1
$\begingroup$

I am solving following Cauchy IVP: $$u_t+uu_x=1,$$ $x$ is real,$t>0$, and initial condition is $$u(t^2/4,t)=t/2$$ and found contradictory results:-Parametrizing the given initial curve as follows $x=s^2/4$, $t=s$, $u=s/2$. Characteristic equations are $dx/u=dt/1=du/1$, solving these and using initial conditions, I got solution as $x=u^2/2+s^2/8$, $t=u+s/2$, eliminating $s$ I got $2u^2-2tu+t^2-2x=0$ giving $2u=t+\sqrt{4x-t^2}$, $2u=t-\sqrt{4x-t^2}$. My question is that inspite of conditions of non-existence being satisfied,viz, $P/x'(s)=Q/t'(s)$ not equal to $R/u'(s)$ at the initial data, non-unique solution exists.How to resolve this situation?

$\endgroup$
  • $\begingroup$ Please use Latex for symbols and equations. $\endgroup$ – Awe Kumar Jha Mar 23 at 5:40
1
$\begingroup$

I agree with your result. This is confirmed below.

$$u_t+uu_x=1$$ Chapit-Lagrange system of ODEs : $\frac{dt}{1}=\frac{dx}{u}=\frac{du}{1}=ds$

A first characteristic equation comes from $\frac{dt}{1}=\frac{du}{1}$ : $$t-u=c_1$$ A second characteristic equation comes from $\frac{dx}{u}=\frac{du}{1}$ : $$2x-u^2=c_2$$ The general solution of the PDE on the form of an implicit equation $c_2=F(c_1)$ is : $$2x-u^2=F(t-u)$$ where $F$ is an arbitrary function, to be determined according to initial condition.

CONDITION : $u(\frac{t^2}{4},t)=\frac{t}{2}$ $$2(\frac{t^2}{4})-(\frac{t}{2})^2=F\left(t-\frac{t}{2}\right)$$ $$\frac{t^2}{4}=F\left(\frac{t}{2}\right)$$ The function $F$ is determined : $\quad F(X)=X^2$ .

This shows that $F$ exists and is unique.

We put this function into the above general solution : $$2x-u^2=(t-u)^2$$ $$2u^2-2tu+t^2-2x=0$$ This is the solution of the PDE fitting to the specified condition.

This solution exists and is unique since $F$ exists and is unique.

We can write it on the form : $$u=\frac{t}{2}\pm\sqrt{x-\frac14 t^2}$$ Writing on this form doesn't mean that we now have two solutions. We still have only one solution, the same as before.

This is like the equation of a circle $x^2+y^2=R^2$ written on the form $y=\pm\sqrt{R^2-x^2}$. There is still one circle made of two demi-circles.

On a purely mathematical viewpoint the result is : A solution exists and is unique.

But the conclusion can be different if the PDE is a mathematical model for a physical problem. The use of the variable noted $t$ makes think to "time". If so, we are going outside the pur mathematics to go into physics. In that case some implicit conditions can appear, for example that the terms be real (no complex root for example). Or other constraints such as the time always increasing, or requirement about stability, for examples.

Without knowing the context of the problem (What the PDE is supposed to model, etc) one cannot discuss the existence of solution, in other words if the above mathematical solution is a convenient answer to the question.

$\endgroup$
  • $\begingroup$ To JJacquelin,I disagree with you.Here is a counter example:-consider following IVP-. (t+2ux)u_x-(x+2ut)u_t=(x^2-t^2)/2 with Cauchy data x-t=0,u=0.Solving as usual,we get (x-t)^2=4(u+1/2)^2-1 , from which we obtain 2u=-1+√{(x-t)^2+1},2u=-1-√{(x-t)^2+1}, but only 2u=-1+√{(x-t)^2+1} satisfies initial data.Hence according to you only solution should be combined form solution (x-t)^2=4(u+1/2)^2-1.Any way ,it is half story.You have not answered the existence part.Pleas reread my comments and original question. $\endgroup$ – user463280 Mar 23 at 17:18
  • $\begingroup$ The example that you cite is not comparable to your problem. In the cited example, only one branch of the whole function satisfies the condition. In your case the two branches satisfies the condition. So, there is a unique solution $2u^2 −2ut+t^2 −2x=0$ which is the whole function (artificially cut in two branches by the solving for $u$ ). $\endgroup$ – JJacquelin Mar 23 at 19:04
  • $\begingroup$ My answer is very simple. The solution $2u^2 −2ut+t^2 −2x=0$ satisfies the PDE $u_t +uu_x =1$ and satisfies the condition $u(t^2/4,t)=t/2$ . One can check it directly. The unicity of this solution is proved by the method used in my answer. I have nothing to add. Good luck for the kind of answer that you expect. $\endgroup$ – JJacquelin Mar 23 at 19:06
  • $\begingroup$ U may certainly.But existence part of my question remains unanswered. $\endgroup$ – user463280 Mar 24 at 9:37
1
$\begingroup$

Using the method of characteristics, one writes

  • $\frac{\text d t}{\text d s}=1$. Letting $t(0) = t_0$ gives $t = s + t_0$.
  • $\frac{\text d u}{\text d s}=1$. Letting $u(0) = \frac{1}{2} t_0$ gives $u = s + \frac{1}{2} t_0$.
  • $\frac{\text d x}{\text d s}=u$. Letting $x(0) = \frac{1}{4} {t_0}^2$ gives $x = \frac{1}{2}s^2 + \frac{1}{2} t_0 s + \frac{1}{4} {t_0}^2$.

The parameter $s$ is eliminated by injecting $s = t-t_0$ in the expression $x(s)$ of the characteristic curves. The latter becomes $x = \frac{1}{4}\big( t^2 + (t-t_0)^2\big)$, which is inverted as $t_0 = t \pm \sqrt{4x - t^2}$. If we inject the equation of the characteristic curves in the expression of $t_0$, we get $t_0 = t \pm |t-t_0|$. Therefore, the top sign ($+$) corresponds to times $t\leq t_0$, while the bottom sign ($-$) corresponds to times $t\geq t_0$. The curves in the $x$-$t$ plane are shown below for several values of $t_0$ (the blue line marks the boundary where the data is imposed).

characteristics

For example, let us consider the point $(x,t) = \big({5}/{32}, -{1}/{4}\big)$ of the $x$-$t$ plane. From the expression of $t_0$, we find the value $t_0 = {1}/{2}$ which satisfies $t<t_0$, and the value $t_0 = -1$ which satisfies $t>t_0$. Hence, two characteristic curves are passing through this point. The first one carries the value $u = t - t_0/2 = -1/2$. The second one carries the value $u = t - t_0/2 = 1/4$. Both solutions $$ u(x,t) = \tfrac{1}{2} \big( t \mp \sqrt{4x - t^2} \big) ,\qquad 4x - t^2\geq 0 $$ exist simultaneously. The classical solution deduced from the characteristics is bi-valued everywhere in the region $4x - t^2 > 0$. The problem is not well-posed in the sense of existence and uniqueness.

$\endgroup$
0
$\begingroup$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=1$ , letting $u(0)=u_0$ , we have $u=s+u_0=t+u_0$

$\dfrac{dx}{ds}=u=s+u_0$ , letting $x(0)=f(u_0)$ , we have $x=\dfrac{s^2}{2}+u_0s+f(u_0)=\dfrac{t^2}{2}+(u-t)t+f(u-t)=ut-\dfrac{t^2}{2}+f(u-t)$ , i.e. $u=t+F\left(x-ut+\dfrac{t^2}{2}\right)$

$u\left(\dfrac{t^2}{4},t\right)=\dfrac{t}{2}$ :

$\dfrac{t}{2}=t+F\left(\dfrac{t^2}{4}\right)$

$F\left(\dfrac{t^2}{4}\right)=-\dfrac{t}{2}$

$F(t)=\mp\sqrt t$

$\therefore u=t\mp\sqrt{x-ut+\dfrac{t^2}{2}}$

$(u-t)^2=x-ut+\dfrac{t^2}{2}$

$u^2-2ut+t^2=x-ut+\dfrac{t^2}{2}$

$u^2-ut=x-\dfrac{t^2}{2}$

$u^2-ut+\dfrac{t^2}{4}=x-\dfrac{t^2}{4}$

$\left(u-\dfrac{t}{2}\right)^2=x-\dfrac{t^2}{4}$

$u=\dfrac{t}{2}\pm\sqrt{x-\dfrac{t^2}{4}}$

$\endgroup$
  • $\begingroup$ That solution I have already given.I am unable to understand how to resolve contradictory non-existence of solution and existence of solution (moreover two solutions). $\endgroup$ – user463280 Mar 23 at 7:19
  • $\begingroup$ I mean that when the problem is viewed as existence problem then we get contradictory answers.Existence test (given in my solution) shows that solution should not exist but routine solving comes up with two solutions.My question is how to explain this happening. $\endgroup$ – user463280 Mar 23 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.