Finding explicit solution of PDE $xu_x+tu_t=pu$

Question

Let $p\in \mathbb{R}$ and consider $xu_x+tu_t=pu$.

Find a) Characteristic curves for the equation.

b) Find an explicit solution for $p=4$, where $u=1$ on the unit circle.

What I tried: a) I solved for the characteristic curves and got $x=e^r+c_1$, $t=e^r+c_2$, $u=e^{pr+f(s)}$. Is this correct?

b) I substituted $p=4$ and $u=1$ to the equation giving me the same $x$ and $t$ from (a) i.e $lnx-lnt=c_2-c_1$ implies $ln\frac{x}{t}=s$ where $s=c_2-c_1$ and $u=4r+f(s)$ which gives me

$u(x,t)=4r+f(ln\frac{x}{t})$

But I don't know what value of $r$ should I substitute to $u$ or is this even correct or I am missing something?

Hope you can help me. Thanks.

Answer 1

a) Let us follow the method of characteristics.

• $\frac{\text d}{\text d s}x = x$, letting $x(0) = x_0$ we know $x = x_0e^s$;
• $\frac{\text d}{\text d s}t = t$, letting $t(0) = t_0$ we know $t = t_0e^s$;
• $\frac{\text d}{\text d s}u = pu$, letting $u(0) = u_0$ we know $u = u_0e^{ps}$.

b) Let us assume that $p=4$. The method of characteristics has transformed the PDE into a set of ODEs for $x(s)$, $t(s)$, $u(s)$, which have been solved analytically. The information we have is $u=1$ on the unit circle (in the $x$-$t$ plane). We set up the problem so that the values $x_0$, $t_0$ of $x$, $t$ at $s=0$ are located on the unit circle, and the value $u_0$ of $u$ at $s=0$ is one: $$ {x_0}^2 + {t_0}^2 = 1 \qquad\text{and}\qquad u_0 = 1 . $$ According to the expression of characteristics, we therefore have $x^2 + t^2 = e^{2s}$. Finally, using $u = (e^{2s})^2$, we have $$ u(x,t) = {(x^2 + t^2)}^2 . $$ To check other examples, see these related posts: (1), (2).

Note: The characteristics are the lines $|t| = a |x|$ for $a = \sqrt{{x_0}^{-2}-1}$ in $\Bbb R^+$.