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Okay, so I am given the following Parabolic PDE:


I find the characteristics to be:

$\frac{dy}{dx} = \frac{-x}{y}$ and therefore $\psi(x,y) = x^2+y^2$

Now, I have been told that I need to choose the other constant, $\eta$. In the answer for the question, the lecturer decides to choose $\eta(x,y) = y^2$ but gives no reason as to why he does. I decided to choose $\eta(x,y) = x^2$ but this does not give the same answer.

Is there a method on how to know what value to choose for the other constant?

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2 Answers 2

The change of variables should be non-singular, so you should be able to reconstruct the original independent variables $x$ and $y$ from the new ones. So your choice is also valid! It might lead to a different answer, that's ok, they're both valid, just with different lower order terms. Just make sure you did indeed end up with the (same) canonical form.

The reason you have this degree of freedom is because the characteristic equation has a zero determinant. So, just like when a quadratic equation has a zero determinant there's one ("double") root, so too here, there is only one root, which leaves a degree of freedom in choosing the second variable.

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Thanks for the answer. We've not actually been told about singularity and non-singularity with respect to PDEs, so do you mind telling me how to go about that? –  Keir Simmons May 26 '13 at 21:23
It's not really a PDE issue, it's an attribute of the transformation (i.e. the change of variables). Just like with integration. A non-singular transform is one that can be inverted. An example of a singular transform would be $\xi=xy$, $\eta=x+y$. –  davin May 26 '13 at 21:31
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Your equation is $Du = 0$ where $$ D = \left(y^2\partial_{x}^2 - 2xy\partial_{x}\partial_{y} + x^2\partial_y^2 - \frac{y^2}x\partial_x - \frac{x^2}y\partial_y\right). $$ With some work, we can factor $D$: \begin{align*} D & = (y\partial_x - x\partial_y)^2 + \left(\frac yx - \frac xy\right)x\partial_y + \left(\frac xy - \frac yx\right)y\partial_x \\ & = (y\partial_x - x\partial_y)^2 + \left(\frac xy - \frac yx\right)(y\partial_x - x\partial_y) \\ & = \left(y\partial_x - x\partial_y + \frac xy - \frac yx\right)(y\partial_x - x\partial_y). \end{align*} This suggests solving the PDE in two steps. The first one is to solve for $v$ from $$ \left(y\partial_x - x\partial_y + \frac xy - \frac yx\right)v = 0. $$ The second one is to solve for $u$ from $$ \left(y\partial_x - x\partial_y\right)u = v. $$ In both steps, the characteristics are defined by $\frac{\partial y}{\partial s} = -x$ and $\frac{\partial x}{\partial s} = y$ where $s$ is the curve parameter (or in your notation, $\frac{dy}{dx} = -\frac{x}y$, and the curve parameter is $x$). The equation for $v$ is $\frac{\partial v}{\partial s} = -\left(\frac xy - \frac yx\right)v$ while the equation for $u$ is $\frac{\partial u}{\partial s} = v$.

To solve for characteristics, differentiate equations for $y$ and $x$ once more to get $\frac{\partial^2 y}{\partial s^2} = -y$ and $\frac{\partial^2 x}{\partial s^2} = -x$. The solution is $x = c_1(r)\cos(s) + c_2(r)\sin(s)$, $y = -c_1(r)\sin(s) + c_2(r)\cos(s)$, where $r$ is the index of the characteristic curve. (Graphically, characteristics are circles, as you have seen from $x^2 + y^2 = \psi$. Your $\psi$ is the same as $c_1(r)^2 + c_2(r)^2$ in my notation.)

With two arbitrary functions of $r$, you do have some freedom here. You get to choose where each curve starts, which is the point on the circle that $s = 0$ corresponds to. Usually, you use the initial condition to decide, because making $s = 0$ correspond to the initial solution curve will simplify a lot of calculation. Anyway, since the initial condition is not given, I'll go ahead and do some computation.

Your lecturer's choice makes the positive $y$-axis correspond to $s = 0$ (so $c_1(r) = 0$), while your choice makes the positive $x$-axis correspond to $s = 0$ (so $c_2(r) = 0$). Your lecturer's choice leads to something like \begin{align*} x(r, s) & = r\sin s \\ y(r, s) & = r\cos s \end{align*} while your choice leads to something like \begin{align*} x(r, s) & = r\cos s \\ y(r, s) & = -r\sin s. \end{align*} It should not matter which one you choose. Note that with your lecturer's choice, you have \begin{align*} \frac yx - \frac xy & = \cot s - \tan s \\ & = 2\cot(2s) \end{align*} and with your choice, you have \begin{align*} \frac yx - \frac xy & = -\tan s + \cot s \\ & = 2\cot(2s). \end{align*} In both cases, $\frac yx - \frac xy$ becomes the same $2\cot(2s)$, and that is the expression you'll need to solve for $v$ and $u$: \begin{align*} \frac{\partial v}{\partial s} & = \left(\frac yx - \frac xy\right)v \\ & = 2\cot(2s)v\\ \therefore v & = C(r)\sin(2s) \\ \text{and from} \frac{\partial u}{\partial s} = v \quad \Longrightarrow \quad u & = -\frac 12C(r)\cos(2s) + D(r) \\ \therefore u & = \tilde C(x^2 + y^2) \cdot (x^2 - y^2) + \tilde D(x^2 + y^2). \end{align*} Although $\cos(2s)$ becomes different expressions in different parametrizations, the final expression is the same because the sign difference is subsumed in $\tilde C$. I did get sloppy and ignored the case $\sin(2s) < 0$. However, the final solution works out if we keep in mind that $\tilde C$ and $\tilde D$, once solved from the initial condition, are valid only in the quadrant that contains the initial solution curve. It is natural to expect that the solution does not emanate past the $x$-axis or the $y$-axis because we have $x$ and $y$ in the denominators in the original PDE.

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