# Issuing with ParametricNDSolveValue

I am having a problem with ParametricNDSolveValue[] . let me start with a simple example. Considered following example with constant parameter $a$.

pfun = ParametricNDSolveValue[{y'[x] == y[x] Cos[x + y[x]],
y == a}, y, {x, 0, 30}, {a}]


Plot the solutions for several different values of the parameter gives:

Plot[Evaluate[Table[pfun[a][t], {a, -1, 1, .1}]], {t, 0, 1},
PlotRange -> All] Now, consider the following case:

pfun = ParametricNDSolveValue[{y'[x] == y[x] Cos[x + y[x]],
y[a] == 1}, y, {x, 0, 30}, {a}]


and trying to plot the function I end up with following error "Cannot find starting value for the variable".

Plot[Evaluate[Table[pfun[a][t], {a, -1, 1, .1}]], {t, 0, 1},
PlotRange -> All]

ParametricNDSolveValue::ndsv: Cannot find starting value for the variable y. >>
ParametricNDSolveValue::ndsv: Cannot find starting value for the variable y. >>
ParametricNDSolveValue::ndsv: Cannot find starting value for the variable y. >>
General::stop: Further output of ParametricNDSolveValue::ndsv will be suppressed during this calculation. >>


I will appreciate for any solution.

## 1 Answer

As you have noted placing the parameter on the left hand side of a boundary condition:

y[b] == 1


does not work. You are forced to keep the parameter on the right hand side.

y == a


What can be done is to keep the parameter on the RHS and then use numerical methods to determine the value of b for a particular a parameter (in other words, seek a relationship between a and b).

First solve the ParametricNDSolveValue using a as the parameter.

pfunRHS = ParametricNDSolveValue[
{
y'[x] == y[x] Cos[x + y[x]],
y == a
},
y,
{x, 0, 30},
{a}
]


Next, define a function that produces a number when given a parameter value and an x coordinate.

y[x_?NumericQ, a_?NumericQ] := Evaluate[pfunRHS][a][x]


Here is a plot in the interval zero to one for various a values.

Plot[Evaluate[Table[y[t, a], {a, -1, 1, .1}]], {t, 0, 1},
PlotRange -> All] What we have to do is select a value for the a parameter and than solve for the corresponding b value where y[b,a]=1. Graphically this is where the black line intersects equal contours of a.

Show[
Plot[Evaluate[Table[y[t, a],
{a, 0.5, 1, .05}]], {t, 0, 1},
PlotRange -> All],
ListLinePlot[{{0, 1}, {1.1, 1}},
PlotStyle -> Black]
] Note that below a certain a parameter threshold (approximately 0.85) there is no solution.

Given an a we can use FindRoot to determine the corresponding b. Here are two values.

FindRoot[
{
a - 1,
y[b, a] - 1
},
{{a, 1}, {b, 0.8}}
]
(* {a -> 1., b -> 0.978505} *)

FindRoot[
{a - 0.9,
y[b, a] - 1
},
{{a, 1}, {b, 0.8}}
]
(* {a -> 0.9, b -> 0.855604} *)


If one tries a = 0.8 you get an error.

About the lowest one can go is 0.85 for a.

FindRoot[
{a - 0.85,
y[b, a] - 1
},
{{a, 0.85}, {b, 0.8}}
]
(* {a -> 0.85, b -> 0.735357} *)


Superimpose the three points on the plot

Show[
Plot[Evaluate[Table[y[t, a], {a, 0.5, 1, .05}]], {t, 0, 1},
PlotRange -> All],
ListLinePlot[{{0, 1}, {1.1, 1}}, PlotStyle -> Black],
ListPlot[{{0.975, 1}, {0.8556, 1}, {0.735357, 1}},
PlotStyle -> {PointSize -> Large, Red}]
] Create a function to retrieve the b value given an a value.

getB[aStart_?NumericQ] := Module[
{
sol
},
sol = FindRoot[
{a - aStart,
y[b, a] - 1
},
{{a, aStart}, {b, 0.8}}
];
sol[[2, 2]]
]


Plot b vs a

Show[
Plot[getB[a], {a, 0.85, 5},
PlotStyle -> Black,
PlotRange -> {{0, 4.1}, {0, 2}}
],
ListPlot[{
{0.85, getB[0.85]}, {1, getB},
{2, getB}, {3, getB}, {4, getB}
},
PlotStyle -> {PointSize -> Large, Red}]
] I didn't want to introduce a diversion but the range of a that works has an upper limit as well, around 4.

It may be, depending upon your actual problem, you may be able to develop a functional relationship between a and b from the numerical results.

• LaVigen Thanks a lot for you professional answer. – Emad Jun 21 '16 at 23:02