Method of characteristics Quasilinear PDE I have the following PDE:
$h_t + 3h^2 \theta h_\theta + h^3 = 0$
Therefore the characteristic equations are:
$\frac{dt}{1} = \frac{d\theta}{3h^2\theta} = \frac{dh}{-h^3}$.
Using 2 and 3, we can obtain:
$h^3\, d\theta + 3h^2\theta \, dh = 0 \Rightarrow \phi = h^3\theta$
Using 1 and 2, we can obtain
$dt = \frac{dh}{-h^3} \Rightarrow d\left(t-\frac{1}{2h^2}\right) = 0 \Rightarrow \psi = t - \frac{1}{2h^2}$
Hence, $F = \left(h^3\theta,t - \frac{1}{2h^2}\right) = 0$. Then via the implicit function theorem,
$h = \frac{1}{\theta^{1/3}} f\left(t - \frac{1}{2h^2}\right)^{1/3}$.
However, when I substitute this calculated $h$ solution into my PDE I get $\frac{f'}{3\theta^{1/3}f^{2/3}}$ clearly $\neq 0$.
Clearly I have made a mistake somewhere, but I am not sure where? Everything seems sound to me. Any pointers would be greatly appreciated. :)
 A: Write the characteristic vector field
$$X_{char} = \frac{\partial}{\partial t} + 3 \, h^2 \theta \,\frac{\partial}{\partial \theta} - h^3 \frac{\partial}{\partial h}$$ so the characteristics are given by the system
\begin{align}
\frac{dt}{ds} &= 1 \\
\frac{d\theta}{ds} &= 3 \, h^2 \theta\\
\frac{dh}{ds} & = - h^3
\end{align} which after expressing $ds$ turns into
\begin{align}
ds &= dt \\
ds &=\frac{d\theta}{3 \, h^2 \theta} \\
ds &=-\frac{dh}{h^3} 
\end{align}
First and third, second and third give us
$$dt =-\frac{dh}{h^3} \,\,\,\text{ and }\,\,\, \frac{d\theta}{3 \, h^2 \theta} = -\frac{dh}{h^3} $$ which simplifies to
$$dt =-\frac{dh}{h^3} \,\,\,\text{ and }\,\,\, \frac{d\theta}{3 \,\theta} = -\frac{dh}{h} $$ The first one gives
$$t = \frac{1}{2\, h^2} + C_1$$ and the second
$$\log(\theta) = -3 \log(h) + C_2$$
$$\theta \, h^3 = \tilde{C}_2$$ so two independent first integrals of $X_{char}$ are
\begin{align}
\lambda &= t - \frac{1}{2\, h^2}\\
\mu &= \theta \, h^3
\end{align}
So for any arbitrary smooth function $F(\mu, \lambda)$ one gets the general solution written as an implicit function satisfying the equation
$$F\left(\theta \, h^3, \, t - \frac{1}{2\, h^2} \right) = 0$$ If you want to use function $F(\mu, \lambda) = f(\lambda) - \mu$ then indeed you get
$$\theta \, h^3 = f\left(t - \frac{1}{2\, h^2}\right)$$ which can indeed turn into
$$h = \frac{1}{\theta^{\, 1/3}} f\left(t - \frac{1}{2\, h^2}\right)^{1/3} = \sqrt[3]{\frac{1}{\theta} f\left(t - \frac{1}{2\, h^2}\right)}$$ so your calculations are correct. However, observe that in the last expression, $h$ is still given as an implicit function, it is present in the left side but also in the right side, in the argument of $f$. I suspect you might have forgotten to differentiate with respect to that $h$ when checking whether the obtained expression is indeed a solution. It is.
Let me try to explain how one gets to the characteristic vector field and what the interpretation of the implicit function is. Given your equation, you assume that the function $h$ is given implicitly as $\Phi \big(t,\theta, h(t,\theta)\big) = 0$ which means it is given by the equation $\Phi\big(t,\theta, h\big) = 0$. Now, differentiate the equation with respect to $t$
$$0 = \frac{\partial}{\partial t} \,\Big(\Phi\big(t,\theta, h\big)\Big) = \frac{\partial \Phi}{\partial t} + \frac{\partial \Phi}{\partial h} \, \frac{\partial h}{\partial t}$$
and with respect to $\theta$
$$0 = \frac{\partial}{\partial \theta} \,\Big(\Phi\big(t,\theta, h\big)\Big) = \frac{\partial \Phi}{\partial \theta} + \frac{\partial \Phi}{\partial h} \, \frac{\partial h}{\partial \theta}$$  Thus you can express
\begin{align}
\frac{\partial h}{\partial t} &= - \, \left(\frac{\partial \Phi}{\partial t} \right)\Big/  \left(\frac{\partial \Phi}{\partial h}\right)  = - \left(\frac{\partial \Phi}{\partial h}\right)^{-1} \, \frac{\partial \Phi}{\partial t} \\
\frac{\partial h}{\partial \theta} &= - \,  \left(\frac{\partial \Phi}{\partial \theta} \right)\Big/ \left( \frac{\partial \Phi}{\partial h}\right) = - \, \left(\frac{\partial \Phi}{\partial h}\right)^{-1} \, \frac{\partial \Phi}{\partial \theta} 
\end{align}
So since you want $h$ to solve your equation
$$ \frac{\partial h }{\partial t} + 3 \, h^2 \theta \,\frac{\partial h}{\partial \theta} + h^3 = 0$$ by plugging the expressions for the partial derivatives for $h$ as an implicit function, you get
$$ - \left(\frac{\partial \Phi}{\partial h}\right)^{-1} \, \frac{\partial \Phi}{\partial t} \, - \, 3 \, h^2 \theta \, \left(\frac{\partial \Phi}{\partial h}\right)^{-1} \, \frac{\partial \Phi}{\partial \theta}  \, + \, h^3 = 0$$ Multiply the whole equation by $- \, \frac{\partial \Phi}{\partial h}$ and you arrive at
$$  \frac{\partial \Phi}{\partial t} \, + \, 3 \, h^2 \theta \,  \frac{\partial \Phi}{\partial \theta}  \, - \, h^3 \, \frac{\partial \Phi}{\partial h}= 0$$ This is exactly
$$X_{char}\,  \Phi = \left( \frac{\partial }{\partial t} \, + \, 3 \, h^2 \theta \,  \frac{\partial}{\partial \theta}  \, - \, h^3 \, \frac{\partial}{\partial h} \right) \, \Phi = 0$$ so you are looking for a function $\Phi(t,\theta,h)$ which is constant along the integral curves of the vector field, i.e. $F$ is constant along the solutions of the system
\begin{align}
\frac{dt}{ds} &= 1 \\
\frac{d\theta}{ds} &= 3 \, h^2 \theta\\
\frac{dh}{ds} & = - h^3
\end{align} $\Phi$ is also called a conserved quantity of the system (of the vector field) and also a first integral. Now if $\lambda(t,\theta,h)$ and $\mu(t,\theta,h)$ are two independent conserved quantities, then they satisfy the equation
$$X_{char}\,  \lambda = \left( \frac{\partial }{\partial t} \, + \, 3 \, h^2 \theta \,  \frac{\partial}{\partial \theta}  \, - \, h^3 \, \frac{\partial}{\partial h} \right) \, \lambda = 0$$
$$X_{char}\,  \mu = \left( \frac{\partial }{\partial t} \, + \, 3 \, h^2 \theta \,  \frac{\partial}{\partial \theta}  \, - \, h^3 \, \frac{\partial}{\partial h} \right) \, \mu = 0$$ and one can check, by applying chain rule, that $\Phi(t,\theta,h) = F(\mu, \lambda) = F\big(\mu(t,\theta,h), \, \lambda(t,\theta,h)\big)$ is also a solution, i.e.
\begin{align}
X_{char}\,  F(\mu, \lambda) &= \left( \frac{\partial }{\partial t} \, + \, 3 \, h^2 \theta \,  \frac{\partial}{\partial \theta}  \, - \, h^3 \, \frac{\partial}{\partial h} \right) \, F(\mu, \lambda) \\
&= \frac{\partial F}{\partial \mu} \, X_{char} \mu + \frac{\partial F}{\partial \lambda} \, X_{char} \lambda  = 0\end{align}
In your case, you have $\lambda = \lambda(t,h)$ and $\mu = \mu(\theta, h)$ so
$$X_{char}\,  \lambda = \frac{\partial \lambda}{\partial t} \, - \, h^3 \, \frac{\partial \lambda}{\partial h} = 0$$
$$X_{char}\,  \mu = 3 \, h^2 \theta \,  \frac{\partial \mu}{\partial \theta}  \, - \, h^3 \, \frac{\partial \mu}{\partial h}= 0$$
It is immediate to see that the last two equations hold for the found solutions
\begin{align}
\lambda &= t - \frac{1}{2\, h^2}\\
\mu &= \theta \, h^3
\end{align}
Assume that you have an initial condition $h(0.\theta) = g(\theta)$. Then your general solution, written in an implicit form as $$0=F\big(\mu(t,\theta, h), \lambda(t,\theta, h)\big) =F\big(\mu(\theta, h), \lambda(t, h)\big) = F\left(\theta h^3, \, t- \frac{1}{2h^2}\right)$$ turns into the equation $$0=F\big(\mu(0,\theta, g), \lambda(0,\theta, g)\big) =F\big(\mu(\theta, g), \lambda(0, g)\big) = F\left(\theta g^3, \, - \frac{1}{2g^2}\right).$$ Next step is to think that $F(\mu, \lambda) = f(\mu) - \lambda$ for some smooth function $f(u)$. Then we have the equation
$$0=f\big(\mu(0,\theta, g)\big) - \lambda(0,\theta, g) = f\big(\mu(\theta, g)\big) - \lambda(0, g) = f\left(\theta g^3 \right) - \frac{1}{2g^2}$$ which is basically
$$f\Big(\theta \, g(\theta)^3 \Big) = \frac{1}{2g(\theta)^2}$$ Now set $$u = \theta \, g(\theta)^3 \,\,\,\, \text{ (which expresses $u$ as a function of $\theta$) }$$ and imagine we can invert this expression as $$\theta = q(u)\,\,\,\, \text{ (which expresses $\theta$ as a function of $u$) }$$
Then $$f(u) = \frac{1}{2g\big(q(u)\big)^2}$$ and here you have your formula for $f$. In the special case of $g(\theta) = \theta^2$ we have the equation
$$f\Big(\theta \, (\theta^2)^3 \Big) = \frac{1}{2 (\theta^2)^2}$$ which becomes
$$f\big(\theta^7 \big) = \frac{1}{2 \, \theta^4}$$ Let $u = \theta^7$. Then $\theta = \sqrt[7]{u}$ and thus
$$f\big(u \big) = \frac{1}{2 \, (\sqrt[7]{u})^4} =  \frac{1}{2 \, ({u})^{4/7}} $$ which is the formula for $f$. Finally, our implicit equation for the function $h$ is
$$ \frac{1}{2 \, (\theta \, h^3)^{4/7}} = t- \frac{1}{2 \, h^2}$$ or equivalently
$$ \frac{1}{ \theta^4 \, h^{12}} =2^7 \left(t- \frac{1}{2 \, h^2}\right)^7$$
A: $h_t+3h^2\theta h_\theta+h^3=0$
$\dfrac{h_t}{h^2}+3\theta h_\theta=-h$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dh}{ds}=-h$ , letting $h(0)=1$ , we have $h=e^{-s}$
$\dfrac{d\theta}{ds}=3\theta$ , letting $\theta(0)=\theta_0$ , we have $\theta=\theta_0e^{3s}=\dfrac{\theta_0}{h^3}$
$\dfrac{dt}{ds}=\dfrac{1}{h^2}=e^{2s}$ , we have $t=f(\theta_0)+\dfrac{e^{2s}}{2}=f(h^3\theta)+\dfrac{1}{2h^2}$
