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I'm taking my first course on PDEs and the text used by the professor has next to 0 examples. I was just wondering if I am approaching this question correctly.

$$ e^{x^2}u_x + xu_y = 0 $$

The characteristic curve satisfies the following ODE: $$ \frac{dy}{dx} = \frac{x}{e^{x^2}}$$ Solving this ODE gives: $$ y= -\frac{1}{2e^{x^2}} + C$$ Then we can isolate for $C$ to find the general solution. $$ C = y + \frac{1}{2e^{x^2}} $$ Thus the general solution is: $$u(x,y) = f( y + \frac{1}{2e^{x^2}})$$


Consider the initial value problem:

$$ u_x + xu_y = 0, u(0,y) = sin(y) $$

So using the steps above I get the general solution to be:

$$ u(x,y) = f(y-\frac{x^2}{2})$$

Then using the initial value: $$u(0,y) = siny = f(y)$$

I am not sure where to go from this step

Any guidance would be greatly appreciated.

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    $\begingroup$ Yes, that looks good. Was there a step you are unsure about? Are there additional informations for this task, like initial values? $\endgroup$ – LutzL Jan 30 at 20:59
  • $\begingroup$ @LutzL Okay awesome. Not quite. But I was unsure about an initial value problem, I'll update the main thread with the question. $\endgroup$ – Safder Jan 30 at 21:19
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    $\begingroup$ So you identified $f$ as $\sin$. Now replace backwards, $u(x,y)=\sin(y-x^2/2)$. $\endgroup$ – LutzL Jan 30 at 21:23
  • $\begingroup$ Oh I completely understand that now. Thank you for the clarification. $\endgroup$ – Safder Jan 30 at 21:24
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    $\begingroup$ Then the Lagrange equations are $dx=\frac{dy}x=\frac{dz}{z}$ with $z=u(x,y)$. Leading to $e^{-x}z=c_2=f(c_1)=f(y-x^2/2)$. $\endgroup$ – LutzL Jan 30 at 22:33
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I'm studying the same material from Partial Differential Equations: Methods and Applications by McOwen. Another way to do the second problem is shown below.

$$ u_x + xu_y = 0 $$ $$ u(0,y) = sin(y)$$

One could first write the initial data curve as $\Gamma: (0,s,\sin(s))$. Then, the method of characteristics would produce the following characteristic equations

$$\frac{dx}{dt}=1, ~\frac{dy}{dt}=x, ~\frac{dz}{dt}=0$$

with the initial data

$$x(s,0)=0, ~y(s,0)=s, ~z(s,0)=\sin(s)$$

For the first equation,

$$\frac{dx}{dt}=1 ~\Rightarrow dx = dt ~\Rightarrow x = t + C$$

Then, the initial data of $x(s,0)=0$ forms $x = 0 + C = 0$. Therefore,

$$x=t$$

For the third equation,

$$\frac{dz}{dt}=0 ~\Rightarrow dz = 0 ~\Rightarrow z = C$$

Then, the initial data of $z(s,0)=\sin(s)$ forms $z = C = \sin(s)$. Therefore,

$$z=\sin(s)$$

For the last equation,

$$\frac{dy}{dt}=x ~\Rightarrow dy = {t}dt ~\Rightarrow y = \frac{t^2}{2} + C$$

Then, the initial data of $y(s,0)=s$ forms $y = 0 + C = s$. Therefore,

$$y=\frac{t^2}{2} + s$$

Putting all of our equations together,

$$x=t$$ $$y=\frac{t^2}{2} + s ~\Rightarrow s = y - \frac{t^2}{2} = y - \frac{x^2}{2}$$ $$z=\sin(s) = \sin(y - \frac{x^2}{2})$$

Hence,

$$z=u(x,y)=\sin(y - \frac{x^2}{2})$$

It might be easier to find a constant function $\phi(x,y,z)$ first. Once you find that $\phi(x,y,z)=c=constant$, you could then find a different function $\psi(x,y,z)$ that is constant and is independent of $\phi$. Letting $\phi=f(\psi)$ for an arbitrary $f\in{C^1}$ would produce a solution.

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  • $\begingroup$ I'll give this method a try as well. Thank you $\endgroup$ – Safder Jan 31 at 19:41

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