# Method of Characteristic for PDE

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?

$$tu_x-u_t=tu$$ 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). $$\boxed{u(x,t)=e^xF(t^2-2x)}$$
$$u(x,0)=x=e^xF(-2x)$$ $$F(-2x)=xe^{-x}$$ 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 : $$\boxed{u(x,t)=(x-\frac{t^2}{2})e^{\frac12 t^2}}$$
• This is a well known property for first order quasilinear PDEs without using boundary condition (Using boundary condition is another way if you prefer). Intuitive oversimplified explanation : Since $c_1$ and $c_2$ are arbitrary constants they are an infinity of equations such as $\Phi(c_1,c_2)=0$ with arbitrary function $\Phi$. Or equivalently on the form of any relationship between the two arbitrary constants : $c_2=F(c_1)$ or equivalently $c_1=G(c_2)$ with arbitrary functions $F$ and $G$ one inverse of the other. Nov 9, 2020 at 8:30