# Solve this PDE $\frac{\partial w}{\partial t} +x\frac{\partial w}{\partial x}=1$

Solve this PDE using the characteristic form

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

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

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

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?

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

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:

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

• yep you forgot the function f is given – Isham Oct 18 '18 at 18:39