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I've been having a very hard time understanding how characteristics work in PDEs, so I'm hoping that knowing how to find them for an equation like this would help me understand them better.

How would you find the characteristics through the point (0,1) of the following PDE?

$$ \frac{\partial^2u}{\partial t^2} + 4\frac{\partial^2u}{\partial x \partial t} + 4\frac{\partial^2u}{\partial x^2}=0$$

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First, factor the differential operator on the left. Since $a^2+4ab+4b^2=(a+2b)(a+2b)$, the operator factors as $$\frac{\partial^2}{\partial t^2} + 4\frac{\partial^2}{\partial x \partial t} + 4\frac{\partial^2}{\partial x^2}= \left(\frac{\partial }{\partial t }+2\frac{\partial }{\partial x }\right) \left(\frac{\partial }{\partial t }+2\frac{\partial }{\partial x }\right) $$ The factors are directional derivatives of 1st order. Sadly, they are in the same direction, of the vector $(2,1)$ in the $(x,t)$ plane. This means we have only one characteristic through each point, namely a line of the form $x=2t+C$. The equation is somewhat degenerate, compared to honest hyperbolic equations such as $\frac{\partial^2u}{\partial t^2} + 4\frac{\partial^2 u}{\partial x^2} =0$.

Anyway, we see that along every line of the form $x-2t=C$ the solution is linear (since its second derivative is zero). Hence, we can write down the general solution as $$u(x,t) = f(x-2t)+t\,g(x-2t)$$ where $f$ and $g$ are arbitrary functions. (Check that this works: the restriction of $u$ to any characteristic is linear.)

If you know the initial value and initial velocity at $t=0$, you can find $f$ and $g$, and with them $u$.

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  • $\begingroup$ So, would this PDE be parabolic then, since it only has one characteristic? I thought it is only hyperbolic PDEs that can be solved by the method of characteristics? $\endgroup$ – Time4Tea Nov 17 '18 at 16:59

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