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I recently asked why $u(x, y) = A(x - y)$ is the general solution to this problem:

Solve the PDE $u_x + u_y = 0$ in the domain $y > \phi(x)$, $x \in \mathbb{R}$ given that $u = g(x)$ on the curve $y = \phi(x)$, where $\phi(x) = \frac{x}{1 + |x|}$.

From the generous responses of Batominovski and Mattos, I tried to develop my own understanding of this. I am seeking advice from others as to whether my reasoning is correct. In addition to my reasoning please also critique my language. I am not sure if the way I have thought about everything is correct so this might be reflected in my use of mathematical language (I might have described certain things incorrectly, confused concepts, and so on).

Starting with the easy part, we use separation of variables on two of the characteristic equations:

$\dfrac{dx}{dt} = 1$

$x(t) = t + C_1(\gamma)$

$\dfrac{dy}{dt} = 1$

$y(t) = t + C_2(\gamma)$

$C_1(\gamma)$ and $C_2(\gamma)$ are constants along any specific characteristic curve but this constant is different for different characteristic curves. That is why I set $C_1$ and $C_2$ as being dependent on $\gamma$. I will say that when $t = 0$, then we are on the initial curve and $\gamma$ will change along the initial curve as it passes through different characteristic curves.

If I am correct in my thinking, the $y$ characteristic must inherit its parameterisation from the curve $y = \phi(x) = \frac{x}{1 + |x|}$. However we can parameterise $x$ in a way that is convenient. I will set $x = \gamma$ when $t = 0$ and $y = \phi(x) = \frac{x}{1 + |x|}$ when $t = 0$.

So we have $x(0) = C_1(\gamma) = \gamma$ and $y(0) = C_2(\gamma) = \frac{x}{1 + |x|} = \frac{\gamma}{1 + |\gamma|}$ (since $x = \gamma$ when $t = 0$). This means that our characteristic curves are $x(t) = t + \gamma$ and $y(t) = \frac{\gamma}{1 + |\gamma|}$.

If I am not mistaken, these are our change of variables, which map from $(s, t) \to (x, y)$.

We now try to find the general solution by using separation of variables on the third characteristic equation.

We have $\dfrac{du}{dt} = 0$.

Using separation of variables gives $u(x, y) = A(C_3)$.

Now we come to the part that was of main confusion in the recently asked question: Why is $A$ dependent on $C_3$? My reasoning is this:

The values of $C_1(\gamma)$ and $C_2(\gamma)$ in the characteristic equations change from one characteristic to another but are constant on any single characteristic. In other words, the values of $C_1(s)$ and $C_2(s)$ distinguish the different characteristics. $A$ is also a constant along any single characteristic but will be different for different characteristics. Therefore, $A$ will change depending on $C_1(\gamma)$ and $C_2(\gamma)$ in the characteristic equations. Therefore, let $C_3 = C_1 - C_2$. We then have $x - y = C_1 - C_2 = C_3$. Therefore, we have our result that $u = A(C_3) = A(x - y)$.

Thank you for kindly reviewing.

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Regarding your confusion of using the third characteristic equation $\frac{du}{dt} = 0$. If one writes $\tilde u(s,t)=u(x,y)$, the third characteristic equation might be written out as

$$ \frac d{dt} u(x(t),y(t)) =\frac d{dt} \tilde u(s,t) = 0$$ so $\tilde u(s,t) = A(s)$ is constant in $t$, only depending on the choice of characteristic curve i.e. choice of initial condition $x(0),y(0)$, which is parameterised by $s$. Undoing the change of coordinates $(x,y)\mapsto (s,t)$ will lead you to the stated result.

Regarding language-

  • I believe separation of variables is a distinct concept, namely when you assume that the solution to some PDE splits as a product $u(x,y)=u_1(x)u_2(y)$.
  • It may or may not be clearer to you to fix the initial curve parameterisation first, say $\mathbf\Gamma$, i.e. $\mathbf\Gamma(s) = (s,\phi(s))$ for $s\in\mathbb R$. This allows you to sidestep defining constants $C_1,C_2,C_3$ with implicit $s$ dependence: you can write that you are considering initial conditions $(x_0,y_0)$ in $\mathbf\Gamma$ , i.e. the initial conditions you want to use are precisely all the points $\mathbf\Gamma(s)$. Also, the solution to the first two characteristic equations are explicitly functions of $s$(and $t$), say $\mathbf x(s,t) = (x(s,t),y(s,t))$, and the two equations are explicitly $$ \frac \partial{\partial t} \mathbf x(s,t) =(1,1) ,\quad \mathbf x(s,0) = \mathbf \Gamma(s)$$ which has the easily spotted unique solution $\mathbf x(s,t) = \mathbf\Gamma(s) + (t,t)$. A graph of what is happening: enter image description here Then as explained above the solution is $\tilde u(s,t) = A(s)$. Taking $t=0$, $$A(s) = \tilde u(s,0) = u(\mathbf x(s,0)) = u(\mathbf \Gamma(s)) = g(s).$$

For completeness- to undo the change of coordinates, \begin{align} x &= s + t, \\y &= \phi(s) + t \\ \implies x-y &= s-\phi(s) = \xi(s)\end{align} Here, it is important that $\xi$ is an invertible function, which is not true for arbitrary initial curves. Then $s = \xi^{-1}(x-y)$, so $$ u(x,y) = \tilde u(\xi^{-1}(x-y),0) = g(\xi^{-1}(x-y)) = : B(x-y)$$ which is the result.

Finally, I have also answered a similar question here which may be of use to you. There, I allowed the characteristic to be solved from a time $t_0=t_0(s)\in\mathbb R$ which deals with the awkwardness of $\xi$ earlier on in the procedure, but is otherwise the same.

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  • $\begingroup$ This is excellent! Thank you very much! $\endgroup$ – Wyuw Aug 8 '18 at 11:45
  • $\begingroup$ @Wyuw (the graph was wrong earlier, my apologies) $\endgroup$ – Calvin Khor Aug 8 '18 at 11:46
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    $\begingroup$ @Wyuw (1) I don't think that's called separation of variables, please have a look at en.wikipedia.org/wiki/Separation_of_variables . (2) The two notations $d/dt$ and $\partial/\partial t$ mean exactly the same thing, the latter is just used if you want to remind yourself that there is a second parameter. So you are 'always' integrating an 'ordinary' derivative, and who's to say integration in $t$ will have an answer independent of $s$? $\endgroup$ – Calvin Khor Aug 8 '18 at 11:54
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    $\begingroup$ @Wyuw This is the definition of invertible- if $a = F(b)$ for an invertible function $F$ then $b = F^{-1}(a)$. $\endgroup$ – Calvin Khor Aug 8 '18 at 12:09
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    $\begingroup$ @Wyuw Glad to be of help, good luck with your studies :) $\endgroup$ – Calvin Khor Aug 8 '18 at 12:11

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