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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.

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$$u_t+u_x=u \tag 1$$ 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 $X=x-t$

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

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

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  • $\begingroup$ This is a weak solution, right? as $u$ has no partial derivatives on the $x=t$ , line. $\endgroup$ – dmtri Dec 15 '18 at 19:23
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    $\begingroup$ Yes, we have to integrate separately Eq.$(1)$ above and below the line $x = t$ and use integration by parts. en.wikipedia.org/wiki/Weak_solution $\endgroup$ – JJacquelin Dec 16 '18 at 7:48
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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.

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  • $\begingroup$ Thanks for the answer, actually I do not need to solve that pde. Why do you apply (2) and (3) conditions in your general integral? The domain of (1) has nothing to do with the curves described in (2) and (3)? $\endgroup$ – dmtri Dec 15 '18 at 12:07
  • $\begingroup$ this is a weak solution, right? $\endgroup$ – dmtri Dec 15 '18 at 17:52
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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).

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  • $\begingroup$ So if the domain of solution of (1), we may proceed as the above 2 answers - posts.Is then this solution unique? How we can proceed if the "curves" do not belong to any solution - domain of (1)? Thanks again. $\endgroup$ – dmtri Dec 15 '18 at 15:25
  • $\begingroup$ @dmtri I don't understand your comment, can you ask your question again? $\endgroup$ – Bananach Dec 15 '18 at 18:16
  • $\begingroup$ @dmtri Also, you might consider asking a new question $\endgroup$ – Bananach Dec 15 '18 at 18:16

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