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I have the equation

$$ \partial_t u(t,x) + \sin(t) \partial_x u(t,x) +\cos(t) u(t,x) = 0. $$

with a initial condition $u(0,x)=f(x)$. I want to solve this using the method of characteristics. First I consider just the equation

$$ \partial_t u(t,x) + \sin(t) \partial_x u(t,x)=0, $$ parametrizing by $s$ we get the equations

$$ \frac{\mathrm d t(s)}{\mathrm d s} = 1, \quad \frac{\mathrm d x(s)}{\mathrm d s} = \sin(t(s)), $$ which has the solutions $$ t(s) = s, \quad x(s) = c_1 - \cos(s). $$ with $c_1$ being a constant (I've determined the constant in $t(s)$ to be zero since we have the initial conditions at $t=0$). Thus the solution is $u(t,x) = f(x + \cos(t)).$ Now along the characteristic curves the PDE reduces to an ODE

$$ \frac{\mathrm d u(s)}{\mathrm d s} + \cos(s) u(s)=0. $$

the argument for this is that $$ \frac{\mathrm d u(s)}{\mathrm d s} = \frac{\mathrm d u(t(s),x(s))}{\mathrm ds} = \partial_{t(s)} u(t(s),x(s)) + \sin(t(s)) \partial_{x(s)} u(t(s),x(s)). $$

the ODE has the solution

$$ u(s) = e^{-\sin(s)} c_2. $$

It's not now obvious to me where I went wrong, I assume that probrably we cannot take $\cos(t) = \cos(s)$ when we are along the characteristic curves. But how do I solve the equation correctly?

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2 Answers 2

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The function $u(s)$ satisfies the differential equation $$ \frac{du(s)}{ds}=-u(s)\cos(t(s))=-u(s)\cos(s), \tag{1} $$ where the second equality follows from $t(s)=s$. The solution to $(1)$ is $$ u(s)=u(0)\,e^{-\sin(s)}. \tag{2} $$ Now $$ u(0)=u(t(0),x(0))=u(0,c_1-1)=f(c_1-1). \tag{3} $$ Solving $$ t(s) = s, \qquad x(s) = c_1 - \cos(s) \tag{4} $$ for $c_1$ and $s$, we get $$ s = t, \qquad c_1 = x+\cos(t). \tag{5} $$ Combining $(2), (3)$ and $(5)$, we finally obtain $$ u(t,x)=f(x+\cos(t)-1)\,e^{-\sin(t)}. \tag{6} $$

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$$\frac{\partial u}{\partial t}+\sin(t)\frac{\partial u}{\partial x}=-\cos(t)u$$ Characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{\sin(t)}=\frac{du}{-\cos(t)u}=ds$$ A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{\sin(t)}$ : $$x+\cos(t)=c_1$$ A second characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{-\cos(t)u}$ : $$u\:e^{\sin(t)}=c_2$$ The general solution of the PDE on the form of implicit equation $c_2=\Phi(c_1)$ is :

$$u\:e^{\sin(t)}=\Phi\big(x+\cos(t)\big)$$ $\Phi$ is an arbitrary function (to be determined according to the initial condition). $$u(x,t)=e^{-\sin(t)}\Phi\big(x+\cos(t)\big)$$ With condition $u(0,x)=f(x)$ : $$u(0,x)=e^{-\sin(0)}\Phi\big(x+\cos(0)\big)=\Phi\big(x+1\big)=f(x)$$ This implies $\Phi(X)=f(X-1)$ .

The function $\Phi$ is determined. We put it into the above general solution where $X=x+\cos(t)$ $$u(x,t)=e^{-\sin(t)}f\big(x+\cos(t)-1\big)$$ This is the particular solution of the PDE which satisfies the initial solution.

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