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Partial Differential.Equation
發問:
4. Solve the equation p : ?u = 0, with u = u(p) only. ( Recall : p = (x^2 + y^2 + z^2.)^(1/2) )5. Find a transformation of the form :U(x, t) = V(x, t)e^(ax?bt),where a and b have to be determined, that would bring or transform thefollowing equation for u(x, t) :Ut ? φUxx ? cUx ? ?U = 0,where φ , c,... 顯示更多 4. Solve the equation p : ?u = 0, with u = u(p) only. ( Recall : p = (x^2 + y^2 + z^2.)^(1/2) ) 5. Find a transformation of the form : U(x, t) = V(x, t)e^(ax?bt), where a and b have to be determined, that would bring or transform the following equation for u(x, t) : Ut ? φUxx ? cUx ? ?U = 0, where φ , c, ? are real nonzero constants, to a diffusion (only) partial differential equation (without sink or source term) for V(x, t).
4. note ?p/?x=x/p, ?p/?y=y/p, ?p/?z=z/p ux=(du/dp)(?p/?x)=(du/dp)(x/p) uxx=(d^2 u /dp^2)(x/p)^2 + (du/dp)(1/p - x^2 /p^3) = (d^2 u /dp^2)(x/p)^2 + (1/p)(du/dp)(1-x^2 / p^2) similarly, uyy=(d^2 u /dp^2)(y/p)^2 + (1/p)(du/dp)(1-y^2 / p^2) uzz=(d^2 u /dp^2)(z/p)^2 + (1/p)(du/dp)(1-z^2 / p^2) Therefore ?u = (d^2 u /dp^2)((x^2 + y^2 + z^2)/p^2) + (1/p)(du/dp)(3-(x^2 + y^2 + z^2)/p^2) =(d^2 u /dp^2) + (2/p)(du/dp) ?u = 0 (d^2 u /dp^2) + (2/p)(du/dp) = 0 d(u')/dp = -2/p u' d(u')/u' = -2dp / p ln (u') = -2ln(p) +c u' = Cp^(-2) u(p) = -Cp^(-1) + B = Ap^(-1)+B where A, B are constants. 5. U(x, t) = V(x, t)e^(ax?bt), Ut = Vt e^(ax?bt) -bV e^(ax?bt) Ux = Vx e^(ax?bt) + aV e^(ax?bt) Uxx = Vxx e^(ax?bt) +2aVx e^(ax?bt) + a^2 V e^(ax?bt) Ut ? φUxx ? cUx ?λU =0 (Vt e^(ax?bt) -bV e^(ax?bt))-φ(Vxx e^(ax?bt) +2aVx e^(ax?bt) + a^2 V e^(ax?bt)) -c( Vx e^(ax?bt) + aV e^(ax?bt)) - λU = 0 e^(ax?bt) (Vt -φVxx -(2aφ+c)Vx -(b+φa^2+ac+λ)V)=0. Vt -φVxx -(2aφ+c)Vx -(b+φa^2+ac+λ)V=0 We need (2aφ+c)=0 and (b+φa^2+ac+λ)=0 Hence a= -c/2φ b=-(φa^2+ac+λ)=-(c^2/4φ - c^2/2φ + λ) = c^2/4φ - λ.
其他解答:
Partial Differential.Equation
發問:
4. Solve the equation p : ?u = 0, with u = u(p) only. ( Recall : p = (x^2 + y^2 + z^2.)^(1/2) )5. Find a transformation of the form :U(x, t) = V(x, t)e^(ax?bt),where a and b have to be determined, that would bring or transform thefollowing equation for u(x, t) :Ut ? φUxx ? cUx ? ?U = 0,where φ , c,... 顯示更多 4. Solve the equation p : ?u = 0, with u = u(p) only. ( Recall : p = (x^2 + y^2 + z^2.)^(1/2) ) 5. Find a transformation of the form : U(x, t) = V(x, t)e^(ax?bt), where a and b have to be determined, that would bring or transform the following equation for u(x, t) : Ut ? φUxx ? cUx ? ?U = 0, where φ , c, ? are real nonzero constants, to a diffusion (only) partial differential equation (without sink or source term) for V(x, t).
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最佳解答:4. note ?p/?x=x/p, ?p/?y=y/p, ?p/?z=z/p ux=(du/dp)(?p/?x)=(du/dp)(x/p) uxx=(d^2 u /dp^2)(x/p)^2 + (du/dp)(1/p - x^2 /p^3) = (d^2 u /dp^2)(x/p)^2 + (1/p)(du/dp)(1-x^2 / p^2) similarly, uyy=(d^2 u /dp^2)(y/p)^2 + (1/p)(du/dp)(1-y^2 / p^2) uzz=(d^2 u /dp^2)(z/p)^2 + (1/p)(du/dp)(1-z^2 / p^2) Therefore ?u = (d^2 u /dp^2)((x^2 + y^2 + z^2)/p^2) + (1/p)(du/dp)(3-(x^2 + y^2 + z^2)/p^2) =(d^2 u /dp^2) + (2/p)(du/dp) ?u = 0 (d^2 u /dp^2) + (2/p)(du/dp) = 0 d(u')/dp = -2/p u' d(u')/u' = -2dp / p ln (u') = -2ln(p) +c u' = Cp^(-2) u(p) = -Cp^(-1) + B = Ap^(-1)+B where A, B are constants. 5. U(x, t) = V(x, t)e^(ax?bt), Ut = Vt e^(ax?bt) -bV e^(ax?bt) Ux = Vx e^(ax?bt) + aV e^(ax?bt) Uxx = Vxx e^(ax?bt) +2aVx e^(ax?bt) + a^2 V e^(ax?bt) Ut ? φUxx ? cUx ?λU =0 (Vt e^(ax?bt) -bV e^(ax?bt))-φ(Vxx e^(ax?bt) +2aVx e^(ax?bt) + a^2 V e^(ax?bt)) -c( Vx e^(ax?bt) + aV e^(ax?bt)) - λU = 0 e^(ax?bt) (Vt -φVxx -(2aφ+c)Vx -(b+φa^2+ac+λ)V)=0. Vt -φVxx -(2aφ+c)Vx -(b+φa^2+ac+λ)V=0 We need (2aφ+c)=0 and (b+φa^2+ac+λ)=0 Hence a= -c/2φ b=-(φa^2+ac+λ)=-(c^2/4φ - c^2/2φ + λ) = c^2/4φ - λ.
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