Cfrtjryk

I cannot understand what we are looking to find in such a problem... For example consider the pde

$u_t+u_x=u$, with $x,t>0$ (1)

and initial and boundary conditions:

$u(x,0)=1$, for $xge0$ (2)

$u(0,t)=1$, for $tge0$ (3)

Are we looking for a solution of (1) which can be extended (continously?)in order to take values implemented by (2), (3)?

Remark: I do not need a solution of the above problem, just a proper defintion or refference. Thanks in advance.

$$u_t+u_x=u tag 1$$
Charpit-Legendre equations :
$$frac{dt}{1}=frac{dx}{1}=frac{du}{u}$$
First characteristic curves equation from $frac{dt}{1}=frac{dx}{1}$ :
$$x-t=c_1$$
Second characteristic curves equation from $frac{dt}{1}=frac{du}{u}$ :
$$ue^{-t}=c_2$$
General solution of the PDE :
$$ue^{-t}=F(x-t)$$
where $F$ is an arbitrary function (to be determined according to boundary conditions).
$$u(x,t)=e^tF(x-t)$$
Condition $u(X,0)=1=F(X)$ for $Xgeq 0$.

Condition $u(0,t)=1=e^tF(-t)$ for $tgeq 0$. Thus $1=e^{-X}F(X)$ for $-Xgeq 0$.

Altogether :
$$F(X)=
begin{cases}1quadtext{for}quad Xgeq 0.\
e^X quadtext{for}quad Xleq 0.
end{cases}$$

Now the function $F$ is determined. We put it into the general solution where $X=x-t$

$$u(x,t)=begin{cases}
e^t1quadtext{for}quad x-tgeq 0.\
e^te^{x-t} quadtext{for}quad x-tleq 0.
end{cases}$$

$$u(x,t)=begin{cases}
e^tquadtext{for}quad xgeq t.\
e^x quadtext{for}quad xleq t.
end{cases}$$

Consider the characteristics problem :

$$frac{mathrm{d}t}{1}=frac{mathrm{d}x}{1} = frac{mathrm{d}u}{u}$$

Taking the first pair, yields :

$$frac{mathrm{d}t}{1}=frac{mathrm{d}x}{1} Leftrightarrow intmathrm{d}t = int mathrm{d}x implies u_1 = x-t $$

Now, the second pair, yields :

$$frac{mathrm{d}x}{1} = frac{mathrm{d}u}{u} Leftrightarrow intmathrm{d}x = intfrac{1}{u}mathrm{d}u implies u_2 = x - ln(u) $$

Since $u_1$ is not dependent on $u$ and $u_2$ is, the solution of the PDE can be written as

$$u_2 = F(u_1) Rightarrow ln u = x - F(x-t) Leftrightarrow u(x,t) = expleft(x-F(x-t)right)$$
$$Leftrightarrow$$
$$u(x,t) = frac{e^x}{F(x-t)} equiv e^xF(x-t)$$

where $F$ is an arbitrary function $in C^1$.

Now, applying the initial values, we get :

$$u(x,0) = 1 implies e^xF(x) = 1 Leftrightarrow F(x) = e^{-x}$$

$$u(0,t) = 1 implies F(-t) = 1$$

It is sufficient then to say that a solution $u(x,t)$ of the given Boundary Value Problem, is the function defined as such :

$$u(x,t) = e^xF(x-t) quad text{where} quad begin{cases} F(x) = e^{-x} \ F(-t) = 1end{cases}$$

To be more precise, consider letting $x := x-t$ and $t := t-x$ in the case of the boundary values. Then :

$$F(x-t) = e^{x-t} quad text{and} quad F(x-t) = 1quad$$

But, that implies that :

$$e^{x-t} = 1 Leftrightarrow x = t$$

Finally, this means that the solution of the given BVP can be written as :

$$u(x,t) = e^xF(0) equiv c_1e^x quad text{or} quad u(x,t) = c_2e^t$$

But note that the first one holds in the case of $x - t leq 0$ thus $x leq t$ and the second one holds in the case of $x-t geq 0$ thus $t geq x$, which stems from your Boundary Value cases for the PDE variables.

Thus, finally, the solution $u(x,t)$ can be written as :

$$u(x,t) = begin{cases}e^t & x geq t \ e^x & x leq t end{cases}$$

Simply substituting confirms that both of them are solutions to the initial PDE BVP.

Are we looking for a solution of (1) which can be extended
(continously?)in order to take values implemented by (2), (3)?

Yes (and yes, continously).