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Math Help - Particular solutions of 3 linear PDEs with nonstandart right-hand side

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation Particular solutions of 3 linear PDEs with nonstandart right-hand side

    Dear MHF members,

    I have the following 3 PDEs, whose solutions I know but I am looking for nice explanary solutions. Please help me.

    1. (D_{x}^{2}+D_{x}D_{y}-2D_{y}^{2})z=8\big(\sec(2x+y)\big)^{2}\tan(2x+y) [Solution: z_{p}=\tan(2x+y)]
    2. (D_{x}^{2}+D_{x}D_{y}-D_{x}-D_{y})z=1/y^{2} [Solution: z_{p}=1/y]
    3. (D_{x}^{2}-D_{y}^{2}-D_{x}+D_{y})z=(x+1)/x^{2} [Solution: z_{p}=-\ln(x)]

    Thanks.
    bkarpuz
    Last edited by bkarpuz; January 17th 2012 at 01:23 AM.
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    Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

    What do you mean by "nice explanary solutions"? Do you mean how they came up with z_p? If so, I think I can explain this.
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    Senior Member bkarpuz's Avatar
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    Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

    Quote Originally Posted by Danny View Post
    What do you mean by "nice explanary solutions"? Do you mean how they came up with z_p? If so, I think I can explain this.
    Exactly!
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    Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

    The second and third are rather straight forward. For the second, the RHS is only a function of y so sek a particular solution in the form z_p = f(y).  You'll get an ODE for f (and it's linear!) Similarly for the third. For the first, you'll notice that the RHS is a function of 2x+y so seek z_p = f(2x+y).  Since the LHS of the PDEs are independent of both independent variables, this method works.

    Hope it helps.
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    Senior Member bkarpuz's Avatar
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    Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

    Quote Originally Posted by Danny View Post
    The second and third are rather straight forward. For the second, the RHS is only a function of y so sek a particular solution in the form z_p = f(y).  You'll get an ODE for f (and it's linear!) Similarly for the third. For the first, you'll notice that the RHS is a function of 2x+y so seek z_p = f(2x+y).  Since the LHS of the PDEs are independent of both independent variables, this method works.

    Hope it helps.
    For the first one, I also noticed that \frac{\mathrm{d}^{2}\tan(u)}{\mathrm{d}u^{2}}=\big  (\sec(u)\big)^{2}\tan(u) as the left-hand side is involves only 2nd order derivatives.
    Dor the 2nd and the 3rd ones, I rewrote left-hand side as (D_{x}-1)(D_{x}+D_{y}) and D_{x}(D_{x}-1)-D_{y}(D_{y}-1) or (D_{x}+D_{y}-1)(D_{x}-D_{y}) but I could not see any nice things here...

    Is it possible to make change of variables to reduce the equation into first-order equation as in the ordinary case?
    Thanks...
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    Re: Particular solutions of 3 linear PDEs with nonstandart right-hand side

    As you've said, for the first, you could made a change of variables where u = 2x + y and pick another independent variable.

    For the second and third, since you've noticed that the operators factor, you can introduced a new variable as to make your problem two linear first order PDEs. For example, for the second one, if you let

    w = z_x + z_y then w_x - w = \frac{1}{y^2}

    Solve the second, then use this in the first.
    Thanks from bkarpuz
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