PDE $(c_1t+c_2)\frac{\partial u}{\partial t}-\left(c_1x+c_3\right)\frac{\partial u}{\partial x}=c_1u$ by method of characteristics

Question

I'm trying to solve $$(c_1t+c_2)\frac{\partial u}{\partial t}-\left(c_1x+c_3\right)\frac{\partial u}{\partial x}=c_1u$$ for $c_1,c_2,c_3\in\mathbb{R}$.

Here is what I am trying (but somthing is off):

Using the Method of characteristics I get $$-\frac{dx}{(c_1x+c_3)}=\frac{dt}{(c_1t+c_2)}=\frac{du}{uc_1}.$$

I'm calling $r=c_1t+c_2$ and $s= c_1x+c_3$ so that we have $$-\frac{ds}{c_1s}=\frac{dr}{c_1r}=\frac{du}{uc_1}.$$ Or equivalently $$-\frac{ds}{s}=\frac{dr}{r}=\frac{du}{u}.$$

From $$\frac{dr}{r}=\frac{du}{u}$$ we get $$\log(u)=log(r)+log(A),$$ for $A$ constant wrt $r$. Thus $$(Eq 1)\quad u=Ar\qquad $$

From $$-\frac{ds}{s}=\frac{du}{u}$$ we get $$(Eq 2)\quad u=\frac{B}{s}.$$ By comparing (Eq 1) and (Eq 2) we get $$u=k\frac{r}{s}$$ for $k\in\mathbb{R}$.

But this doesn't solve the PD: using $$\frac{\partial u}{\partial t}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}=\frac{k}{s}c_1=\frac{kc_1}{c_1t+c_2}$$ and $$\frac{\partial u}{\partial x}= \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}=-\frac{kr}{s^2}c_1=-\frac{k(c_1t+c_2)}{(c_1x+c_3)^2}c_1$$ Then $$(c_1t+c_2)\frac{\partial u}{\partial t}-(c_1x+c_3)\frac{\partial u}{\partial x}=kc_1+c_1\frac{k(c_1t+c_2)}{(c_1x+c_2)}=kc_1+c_1(u).$$

What I am doing wrong?

Answer 1

$$-\frac{ds}{s}=\frac{dr}{r}=\frac{du}{u}\quad\text{is OK.}$$ I agree with $$(Eq 1)\quad u=Ar$$ $$(Eq 2)\quad u=\frac{B}{s}$$ One can also solve $-\frac{ds}{s}=\frac{dr}{r}$ $$(Eq3)\quad r=\frac{C}{s}$$

So one found three characteristic equations : $$\frac{u}{r}=A$$ $$u\:s=B$$ $$r\:s=C$$ These three equations are not independant since $AC=B$.

$A,B,C$ are constant on the respective characteric curves, but not elsewere. Elsewhere $A,B,C$ depend on $r$ and $s$.

Two independant characteristic equations are necessary and sufficient to find the general solution of the PDE. So one can write the general solution on various equivalent forms.

With Eqs.1 and 2 the general solution of the PDE expressed on the form of implicit equation $F(A,B)=0$ is : $$F(\frac{u}{r}\:,\:u\:s)=0$$ $F$ is an arbitrary function of two variables.

With Eqs.1 and 3 the general solution of the PDE expressed on the form of implicit equation $G(A,C)=0$ is : $$G(\frac{u}{r}\:,\:r\:s)=0$$ $G$ is an arbitrary function of two variables (Different from $F$ , but related). Or equivalently : $$\frac{u}{r}=\Phi(r\:s)$$ $\Phi$ is an arbitrary function. $$\boxed{u(x,t)=(c_1x+c_3)\Phi\big((c_1t+c_2)(c_1x+c_3)\big)}$$

With Eqs.2 and 3 the general solution of the PDE expressed on the form of implicit equation $H(B,C)=0$ is : $$H(u\:s\:,\:r\:s)=0$$ $H$ is an arbitrary function of two variables (Different from $F$ and $G$ , but related). Or equivalently : $$u\:s=\Psi(r\:s)$$ $\Psi$ is an arbitrary function. $$\boxed{u(x,t)=\frac{1}{c_1t+c_2}\Psi\big((c_1t+c_2)(c_1x+c_3)\big)}$$

The functions $\Phi$ and $\Psi$ are both arbitrary but related : $$\Psi\big((c_1t+c_2)(c_1x+c_3)\big)=(c_1t+c_2)(c_1x+c_3)\Phi\big((c_1t+c_2)(c_1x+c_3)\big)$$ Of course all those arbitrary functions could be determined if some valid boundary condition was specified.

Answer 2

From the equations you got,

$u=Ar$, $A=\dfrac{u}{r}$

$u=\dfrac{B}{s}$, $B=us$

the general solution is obtained setting $A=f(B)$

With $f(x)$ some one variable differentiable function to be determined by the initial conditions.

So, $u=rf(us)$ is the function you have to check.

Cannot be deduced $u=k\dfrac{r}{s}$ because $A$ and $B$ are not in fact constants (they are so along each characteristic curve, but not over the surface solution).

Added

With $r=c_1t+c_2$, $s=c_1x+c_3$, the PDE is

$r\dfrac{\partial u}{\partial r}-s\dfrac{\partial u}{\partial s}=u$

The solution to check is $u=rf(us)$

$\dfrac{\partial u}{\partial r}=f(us)+rsf'(us)\dfrac{\partial u}{\partial r}$; $\dfrac{\partial u}{\partial r}\left(1-rsf'(us)\right)=f(us)$; $\dfrac{\partial u}{\partial r}=\dfrac{f(us)}{1-rsf'(us)}$

$\dfrac{\partial u}{\partial s}=rf'(us)\left(u+s\dfrac{\partial u}{\partial s}\right)$; $\dfrac{\partial u}{\partial s}=rf'(us)u+rsf'(us)\dfrac{\partial u}{\partial s}$; $\dfrac{\partial u}{\partial s}=\dfrac{rf'(us)u}{1-rsf'(us)}$

Then,

$r\dfrac{\partial u}{\partial r}-s\dfrac{\partial u}{\partial s}=\dfrac{rf(us)}{1-rsf'(us)}-\dfrac{rsf'(us)u}{1-rsf'(us)}=\dfrac{rf(us)-rsf'(us)u}{1-rsf'(us)}=$

$=\dfrac{u-rsf'(us)u}{1-rsf'(us)}=\dfrac{u(1-rsf'(us))}{1-rsf'(us)}=u$ $\LARGE\checkmark$