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Given the PDE, find the general solution.

$$ -4yu_x + u_y-yu=0 $$

So,

$$ \frac{dy}{dx} = \frac{1}{-4y} $$

$ x+2y^2=k $ is a characteristic curve. If I do the parametrization

$$ y=s \\ x=k-2y^2=k-2s^2 $$

The PDE becomes the ODE

$$ \frac{du}{ds} -yu = 0 \rightarrow \frac{du}{ds}-su = 0 \\ u=f(k)e^{\frac{s^2}{2}} $$

Which gives me the answer

$$ u(x,y)=f(x+2y^2)e^{\frac{1}{2}y^2} $$

However, if I divide the original PDE by $-4y$, then I get: $$ u_x + \frac{1}{-4y}u_y+\frac{1}{4}u=0 $$

with the same characteristic curve. But now I can do the following parametrization:

$$ \frac{dx}{ds}=1 \rightarrow x=s\\ \frac{dy}{ds}=-\frac{1}{4y}\rightarrow -2y^2=s+k $$

The PDE then becomes:

$$ \frac{du}{ds} + \frac{1}{4}u = 0 \Rightarrow u=f(k)e^{-\frac{s}{4}} $$

The final answer this time is:

$$ u(x,y) = f(x+2y^2)e^{-\frac{x}{4}} $$

My question is: are both answers the same? WolframAlpha gives as the answer the second option. If I say $x+2y^2=0 \rightarrow y^2=-x/2$, I can change the $exp$ term from one to another, but is this right to do? And why should I get $x+2y^2=0$, and not any other value of the characteristic $k$? Is it somewhat preferably to write the solution in terms of $x$ instead of $y$?

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  • $\begingroup$ You still need to have $$ u(x,0) = f_2(x)\mathrm{e}^{-x/4} = f_1(0) $$ So it may work out in the end - but the function is determined by conditions. $\endgroup$ – Chinny84 Feb 14 '17 at 19:48
  • $\begingroup$ The exercise proposes to try to obtain the solutions to (i) $ u(x,y)=x^3 $ along $ x+2y=3 $, (ii) $ u(x,y)=-y $ along $ y^2=x $ and (iii) $ u(x,y)=2 $ along $ x+2y^2=1 $. Not exactly initial conditions like $ u(x,0)=u_0(x) $, so I was wondering why should I stick with the solution in terms of $ x $. $\endgroup$ – Thales Feb 14 '17 at 19:54
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The solutions are indeed the same:

$$u(x,y)=f(x+2y^2)e^{y^2/2}=g(x+y^2)e^{-(1/4)(x+2y^2)}e^{y^2/2}=g(x+2y^2)e^{-x/4}$$

You still have a function with $x-y^2$ as argument, but not the same as in the first case.

$$f(x+2y^2)=g(x+2y^2)h(x+2y^2)\;\mathbb{with}\;h(x+2y^2)=e^{-(1/4)(x+2y^2)}$$

Added You can get now any of the particular solutions you are requested to give. I solve here the second one in your comment. The others can be solved in the same fashion.

It can be write "$u(x,y)=−y$ along $y^2=x$" this way:

$$u(y^2,y)=-y$$

Substituting into the general solution,

$$u(x,y)=f(x+2y^2)e^{y^2/2}$$

gives:

$$u(y^2,y)=f(y^2+2y^2)e^{y^2/2}=-y$$

Rearranging:

$$f(3y^2)=-ye^{-y^2/2}$$

Now, the variable change $z=3y^2$ drives to the function $f$ corresponding to these initial conditions.

$$f(z)=\pm\sqrt{z/3}e^{-z/6}$$

Now, the argument of $f$ have to be $x+2y^2$, and substituted into $u$ is

$$u(x,y)=\pm\sqrt{\frac{x+2y^2}{3}}e^{-(x+2y^2)/6}e^{y^2/2}=\pm\sqrt{\frac{x+2y^2}{3}}e^{(y^2-x)/6}$$

As surface in $\mathbb R^3$ the two halves fit smoothly together as to form a unique surface. Nevertheless, due to the introduction of the square in the varible change, it's necessary to "tune" them to fit to the intial conditions. We use both them in a suitable domain to get:

$$u(x,y)=\begin{cases} -\sqrt{\frac{x+2y^2}{3}}e^{(y^2-x)/6}\;\mathbb{if}\;y>0\\ \sqrt{\frac{x+2y^2}{3}}e^{(y^2-x)/6}\;\mathbb{if}\;y\le 0 \end{cases}$$

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