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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?

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1 Answer 1

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$$tu_x-u_t=tu$$ 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 : $$\boxed{u(x,t)=(x-\frac{t^2}{2})e^{\frac12 t^2}}$$

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  • $\begingroup$ Sorry, Why we could write C2 = F(C1)? $\endgroup$
    – Snow Max
    Nov 9, 2020 at 4:46
  • $\begingroup$ 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. $\endgroup$
    – JJacquelin
    Nov 9, 2020 at 8:30

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