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    Hi,

    I have a couple of questions remaining on a differential equations example sheet that I can't seem to crack. They have a common theme -- changing variables in a PDE.

    Here's the first part of the first one. I'm hoping that with a gentle nudge in the right direction the rest of it should fall into place

    Consider the change of variables
    x=e^{-s} \sin t, \; y=e^{-s}\cos t such that u(x,y) = v(s,t).

    Use the chain rule to express \partial v / \partial s and \partial v / \partial t in terms of x,y,\partial u/\partial x and \partial u / \partial y.
    What I don't see is how to use the chain rule to get the required expression for the derivatives. I'm also not convinced I understand the meaning of the condition u(x,y) = v(s,t).

    Thanks,
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    (Original post by iainfs)
    Hi,

    I have a couple of questions remaining on a differential equations example sheet that I can't seem to crack. They have a common theme -- changing variables in a PDE.
    u(x,y) = v(s,t) means that a function u of the variables x and y, after using the given substitution, becomes a (probably) very different function v, which now has variables s and t. Do you see why you can't necessarily write u(x,y) = u(s,t)?

    For the partial derivative, as s depends on both x and y, you will have both \frac{\partial x}{\partial s} and \frac{\partial y}{\partial s} terms. I will just state the chain rule first:

    \displaystyle \frac{\partial v}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}

    Someone on here may be able to give you a more rigorous argument for the why the above is true, but the informal way I think of it is: the (du/dx)(dx/ds) sort of 'cancels' to give a (du/ds) term, although we really can't write it this way since u isn't really a function of s. Same with the y part. So we have to split it up into the x and y components, if you will, to get an expression for (dv/ds).
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    (Original post by Dream Eater)
    u(x,y) = v(s,t) means that a function u of the variables x and y, after using the given substitution, becomes a (probably) very different function v, which now has variables s and t.
    Thanks. I think I was trying to read too much into that statement.

    I think I've done the first two parts; I'd appreciate it if someone could check it over because I've probably ballsed up somewhere!

    \frac{\partial x}{\partial s} = -x, \; \frac{\partial x}{\partial t} = y, \; \frac{\partial y}{\partial s} = -y, \; \frac{\partial y}{\partial t} = -x

    Then we have
    \frac{\partial v}{\partial s} = -x\frac{\partial u}{\partial x}  - y\frac{\partial u}{\partial y}

    \frac{\partial v}{\partial t} = y\frac{\partial u}{\partial x}  - x\frac{\partial u}{\partial y}.

    (ii) Find, similarly, an expression for \partial^2 v/\partial t^2.
    Rewriting the operator \frac{\partial}{\partial t} =  y\frac{\partial}{\partial x}  - x\frac{\partial}{\partial y}, so we find \frac{\partial^2 v}{\partial t^2} = y\frac{\partial}{\partial x} \left( y\frac{\partial u}{\partial x}  - x\frac{\partial u}{\partial y} \right)  - x\frac{\partial}{\partial y} \left( y\frac{\partial u}{\partial x}  - x\frac{\partial u}{\partial y} \right)
    =y \left[ \frac{\partial y}{\partial x}\frac{\partial u}{\partial x} + y \frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} - x \frac{\partial^2 u}{\partial x \partial y} \right] - x\left[ \frac{\partial u}{\partial x} + y\frac{\partial^2 u}{\partial y \partial x} - \frac{\partial x}{\partial y}\frac{\partial u}{\partial y} - x\frac{\partial^2 u}{\partial y^2}\right]

    \Rightarrow \frac{\partial^2 v}{\partial t^2} = y^2 \frac{\partial^2 u}{\partial x^2} - 2xy\frac{\partial^2 u}{\partial x \partial y}  + x^2 \frac{\partial^2 u}{\partial y^2} + y\frac{\partial y}{\partial x}\frac{\partial u}{\partial x} + x\frac{\partial x}{\partial y}\frac{\partial u}{\partial y} - y\frac{\partial u}{\partial y} - x \frac{\partial u}{\partial x}

    Further, \frac{x}{y}=\tan t, so \frac{\partial x}{\partial y} = \tan t = \frac{x}{y}, and similarly \frac{\partial y}{\partial x} = \frac{y}{x}.

    Then we have
    \frac{\partial^2 v}{\partial t^2} = y^2 \frac{\partial^2 u}{\partial x^2} - 2xy\frac{\partial^2 u}{\partial x \partial y}  + x^2 \frac{\partial^2 u}{\partial y^2} + \frac{y^2}{x}\frac{\partial u}{\partial x} + \frac{x^2}{y}\frac{\partial u}{\partial y} - y\frac{\partial u}{\partial y} - x \frac{\partial u}{\partial x}

    (iii) Hence transform the equation y^2 \frac{\partial^2 u}{\partial x^2} - 2xy\frac{\partial^2 u}{\partial x \partial y}  + x^2 \frac{\partial^2 u}{\partial y^2}=0 into a partial differential equation for v.
    We have
    \frac{\partial^2 v}{\partial t^2} - \frac{y^2}{x}\frac{\partial u}{\partial x} - \frac{x^2}{y}\frac{\partial u}{\partial y} + y\frac{\partial u}{\partial y} + x \frac{\partial u}{\partial x}=0

    \Rightarrow \frac{\partial^2 v}{\partial t^2} - \frac{\partial v}{\partial s}  - \frac{y^2}{x}\frac{\partial u}{\partial x} - \frac{x^2}{y}\frac{\partial u}{\partial y}=0

    ...but I can't see how to get rid of that last pair of terms.
 
 
 
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