Consider the following initial-value problem:



I've come to the follow characteristic equations:

$$\frac{\mathrm{d}x}{\mathrm{d}\tau }=x,\,\,\,\frac{\mathrm{d}t}{\mathrm{d}\tau }=-u,\,\,\,\frac{\mathrm{d}u}{\mathrm{d}\tau }=t$$

I can solve the first easily;

$$\int \frac{1}{x}\mathrm{d}x=\int \mathrm{d}\tau $$

$$\therefore x=e^\tau $$

The second and third applying the chain rule;


$$\int u \mathrm{d}u=-\int t\mathrm{d}t$$ $$\therefore u^2=F(\xi )-t^2$$

Putting back in the initial conditions, I wind up with the following;

$$F(\xi )=2\xi ^2$$

$$\therefore u=\sqrt{2\xi ^2-t^2}$$

Now, from here I cannot find a way to evaluate what $\xi$ is in terms of the natural variables $x$ and $t$.

According to Maple, the solution is as follows:


How do I find this solution algebraically?


$$xu_x-uu_t=t $$

Your system of ODEs is correct. Written equivalently as : $$\frac{dx}{x}=\frac{dt}{-u}=\frac{du}{t}=d\tau$$ A first characteristic equation comes from $\frac{dt}{-u}=\frac{du}{t}$ : $$u^2+t^2=c_1 \tag 1$$

A second characteristic equation comes from $\frac{dx}{x}=\frac{dt}{-u}=\frac{dt}{-\sqrt{c_1-t^2}}$

$\ln|x|=-\int\frac{dt}{\sqrt{c_1-t^2}}=-\tan^{-1}\left(\frac{t}{\sqrt{c_1-t^2}}\right)+$constant. $$\ln|x|+\tan^{-1}\left(\frac{t}{\sqrt{c_1-t^2}}\right)=c_2$$ $$\ln|x|+\tan^{-1}\left(\frac{t}{\sqrt{(u^2+t^2)-t^2}}\right)=c_2$$ $$\ln|x|+\tan^{-1}\left(\frac{t}{u}\right)=c_2 \tag 2$$

The general solution of the PDE expressed on the form of implicit equation $c_2=\Phi(c_1)$ is : $$\ln|x|+\tan^{-1}\left(\frac{t}{u}\right)=\Phi(u^2+t^2) \tag 3$$ where $\Phi$ is an arbitrary function, to be determined according to the boundary condition.

CONDITION : $u(1,t)=t$ $$\ln|1|+\tan^{-1}\left(\frac{t}{t}\right)=\Phi(t^2+t^2)$$ $$\Phi(2t^2)=\frac{\pi}{4}$$ This implies that the function $\Phi$ is a constant function $=\frac{\pi}{4}$.

Such case can be encountered when the boundary condition is specified on a characteristic curve, which is the present case.

Nevertheless the function $\Phi$ is determined. We put it into the general solution $(3)$. $$\ln|x|+\tan^{-1}\left(\frac{t}{u}\right)=\frac{\pi}{4}$$

$$\tan^{-1}\left(\frac{t}{u}\right)=-\ln|x|+ \frac{\pi}{4}$$ $$u(x,t)=\frac{t}{\tan\left(-\ln|x|+\frac{\pi}{4}\right)}$$ This is the same result than Maple since $\ln(\frac{e^{\pi/4}}{x})=\frac{\pi}{4}-\ln(x)$ .

  • $\begingroup$ Thank-you so much!!! Makes sense! $\endgroup$ – Andrew Sullivan Mar 18 at 20:40

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