Solving PDE by variable coefficient equation method

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.

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

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.

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