Laplace equation with initial conditions

Question

So I have Laplace equation:

$$ u_{xx}+u_{yy}= 0 $$

and initial conditions

$$ u(0,y)=0, \;\: u_x(0,y)=y $$

And I have to solve it.

My solution:

If we assume that the solution is of the form: $$ u(x,y)=X(x)Y(y) $$

Then follows: $$ u(0,y)=X(0)Y(y)=0; \:\; u_x(0,y)=X'(0)Y(y)=y.$$ Furthermore: $$ u_{xx}(x,y)=X''(x)Y(y); \:\; u_{yy}(x,y)=X(x)Y''(y) $$ From that, we get: $$ \begin{aligned}&\;u_{xx}+u_{yy}=X''(x)Y(y)+X(x)Y''(y)=0 \\ \\\Leftrightarrow& \; X''(x)Y(y)=-X(x)Y''(y) \\\\ \Leftrightarrow& \; -\frac{X(x)}{X''(x)}=\frac{Y(y)}{Y''(y)} \end{aligned}$$

My problem: Every time I try to go further I get: $X(0)=0$, which then implies, that $X'(0)=0$, which furthermore is in contradiction with $X'(0)Y(y)=y$

This is a part where I didn't know where to go further, everything I try to insert doesn't end well and I am not sure, whether my approach was good at all. I would appreciate any kind of help!

Answer 1

From your calculation, we have to solve $$(1)\left\{\begin{aligned}&X''+\lambda X=0\\ &X(0)=0\end{aligned}\right.\qquad\text{and}\qquad (2)\left\{\begin{aligned}&Y''-\lambda Y=0\\ &Y(y)=ky \end{aligned}\right.$$ where $\lambda$ and $k=(X'(0))^{-1}$ are constants.

The nonzero solutions of $(1)$ are

$$X(x)=\left\{\begin{aligned} &c_1\sin(\sqrt{\lambda} x),\quad&&\text{if } \lambda>0\\ &c_1e^{\sqrt{-\lambda} x}-c_1e^{-\sqrt{-\lambda} x},\quad&&\text{if } \lambda<0\\ & c_1 x, \quad&&\text{if } \lambda=0 \end{aligned}\right.\tag{3}$$

with $c_1\neq 0$. From system $(2)$, we have $$Y(y)=ky\quad \Rightarrow \quad Y'(y)=k\quad \Rightarrow \quad Y''(y)=0\quad \Rightarrow \quad \lambda=0.$$

Thus, the only possibility is the third case of $(3)$: $$X(x)=c_1 x,\qquad Y(y)=ky=(X'(0))^{-1}y=\frac{1}{c_1} y.$$ and the desired solution is $$u(x,y)=X(x)Y(y)=(c_1 x)\left(\frac{1}{c_1}y\right)=xy.$$

Answer 2

After a little tinkering with multilinear functions, I found a solution to be $u(x,y)=xy$.

Answer 3

Of course, only inspection is sufficient to find the solution $u(x,y)=xy$ ( through trial and error). I suppose that the question is how analytically ? They are several methods. For example :

It is well known that the general solution of $$u_{xx}+u_{yy}=0\tag 1$$ can be expressed on the form : $$u(x,y)=F(y+ix)+G(y-ix)\tag 2$$ in which $F$ and $G$ are arbitrary complex functions in the most general case.

The conditions considered here are : $$u(0,y)=F(y)+G(y)=0\quad\implies\quad F'(y)+G'(y)=0$$ $$u_x(0,y)=iF'(y)-iG'(y))=y$$ Solving for $F'$ leads to : $$F'(y)=-\frac{i}{2}y$$ Integrating wrt $y$ leads to : $$F(y)=-\frac{i}{4}y^2+c$$ $$G(y)=-F(y)=\frac{i}{4}y^2-c$$ Now the functions $F$ and $G$ are known. We put them into Eq.$(2)$ where the variables are not $y$ but are $(y+ix)$ and $(y-ix)$ respectively. $$u=\left(-\frac{i}{4}(y+ix)^2+c\right)+\left(\frac{i}{4}(y-ix)^2-c\right)$$ Expend and simplify. The result is : $$u(x,y)=xy$$