Given the following partial differential equation with initial condition
$$tu_x+u_t=2t\cdot u,\;IV: u(x,0)=x.$$
So I apply the method of characteristic to solve this problem as following
Forming three characteristic equation
$$\frac{dx}{ds}=t,\;\frac{dt}{ds}=1,\; and\;\frac{du}{ds}=2t\cdot u $$
$$\frac{dt}{ds}=1\rightarrow dt = ds\rightarrow t = s+t_0\rightarrow t=s\;since\; t_0=0\; from\; IV$$
then
$$\frac{dx}{ds}=t\rightarrow \frac{dx}{ds}=s\rightarrow x=\frac{s^2}{2}+x_0$$
also for the last equation we have
$$\frac{du}{ds}=2t\cdot u\rightarrow \frac{du}{u}=2s\;dt\rightarrow u=e^{s^2+u_0}$$
Then we substitute the $s=t$, and $x_0=x-\frac{s^2}{2}$ to initial condition
$$ u(x_0,0)=e^{0+u_0}=x\rightarrow u_0=lnx$$
Thus the finial solution I have should be $u(x,t)=e^{t^2+lnx}=xe^{t^2}$
However this is not same as the answer given, could someone tell me where I make the mistake in this problem?
Best Answer
$$tu_x-u_t=tu$$ Charpit-Lagrange system of characteristic ODEs : $$\frac{dx}{t}=\frac{dt}{1}=\frac{du}{tu}=ds$$ This is what you correctly found, but presented on an equivalent form.
A first characteristic equation comes from solving $\frac{dx}{t}=\frac{dt}{1}$ : $$t^2-2x=c_1$$ A second characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{tu}$ : $$ue^{-x}=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is : $$ue^{-x}=F(t^2-2x)$$ $F$ is an arbitrary function (to be determined later in order to satisfy the initial condition). $$\boxed{u(x,t)=e^xF(t^2-2x)}$$
CONDITION :
$$u(x,0)=x=e^xF(-2x)$$ $$F(-2x)=xe^{-x}$$ Let $x=-\frac12 X$ $$F(X)=-\frac12 X e^{\frac12 X}$$ Now the function $F(X)$ is known. We put it into the above general solution where $X=t^2-2x$
$F(t^2-2x)=-\frac12 (t^2-2x) e^{\frac12 (t^2-2x)}$ $$u(x,t)=e^x\left(-\frac12 (t^2-2x) e^{\frac12 (t^2-2x)}\right)$$ After simplification, the particular solution which satisfies both the PDE and the condition is : $$\boxed{u(x,t)=(x-\frac{t^2}{2})e^{\frac12 t^2}}$$