2nd order PDE

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Hi there,

I have

f(x,y)=x^2y^2

I have to find $\frac{df}{dt}$ and $\frac{d^2f}{dt^2}$

The first part by chain rule:

$\frac{df}{dt}=\frac{\partial f}{dx}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$

I am not sure how to find $\frac{d^2f}{dt^2}$ from this though?

$\frac{\partial^2 f}{dx^2}=\frac{\partial}{\partial x}(\frac{\partial f}{dx}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t})$
$\frac{\partial^2 f}{dx^2}=(\frac{\partial^2 f}{dx^2}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x\partial y}\frac{\partial y}{\partial t})$

similarly

$\frac{\partial^2 f}{dy^2}=\frac{\partial}{\partial y}(\frac{\partial f}{dx}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t})$
$\frac{\partial^2 f}{dy^2}=(\frac{\partial^2 f}{\partial y \partial x}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial t})$

How do I go from this to the expression for $\frac{d^2y}{dx^2}$ ?

outermeasure · **Posted:** Wed, 28 Mar 2012 17:03:28 UTC

Bazman wrote:

Hi there,

I have

f(x,y)=x^2y^2

I have to find $\frac{df}{dt}$ and $\frac{d^2f}{dt^2}$

The first part by chain rule:

$\frac{df}{dt}=\frac{\partial f}{dx}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$

I am not sure how to find $\frac{d^2f}{dt^2}$ from this though?

$\frac{\partial^2 f}{dx^2}=\frac{\partial}{\partial x}(\frac{\partial f}{dx}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t})$
$\frac{\partial^2 f}{dx^2}=(\frac{\partial^2 f}{dx^2}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial x\partial y}\frac{\partial y}{\partial t})$

similarly

$\frac{\partial^2 f}{dy^2}=\frac{\partial}{\partial y}(\frac{\partial f}{dx}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t})$
$\frac{\partial^2 f}{dy^2}=(\frac{\partial^2 f}{\partial y \partial x}\frac{\partial x}{\partial t}+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial t})$

How do I go from this to the expression for $\frac{d^2y}{dx^2}$ ?

Be careful! Do you really mean $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in the last line?

Also, unless you do additional manipulations, $\dfrac{\mathrm{d}f}{\mathrm{d}t}$ you found is an explicit function of x,y,t

, not

alone. So when you apply the chain rule you need to use
$\displaystyle \frac{\mathrm{d}^2f}{\mathrm{d}t^2} = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\frac{\partial}{\partial x}+\frac{\mathrm{d}y}{\mathrm{d}t}\frac{\partial}{\partial y}+\frac{\partial}{\partial t}\right)\frac{\mathrm{d}f}{\mathrm{d}t}.$

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

thanks OM and yes should have been $\frac{d^2}f{dt^2}$

Shadow · **Posted:** Wed, 28 Mar 2012 17:26:42 UTC

Bazman wrote:

thanks OM and yes should have been $\frac{d^2}f{dt^2}$

??

I'm confused, why not just make the substitutions?

Then $f(x(t),y(t))=\sin^2t\cos^2t={1\over 4}\sin^2(2t)$ and use one variable calculus?

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

umm its a good point, but the unit we are working on is PDE's and it explicitly reuqests you use PSE's to solve it.

Shadow · **Posted:** Wed, 28 Mar 2012 18:33:41 UTC

Bazman wrote:

umm its a good point, but the unit we are working on is PDE's and it explicitly reuqests you use PSE's to solve it.

I suppose. I guess I always think of ODEs as a subset of PDEs. In any case, this is a good way to check your work.

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Thanks Shadow, your method is clearly better.

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2nd order PDE

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