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chihiroasleaf

  • 3 years ago

Partial Differential Equation Solve the following boundary value problems \[\Large \frac{\partial^{2} u}{\partial x \partial y} (x,y) =3x^{2} , u(x,0) = x^n (n > 0) , u(0,y) = 0 \]

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  1. chihiroasleaf
    • 3 years ago
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    I've tried solve it \[ \Large \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial y} (x,y) \right) = 3x^{2} \] Integrate respect to \(x\) \[ \Large \frac{\partial u}{\partial y} (x,y) = x^3 + f(y) \] Integrate respect to \(y\) \[ \Large u(x,y) = x^{3} y + F(y) + g(x) ; \frac{\partial}{\partial y} F(y) = f(y) \] \[\Large u(x,0) = x^n \implies x^{3} \cdot 0 + F(0) + g(x) = x^{n} \implies F(0) + g(x) = x^n\] \[\Large u(0,y) = 0 \implies 0 \cdot y + F(y) + g(0) = 0 \implies F(y) + g(0) = 0\] \[ \Large u(x,y) = x^3y - g(0) + x^n - F(0) \] \[ \Large u(0,y) = 0 \implies 0 - g(0) + 0 -F(0) = 0 \implies g(0) = - F(0) \] so.., \[ \Large u(x,y) = x^3y + x^n \] is this correct?

  2. AccessDenied
    • 3 years ago
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    While I am not learning this currently, the work seems to be correct, and the answer seems to agree with your initial problem (checking by taking partial derivative).

  3. chihiroasleaf
    • 3 years ago
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    thank you...

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