Boundary and initial conditions in quasi linear first order pde

Question

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 $x\ge0$ (2)

$u(0,t)=1$, for $t\ge0$ (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.

Answer 1

$$u_t+u_x=u \tag 1$$ 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 $X\geq 0$.

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

Altogether : $$F(X)= \begin{cases}1\quad\text{for}\quad X\geq 0.\\ e^X \quad\text{for}\quad X\leq 0. \end{cases}$$

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

$$u(x,t)=\begin{cases} e^t1\quad\text{for}\quad x-t\geq 0.\\ e^te^{x-t} \quad\text{for}\quad x-t\leq 0. \end{cases}$$

$$u(x,t)=\begin{cases} e^t\quad\text{for}\quad x\geq t.\\ e^x \quad\text{for}\quad x\leq t. \end{cases}$$

Answer 2

Consider the characteristics problem :

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

Taking the first pair, yields :

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

Now, the second pair, yields :

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

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

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

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 :

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

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

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

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.

Answer 3

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).