmon, crayfish (lobster), Bluff oysters, paua (abalone), mussels, scallops, pipis and tuatua (both are types of New Zealand shellfish); kumara (sweet potato); kiwifruit, tamarillo and feijoa; and pavlova, the national dessert.

Perhaps even better known than New Zealand lamb, mussels and other foods are the country’s wines. There are 10 major wine-producing areas in New Zealand, with Marlborough famed for its sauvignon blanc, Gisborne for its chardonnay, and Central Otago and Martinborough building a reputation for pinot noir and pinot gris. Hawke’s Bay is known for its bold cabernets and Auckland’s Waiheke Island is home to one of the top 20 cabernet blends in the world. Marlborough and Hawke's Bay are New Zealand’s two premium wine-growing regions.

Casual Kiwi cuisine
The New Zealand summer sees many eat outdoors, often in a barbecue setting. Kiwi barbecues - featuring New Zealand beef, lamb and seafood - are a big part of the culture, and typify the laid-back nature of the New Zealand people.

A Maori specialty is the hangi (pronounced hung-ee), a pit in which meats or fish are cooked with vegetables. A deep hole is dug in the ground, lined with red-hot stones and covered with vegetation. The food is then placed on top. The whole oven is sprinkled with water and sealed with more vegetation. The hole is then filled with earth and left to steam for several hours. Traditionally, men dig and prepare the hole, and women prepare the food to go in it. All members of an extended family (whanau) help out for such a feast. The occasion is relaxed, friendly and fun, with people often eating the meal under a marquee.

New Zealand's cuisine experience is incomplete without enjoying traditional fish and chips served wrapped in paper.

New Zealand dishes include:

 * Fish and Chips
 * Colonial Goose (a surprisingly effective preparation of roast leg of lamb)

Desserts:

Drinks:
 * Pavlova (a light and fluffy meringue dessert named after the ballet dancer, Anna Pavlova)
 * Lamingtons (a type of sponge cake (or more traditionally, butter cake) squares, coated in a layer of chocolate icing or strawberry jam, then desiccated coconut)
 * Anzac biscuits

 * L&P
 * Flat white

title: Cookbook:Cuisine of New Zealand


text:
User:Christiaan~enwikibooks

Articles I've written:
 * Cookbook:Cuisine_of_New_Zealand

title: User:Christiaan~enwikibooks


text:
Cookbook talk:Cuisine of New Zealand

Well, that's a first effort, I'm sure it can be much improved upon. -Christiaan

Kiwi barbecues
To my mind, this involves cooking small fuzzy fruits. I hear that "kiwi" can also mean a human from New Zealand or an endangered bird. Uh, the bird sounds tasty, but... AlbertCahalan 19:38, 4 May 2005 (UTC)

The Picture
Couldn't we take the Vegemite picture down and replace it with a Marmite one? —The preceding unsigned comment was added by 222.154.5.87 (talk • contribs) 2007-01-07T14:53:33.

 * G'day, I couldn't find any Kiwi Marmite photos in Commons, so I replaced the Vegemite with a photo of the big L&P bottle at Paeroa. I miss that stuff :( Webaware talk 10:01, 17 January 2007 (UTC)

title: Cookbook talk:Cuisine of New Zealand


text:
Cookbook talk:Vegetable

why has everything got "CookBook" in front of it?

 * To answer Christiaan's question above, we use the "Cookbook:" prefix to avoid possible nameing conflicts. While many pages in the cookbook have unique names without the prefix, some would not.  For example, the cookbook module on salt, at Cookbook:Salt, should have different information than Salt (general chemistry), if it's there yet (I don't really know what naming convention they're using over there), while our module on oysters is very different than what you'd expect to find in Marine Biology:Oyster (which doesn't exist, but might one day). Gentgeen 17:37, 11 Apr 2004 (UTC)

title: Cookbook talk:Vegetable


text:
Guide to Unix/Commands

The Unix command line is often considered difficult to learn. This book aims to help beginners by introducing various commands in lucid and simple language. Unlike most command references, this book is designed to be a self-study guide.

title: Guide to Unix/Commands


text:
First Aid/Appendix D: Notes for First Aid Instructors

Education
The teaching of first aid is in itself a valuable safety measure. Educating even a few people in first aid improves the safety of everyone they come into contact with.

Teaching first aid also provides an opportunity for teaching safety, prevention and risk reduction measures at the same time, as part of the first aid curriculum.

A person trained in first aid is more motivated to avoid personal injury, because their first aid training gives them a greater appreciation of the potential serious consequences.

Last but not least, someone trained in first aid can be taught to appreciate the importance of safety, prevention and risk reduction. This makes them an ideal advocate for spreading safety awareness to others in their organization or family. This "peer to peer" education can be very valuable.


text:
User:Vuara/PINXXIEE Method

 * http://www.tienstrokes.com/english/Technology.htm
 * http://www.tienstrokes.com/english/company.htm
 * http://www.tienstrokes.com/english/beijing_px_report.htm
 * http://www.tenstrokes.com/english/Global%20Alphabet%20Chart.htm

 * maxx 	=	mother
 * ccaaott =	grass
 * muutv 	=	tree
 * fentt	=	scent
 * xiaaokk	=	smile
 * gguoq	=	nation

 * liannthht + hhejy + gguoq => Lianheguo (United Nations)

 * muu 	=	eye

CHINESE CHARACTER PINXXIEE METHOD

 * Introduction
 * What is PX or Modernized Internet-Chinese Script
 * The PX Rules

Introduction:

Chinese Character Pinxxiee Method is based on the Chinese Language Pinyin Scheme, the official Chinese phonetic standard, as promulgated and published by China and recognized by the United Nations.

What is PX or Modernized Internet-Chinese script?

If the user wishes to learn the complete Pinxxiee language input method, Chinese Computer Communication, Inc. has a DOS version of PX 2001 Tutor#, which serves as a Chinese word processor and also instructs the user how to input Chinese characters using the complete Pinxxiee language technology.

Pinxxiee language technology (Æ´Ð´ÓïÑÔ¼¼Êõ) has advanced to the point that it will complement the other input methods (ÊäÈë•½•¨)), popular on the market, such as Cangjie Code (²Öò¡), Wubizi Code (Îå ±Ê×Ö), Pinyin (Æ´Òô) or Zhuyin (×¢Òô) spellings, and can also facilitate the use of telegraphic code (µç±¨ Âë), Three-corner Code (Èý½Ç), Four-corner Code (ËÄ½Ç), or Simple Stroke (±Ê») or other graphic-phonetic (ÐÎÉù) hybrid methods, with or without AI or abbreviations, for speedy Chinese word-processing . However, in the long run, speed of input should not be the only criterion of economy. We need to consider the meaning of the input language in word-processing and writing manuscripts for the users’ true productivity and creativity.

The meaning of an input language is highly significant for the average Chinese user, as it is for the ordinary English user of the typewriter or word processor. Stenography and shorthand are high-speed tools of English language specialists, but ordinary writers of English do not learn stenography or shorthand to do typing or word-processing. In fact, ordinary writers need a meaningful language to learn, to read, and to type. What’s typed should be readable and memorable. There are hundreds of Chinese input methods but they are mostly not readable or easily memorable by the average person. Hence, the need for this Pinyin+Pinxxiee Computer-Chinese language.

In brief, Pinxxiee language is Computer-Chinese, electronic Chinese, or e-Chinese for the Internet. Its time has come. It is the linear transformation of the square Chinese character in the computer of Information Age. Pinxxiee Æ´Ð´ integrates, explains, and aids the other input methods in their functions as speed tools, just like the English language must be the foundation of English shorthand, stenography or abbreviations.

And similarly, Pinxxiee completely solves the Chinese homophone problem and also serves as a computer-compatible language. As with any languages, you, of course, have to learn it. On the average, it takes 15 hours to learn Pinxxiee, if you are familiar with Pinyin and Chinese characters. Furthermore, the user can learn Pinyin+ Pinxxiee language with PX 2001 Tutor, independently. Now, let’s learn step 1: Pinyin+Tones, so that we can apply Pinyin+Tones to Chinese input for Palm-size personal computers, effectively and efficiently.

The CAIS’99 Dictionaries within the Manual serves you as a standard for Chinese and Pinyin+ spellings of Chinese characters. Pinyin+ gives you faster and more efficient graphic input tools with a future. And this CAIS’99 Dictionaries helps you now with proper spelling and writing of any lanugage as the PX 2001 Tutor will give you a glimpse into the bright future of Chinese/English and bilingual word-processing and multilingual data management.

The PX Rules

1. The Pinyin Rule: Use the Pinyin syllable as the phonetic base. Xinhhua ZiidiaannÑ [New China Dictionary] serves as this Manual’s standard reference for Pinyin syllabary, character fonts and stroke counts.

2. The Pinyin+Tone Rule: Replace the Pinyin diacritic tone marks with tonal letters, according to a linguistic consonant-vowel principle: that is, repeat a consonant (to raise the tone) or repeat a vowel (to lower the tone). For example:

1st tone: ma = £í ¨¡= ma (Âè )

2nd tone: mma = £í ¨¢= mma (Âé )

3rd tone: mmaa= £í ¨£= mmaa (Âí )

4th tone: maa = £í ¨¤= maa ( Âî ) so that,

1st tone: The Pinyin spelling base remains unchanged. Remove the level tone mark from the principal vowel (¨¡), but add no letters. e.g., ma (Âè ), ba, ci, de, er, etc.

2nd tone: Remove the tone-raising mark from the principal vowel. Replace it with the tone-raising consonant by doubling the consonant. e.g., mma (Âé ), bba, cci, dde, err, etc.

3rd tone: Remove the tone-lowering-and-raising mark from the principal vowel. R eplace it with two letters, by doubling both the associated consonant and the principal vowel a. e.g., mmaa (Âí ), bbaa, ccii, ddee, eerr, etc.

4th tone: Remove the tone-lowering mark from the principal vowel. Replace it with the tone-lowering letter, by doubling the principal vowel. e.g., maa ( Âî ), baa, cii, dee, eer, etc.

What if the Pinyin syllable has more than one consonant?

2.1 The Pinyin+tone nth Rule. Whenever the consonant letter n is present in any syllable e.g. man, n is the consonant to be doubled to signal the tone (2nd or 3rd). The nth Rule’s power and its ease to master becomes more obvious with the following examples:

1st tone 2nd tone 3rd tone 4th tone

an °² ann [ÎÞ ] aann Ûû aan °µ

ban °ã bann ÙØ baann °å baan °ë

chan ²ó chann µ¥ chaann Ùæ chaan âã

…

man ò© mann Âù maann Âú maan Âý

nan àî nann ÄÐ naann [ÎÞ ] naan ÄÑ

ren [ÎÞ ] renn ÈË reenn ÈÌ reen ÈÏ

zheng Õù zhenng [ÎÞ ] zheenng Õû zheeng Õý

What if the Pinyin syllable has two or more consonants and no letter n?

The following 2.3 Rule is applied:

2.3. The h-rule: Whenever h is present without n (e.g. cha), h is the consonant to be doubled. Examples of the two rules, respectively:

1st tone cha = cha (²æ) 2nd tone cha = chha (²è) 3rd tone cha = chhaa (ïï) ) 4th tone cha = chaa (²î )

What if a syllable has no consonants at all, but only vowels?

The following Rules are applied:

2.4. The single-vowel rule: For single-vowel Pinyin syllables (e.g. "a" or "e"), add an h at the end, and then follow the 2.3.h-rule:

1st tone a = ah (°¢) 2nd tone a = ahh (àÄ) 3rd tone a = aahh (ºÇ) 4th tone a = aah (ºÇ )

2.5. The double-vowel rule: For Pinyin syllables with two vowels and no consonants (e.g. "ao"), have the second vowel act as a consonant:

1st tone ao = ao (°¼) 2nd tone ao = aoo (°½) 3rd tone ao = aaoo (°À) 4th tone ao = aao (°Â)

What about homotonic and homophonic Ideographic characters?

Please, once again, recall the Pinxxiee Formula to our aid:

Pinxxiee = Pinyin + Tone + Radical = Ideograph Px = Py + T + R = Hz Æ´Ð´ = Æ´Òô + Éùµ÷ + ²¿ Ê× = ºº ×Ö Pinxxiee = Pinyin + Shengdiaao + Buushhaao = Haanzii

That’s when we add the needed radical stroke letter(s)to the Pinyin+toned syllable to complete the spelling.

3. The Pinxxiee Radical Rule: Change the ideographic radicals to look-like alphabetic letter suffixes. Chinese characters, through a process of TiENSTROKES Pinxxiee transformation, are changing stroke(s) into a letter or letters, and then make the stroke letters into a radical suffix. For instance, the t-radical is condensed from the H-stroke Ò» and L-stroke l, which, combined, form the t letter, or the -t suffix. A Pinyin+ syllable plus a radical suffix produce a Pinxxiee word, similar to an English alphabetic spelling, e.g., ma + xx = maxx (mother). The following 4 examples show how spellings of Pinxxiee words are naturally derived from Chinese characters themselves, thus retaining their ancient historical roots:

Hanzi Tenstrokes Radical Script Pinyin+Plus Radical Pinxxiee English

Âè CJH-PZH cjh-pzh ma -xx maxx mother

Ä¿ LP-HH-H d-hh muu -dhh muu* eye

²Ý HLL-LPHH-HL tt-lphh-t ccaao -tt ccaaott grass

Ä¾ HL-JY tv muu -tv muutv tree

 * muu Ä¿ can also be spelled as muudhh with itself as radical.

These spellings, maxx Âè, muudhhÄ¿, ccaaott ²Ý, and muutv Ä¾, contain silent Pinxxiee suffixes, serving as the semantic radicals in the original square characters. These radicals must be added to Pinyin+Plus spellings in order to be differentiated from their homophones, e.g. muu Ä¿ must be spelled differently from muutv Ä¾. And so, each Chinese character (square ideograph) has been transformed into a unique monosyllabic Pinxxiee word (linear ideograph) as its original in the computer Internet world.

However, you only need to learn TiENSTROKES Alphabet VHLJY TGPCZ and then about 50 of the most-frequently-used radical-suffixes (like -xx, -dhh, -tt, -tv…) in order to process 95% or more of the ordinary Chinese text. A set of 250+ silent Pinxxiee radical-suffixes has been created to spell differentially all homophones for both complex and simplified characters. This applies to all 60,000+ Chinese ideographs, and even more characters, if created. These suffixes are easy to remember, because they are created to look like the original radicals, by neuro-association, based on a child’s game: search for English letters in Chinese characters! Study the following characters and you will see what we mean:

ccaaott = ²Ý = grass; fentt = •Ò = scent; zhhukk = óÃ; xiaaokk = Ð¦ smile and others.

4. The Two-Syllable Compound Rule: Omit the silent radical suffixes when entering a two-character compound word as follows:

Âè + Âè => ÂèÂè

maxx + maxx => mama

5. The Three-Syllable Compound Rule: Omit the silent suffixes and the tone letters when entering a three-character compound word as follows:

Áª + ºÏ + ¹ú => ÁªºÏ ¹ú

liannthht + hhejy + gguoq => Lianheguo (United Nations)

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Conclusion Report On Dr. H. C. Tien’s [Ìï Èý ÎÄ ] PINXXIEE Technology

by Experts at Standardization Conference

AUTHENTICATION OF CERTIFICATE

The China Standardization Association hereby decrees that the panel of experts were indeed authorized to review Dr. H. C. Tien’s (Tiann San Wenn, Ìï Èý ÎÄ ²© Ê¿ ) presentation and demonstration in defense of his Pinxxiee Language Technology. Furthermore, this conference was declared on public record that the expert panel’s conclusions are true and valid.

On November 3, 1999, a conference was held in Beijing, China, at which Dr. H. C. Tien demonstrated and defended his Pinxxiee Language Technology. The technology was subsequently validated by the China Standardization Association. The CSA invited China’s experts in the following fields to participate in the review: information technology, standardization, ergonomics, software evaluation, software design, language coding analysis, psychology, internet software analysis, et al.

The following conclusions were reached by the commission of panel experts.

Pinxxiee Language Technology was developed based on the Chinese Government’s official standards of spoken Mandarin and the Chinese ("Pinyin") phonetic spelling method. Moreover, this technology adheres to the contents of China’s official Mandarin language Xinhua Dictionary. In conclusion this language technology’s overall design is in accordance with China’s national Mandarin language policies. This technology is an effective system of writing Chinese characters because the problem of duplicate Chinese character spellings is rendered small. All aspects of a Chinese character -- i.e. the phonetic spelling, tone, and semantic root -- are expressed in its Pinxxiee language spelling. This technology is a breakthrough in solving the problem of homophone ambiguity in Mandarin because Pinxxiee spells out all aspects of the Chinese character (see above point # 2). For all practical purposes, the homophone problem has been solved. Pinxxiee Language Technology is in ASCII, so it can be used on the Internet to send Chinese Mandarin email messages (as easily as in English) without the ambiguity that arises when the phonetic "Pinyin" spelling method is used. Thus, fluent speakers of Mandarin, who do not know English and do not have a Chinese character platform on their PC, now have a convenient method for sending and receiving email messages in Mandarin Chinese. This is especially true for international communication, and will have ever increasing usefulness in the future. There is sufficient education software available for Pinxxiee Language Technology, and the PX-2001 software package (PX Toolbox) makes learning this technology easy. In summary, Pinxxiee Language Technology’s method of spelling Chinese characters is technically sound; it has simple rules and very little memorizing needed for Mandarin’s inflection tones. It is suitable for being introduced for learning as early as the elementary and junior high school grade levels by students who have knowledge of the phonetic "Pinyin" spelling method.

The Commission’s panel of experts raised the following relevant issues regarding Pinxxiee Language Technology that it felt must be considered and addressed:

China is a huge country with many different ethnic dialects; and although the Chinese government has extensively taught Mandarin Chinese for more than 30 years, the vast majority of Chinese people still find difficulty in speaking Mandarin properly. They find even greater difficulty in writing Chinese phonetic "Pinyin" spellings. Since teaching Pinxxiee Language Technology is based on both the correct speaking of Mandarin and the spelling of "Pinyin" sounds, Pinxxiee Language Technology will be very difficult to promulgate. Because Pinxxiee Language Technology has all the aspects of the Chinese character contained in its spelling system, its Chinese character representations are much longer than all the other systems used in China today. In addition, the Commission stated that any further developments or improvements of Pinxxiee Language Technology must be in accordance with the Chinese language standards set down by the Government’s relevant official bodies. Up until now the phonetic "Pinyin" spelling method has been primarily used for learning Mandarin Chinese. It has also been a beneficial tool in Chinese character display for software. With the introduction of Pinxxiee Language Technology, we now have what can be called "the Chinese writing script" for the PC and Internet environments. In fact, it has become Chinese language technology for international communication. Needed now is for the general populace to master it while the relevant government bodies support its use. The China Standardization Association hereby decrees that the panel of experts were indeed authorized to review Dr. H. C. Tien’s demonstration and defense of his Pinxxiee Language Technology. Furthermore, this conference was on public record and the expert panel’s conclusions valid.

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1999 EXECUTIVE SUMMARY 2001

1. COMPANY OVERVIEW: In 1960, while working as a practicing physician with dyslexic children, H. C. Tien, M.D.[UMICH.MED] discovered a historic and scientific link between the two major languages of the world, English letters and Chinese characters. In 1983,

Dr. Tien, President and CEO founded CCC, Inc. By 1997, after having received 3 patents for his inventions, he closed his private practice of medicine and decided to launch CCC on the Internet supermarket.

2. DISCOVERY: What he discovered in 1960 turned out to be the "genetic code" of Englishó Chinese transcription in TiENSTROKES ALPHABET. Using TiENSTROKES®, CCC can speed up U.S. trade & communication with China.

3. RESEARCH & DEVELOPMENT: After more than a quarter of a century (1961-1999) of R&D with considerable equity investment from family’s savings and a number of private investors, Dr. TiEN perfected the PINXXIEE theory and its language technology TREE model:

HLJY ó Ä¾ ó MUUH ó TREE.

CCC has copyrighted, trademarked and patented [1960-99] the intellectual properties of a global TiENSTROKES ® Alphabet, and its derived Pinxxiee ComputerChinese™ dictionaries, InternetChinese scripts and language technologies.

4. BUSINESS MISSION: To achieve this mission, CCC seeks strategic investors and experienced technology partners to help plan an IPO by 2001. Today, almost one-forth (1/4) of the world’s population is Chinese, the vast majority can’t touch-type Chinese characters directly on the computer keyboard in their own language (ºº ×Ö ) ! The remaining three-quarters (3/4) including Americans, who are trading with China can’t read Chinese ºº ×Ö. However, the basic invention and the applications of TiENSTROKES to Chinese-English translations has readily enhanced the commercialization of products and services on the market. CCC supplies accurate and rapid Englishó Chinese translation service, using its own products, the 3 InternetChinese™ word-processing and translation tools.

5. PRODUCT & SERVICE:

Ìï ™ TiENSTROKES® INTERNET-CHINESE™ TOOLBOX™, retailing @ $775/box: contains the 3 key tools used by CCC over the years (1983-date) in its Translation Services:

[i] Pinxxiee® Internet-Chinese AutoInput System’98 (CAIS’98), retailing @ $375/ per tool; a Chinese character word-processor that works like English.

[ii] TiENSTROKES® Pinxxiee® e-Chinese SpellChecker’99, retailing@ $350/per tool; for checking the spellings of linearized Chinese characters!

[iii] TiENSTROKES® MindMasterTutor2000™, retailing $300/per tool; a hand-held computer educational tool for students and language workers in translating Chinese ºº ×Ö ó Pinxxiee ó English documents and database.

6. FOCUS: Washington Post has forecast that CCC will eventually become the leader in Pinxxiee Chinese script technology. CCC is ready to enter the largest and fastest growing Internet markets with its Toolbox™99 for US-China & global trade. Although already well known to some important governmental customers and users, e.g., the State of Michigan; GM and EDS, and other fortune companies, CCC aims to fulfill its destiny as a leading corporation in the multilingual information industry, championing for the Golden Age of East-West trade in Peace with Prosperity for the 21st Century.

Æ´ Ð´ PINXXIEE (pin+shay = phonics+graphics) CORPORATION ¹« Ë¾

801 Glenhaven Avenue, East Lansing, Michigan 48823 USA

e-mail contact: dr.tien@pinxxiee.com; Phone:517 336-9500; Fax: 517 336-1212

title: User:Vuara/PINXXIEE Method