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Solve this PDE using the characteristic form

$\begin{equation} \frac{\partial w}{\partial t} +x\frac{\partial w}{\partial x}=1 \\ w(x,0)=f(x) \end{equation}$

My attempt Let's go to rewrite the PDE.

$\frac{\partial w}{\partial t} +x\frac{\partial w}{\partial x}-1=0$

For other way We know $w(t)=w(x(t),t)$

Then by chain rule

$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\times \frac{\partial x}{\partial t} + \frac{\partial w}{\partial t}$

This implies

$\begin{equation} \frac{\partial x}{\partial t}=x \\ \frac{\partial w}{\partial t}=0 \end{equation}$

Then:

$x=e^{t+x_0}$

This implies

$w(x,t)=c=w(x_0,0)=f(x_0)=f(ln(x)-t)$

Is correct this?

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    $\begingroup$ I got $w(x,t)=t+f(xe^{-t})$ $\endgroup$ – Isham Oct 18 '18 at 18:37
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    $\begingroup$ frogeye got the same answer you didnt take account of the function f that is given to you ... $\endgroup$ – Isham Oct 18 '18 at 19:00
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    $\begingroup$ Yes, you have reason. I didn't consider that. Thanks $\endgroup$ – Bvss12 Oct 18 '18 at 19:01
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    $\begingroup$ you're welcome..once you take account of the function f you will end with the same answer as frog's one $\endgroup$ – Isham Oct 18 '18 at 19:02
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    $\begingroup$ yeah you have a mistake there also its 1 not 0 $\endgroup$ – Isham Oct 18 '18 at 19:04
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By the principle of superposition, a single solution to $$ \frac{\partial w}{\partial t} + x \frac{\partial w}{\partial x} = 1$$

Combined with the general solution to

$$ \frac{\partial w}{\partial t} + x \frac{\partial w}{\partial x} = 0 $$ Will form the general solution of:

$$ \frac{\partial w}{\partial t} + x \frac{\partial w}{\partial x} = 1$$

So we now focus our attention to:

$$ \frac{\partial w}{\partial t} + x \frac{\partial w}{\partial x} = 0$$

As you noted correctly for any differentiable function $f$ we have that $f(\ln(x) - t)$ is a $w$ that obeys the above equation.

So what remains is to find a single solution to:

$$ \Omega[w] = \frac{\partial w}{\partial t} + x \frac{\partial w}{\partial x} = 1$$

We can use a power-series approach. Let $w_0 = t$. then $\Omega[w_0] = 1$ [We got lucky with our first guess, but if it weren't so we would continue adding terms to create a series solution] so we have that:

$$w(x,t) = t + f(\ln(x) - t)$$

Is the general solution.

Note:

To support $w(x,0) = h(x)$ as an initial value problem for some $h(x)$ given ahead of time you have that:

$$ f(\ln (x)) = h(x) \rightarrow f(x) = h(e^x) \rightarrow w(x,t) = t+ h(e^{-ln(x)-t}) \rightarrow t + h(xe^{-t}) $$

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    $\begingroup$ yep you forgot the function f is given $\endgroup$ – Isham Oct 18 '18 at 18:39

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