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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 Answers2

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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

Asaf Karagila
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Paul Carter
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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? –  May 18 '12 at 17:14
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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.

user26872
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  • @morphism: Is there some way I can make this more clear? – user26872 May 22 '12 at 18:10
  • Slightly; we're performing $\frac{\partial}{\partial t}s(x+b(\tau-t),\tau) d\tau$ and showing that it's equal to $-b \frac{\partial}{\partial x}s(x+b(\tau-t),\tau) d\tau $ and so we say $\frac{\partial}{\partial t}s(x+b(\tau-t),\tau) d\tau = \frac{\partial}{\partial X}\ s(X,\tau)\frac{\partial X}{\partial t}$ where $X=x+b(\tau-t)$ and so this gives $-b\frac{\partial}{\partial X}\ s(X,\tau) = -b\frac{\partial}{\partial X}\ s(x+b(\tau-t),\tau)$ but then I don't see how that equals $-b \frac{\partial}{\partial x}s(x+b(\tau-t),\tau)$ because we're differentiating with respect to $X$ and not $x$? –  May 23 '12 at 14:43
  • @morphism: Ah, okay. Notice that $$\begin{eqnarray} \frac{\partial}{\partial x} s(x+b(\tau-t),\tau) &=& \frac{\partial}{\partial X} s(X,\tau)\frac{\partial X}{\partial x} \ &=& \frac{\partial}{\partial X} s(X,\tau) \end{eqnarray}$$ since $\frac{\partial X}{\partial x} = \frac{\partial }{\partial x}(x+b(\tau-t)) = 1$. – user26872 May 23 '12 at 17:00
  • Ah perfect! Thank you! –  May 23 '12 at 17:25
  • @morphism: Glad to help! – user26872 May 23 '12 at 17:48