# Method of characteristics non-linear PDE

Consider the following initial-value problem:

$$xu_x-uu_t=t$$

$$u(1,t)=t$$

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;

$$\frac{\mathrm{d}u}{\mathrm{d}t}=-\frac{t}{u}$$

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

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:

$$u(x,t)=\frac{t}{\tan{\left(\ln{\left(\frac{e^{\frac{\pi}{4}}}{x}\right)}\right)}}$$

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 $$\ln(\frac{e^{\pi/4}}{x})=\frac{\pi}{4}-\ln(x)$$ .

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