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I know this question is quite maths-focused; however, it relates closely to numerical methods that are used to solve Physics problems (for example in Fluid Dynamics/CFD). I asked the same question here on Mathematics SE a few days ago, but no-one there has answered it and it was suggested that I might get a better answer here, since PDEs are core to many physical theories.

I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.

They give a generalized second-order PDE as:

$$A\frac{\partial^2u}{\partial x^2}+B\frac{\partial^2u}{\partial x\partial y}+C\frac{\partial^2u}{\partial y^2}+H=0\tag{1}$$

where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:

$$P=\frac{\partial u}{\partial x}, Q=\frac{\partial u}{\partial y}, R=\frac{\partial^2 u}{\partial x^2}, S=\frac{\partial^2 u}{\partial x\partial y}, T=\frac{\partial^2 u}{\partial y^2}$$

They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:

$$dP=Rdx+Sdy\tag{2}$$

$$dQ=Sdx+Tdy\tag{3}$$

Using the substitutions above, the original PDE can be written:

$$AR+BS+CT+H=0$$

Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:

$$S\Bigl[A\Bigl(\frac{dy}{dx}\Bigr)^2-B\Bigl(\frac{dy}{dx}\Bigr)+C\Bigr]-\Bigl\{\Bigl[A\Bigl(\frac{dP}{dx}\Bigr)+H\Bigr]\frac{dy}{dx}+C\frac{dQ}{dx}\Bigr\}=0\tag{4}$$

It then states that if:

$$A\Bigl(\frac{dy}{dx}\Bigr)^2-B\Bigl(\frac{dy}{dx}\Bigr)+C=0\tag{5}$$

then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $\frac{dP}{dx}$ and $\frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.

So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?

Thanks in advance!

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  • $\begingroup$ Hi! You said, "...when the terms P, Q and H on the right-hand side of equation (4) still contain partial differentials?" I'm wondering what makes you think this since there is no such dependence explicity? Do you mean that since $P$ and $Q$ are themselves defined as partial derivatives? $\endgroup$ – N. Steinle Nov 28 '18 at 19:21
  • $\begingroup$ @N.Steinle yes. Since $P$ and $Q$ are both partial derivatives of $u$, I fail to see how the rearrangement has reduced the equation to total derivatives, or how it is helping in the ultimate 'quest' to find the solution to the original pde ($u(x,y)$). $\endgroup$ – Time4Tea Nov 28 '18 at 19:25
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Having looked at the book you based this question on I will provide answers to your questions which is partly paraphrased from the book. In my opinion your questions are answered in the book if you consider the section on which you have question in context of the whole chapter.

First some definitions; A second-order PDE of the form of (1) is classified according to its discriminant $\Delta = B^2-4AC$:

  • elliptic PDE: $\Delta<0$
  • parabolic PDE: $\Delta=0$
  • hyperbolic PDE: $\Delta>0$

To classify a PDE into the above categories, we first seek characteristic direction along which the governing equations only involve total differentials.

To understand why (5) gives the characteristic direction, lets first find it for a first-order PDE (taken from the book but with slightly different symbols): $$A\frac{\partial u}{\partial x} + B\frac{\partial u}{\partial y} = C$$

The characteristic direction is defined as: $$\frac{dy}{dx}=\frac{B}{A}$$ Along this direction, the original PDE reduces to: $$\frac{du}{dx}=\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\frac{dy}{dx} = \frac{C}{A}$$ and $$\frac{du}{dy}=\frac{\partial u}{\partial y} + \frac{\partial u}{\partial x}\frac{dx}{dy} = \frac{C}{B}$$ This allow one to solve it as two ordinary differential equations in $x$ and $y$ along a grid defined by the characteristic direction as long as the initial conditions are given on a non-characteristic line.

Now considering your questions:

By subtituting (2) and (3) you have eliminated $R$ and $T$, what is left of the partial derivatives is $S$ which only appears in the first term in (4); by making sure (5) applies, you effectively also remove $S$ from (4). You are now left with a differential equation with total derivatives only involving $P$, $Q$ and $H$: $$\left[A\left(\frac{dP}{dx}\right)+H\right]\frac{dy}{dx}+C\frac{dQ}{dx}=0\tag{6}$$

where the characteristic direction, $\frac{dy}{dx}$, is defined by the quadratic equation (5) with solution: $$\frac{dy}{dx}=\frac{B\pm\sqrt{\Delta}}{2A}\tag{7}$$

Consideration of the discriminant determines both the type of the PDE and the nature of the characteristic:

  • elliptic PDE, two omplex characteristics exist
  • parabolic PDE, one real characteristic exists
  • hyperbolic PDE, two real characteristics exist

The new differential equation (6) only involves total differentials as it has combined the partial derivatives using the characteristic direction (7): $$\frac{dP}{dx}=\frac{\partial P}{\partial x} + \frac{\partial P}{\partial y}\frac{dy}{dx}$$ $$\frac{dP}{dy}=\frac{\partial P}{\partial y} + \frac{\partial P}{\partial x}\frac{dx}{dy}$$ and similar for $Q$ such that it can be solved in a similar manner to the first-order PDE.

The usefulness of these characteristics in terms of solutions to the PDEs is up for debate, per classification (taken from sections on Interpretation by Characteristics):

  • elliptic PDE: Characteristics are complex and cannot be displayed in the (real) computational domain. For problems in fluid dynamics, identification of characteristic directions serves no useful purpose.
  • parabolic PDE: Characteristics do not play such a significant role as for hyperbolic PDEs. There is no equivalent method of characteristics.
  • hyperbolic PDE: it is possible to use characteristic directions to develop a computational grid on which the compatibility conditions hold. This is the strategy behind the method of characteristics, Sect. 2.5.1.
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