method of characteristics?

i want to solve the followning second order non linear PDE :

$$\frac{\partial V}{\partial t}(t,x) + g(x)\frac{\partial V}{\partial x}(t,x) + q(x)\frac{\partial^{2} V}{\partial x^{2}}(t,x) + h(x) =0$$

with boundary condition $V(T,x) = \Phi(x)$ where $t\le T$.

I heard that the method of characteristics can work in order to determine $V(t,x)$. However, i have never met this method before. Could someone give me a quick quide ?

thanks

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Yes, someone could: Method of characteristics. – Did Jan 28 '12 at 8:26
@Didier: Dear Didier, That article only really describes the application of the method to first order equations, though. Regards, – Matt E Jan 28 '12 at 12:25
Dear peter, You describe your PDE as non-linear, but looking at it, it seems linear to me. What am I missing? Regards, – Matt E Jan 28 '12 at 12:27
Firstly, thanks for the immediate responce. @Matt E: Dear Matt it is a non linear PDE, probably my fault that i didnt write the functions g,q,h in a better way. Dear Didier, i have seen this article too, but i am looking for second orfer PDEs. In fact i have to solve a system of two PDEs of this forms, but i suppose that if i manage to solve one of them i would probably solve the system too. I was thinking to give $V(t,x)$ the form of a Cole-Hopf factorization , i.e. $V(t,x)=e^{\alpha f(t)}$ and then get a Burgers-like PDE. However, i coudn't find simething instructive in order to begin my wor – peter Jan 28 '12 at 12:41
Your PDE is linear. In fact, linearity is meant w.r.t. the unknown $V$, not w.r.t. the variables $(x,t)$: this means that a differential operator $\mathcal{L}$ is said to be linear iff $\mathcal{L}[\alpha V_1+\beta V_2]=\alpha \mathcal{L}[V_1] + \beta \mathcal{L}[V_2]$, and this is the case for your operator $\mathcal{L}=q \frac{\partial^2}{\partial x^2} + g\frac{\partial}{\partial x} + \frac{\partial}{\partial t}$, where $q,g,h$ depend on $(x,t)$. Neverthless, your PDE is inhomogeneous because of the term $h$. Hence, you have a second order, inhomogeneous linear PDE of parabolic type. – Pacciu Mar 28 '12 at 16:11

Case $1$: $g(x)=q(x)=0$

Then $\dfrac{\partial V(t,x)}{\partial t}+h(x)=0$

$\dfrac{\partial V(t,x)}{\partial t}=-h(x)$

$V(t,x)=C(x)-h(x)t$

$V(T,x)=\Phi(x)$ :

$C(x)-h(x)T=\Phi(x)$

$C(x)=\Phi(x)+h(x)T$

$\therefore V(t,x)=\Phi(x)+h(x)T-h(x)t=\Phi(x)+h(x)(T-t)$

Case $2$: $g(x)\neq0$ and $q(x)=0$

Then $\dfrac{\partial V(t,x)}{\partial t}+g(x)\dfrac{\partial V(t,x)}{\partial x}+h(x)=0$

$\dfrac{\partial V(t,x)}{\partial t}+g(x)\dfrac{\partial V(t,x)}{\partial x}=-h(x)$

This belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1212.pdf

The general solution is $V(t,x)=C\biggl(\int^x\dfrac{1}{g(x)}dx-t\biggr)-\int^x\dfrac{h(x)}{g(x)}dx$

$V(T,x)=\Phi(x)$ :

$C\biggl(\int^x\dfrac{1}{g(x)}dx-T\biggr)-\int^x\dfrac{h(x)}{g(x)}dx=\Phi(x)$

$C\biggl(\int^x\dfrac{1}{g(x)}dx-T\biggr)=\Phi(x)+\int^x\dfrac{h(x)}{g(x)}dx$

$C(x)=\Phi\biggl(\biggl(\int^x\dfrac{1}{g(x)}dx-T\biggr)^{-1}(x)\biggr)+\int^{\left(\int^x\frac{1}{g(x)}dx-T\right)^{-1}(x)}\dfrac{h(x)}{g(x)}dx$

$\therefore V(t,x)=\Phi\biggl(\biggl(\int^x\dfrac{1}{g(x)}dx-T\biggr)^{-1}\biggl(\int^x\dfrac{1}{g(x)}dx-t\biggr)\biggr)+\int^{\left(\int^x\frac{1}{g(x)}dx-T\right)^{-1}\left(\int^x\frac{1}{g(x)}dx-t\right)}\dfrac{h(x)}{g(x)}dx-\int^x\dfrac{h(x)}{g(x)}dx$

Case $3$: $q(x)\neq0$

Then $\dfrac{\partial V(t,x)}{\partial t}+g(x)\dfrac{\partial V(t,x)}{\partial x}+q(x)\dfrac{\partial^2V(t,x)}{\partial x^2}+h(x)=0$

It is possible that the subsititution $V(t,x)=V_c(t,x)+V_p(t,x)$ can make the inhomogeneous linear PDE becomes homogeneous linear PDE if $V_p(t,x)$ can be found.

For this question, the form of $V_p(t,x)$ is not difficult to guess, just the particular solution of the ODE $g(x)\dfrac{dV(x)}{dx}+q(x)\dfrac{d^2V(x)}{dx^2}+h(x)=0$ , i.e. $V_p(t,x)=-\int^xe^{-\int^x\frac{g(x)}{q(x)}dx}\int^x\dfrac{h(x)}{q(x)}e^{\int^x\frac{g(x)}{q(x)}dx}~dx~dx$

So let $V(t,x)=V_c(t,x)-\int^xe^{-\int^x\frac{g(x)}{q(x)}dx}\int^x\dfrac{h(x)}{q(x)}e^{\int^x\frac{g(x)}{q(x)}dx}~dx~dx$ ,

Then $\dfrac{\partial V_c(t,x)}{\partial t}+g(x)\dfrac{\partial V_c(t,x)}{\partial x}+q(x)\dfrac{\partial^2V_c(t,x)}{\partial x^2}=0$

Of course we use separation of variables:

Let $V_c(t,x)=T(t)X(x)$ ,

Then $T'(t)X(x)+g(x)T(t)X'(x)+q(x)T(t)X''(x)=0$

$T'(t)X(x)=-(q(x)X''(x)+g(x)X'(x))T(t)$

$\dfrac{T'(t)}{T(t)}=-\dfrac{q(x)X''(x)+g(x)X'(x)}{X(x)}=f(s)$

$\begin{cases}\dfrac{T'(t)}{T(t)}=f(s)\\q(x)X''(x)+g(x)X'(x)+f(s)X(x)=0\end{cases}$

$\therefore V(t,x)=\sum_sC_1(s)e^{tf_s(s)}X_{s,1}(x,s)+\sum_sC_2(s)e^{tf_s(s)}X_{s,2}(x,s)+\int_sC_3(s)e^{tf_g(s)}X_{g,1}(x,s)~ds+\int_sC_4(s)e^{tf_g(s)}X_{g,2}(x,s)~ds-\int^xe^{-\int^x\frac{g(x)}{q(x)}dx}\int^x\dfrac{h(x)}{q(x)}e^{\int^x\frac{g(x)}{q(x)}dx}~dx~dx$

or $V(t,x)=\sum_sC_1(s)e^{tf_s(s)}X_{s,1}(x,s)+\sum_sC_2(s)e^{tf_s(s)}X_{s,2}(x,s)+\sum_sC_3(s)e^{tf_g(s)}X_{g,1}(x,s)+\sum_sC_4(s)e^{tf_g(s)}X_{g,2}(x,s)-\int^xe^{-\int^x\frac{g(x)}{q(x)}dx}\int^x\dfrac{h(x)}{q(x)}e^{\int^x\frac{g(x)}{q(x)}dx}~dx~dx$

Then subsititute $V(T,x)=\Phi(x)$ into the above general solutions to eliminate part of the arbitrary parts to get the final answer.

The suffix $s$ stand for special cases while suffix $g$ stand for general cases. The special cases means that the values of $s$ that lead $q(x)X''(x)+g(x)X'(x)+f(s)X(x)=0$ will have the forms of their general solution differ from the current form of the general solution of most values of $s$ in $q(x)X''(x)+g(x)X'(x)+f(s)X(x)=0$ .

Note that different $q(x)$ , $g(x)$ and $f(s)$ may result the special cases appearing in the different values of $s$ . Like the previous question Indication on how to solve the heat equations with nonconstant coefficients $(q(x)=-x^2 , g(x)=-2x)$ , choose $f(s)=-\dfrac{4\pi^2s^2+1}{4}$ will result the only special case appearing in $s=0$ . We should choose the form of $f(s)$ wisely so that the special cases appear in the nice values of $s$ .

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