Runge-Kutta 4 failure?

Say we want to solve numerically $y'(x) = f(x) \cdot y$, with $y_0 = y(x=0) = 0$ and applying RK4 method with step $dx = h$:

\begin{align} k_1 &= f(0) \cdot y(0) \cdot h = 0\\ k_2 &= f(0+h/2) \cdot (y_0 + k_1/2) \cdot h = f(h/2) \cdot 0 = 0\\ k_3 &= f(h/2) \cdot (y_0 + k_2/2) \cdot h = 0\\ k_4 &= f(c+h) \cdot (y_0 + k_3) \cdot h = 0 \end{align}

Hence: $y(h) = 0 + (k_1+2 k_2+2 k_3+k_3)/6 = 0...$

In the end: $y_i = y(0+i\cdot h) = 0$, no matter what $f(x)$ looks like!

Any suggestions on this?

• Why would you make $y'(x)$ depend on $y$ that way? Since $y(0)=0$ and $y'(x) = f(x)\cdot y$, you have $y'(0)=0$ (because $y=0$) no matter what $f(x)$ is. Are you sure that's the equation you're trying to solve? – David K Feb 10 '16 at 3:44
• thanks, @DavidK! – Chip Feb 10 '16 at 5:02
• I realized that in the form I wrote it, the unique solution is zero as @runaround pointed out. However, the long story is the following: in solving the hydrogen radial Schr$\:o$dinger equation (with r the radial coordinate) for angular momentum $L=1$ and the modified radial wave function $P(r) = rR(r), P(r)$ satisfies: $P"(r) = 2(-1/r-E+1/r^2)$, where $E$ is the eigen-energy to be found. One can re-write this as first order system of equations: \begin{align} dP/dr(r) &= Q(r) \\ dQ/dr(r)&= 2(-1/r-E+1/r^2) P(r)\\ \end{align}, $P(0)=0$ and $Q(0)=0$. Hereby, RK4 will output zero for all r > 0. – Chip Feb 10 '16 at 5:15
• Looks like a two-parameter version of the same thing. Maybe the fault is in the initial conditions. Wolfram Alpha gives as a solution $P(r) = c_2 r+c_1-e r^2-2 r log(r)-2 log(r)$, which is not defined at $r=0$. It seems the thing to do is (a) start this at a positive $r=r_0$ and (b) make at least one of $P(r_0)$ and $Q(r_0)$ non-zero. – David K Feb 10 '16 at 6:42

y = 0 is the solution of the equation with $y(0) = 0$. You get exact solution for this.