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Solve the following:

$$u_t + \frac{1}{1+0.5\cos x} u_x = 0$$

where $u(x,0) = \cos (x-1+0.5\sin (x+1)).$

My attempt:

I applied method of characteristics directly.

$$\frac {dt}{1}=(1+0.5\cos x)dx, \frac {du}{ds} = 0$$

From the equations, I obtain

$$x + 0.5\sin x - t = c$$ $$\Rightarrow u(x,t) = g(x+0.5 sin x - t)$$ where g is an arbitrary function to be determined.

By $u(x,0)$, $$\cos (x-1+0.5\sin (x+1)) = g(x+0.5\sin x)$$

At this point, I can already feel that things may get hairy. Nevertheless, I tried to expand the sine term on the left hand side using trigonometric identities, but to no avail.

How do I proceed from here?

Air Christmas
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  • Your general solution $g(x+0.5\sin(x)-t) $ is correct. Are you sure that there is no typo in the condition ? A wrong bracket ? Where this condition is coming from ? – JJacquelin Mar 09 '17 at 12:38
  • The initial condition is given as it is (as in, as part of the question). Unfortunately, there are no mistakes. – Air Christmas Mar 09 '17 at 13:26

1 Answers1

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The function $$ \alpha(x)=x+0.5\sin x $$ has a positive derivative everywhere on $\mathbf R$ and hence it is invertible. Then starting with $$ g(x+0.5\sin x)=g(\alpha(x))=\cos (x-1+0.5\sin (x+1))=\cos(\alpha(x+1)-2) $$ we'll have that $$ g(x)=\cos(\, \alpha(\alpha^{-1}(x)+1)-2\,) $$ for all $x \in \mathbf R.$ Maybe $$ \alpha(\alpha^{-1}(x)+1)-2 $$ can be simplified somehow (doubtful); if not, leave the answer as it is.

Olod
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  • Your observation of $\alpha$ being invertible has been useful. While the exact answer is not a nice expression, at the very least the answer is exact. Thanks! – Air Christmas Mar 09 '17 at 13:28