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$$2x^2\phi_{xx}-5xy\phi_{xy}+2y^2\phi_{yy}+2x\phi_x+2y\phi_y=0$$

I know how to get general solution which is $$\phi(x,y)=F(x^2y)+G(xy^2)$$

now my question is if $$\phi(1,y)=10+y-2y^4$$ and $$\phi_x(1,y)=6+2y-4y^4$$ show that the unique solution is given by $$\phi(x,y)=10-2x^2y^4+yx^2+6lnx$$

General solution is not a problem.

Approach 1 : I couldn't get the particular solution directly from the initial conditions.

Approach 2 : I can prove the given solution indeed satisfies the PDE and the initial conditions but then i cant prove the uniqueness of it.

NOTE: Existence and uniqueness of PDE is not covered in the module.

I would be grateful for any help Thanks. couldn't find similar question on MSE

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"Approach 1" : The solution can be obtained directly from the initial conditions on a straightforward calculus which introduces no unexpected arbitrary parameter :

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  • $\begingroup$ thanks for answer however $F'(y)$ in equation 5 is same as in equation 4 for you to subtract i feel like $F'$ is derivation w.r.t $y$ in equation 5 but in 2 & 4 it is w.r.t $x$. $\endgroup$ – MRK Aug 28 '16 at 18:44
  • $\begingroup$ @MRK : It seems that you don't understand. Try to use dummy variables for example $F(z)$ and $F'(z)=\frac{dF}{dz}$ Sometimes $z=x^2y$ and one can differentiate with respect to $x$ or to $y$. Sometimes $z=y$ and one can differentiate only with respect to $y$ All these with the same fonction $F$ . If you are not well experienced with this kind of calculus, this will be not easy for you to understand. Moreover, I am on the verge to leave and I will not be available all next week. Sorry, I have not enough time for more explanation. I hope that other people will help you. $\endgroup$ – JJacquelin Aug 28 '16 at 21:11

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