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Solve the IVP

$$u_{xx} + y u_{xy} = 0 \text{ for } u(0,y)=y^3$$

Is the solution unique?

My attempt:

Let $u_x = v$. Substituting on equation:

$$v_x + y v_y = 0$$

We solve this using the characteristics method: $x=x(s), \; y=y(s),\; w(s)=v(x,y)$

$$\Rightarrow \frac{dx}{ds}=1, \; \frac{dy}{ds}=y, \; \frac{dw}{ds}=0$$

Doing all the calculation,

$$v(x,y) = f(y e^{-x})$$ where $f: R \rightarrow R$ is an arbitrary function. Now,

$$u_x = v \Rightarrow u(x,y) = \int_?^x f(y e^{-\eta}) d \eta + C(y) $$

My doubt is:

  1. What are the integration limits? Usually, I use 0 to $x$ but I don't know why.
  2. Using the initial conditions: $$u(0,y) = \int_?^0 f(y e^{-\eta}) d \eta + C(y) = y^3 $$. If I use the limits that I am used to, we must have $C(y)=y^3$ and $f$ still arbitrary (and integrable). But I don't know if it is rigorous. In this case, the solution won't be unique (there are many for any $f$ integrable)

Thanks in advance!

Giiovanna
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3 Answers3

2

Hint: let $$u=XY$$ so $$u_{xx}=X''Y$$ $$u_{xy}=X'Y'$$ then you get $$X''Y+yX'Y'=0$$ divide by $$X'Y$$

hence $$\frac{X''}{X'}+y\frac{Y'}{Y}=0$$

now you can let $$\frac{X''}{X'}=\lambda $$ and $$y\frac{Y'}{Y}=-\lambda $$ you can solve this equation by euler method as follow $$Y=c_1y^m$$ $$Y'=c_1my^{m-1}$$ substitute to get $$y(c_1my^{m-1})=-\lambda (c_1y^m)$$ $$m=-\lambda $$ $$Y=c_1y^{-\lambda }$$ now we will solve the $\frac{X''}{X'}=\lambda$ The general solution is $$X=c_2e^{\lambda x}-\frac{c_3}{\lambda}$$

hence $$u=(c_1y^{-\lambda})(c_2e^{\lambda x}-\frac{c_3}{\lambda})$$ when the $u(0,y)=y^3$ $$(c_1y^{-\lambda})(c_2-\frac{c_3}{\lambda})=y^3$$ $$\lambda=-3$$ and $$c_1(c_2-\frac{c_3}{\lambda})=1$$ or $$(k_1-\frac{k_2}{\lambda})=1$$

then $$u=(c_1y^{3})(c_2e^{-3 x}+\frac{c_3}{3})$$ or you can write it as $$u=(y^{3})(k_1e^{-3 x}+\frac{k_2}{3})$$

E.H.E
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  • Sorry, I don't understand what you are trying to do and I would be glad if you could point what is wrong with my solution. As you can see, I know how to solve it, except for the integration limits. – Giiovanna Sep 27 '15 at 18:07
  • Thanks for your edit and patience, but I really don't understand your method. – Giiovanna Sep 27 '15 at 18:18
  • This called separation method for partial differential equations – E.H.E Sep 27 '15 at 18:19
  • @Essam The OP is asking specifically for a review of the application of the method of characteristics. ;-) – Mark Viola Sep 27 '15 at 18:27
  • @Essam I never heard it before, but won't you have to integrate two times to find $X$? If so, what are the limits of the last integration, since it will give an arbitrary function? – Giiovanna Sep 27 '15 at 18:28
  • @Giiovanna see the more details – E.H.E Sep 27 '15 at 18:53
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    Thanks for your effort but I need to use the characteristics method :) – Giiovanna Sep 27 '15 at 19:18
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Since you know $u(0,y)$ and $u_x$, you can get $u(x,y)$ by $$ u(x,y) = u(0,y) + \int\limits_{0}^{x}{u_x(\eta,y)\text{ d}\eta} = y^3 + \int\limits_{0}^{x}{f(ye^{-\eta})\text{ d}\eta}. $$ Also, you should double check exactly how "arbitrary" $f$ can be (in particular, it probably should be at least differentiable on $(0,\infty)$).

Joey Zou
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1
  1. Use the fundamental theorem of calculus: $$u(x,y) - u(0,y) = \int_0^xu_x(\eta,y)\,d\eta \Longrightarrow u(x,y) = \int_0^xf(ye^{-\eta})\,d\eta + y^3.$$
  2. You are correct, you don't get uniqueness, indeed for any integrable and differentiable function $f$ you have that $u(x,y)$ defined as above solves the system. This follows by construction but you can easily verify it by noticing that the initial condition is trivially satisfied and that \begin{align} u_{xx}(x,y) = &\ -f'(ye^{-x})ye^{-x} \\ u_{xy}(x,y) = &\ f'(ye^{-x})e^{-x} \end{align}
Giovanni
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