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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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    $\begingroup$ The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics. $\endgroup$ – Bedovlat Nov 27 '18 at 22:41
  • $\begingroup$ @Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $\frac{dP}{dx}$ and $\frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding. $\endgroup$ – Time4Tea Nov 28 '18 at 0:25
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    $\begingroup$ Crossposted to physics.stackexchange.com/q/443899/2451 $\endgroup$ – Qmechanic Nov 28 '18 at 19:24
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If $A\neq0$ then the equation (5) is a quadratic equation, and unless your PDE is elliptic ($B^2-4AC<0$), locally it has two solutions $y_1(x)$ and $y_2(x)$ which satisfy Vieta's formula, $$ A\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)=B,\quad A\frac{dy_1}{dx}\frac{dy_2}{dx}=C. $$ If your PDE is parabolic ($B^2-4AC=0$) then $y_1=y_2$. It follows that $$ A\partial_x^2+B\partial_x\partial_y+C\partial_y^2=A\left(\partial_x^2+\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)\partial_x\partial_y+\frac{dy_1}{dx}\frac{dy_2}{dx}\partial_y^2\right) $$ $$ =A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)+\mbox{first order terms}. $$ Therefore (1) now becomes $$ A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)u+\mbox{lower order terms}=0, $$ which is the expected simplification.

This is the high school level answer to the question why characteristics matter. The scientific answer is much longer and involves microlocal analysis (propagation of singularities), quantum field theory (trajectories of classical point particles corresponding to a linear quantum field), etc.

This is, however, not a complete answer to the question posted, for I have no suggestion as to why (4) is relevant.

Edit: By no means could I read the author's mind when they wrote equation (4), but let me elaborate on what could be done along those lines (at OP's request).

(5) is a first order ODE, of which the solutions depend on one parameter (initial data). For every $(x_0,y_0)$ in the domain let $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ be the two solutions of (5) satisfying $$ y_1(x_0;x_0,y_0)=y_0,\quad y_2(x_0;x_0,y_0)=y_0. $$ In other words, $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ are the two characteristics through the point $(x_0,y_0)$. Fix $(x_0,y_0)$ for a moment, and omit their mention for brevity. Let us do with the characteristic $y_1(x)$ first. Consider $P$ and $Q$ as independent functions, then their restriction to the characteristic $P(x,y_1(x))$ and $Q(x,y_1(x))$ as unknown functions of $x$. If $H$ contains no zero order terms then $H(P,Q,x,y_1(x))$ is a function of $P$,$Q$ and $x$. This renders (4) a first order ODE relating two unknown functions $P$ and $Q$ of one variable $x$, $$ \left[A(x,y_1(x))\frac{dP}{dx}+H(P,Q,x,y_1(x))\right]y_1'(x)+C\frac{dQ}{dx}=0. $$ If you can solve this equation in one way or another then you obtain $$ F_1(P,Q,x)=C_1, $$ where $F_1$ is a function and $C_1$ is a number, both depending on the choice of the characteristics, i.e., $(x_0,y_0)$. Now if you do the same thing with the characteristic $y_2(x)$ you end up with $$ F_2(P,Q,x)=C_2. $$ On the other hand, if your $H$ contains zero ordet terms, $H=H(P,Q,x,y,u)$, then $$ u(x,y_1(x))=\int^x\left[P+y_1'(t)Q\right]dt, $$ and (4) becomes an integral-differential equation. Good luck solving it.

Now the final solution of (1) depends on the specification of the boundary (Cauchy) data. Suppose that you have a Cauchy curve $\Sigma$ that intersects every characteristic exactly once, and suppose that $P$, $Q$, $R$, $S$, $T$ and $u$ are known along $\Sigma$. Consider only those $(x_0,y_0)\in\Sigma$, and let $p(x_0,y_0)$, $q(x_0,y_0)$ be the values of $P$ and $Q$ at $(x_0,y_0)$. We now recall all dependences upon $(x_0,y_0)$ above. If $(x,y)$ is any point in the domain then let $(x_1,y_1)$ and $(x_2,y_2)$ be the intersections of $y_1(x)$ and $y_2(x)$ with $\Sigma$, respectively, $$ y=y_1(x;x_1,y_1),\quad y=y_2(x;x_2,y_2). $$ This brings us to the system $$ \begin{cases} y=y_1(x;x_1,y_1)\\ y=y_2(x;x_2,y_2)\\ F_1(P,Q,x;x_1,y_1)=F_1(p(x_1,y_1),q(x_1,y_1),x_1;x_1,y_1)\\ F_2(P,Q,x;x_2,y_2)=F_2(p(x_2,y_2),q(x_2,y_2),x_2;x_2,y_2) \end{cases} $$ This is the parametric solution of the problem (1) with given boundary data $p$ and $q$.

If we want an explicit solution, we make a few more steps. The curve $\Sigma$ can be locally parameterized by one variable $s$, so that $(x_1,y_1)=\eta(s_1)$ and $(x_2,y_2)=\eta(s_2)$. Therefore the above is a system of 4 equation for 4 variables $P$, $Q$, $s_1$ and $s_2$. Eliminating $s_1$ and $s_2$ we find the desired solution $$ P(x,y),\quad Q(x,y). $$ Finally, $$ u(x,y)=u(x_0,y_0)+\int_{(x_0,y_0)}^{(x,y)}\left[Pdx+Qdy\right] $$ along your favourite curve joining $(x_0,y_0)$ and $(x,y)$. The point $(x_0,y_0)\in\Sigma$ therefore we know $u(x_0,y_0)$ from the boundary data.

This is how, I think, you would solve (1) if you want to start from (4). Generally, the method of characteristics always ends with a parametric solution, and for an explicit solution you need to exliminate the dependence upon the characteristic.

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    $\begingroup$ The benefit is that, if $H$ doesn't contain zero order terms then it is a function $H(P,Q,x,y(x))$ and (4) is a quasilinear first order ODE for two functions$P$ and $Q$. As in Hamiltonian dynamics, you forget that $P$ and $Q$ are actually partial differentials, and consider them as independent functions, which are then coupled by a further equation. If you manage to find an integrating coefficient for (4) then you can solve it as $F(P,Q,x)=0$. $\endgroup$ – Bedovlat Nov 29 '18 at 16:44
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    $\begingroup$ If $H$ contains zero order terms, i.e., $H=H(P,Q,x,y(x),u)$, then you do not get an equation involving $P,Q,x$ alone, since you don't still know the function $u(x,y(x))$ along the characteristics. If you are able to solve for $u(x,y(x))$ in the first place though, then zero order terms work as well. Concerning first order methods, its not that quick, but eventually you come to that. The function $F(P,Q,x)=0$ I wrote above depends on the characteristic, so it is really $F(P,Q,x,y)=0$. Generally speaking, the method of characteristics works only in that sense - eliminating the initial data. $\endgroup$ – Bedovlat Dec 1 '18 at 3:36
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    $\begingroup$ I will be glad to know this helps your research. In linear quantum field theory, quantum fields are assumed to weakly satisfy a hyperbolic PDE (e.g., Klein-Gordon). It is known in microlocal analysis that singularities of solutions to hyperbolic PDE travel along characteristics (lightlike trajectories). So if your field has a delta function at $t=0$ then further in time your delta function will travel along the characteristics. This is a multifaceted and interesting interplay between maths and physics. $\endgroup$ – Bedovlat Dec 1 '18 at 23:31
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    $\begingroup$ Congrats, @Bedovlat!! Nice work! $\endgroup$ – Namaste Dec 3 '18 at 19:28
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    $\begingroup$ Thank you, @amWhy, for the bounty. I am glad to see that 6 out of 34 people who liked the question also liked my answer. The rest maybe expected something more spectacular :) $\endgroup$ – Bedovlat Dec 4 '18 at 1:17

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