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Background: I've solved this linear PDE: $$\frac{\partial H_a}{\partial p_r}-g_0({p_r})\frac{\partial H_a}{\partial z}=0$$ Like this (and this is true) that I used method of characteristics: $$\frac{dp_r}{ds}=1 \And (p_r(0)=0)\implies p_r=s$$ $$\frac{dz}{ds}=g_0({p_r}) \And(z(0)=\aleph)\implies z=-\int_{0}^{p_r} g_0(s) ds+\aleph$$ $$\frac{dH_a}{ds}=0 \And (H_a(0)=\phi(\aleph))\implies H_a(z,p_r)=\phi(z+\int_{0}^{p_r} g_0(s) ds)$$

Question: Now, I should solve this quasilinear PDE:

$$\frac{\partial H_a}{\partial p_r}-2g_0({p_r})\frac{\partial H_a}{\partial z}=g_0({p_r})k_0z$$

Although I like to use method of characteristics again, I don't know hot to do. So I do this:

Forming this: $$\frac{dp_r}{1}=\frac{dz}{-2g_0({p_r})}=\frac{dH_a}{g_0({p_r})k_0z}$$ Taking first two fractions integral leads: $$dz=-2g_0({p_r})dp_r \implies z=-2\int_{0}^{p_r} g_0(s) ds+c_1$$ And for last two fractions: $$\frac{k_0}{-2}zdz=dH_a \implies H_a=\frac{k_0}{-2}z^2+c_2$$

And I'm sure I miss some points (for example $c_1$ and $c_2$ are not necessarily constants). Could anyone help me?

1 Answers 1

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Without checking the preliminary calculus above.

You obtained the equations of two families of characteristic curves : $$z+2\int_{0}^{p_r} g_0(s) ds=c_1$$ $$ H_a-\frac{k_0}{-2}z^2=c_2$$ Those equations are valid on the respective characteristic curves for any $c_1$ and $c_2$. But this is not true everywhere on the surface linking the characteristic curves : $c_1$ and $c_2$ are no longer independent. The relationship can be expressed on the form of an implicit equation : $\Phi(c_1,c_2)=0$. This leads to general solution of the PDE on the form : $$\Phi\left((z+2\int_{0}^{p_r} g_0(s) ds)\:,\:(H_a-\frac{k_0}{-2}z^2)\right)=0$$ where $\Phi$ is any differentiable function of two variables.

An equivalent way to express the above relationship consists in expressing one variable as a function of the other : $c_2=F(c_1)$ or $c_1=G(c_2)$ where $F$ and $G$ are any differentiable functions, but related (one is the inverse of the other).

For example with the first : $$\left(H_a-\frac{k_0}{-2}z^2\right)=F\left(z+2\int_{0}^{p_r} g_0(s) ds\right)$$ So, an explicit form of the general solution is : $$H_a=-\frac{k_0}{2}z^2+F\left(z+2\int_{0}^{p_r} g_0(s) ds\right)$$

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    I think I got my answer. Thanks. – 2017-01-25
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    By the way, Is last part of my calculus true? (I mean finding $c_1$ and $c_2$ part). Thanks in advance. – 2017-01-25
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    Well, these two equations in general define separately a family of surfaces (in general they are of the form $f(x,y,z,c_1)=0$). It's its intersection giving a specific value for $c_1$ and $c_2$, a curve, one of the characteristics. The particular selection for the arbitrary function in the general solution tell us how these surfaces are pasted together to form the family of characteristics leading to a new surface, a solution of the PDE. – 2017-01-25
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    @Abraham Two first order ODE, each one with its own arbitrary constant: All ok. – 2017-01-25
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    @RafaBudría Thanks. – 2017-01-25
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    @Rafa Budría. OK. Your comment is right and more clear than mine. I appreciate it. – 2017-01-25
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    @ Abraham @ JJacquelin. You are welcome :) – 2017-01-25
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    @Abraham Oh, oh. I am very sorry. There is a minor mistake in your calculations. The factor $\frac{1}{-2}$ has to be $\frac{1}{-4}$, all from your second integral. Well, the trick is "always try to differenciate your integral". – 2017-01-25
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    @JJacquelin thanks to you. I've checked my questions and you were my teacher about that topic! :) – 2017-01-25