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P. Dravek and G. Holubova, Elements of Partial Differential Equations, Section 3.4 Exercise 22:

Show that the initial value problem $$u_t + u_x = 0,\; u(x,t) = x \;\text{ on }\; x^2+t^2=1.$$

has no solution. However, if the initial data are given only over the semicircle that lies in the half-plane $x + t \leq 0$, a solution exists but is not differentiable along the characteristics coming from the two end points of the semicircle.

I managed to show the first part. As for the second part, any hints?

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A graphical representation of the problem in the $x$-$t$ plane may be relevant (see this post). The method of characteristic provides the set of lines $x=t-t_0+x_0$ along which $u$ is constant. Here, the value $u=x_0$ is specified along the unit circle ${x_0}^2+{t_0}^2=1$. Characteristics coming from the region $x_0+t_0< 0$ of the unit circle will cross the circle again in the region $x_0+t_0 >0$ where another boundary-value is specified. Hence, the problem is ill-posed.

To avoid this, we restrict the boundary data to the region $x_0+t_0\leq 0$ of the unit circle. One should note that characteristic curves become parallel to the unit circle at the extremities of this half-circle. To see what happens, we express the characteristic curves as $x-t-x_0=-t_0$ with $t_0=\pm\sqrt{1-{x_0}^2}$. Squaring this identity, we have $$ (x-t)^2 -2x_0(x-t) + {x_0}^2=1-{x_0}^2 $$ so that $$ x_0 = \frac{x-t \pm \sqrt{2-(x-t)^2} }{2} = u(x,t), $$ for $t-\sqrt{2} \leq x\leq t+ \sqrt{2}$. At the extremities of the domain, we have $x-t = \pm\sqrt{2}$. Differentiating the expression of $u$ along these curves would require to differentiate a square root in the vicinity of zero, which is not possible.

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Solve the pde, calculate the partials and check them for the endpoints (as a hint) and for the characteristics passing by them. Probably you know that the general solution is $u(x,t)=f(x-t)$ with $f$ some function. Incidentally, the projection of the characteristics are lines of the form $x=t+K$, so, those with $K>\sqrt{2}$ have not assigned the initial data and the solution is not unique.

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From the boundary conditions: $u(x,\pm\sqrt{1-x^2})=f(x\pm\sqrt{1-x^2})=x$ we can find $f$ Call $w$ to $f$'s argument to isolate $x$ $$w=x\pm\sqrt{1-x^2}\text{ and }w^2-2xw+2x^2-1=0$$ $$x=(w\mp\sqrt{2-w^2})/2$$ and we get $f$ $$f(w)=(w\mp\sqrt{2-w^2})/2$$ and $u(x,t)$ $$u(x,t)=f(x-y)=\left(x-y\pm\sqrt{2-(x-y)^2}\right)/2$$ There is some work to do with the signs for the square root, but we let it aside as it is not relevant for the question about derivability. Now, the derivatives $$\dfrac{\partial u}{\partial x}=\dfrac12\left(1\mp\dfrac{x-t}{\sqrt{2-(x-t)^2}}\right)$$ The end points are $P(-\sqrt{2}/2,\sqrt{2}/2)$ and $Q(\sqrt{2}/2,-\sqrt{2}/2)$ and the projections of the characteristics passing by those points are $x-t=\sqrt{2}$ and $x-t=-\sqrt{2}$

It's easy to check that the partial has a division by zero for every point in those characrteristics and hence, is not derivable.

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