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Use the method of characteristics to find the general solution of the following PDE $$ e^xu_x + u_y = xu. $$ Show explicitly that your result is indeed a solution of the PDE.

So I believe I have found the general solution as follows:

$\frac{dy}{dx} = e^{-x},$ $\frac{du}{dx} = xue^{-x}$, $\frac{du}{dy} = xu$.

Then, the first eq'n gives $C_1 = y + e^{-x}$. And, the second equation gives $C_2 = \ln{u} + (x+1)e^{-x}$. So, the general solution is given by $\ln{u} + (x+1)e^{-x} = \omega(y + e^{-x})$, where $\omega$ is an arbitrary function. We can rearrange to give an explicit general sol'n: $$u(x,y) = \exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\}.$$

Now, I'm struggling to, as the question asks, show explicitly that this is a solution. I'm assuming this is wanting me to substitute my solution into the PDE. But, because I have this arbitrary function it doesn't seem to be working.. (?) I can't get my expression for $u(x,y)$ to give me the PDE, i.e., taking partial derivatives and subbing into the PDE I don't get back an expression for $xu$. This is what I have: \begin{align} u_x &= \frac{\partial}{\partial x} \exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \left( \frac{\partial}{\partial x} [\omega(y + e^{-x})] - \frac{\partial}{\partial x}[(x+1)e^{-x}] \right) \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \left( \frac{\partial}{\partial x} [\ln{u} + (x+1)e^{-x}] - [-xe^{-x}] \right) \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \left( \frac{\partial}{\partial x} [\ln{u}] + \frac{\partial}{\partial x}[(x+1)e^{-x}] + [xe^{-x}] \right) \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \frac{u_x}{u} \\ &= u \cdot \frac{u_x}{u} \\ &= u_x \\ \end{align}

\begin{align} u_y &= \frac{\partial}{\partial y} \exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \left( \frac{\partial}{\partial y} [\omega(y + e^{-x})] - \frac{\partial}{\partial y}[(x+1)e^{-x}] \right) \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \frac{\partial}{\partial y} [\ln{u} + (x+1)e^{-x}] \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \frac{\partial}{\partial y} [\ln{u}] \\ &=\exp\{\omega(y + e^{-x}) - (x+1)e^{-x}\} \cdot \frac{u_y}{u} \\ &= u \cdot \frac{u_y}{u} \\ &= u_y \\ \end{align}

Is this enough to show that it's a solution? What else can I do with this? Have I missed something?

spooleey
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  • Is your second line meant to be your characteristics ODEs, if so it's not right. You also shown that $u_x=u_x$, $u_y=u_y$ in the bottom which is trivially true and doesn't shown anything. – Finn Pillar Mar 30 '24 at 13:48
  • @FinnPillar Yes, they are supposed to be the characteristic ODEs. In my lectures we were told that for $a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)$ the general solution is $v_1(x,y,u) = \omega(v_2(x,y,u))$ for arbitrary $\omega$ and $c_1 = v_1(x,y,u), c_2 = v_2(x,y,u)$ are any solutions of the characteristic ODEs $\frac{dy}{dx} = \frac{b}{a}, \frac{du}{dx} = \frac{c}{a}, \frac{du}{dy} = \frac{c}{b}$. Is this wrong? – spooleey Mar 30 '24 at 14:03
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    I don't know about the validity of your version but it is definitely not the one I was taught, nor the one used in most PDE textbooks (standard one is usually Evans PDEs). – Finn Pillar Mar 30 '24 at 14:55
  • @FinnPillar It's used here: https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/01%3A_First_Order_Partial_Differential_Equations/1.03%3A_Quasilinear_Equations-_The_Method_of_Characteristics -- I don't understand why it is giving me the wrong results? – spooleey Mar 30 '24 at 16:03
  • (1.3.2) from the link you posted is usually the standard characteristic ODEs. Nonetheless, I haven't actually checked your solution. Maybe it's just some error with your computations. – Finn Pillar Mar 30 '24 at 16:32

1 Answers1

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Rewrite the third line of the equation $u_x=\ldots$ as $$ u_x=u\left(-e^{-x}\omega'(\xi)+xe^{-x}\right), \tag{1} $$ where $\xi:=y+e^{-x}$, and the third line of the equation $u_y=\ldots$ as $$ u_y=u\, \omega'(\xi). \tag{2} $$ From $(1)$ and $(2)$ it follows that $$ e^{x}u_x+u_y=xu.\quad{\square} \tag{3} $$

Gonçalo
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