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The question (stewart 14.5.50) reads:

If $u = f(x,y)$, where $x=e^scos(t)$ and $y=e^ssin(t)$, show that $$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = e^{-2s} \left[\frac{\partial^2u}{\partial s^2} + \frac{\partial^2u}{\partial t^2}\right] $$

Moving $e^{-2s}$ over, the LHS becomes $\left(\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}\right)e^{2s}\left(sin^2t + cos^2t\right)$ . so the LHS becomes $$y^2\frac{\partial^2u}{\partial x^2} + x^2\frac{\partial^2u}{\partial y^2}$$

Then I'm stuck.

kel c.
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1 Answers1

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Note that

$$\frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}$$

and

$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial t}.$$

Then it follows that

$$\begin{aligned}\frac{\partial^2u}{\partial s^2} &= \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial s}\right) \\ &= \frac{1}{2}\frac{\partial x}{\partial s}\cdot\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial s}\right) + \frac{1}{2}\frac{\partial y}{\partial s}\cdot\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial s}\right)\\ &= \frac{1}{2}\left[\frac{\partial^2 u}{\partial x^2}\left(\frac{\partial x}{\partial s}\right)^2+\frac{\partial^2u}{\partial x\partial y}\frac{\partial y}{\partial s}\frac{\partial x}{\partial s}\right] + \frac{1}{2}\left[\frac{\partial^2 u}{\partial y\partial x}\frac{\partial x}{\partial s}\frac{\partial y}{\partial s} +\frac{\partial^2u}{\partial y^2}\left(\frac{\partial y}{\partial s}\right)^2\right] \\ &= \frac{1}{2}\left[\frac{\partial^2u}{\partial x^2}\left(\frac{\partial x}{\partial s}\right)^2+\frac{\partial^2u}{\partial y^2}\left(\frac{\partial y}{\partial s}\right)^2\right]+\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial s}\frac{\partial y}{\partial s}\end{aligned}$$

Likewise, we have that

$$\frac{\partial^2u}{\partial t^2} = \frac{1}{2}\left[\frac{\partial^2u}{\partial x^2}\left(\frac{\partial x}{\partial t}\right)^2+\frac{\partial^2u}{\partial y^2}\left(\frac{\partial y}{\partial t}\right)^2\right]+\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial t}\frac{\partial y}{\partial t}$$

Since we know that $x(s,t) = e^s\cos t$ and $y(s,t)=e^s\sin t$, it follows that

$$\frac{\partial x}{\partial s} = e^s\cos t,\quad \frac{\partial x}{\partial t}=-e^s\sin t,\quad \frac{\partial y}{\partial s}=e^s\sin t,\quad\text{and}\quad \frac{\partial y}{\partial t}=e^s\cos t$$

We now note that

$$\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial s}\frac{\partial y}{\partial s} = e^{2s}\cos t\sin t\frac{\partial^2u}{\partial x\partial y}$$

and

$$\frac{\partial^2u}{\partial x\partial y}\frac{\partial x}{\partial t}\frac{\partial y}{\partial t} = -e^{2s}\cos t\sin t\frac{\partial^2u}{\partial x\partial y}$$

hence the mixed partial terms drop out when we look at $\dfrac{\partial^2u}{\partial s^2}+\dfrac{\partial^2u}{\partial t^2}$. Therefore, we now see that

$$\begin{aligned}\frac{\partial^2u}{\partial s^2}+\frac{\partial^2 u}{\partial t^2} &= \frac{1}{2}\left[e^{2s}\cos^2t\frac{\partial^2u}{\partial x^2}+e^{2s}\sin^2t\frac{\partial^2u}{\partial y^2}\right] + \frac{1}{2}\left[e^{2s}\sin^2t\frac{\partial^2u}{\partial x^2}+e^{2s}\cos^2t\frac{\partial^2u}{\partial y^2}\right] \\ &= e^{2s}\left(\cos^2t+\sin^2t\right)\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)\\ &= e^{2s}\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)\end{aligned}$$

Therefore, $\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2} = e^{-2s}\left(\dfrac{\partial^2u}{\partial s^2}+\dfrac{\partial^2u}{\partial t^2}\right)$.

Christopher Toni
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  • can you explained a little bit on the second equality on $\frac{\partial^2 u}{\partial s^2}$? where does the half come from? – kel c. Oct 25 '13 at 02:03
  • Let me denote $\frac{\partial}{\partial s}$ by $\partial_s$. Since the objective is to get $u_{xx}$ and $u_{yy}$ to appear when we compute $u_{ss}$, we need to apply chain rule again to partially differentiate $u_s$ with respect to $s$. To do this properly though, we need to introduce $\partial_x$ and $\partial_y$ somehow into this process, so the best way is to note that $\partial_s = x_s\partial_x$ and $\partial_s = y_s\partial y$ and hence $$2\partial_s = x_s\partial_x + y_s\partial_y \implies \partial_s = \frac{1}{2}(x_s\partial_x+y_s\partial_y)$$ Does this clarify things? – Christopher Toni Oct 25 '13 at 07:28
  • yeah. it makes a lot more sense. thanks for the detailed answer. – kel c. Oct 26 '13 at 03:40