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I'm trying to show that the function $u(x,t) = \int^t_0 s(x + b(\tau - t), \tau) d\tau$ satisfies the partial differential equation $u_t + bu_x = s(x,t).$

I start by finding $u_t(x,t) = \frac{\partial}{\partial t}\int^t_0 s(x + b(\tau - t), \tau) \, d\tau =s(x,t)$ and then $u_x(x,t) = \frac{\partial}{\partial x}\int^t_0 s(x + b(\tau - t), \tau) \, d\tau$ $= \int^t_0 \frac{\partial}{\partial x}s(x + b(\tau - t),\tau) \, d\tau$ and this is where I get stuck.

Am I on the right track?

2 Answers 2

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As @PaulCarter mentions, your trouble is in the calculation of $u_t$. In gory detail, according to Leibniz's integral rule, $\begin{eqnarray*} u_t(x,t) &=& \frac{\partial}{\partial t} \int_0^t s(x+b(\tau-t),\tau) d\tau \\ &=& \frac{\partial t}{\partial t} s(x+b(\tau-t),\tau)|_{\tau=t} + \int_0^t \frac{\partial}{\partial t}s(x+b(\tau-t),\tau) d\tau \\ &=& s(x,t) -b \int_0^t s^{(1,0)}(x+b(\tau-t),\tau) d\tau, \end{eqnarray*}$ where $s^{(1,0)}(x+b(\tau-t),\tau) = \frac{\partial}{\partial X} s(X,\tau)|_{X=x+b(\tau-t)}$. Chain rule brings out the factor $-b$. Notice that the derivative with respect to $t$ acts on the first argument and that $s^{(1,0)}(x+b(\tau-t),\tau) = s_x(x+b(\tau-t),\tau)$.

Likewise $\begin{eqnarray*} u_x(x,t) &=& \int_0^t s^{(1,0)}(x+b(\tau-t),\tau) d\tau. \end{eqnarray*}$ Therefore, $u_t + b u_x = s(x,t)$, as claimed.

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    @morphism: Glad to help!2012-05-23
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Your calculation of $u_t(x,t)$ is incorrect. Should be

$u_t(x,t) = \frac{\partial}{\partial t}\int^t_0 s(x + b(\tau - t), \tau) \, d\tau =s(x,t) - b\int^t_0 s_x(x + b(\tau - t),\tau) \, d\tau$

as the integrand also depends on t in the first argument of $s$. See the Leibniz rule: http://en.wikipedia.org/wiki/Leibniz_integral_rule

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    Hang on how do you get the $b$ to come out and $s_x$ instead of $s_t$ like in Leibniz?2012-05-18