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I’m currently a year 10 in high school and I’ve recently been really interested in partial differential equations. I can only solve really simple ones though due to gaps in my knowledge. Anyway, one day while waiting for my teacher to print out a practice test I decided to come up with a simple looking PDE and try to solve it. I tried solving it but it turned out to be more complex than I expected so I was wondering if someone could help me solve it if it is solvable. Anyway, the question is

$$UU_{x}=(U+1)U_{t}.$$

I managed to reduce it to a system of ordinary differential equations but I don’t know how to solve systems of ordinary differential equations or even if that was the right way to attempt the partial differential equation.

Just_A_User
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Ovin Herath
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1 Answers1

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The reason why this problem is so difficult is because it is fully nonlinear, in the sense that the highest derivatives in the equation (both $u_{t}, u_{x}$) are nonlinear. To solve this we use the method of characteristics. Whilst it may be too advanced for you, I'm not sure of another simpler way to solve the problem. So hopefully you can follow.

You have

$$(u+1)u_{t} = uu_{x}$$

The characteristic curves are determined from the parametrization invariant form of the Lagrange-Charpit equations

$$\frac{dt}{u+1} = \frac{dx}{- u} = \frac{du}{0}$$

From the final characteristic, we have

$$u = c_{2}, \quad \text{$c_{2}$ a constant}$$

From the first equality, substituting $u = c_{2}$ and integrating, we have

\begin{align} dx &= - \left( \frac{c_{2}}{c_{2}+1} \right) dt \\ \implies x + \left( \frac{c_{2}}{c_{2}+1} \right) t &= x + \left( \frac{u}{u+1} \right) t \\ &= c_{1} \end{align}

Using the functional relationship $c_{2} = f(c_{1})$, we have

\begin{align} u &= c_{2} \\ &= f(c_{1}) \\ &= f \left(x + \left( \frac{u}{u+1} \right) t \right) \end{align}

which you can check via differentiation satisfies the PDE.

Matthew Cassell
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