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I have a boundary-value problem:

$$ xu_x - uu_t = t $$

with boundary conditions:

$$ u(1, t)= t, -\infty<t<\infty $$

Finding the characteristic equations is no problem, and I get a general solution of:

$$ u(x, t) = \frac{\cos(\ln x) + \sin(\ln x)}{\cos(\ln x)-\sin(\ln x)}t $$

I have base characteristics that look like this:

enter image description here

And I know that my solution is defined only when $x>0$.

However, I am having trouble gaining any intuition on how to interpret these characteristic plots, and determining well-posedness.

So my main questions are:

  1. For this example, is the initial curve the line $t=0$?

  2. Is the solution defined everywhere for $x>0$? I would say yes.

  3. Is the solution unique? I would say no, because of the intersection of all of the base characteristics at that point on the $x$-axis implies that I can have the same solution for different characteristics.

  4. Are there restrictions on my parameters? For example can I allow my parameters to run over arbitrary intervals?

  5. Although I know the criteria for a well-posed problem, I feel like I still cannot determine whether this problem is well posed. I would say no, because it fails uniqueness.

  6. Is the solution single valued everywhere? I have no idea.

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For this problem, the boundary data is located along the line $x=1$. Let us apply the method of characteristics:

  • $\frac{\text d x}{\text d s} = x$. Letting $x(0)=1$ we know $x(s) = e^s$.
  • $\frac{\text d t}{\text d s} = -u$ and $\frac{\text d u}{\text d s} = t$. Letting $t(0)=t_0$ and $u(0)=t_0$, we know $t(s) = t_0 \left(\cos s - \sin s\right)$ and $u(s) = t_0 \left(\cos s + \sin s\right)$.

The expression of $x(s)$ gives $s = \ln x$ for positive $x$. Combining the expressions of $t(s)$ and $x(s)$, one obtains indeed the expression of the solution $u(x,t)$ in OP.

Now, let us analyze the expression of characteristic curves, which tell how the boundary information located at $x=1$ propagates in the $x$-$t$ plane. These curves are expressed by the equations $t = t_0 \left(\cos\ln x - \sin\ln x\right)$ where $t_0 \in \Bbb R$, and they carry the information $u = t_0 \left(\cos\ln x + \sin\ln x\right)$. All these curves intersect at the roots of the function $x \mapsto \cos\ln x - \sin\ln x$, i.e., at the abscissas $x_n = e^{\pi/4 + n\pi}$ with $n\in \Bbb Z$. At the abscissas $x_n$, the characteristic curves carry the value $u = \pm t_0\sqrt{2}$. Therefore, the classical solution $u(x,t)$ deduced from the characteristics is multivalued at the abscissas $x_n$. Since the boundary-value problem sets the data at $x=1$, we can increase $x$ until the solution becomes multivalued at $x_0 = e^{5\pi/4} \approx 2.19$, and we can decrease $x$ until $x_{-1} = e^{-3\pi/4} \approx 0.095$ only. The classical solution is only valid for $s$ within the interval $]{-3\pi}/{4}, {5\pi}/{4}[$, such that $x$ belongs to $]x_{-1}, x_0[$.

The boundary-value problem is well-posed, but the solution blows up at some point. With the present boundary data located at $x=1$, it does not make sense to look for values of $s < {-3\pi}/{4}$ or $s > {5\pi}/{4}$.

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  • $\begingroup$ Thanks, that clears up a few things. So is this not a well-posed problem? $\endgroup$ – monkeyofscience Apr 2 at 17:57
  • $\begingroup$ What about those values above $5\pi/4$? For example, if I expand the $x$-axis, the curves cross the $t=0$ line again. So is the solution single-valued everywhere within these intervals? And since each characteristic intersects the plane at more than one point, ie different values of x, and the intersection are distinct, does this not mean it is ill-posed? $\endgroup$ – monkeyofscience Apr 3 at 2:06
  • $\begingroup$ @monkeyofscience After the solution is blown up (e.g., $s > 5\pi/4$), there is no more solution. The picture is similar to the Riccati ODE $y' = y^2$ with $y(0) = 1$, which solution $t\mapsto 1/(1-t)$ blows up at $t=1$. We can indeed plot the function $t\mapsto 1/(1-t)$ for $t>1$, but this part of the solution is no longer related to the original IVP. $\endgroup$ – Harry49 Apr 3 at 8:09

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