# Is it possible to find a non-zero solution of an ODE?

Is it possible to solve the 4th-order ode analytically?

op[r_] = (D[#, {r, 2}] - 1/r*D[#, r]) &;

DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0, (f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. r -> (1 + b)) == 0, (f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b)) == 0}, f[r], r]


I am using v11 in which DSolve returns a trivial solution f[r] -> 0 only.

Any suggestion is welcome. Thank you!

For general values of $$b$$ and $$c$$, the trivial solution is indeed the only solution. However, there are special values of $$b$$ and $$c$$ which can yield non-trivial solutions.

To find these, we can start by telling Mathematica to find the general solution to the ODE:

soln = DSolve[{op[r][op[r][f[r]]] == 0}, f, r]

(* {f -> Function[{r}, (r^2 C)/2 - (r^2 C)/4 + (r^4 C)/4 + C + 1/2 r^2 C Log[r]]} *)


We can then look at the boundary conditions and see what they imply about the coefficients C[i]. We do this by storing the vanishing quantities in a list BCs; the actual equations are then BCs == 0.

BCs = { f[b],
f'[b],
(f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. r -> (1 + b)),
(f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b))};
quants = BCs /. First[soln]


The quantities in quants are four linear combinations of the coefficients C, C, C, and C, all of which must vanish. We can think of this set of simultaneous equations as the result of a matrix $$M$$ multiplying the vector $$\vec{v} = \{C_1, C_2, C_3, C_4\}$$. From basic results in linear algebra, we know that the only way for there to be a non-trivial solution for $$\vec{v}$$ is for the matrix $$M$$ to have a non-zero determinant. So we construct this matrix and take its determinant:

mat = Outer[Coefficient, quants, {C, C, C, C}];
Simplify[Det[mat]]

(* -((b (1 + b) (-3 + 2 b + 4 c + 2 (1 + b)^2 Log[b] - 2 (1 + b)^2 Log[1 + b]))/(-1 + c)) *)


This implies that the trivial solution is the only solution to the ODE unless this last quantity (in terms of the constants $$b$$ and $$c$$) is zero, which occurs when $$b = 0$$, $$b = -1$$, or $$c = \frac{3 - 2b + 2 (1+b)^2 \ln( (1+b)/b )}{4}.$$

• thank you very much, but the question is actually not solved completely... I tried csoln = Solve[Det[mat] == 0, c] // Simplify and substituted the special c in the original system but the solution still includes one unknown constant. Please see DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0, (f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. {r -> (1 + b), csoln[[1, 1]]}) == 0, (f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b)) == 0}, f[r], r], it also gives a warning _Unable to resolve some of the arbitrary constants _ Jan 20 at 9:20
• Yes, there will be unknown constants in the solution. They should correspond to the overall normalization of the solution; if $f(x)$ is a solution then so is $\alpha f(x)$ for any $\alpha$, because the equation is linear. Jan 20 at 12:13
• Prof. Seifert thanks a lot! Your comment is plausible. But with Det[mat]==0 we should have non-zero solution for C[i], however, when substituting the eigenvalue c back to mat, which represents vanishing quantities for the b.c.s, I got zero solution... Please try csoln = Solve[Det[mat] == 0, c] and LinearSolve[mat/.First[csoln], {0, 0, 0, 0}] Jan 21 at 2:41

A somewhat different way is as follows.

s = DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0}, f, r]


{{f -> Function[{r}, (1/( 4 b^2))(-b^4 C + 2 b^2 r^2 C - r^4 C + b^4 C - b^2 r^2 C - b^4 C Log[b] - r^4 C Log[b] + 2 b^2 r^2 C Log[r])]}}

Resolve[Exists[{C, C},Simplify[f'[1 + b]/(1 + b)^2 - f''[1 + b]/(1 + b) + f'''[1 + b] /.
s[]]==0 && Simplify[-(f[1 + b]/(-1 + c)) - f'[1 + b]/(1 + b) + f''[1 + b] /.
s[]]==0&& C^2 + C^2 != 0], Reals]


b > 0 && c == 1/4 (3 - 2 b - 2 Log[b] - 4 b Log[b] - 2 b^2 Log[b] + 2 Log[1 + b] + 4 b Log[1 + b] + 2 b^2 Log[1 + b])

The main difference from Michael Seifert's answer consists in the use of quantifiers instead of linear algebra. This way is more automatical.

• One could also start from DSolve[{op[r][op[r][f[r]]] == 0, f, r]. Jan 16 at 17:12