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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](http://en.wikipedia.org/wiki/Method_of_characteristics).2012-01-28
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    @Didier: Dear Didier, That article only really describes the application of the method to first order equations, though. Regards,2012-01-28
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    Dear peter, You describe your PDE as non-linear, but looking at it, it seems linear to me. What am I missing? Regards,2012-01-28
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    This is a linear equation.2012-08-01
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    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 wor2012-01-28
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    any suggestions on the above matter ?2012-01-28
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    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.2012-03-28
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    This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date.2012-09-10

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