Existence and uniqueness of solution to Cauchy problem for $xu_x + u_y+ yu =0$

Question

Consider the PDE

$$xu_x+u_y+yu=0$$

a) Solve the IVP for $ u(x,0)=g(x)$. In which region the solution exists and is unique?

b) Solve the IVP for $u(0,y)=h(y)$. In which region the solution exists and is unique?

My attempt: as we can check here the solution has the form $$u(x,y) =K(\ln|x|-y)e^{-y^2/2}$$ Or $$ u(x,y)= K(xe^{-y})e^{-y^2/2}$$

a) clearly it is not defined for x=0. If $x \neq 0$, $$u(x,y)=g(xe^{-y})e^{-y^2/2}$$ And the solution is unique

b) it looks like there is no solution satisfying this (we should have $K(0)=h(y)$), but I dont know how to argue this formally.

Answer 1

The characteristics $(x(t), y(t), z(t))$ of the system, where $z(t) := u(x(t), y(t))$ , are defined by the ODEs \begin{align*} \dot x &= x & \dot y &= 1 & \dot z &= -zy\\ x(0) &=x_0 & y(0)&=y_0 & z(0) &= u(x_0,y_0) = z_0, \end{align*} where $p_0 = (x_0, y_0)$ is a point in the domain with a known value for $u$, which we call $z_0$. The solution to this system is \begin{align*} x &= x_0 e^t & y &= t+y_0 & z &= z_0e^{-(t^2/2+y_0t)}. \end{align*} Moreover, these solutions are unique by the uniqueness of solutions to systems of ODEs with prescribed initial values.

Notice that once you fix the value of $u$ at some point $p_0 = (x_0,y_0)$ of the domain, the values of $u$ along the characteristic passing through $p_0$ are immediately determined by the solutions to these ODEs. Consequently, it is not possible to prescribe arbitrary values of $u$ at more than one point of a characteristic.

In particular, the characteristic passing through $p_0=(0,0)$ is \begin{align*} x &= 0 & y &= t & z &= z_0e^{-t^2/2}, \end{align*} therefore it is not possible to prescribe values on more than one point of the line $(0,t)$.

Conclusion:

Part a) Has a unique solution defined for all points $x,y$ given by the formula you found: $u(x,y) = g(xe^{-y})e^{-y^2/2}$. This is obtained from the general solution to the ODEs by setting $y_0=0, \, z_0=g(x_0)$ and writing $x_0$ in terms of $x$.

Part b) Has no solutions, except for the case where $h$ agrees with the system of ODEs, in which there are an infinite number of solutions; one for each possible $g(x)$ defined in the $x$-axis agreeing with $h$ in the origin ($g(0)=h(0)$), that is

$$h(y)=h(0)e^{-y^2/2}.$$

Answer 2

let us define the characteristic curve $C$, parametrised bty $t,$ through $(a, b)$ by $$ \frac{dx}{dt} = x, \frac{dy}{dt} = 1 \text{ subject to initial conditions } x = a, y = b \text{ at } t = 0.$$ the solution is $$x = ae^t, y = t+b. $$ along $C,$ we have $$\frac{du}{dt} = -yu = -(t+b)u, u = f(a,b).\tag 1 $$
the solution to $(1)$ is $$u = f(a,b)e^{-tb - t^2/2}.$$

we can specialize for the cases $$f(a, 0) = g(a), f(0,b)=h(b). $$ i will come back to it later if i can.

Answer 3

I was curious to try another method instead of method characteristics for obtaining solutions. I tried the method of separation of variables. Assuming a separable solution $u(x,y) = X(x)Y(y)$ and putting into the PDE and dividing by $X(x)Y(y)$ you get

$$x{{X'} \over X} + {{Y'} \over Y} + y = 0$$

where this can hold only when

$$\left\{ \matrix{ x{{X'} \over X} = \mu \hfill \cr \mu + {{Y'} \over Y} + y = 0 \hfill \cr} \right.$$

and hence you can find $X(x)$ and $Y(y)$ by solving these two ODEs which results in

$$\left\{ \matrix{ X(x) = A{x^\mu } \hfill \cr Y(y) = B{e^{ - \left( {{{{y^2} + 2\mu y} \over 2}} \right)}} \hfill \cr} \right.$$

and

$$u(x,y) = C{x^\mu }{e^{ - \left( {{{{y^2} + 2\mu y} \over 2}} \right)}}$$

which also can be written as

$$u(x,y) = C{\left( {x{e^{ - y}}} \right)^\mu }{e^{ - {{{y^2}} \over 2}}}$$

and hence a special case of your solution. Now we get to satisfaction of the IC's.

Case a. $u(x,0)=g(x)$

Let us try our solution and see what we get. Substitution into the initial condition gives

$$C{x^\mu } = g(x)$$

What this recalls you? That's right! The Taylor series. This equation motivates us to define

$$\left\{ \matrix{ {\mu _n} = n \hfill \cr {C_n} = {{{g^{(n)}}(0)} \over {n!}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n = 0,1,2,...$$

Then by superposing our solutions we finally get

$$\eqalign{ & \sum\limits_{n = 0}^\infty {{1 \over {n!}}{{\left. {{{{d^n}g\left( x \right)} \over {d{x^n}}}} \right|}_{x = 0}}} {e^{ - \left( {{{{y^2} + 2ny} \over 2}} \right)}}{x^n} \cr & = \left( {\sum\limits_{n = 0}^\infty {{1 \over {n!}}\left( {{{\left. {{{{d^n}g\left( x \right)} \over {d{x^n}}}} \right|}_{x = 0}}{e^{ - ny}}} \right){x^n}} } \right){e^{ - {{{y^2}} \over 2}}} \cr & = \left( {\sum\limits_{n = 0}^\infty {{1 \over {n!}}\left( {{{\left. {{{{d^n}g\left( {x{e^{ - y}}} \right)} \over {d{x^n}}}} \right|}_{x = 0}}} \right){x^n}} } \right){e^{ - {{{y^2}} \over 2}}} \cr & = g\left( {x{e^{ - y}}} \right){e^{ - {{{y^2}} \over 2}}} \cr} $$

where I used the chain rule in second equality. This completes the process.

Case b. $u(0,y)=h(y)$

When we substitute our solution in the initial condition we get

$$0=h(y)$$

and it turns out that we fail to satisfy the IC in this case. Existence and uniqueness in this case must also be investigated. I don't have any Idea yet.