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$$u=f(x-(k+u)t)$$

How would I go about finding $\frac{\partial u}{\partial t}$ and $\frac{\partial u}{\partial x}$?

So far I have written that $$du=\left(\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial t}\frac{dt}{dt}\right)dt$$

But I'm not even sure if this is correct, or the right way to go about it. I Would appreciate any hints/tips (especially for how to find partial derivatives using the chain rule in general).

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  • $\begingroup$ @ChristianBlatter The fact that $x=x(t)$ is given in the problem. Also, $f$ is an arbitrary function of a single variable. $\endgroup$ – mrnovice Jan 27 '18 at 19:46
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Since $x$ is not an independent variable, but a function of $t$, we have the following situation: We are given the function $$F(t,u):=f\bigl(x(t)-(k+u)t\bigr)-u=0\ ,$$ assumed of class $C^1$. The equation $$F(u,t)=0$$ then implicitly defines $u$ as a function of $t$ under favorable circumstances. The implicit function theorem has the following to say in this regard: Given a point $(t_0,u_0)$ satisfying $F(t_0,u_0)=0$ there is a $C^1$-function $$\psi: \>t\mapsto u:=\psi(t)\qquad(t_0-h<t<t_0+h),\qquad \psi(t_0)=u_0\ ,$$ such that $$F\bigl(t,\psi(t)\bigr)\equiv0\qquad(t_0-h<t<t_0+h)\ .$$ Furthermore one has the formula $$u'(t_0)=-{F_t(t_0,u_0)\over F_u(t_0,u_0)}={f'\bigl(x(t_0)-(k+u_0)t_0\bigr)\bigl(x'(t_0)-(k+u_0)\bigr)\over f'\bigl(x(t_0)-(k+u_0)t_0\bigr)t_0+1}\ .\tag{1}$$ The all important assumption for all this to hold is that $F_u(t_0,u_0)\ne0$, i.e., that the denominator in $(1)$ does not vanish.

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by the chain rule we obtain $$\frac{\partial u}{\partial x}=f'(x-(k+u)t)(1-\frac{\partial u}{\partial x}t)$$ and $$\frac{\partial u}{\partial t}=f'(x-(k+u)t)(-(k+u)-(k+\frac{\partial u}{\partial t})t)$$

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