2.1.78 \(u_x u_y = u\) with \(u(x,0)=0,u(0,y)=0\)

problem number 78

Added May 27, 2019.

From UMN Math 5587 HW2, Fall 2016, problem 5(b).

Solve for \(u(x,y)\) \begin {align*} u_x u_y = u \end {align*}

With \(u(x,0)=0,u(0,y)=0\).

Mathematica


Failed

Maple


\[u \left (x , y\right ) = x y\]

Hand solution

Solve \(u_{x}u_{y}=u\) with \(u\left ( x,0\right ) =0,u\left ( 0,y\right ) =0\).

Let \(u=X\left ( x\right ) Y\left ( y\right ) \). The pde becomes \(\left ( X^{\prime }Y\right ) \left ( Y^{\prime }X\right ) =XY\) or \(X^{\prime }Y^{\prime }=1\) or \(X^{\prime }=\frac {1}{Y^{\prime }}\). Hence both sides are constant, say \(\lambda \). Therefore we have 2 differential equations to solve\[ X^{\prime }=\lambda \] And\[ \frac {1}{Y^{\prime }}=\lambda \] The solution to the first one is \(X\left ( x\right ) =\lambda x+C_{1}\). The solution to the second equation is \(Y\left ( y\right ) =\frac {1}{\lambda }y+C_{2}\).

Since \(u=X\left ( x\right ) Y\left ( y\right ) \) then the solution is\begin {align*} u\left ( x,y\right ) & =\left ( \lambda x+C_{1}\right ) \left ( \frac {1}{\lambda }y+C_{2}\right ) \\ & =xy+C_{2}\lambda x+\frac {C_{1}}{\lambda }y+C_{1}C_{2} \end {align*}

Using the condition \(u\left ( x,0\right ) =0\) gives\[ 0=C_{2}\lambda x+C_{1}C_{2}\] Differentiating gives \[ 0=C_{2}\lambda \] But \(\lambda \neq 0\) for nontrivial solution, hence \(C_{2}=0\). Therefore (1) reduces to\[ u\left ( x,y\right ) =xy+\frac {C_{1}}{\lambda }y \] Using the condition \(u\left ( 0,y\right ) =0\) gives\[ 0=\frac {C_{1}}{\lambda }y \] Hence \(C_{1}=0\). Therefore the solution becomes\[ u\left ( x,y\right ) =xy \]