HARDMath2

How subjects answered

• GPT-4o incorrect
  $$\boxed{y(x) = a \left(1 - \frac{1}{16} x^8 \right)}$$
• GPT-4.1 incorrect
  $$\boxed{ y(x) \approx \frac{1}{3} \exp\left(\frac{2}{3 x^2}\right) + \frac{2}{3} \cos\left(\frac{2}{3 x^2}\right) }$$
• GPT-o1 incorrect

  A convenient way to see why the solution is exponentially small as x → 0 is to make the change of variables t = 1/x so that y(x) = Y(t) with t → ∞. One finds that, to leading orders in 1/t, the differential equation forces Y(t) to behave like

      -Y‴(t) + Y(t) = 0

  which has characteristic equation λ³ = 1. The three roots are

  1. λ = 1 (which gives a solution that blows up like e^(t) as t → ∞),
  2. λ = -1/2 ± i(√3)/2 (which give solutions that decay like e^(-t/2) times oscillations).

  Discarding the exponentially growing root (since we want y(x) finite as x → 0 ⇔ t → ∞) and allowing for the small additional 1/t-terms in the full ODE shows that the overall amplitude must behave like 1/t² for large t. Translating back to x = 1/t then gives, up to the first two “WKB orders” as x → 0,

  y(x) ≈ x² e^(-1/(2x)) [ A cos( (√3)/(2x) ) + B sin( (√3)/(2x) ) ].

  A single-expression real solution with amplitude and phase can also be written as
  y(x) ≈ C x² e^(-1/(2x)) cos( (√3)/(2x) - δ ).

  Either form is the standard “second-order” WKB approximation (the leading exponential factor together with the x² prefactor) for this third-order problem.

  Accordingly, a succinct final answer (with arbitrary real constants A, B) is:

  $$\boxed{y(x)\;\approx\;x^{2}\,e^{-\frac{1}{2x}}\Bigl[A\,\cos\!\bigl(\tfrac{\sqrt{3}}{2x}\bigr)\;+\;B\,\sin\!\bigl(\tfrac{\sqrt{3}}{2x}\bigr)\Bigr]}.$$

How subjects answered

• GPT-4o incorrect

  \boxed{y(x) = (1)/(2) + (1)/(2) (1-tan^2(x) )}

• GPT-4.1 incorrect
  $$\boxed{ y(x) = \frac{\int_{-1}^x \exp\left( \frac{1}{\epsilon} \int_{-1}^s \tan(t) dt \right) ds}{\int_{-1}^1 \exp\left( \frac{1}{\epsilon} \int_{-1}^s \tan(t) dt \right) ds} }$$
• Gemini 2.5 Flash Thinking incorrect

  The given differential equation is $\epsilon y'' - 2 \tan(x) y' + y = 0$ with boundary conditions $y(-1)=0$ and $y(1)=1$. This is a singular perturbation problem as $\epsilon \to 0$.

  Let $y_0(x)$ be the outer solution, obtained by setting $\epsilon=0$: $-2 \tan(x) y_0' + y_0 = 0$. This is a first-order ODE. For $x \neq 0$, we can write $\frac{y_0'}{y_0} = \frac{1}{2\tan x} = \frac{\cos x}{2\sin x}$. Integrating gives $\ln|y_0| = \frac{1}{2}\ln|\sin x| + C_0$, so $y_0(x) = C\sqrt{|\sin x|}$.

  The coefficient $a(x) = -2\tan x$ is $0$ at $x=0$. For $x \in [-1, 0)$, $\tan x < 0$, so $a(x) > 0$. Boundary layers are typically at the left end of such an interval. For $x \in (0, 1]$, $\tan x > 0$, so $a(x) < 0$. Boundary layers are typically at the right end of such an interval.

  We analyze the problem by considering two approaches for the outer solution:

  1. The outer solution $y_0(x)=0$. In this case, $y_0(-1)=0$ satisfies the left boundary condition. The right boundary condition $y(1)=1$ is not satisfied by $y_0(1)=0$. So, a boundary layer is needed at $x=1$. The stretched coordinate at $x=1$ is $X = (1-x)/\epsilon$. The original equation becomes, approximately, $Y''(X) - 2\tan(1) (-1/\epsilon) Y'(X) \approx 0$, which simplifies to $Y''(X) + (2\tan 1)Y'(X) \approx 0$. The solution that decays as $X \to \infty$ (i.e. away from $x=1$) is $Y(X) = C_1 e^{-2\tan(1)X} + C_2$. Matching with the outer solution $y_0(1)=0$: $\lim_{X\to\infty} Y(X) = 0 \implies C_2=0$. Applying the boundary condition $y(1)=1$, which is $Y(0)=1$: $C_1=1$. So the boundary layer solution at $x=1$ is $Y_R(X) = e^{-2\tan(1)X}$. The composite solution is $y(x) \approx y_0(x) + Y_R((1-x)/\epsilon) - \lim_{X\to\infty} Y_R(X) = 0 + e^{-2\tan(1)(1-x)/\epsilon} - 0$. So, $y(x) \approx e^{-2\tan(1)(1-x)/\epsilon}$. This solution satisfies $y(1)=1$. At $x=-1$, $y(-1) \approx e^{-4\tan(1)/\epsilon}$. Since $\tan(1) > 0$, this is exponentially small, so $y(-1) \approx 0$ is satisfied. This solution is uniformly valid if the true solution is indeed exponentially small for $x$ not close to $1$. The vanishing of $a(x)$ at $x=0$ can sometimes lead to more complex behavior, but this exponential form is a common type of leading-order uniform approximation.

  2. Piecewise outer solution based on characteristics. The characteristics of the outer equation point from the boundaries $x=\pm 1$ towards $x=0$ (since $a(x)/x = -2\tan x/x < 0$ near $x=0$). So $y_0(x)$ for $x<0$ is determined by $y(-1)=0$: $C_L \sqrt{-\sin(-1)} = C_L \sqrt{\sin 1} = 0 \implies C_L=0$. So $y_0(x)=0$ for $x \in [-1,0]$. And $y_0(x)$ for $x>0$ is determined by $y(1)=1$: $C_R \sqrt{\sin 1} = 1 \implies C_R=1/\sqrt{\sin 1}$. So $y_0(x)=\sqrt{\sin x/\sin 1}$ for $x \in (0,1]$. This combined outer solution y_0(x) = $$\begin{cases} 0 & x \in [-1,0] \\ \sqrt{\frac{\sin x}{\sin 1}} & x \in (0,1] \end{cases}$$$ is continuous and satisfies both boundary conditions. However, y_0'(0^+) = \lim_{x\to 0^+} \frac{\cos x}{2\sqrt{\sin 1 \sin x}} …

How subjects answered

• Gemini 2.5 Flash Thinking correct

  We are looking for a travelling wave solution of the form $u(x,t) = f(\xi)$, where $\xi = x-ct$. Substituting this into the PDE $u_{tt} - u_{xx} - 6 (u^2)_{xx} - u_{xxxx}=0$: $c^2 f''(\xi) - f''(\xi) - 6 (f(\xi)^2)'' - f''''(\xi) = 0$ $(c^2-1)f'' - 6(f^2)'' - f'''' = 0$. We seek a localized solution, so we assume $f(\xi)$ and its derivatives vanish as $|\xi| \to \infty$. Integrating with respect to $\xi$: $(c^2-1)f' - 6(f^2)' - f''' = K_1$. Since $f, f', f''', (f^2)' = 2ff'$ all go to zero as $|\xi| \to \infty$, the integration constant $K_1=0$. So, $(c^2-1)f' - 12ff' - f''' = 0$. Integrating again with respect to $\xi$: $\int ((c^2-1) - 12f) f' d\xi - f'' = K_2$. This gives $(c^2-1)f - 6f^2 - f'' = K_2$. Again, since $f, f'' \to 0$ as $|\xi| \to \infty$, the integration constant $K_2=0$. Thus, we obtain the ordinary differential equation: $f'' = (c^2-1)f - 6f^2$.

  We look for a solution of the form $f(\xi) = A_0 \text{sech}^2(k\xi)$. The derivatives are: $f'(\xi) = -2kA_0 \text{sech}^2(k\xi)\tanh(k\xi)$. $f''(\xi) = -2kA_0 [-2k\text{sech}^2(k\xi)\tanh^2(k\xi) + k\text{sech}^4(k\xi)]$ $f''(\xi) = 4k^2A_0 \text{sech}^2(k\xi)\tanh^2(k\xi) - 2k^2A_0 \text{sech}^4(k\xi)$. Using $\tanh^2(z) = 1-\text{sech}^2(z)$: $f''(\xi) = 4k^2A_0 \text{sech}^2(k\xi)(1-\text{sech}^2(k\xi)) - 2k^2A_0 \text{sech}^4(k\xi)$ $f''(\xi) = 4k^2A_0 \text{sech}^2(k\xi) - 4k^2A_0 \text{sech}^4(k\xi) - 2k^2A_0 \text{sech}^4(k\xi)$ $f''(\xi) = 4k^2A_0 \text{sech}^2(k\xi) - 6k^2A_0 \text{sech}^4(k\xi)$. Substitute this into the ODE $f'' = (c^2-1)f - 6f^2$: $4k^2A_0 \text{sech}^2(k\xi) - 6k^2A_0 \text{sech}^4(k\xi) = (c^2-1)A_0 \text{sech}^2(k\xi) - 6A_0^2 \text{sech}^4(k\xi)$. Comparing the coefficients of $\text{sech}^2(k\xi)$ and $\text{sech}^4(k\xi)$:

  1. $4k^2A_0 = (c^2-1)A_0 \implies 4k^2 = c^2-1$ (assuming $A_0 \neq 0$).
  2. $-6k^2A_0 = -6A_0^2 \implies k^2 = A_0$ (assuming $A_0 \neq 0$).

  We are given the condition $u(0,0) = \frac{1}{2}$. For $f(\xi) = A_0 \text{sech}^2(k\xi)$, the maximum value is $A_0$ at $\xi=0$. The condition $u(0,0)=\frac{1}{2}$ implies $f(0) = A_0 = \frac{1}{2}$. Using $A_0 = k^2$, we get $k^2 = \frac{1}{2}$. We can choose $k = \frac{1}{\sqrt{2}}$ (the sign of $k$ can be absorbed in $\xi$, as $\text{sech}^2(-z)=\text{sech}^2(z)$). Using $4k^2 = c^2-1$, we get $4\left(\frac{1}{2}\right) = c^2-1 \implies 2 = c^2-1 \implies c^2=3$. So $c = \pm\sqrt{3}$. Thus, the travelling wave solutions are $u(x,t) = \frac{1}{2} \text{sech}^2\left(\frac{x \mp \sqrt{3}t}{\sqrt{2}}\right)$. We can write these two solutions using $\operatorname{sech}(z) = 1/\cosh(z)$.

  $\boxed{\frac{1}{2} \operatorname{sech}^2\left(\frac{x-\sqrt{3}t}{\sqrt{2}}\right); \frac{1}{2} \operatorname{sech}^2\left(\frac{x+\sqrt{3}t}{\sqrt{2}}\right)}$

• Gemini 2.5 Pro Preview correct

  Let the given partial differential equation be $u_{tt} - u_{xx} - 6 (u^2)_{xx} - u_{xxxx}=0$ We seek a travelling wave solution of the form $u(x,t) = f(\xi)$, where $\xi = x-ct$ for some wave speed $c$. The partial derivatives are: $u_t = -c f'(\xi)$ $u_{tt} = c^2 f''(\xi)$ $u_x = f'(\xi)$ $u_{xx} = f''(\xi)$ $u_{xxxx} = f''''(\xi)$ Also, $(u^2)_{xx} = (f^2)''$. Substituting these into the PDE yields an ODE for $f(\xi)$: $c^2 f'' - f'' - 6 (f^2)'' - f'''' = 0$ $(c^2-1)f'' - 6(f^2)'' - f'''' = 0$ For a localized solution (soliton), we assume that $f$ and its derivatives vanish as $|\xi| \to \infty$. Integrating the equation twice with respect to $\xi$ and using these boundary conditions (so the integration constants are zero), we get: $(c^2-1)f - 6f^2 - f'' = 0$ Rearranging, we have: $f'' = (c^2-1)f - 6f^2$ Let $A = c^2-1$ and $B=6$. The equation is $f'' = Af - Bf^2$. To solve this, we multiply by $f'$: $f'f'' = A f f' - B f^2 f'$ Integrating with respect to $\xi$: $\frac{1}{2}(f')^2 = \frac{A}{2}f^2 - \frac{B}{3}f^3 + E$ Using the boundary conditions $f \to 0$ and $f' \to 0$ as $|\xi| \to \infty$, we find $E=0$. So, $(f')^2 = Af^2 - \frac{2B}{3}f^3 = f^2\left(A - \frac{2B}{3}f\right)$. We are given that $u(0,0) = 1/2$ is the maximum value. For $u(x,t)=f(x-ct)$, this means $f(0)=1/2$ is the maximum value. At a maximum, $f'(0)=0$. Since $f(0)=1/2 \ne 0$, we must have: $A - \frac{2B}{3}f(0) = 0$ Substitute $f(0)=1/2$, $A=c^2-1$, and $B=6$: $(c^2-1) - \frac{2(6)}{3}\left(\frac{1}{2}\right) = 0$ $(c^2-1) - 4\left(\frac{1}{2}\right) = 0$ $(c^2-1) - 2 = 0$ So, $c^2-1 = 2$, which implies $c^2=3$, and $c=\pm\sqrt{3}$. The ODE for $f$ is now $f'' = 2f - 6f^2$. The standard solution for $f'' = Af - Bf^2$ that is localized is $f(\xi) = P \mathrm{sech}^2(Q\xi)$. The maximum value is $P = f(0)$. So $P=1/2$. Let's substitute $f(\xi) = P \mathrm{sech}^2(Q\xi)$ into $f'' = Af - Bf^2$: $f'(\xi) = -2PQ \mathrm{sech}^2(Q\xi)\tanh(Q\xi)$. $f''(\xi) = -2PQ [-2Q \mathrm{sech}^2(Q\xi)\tanh^2(Q\xi) + Q \mathrm{sech}^4(Q\xi)]$. Using $\tanh^2(Q\xi) = 1-\mathrm{sech}^2(Q\xi)$: $f''(\xi) = -2PQ^2 [-2 \mathrm{sech}^2(Q\xi)(1-\mathrm{sech}^2(Q\xi)) + \mathrm{sech}^4(Q\xi)] = -2PQ^2 [-2 \mathrm{sech}^2(Q\xi) + 3 \mathrm{sech}^4(Q\xi)]$. $f''(\xi) = 4PQ^2 \mathrm{sech}^2(Q\xi) - 6PQ^2 \mathrm{sech}^4(Q\xi)$. Comparing this with $Af - Bf^2 = AP \mathrm{sech}^2(Q\xi) - BP^2 \mathrm{sech}^4(Q\xi)$: Equating coefficients of $\mathrm{sech}^2(Q\xi)$: $4PQ^2 = AP \implies 4Q^2 = A$. Equating coefficients of $\mathrm{sech}^4(Q\xi)$: $6PQ^2 = BP^2 \implies 6Q^2 = BP$. We have $A=c^2-1=2$, $B=6$, and $P=1/2$. From $4Q^2=A$: $4Q^2=2 \implies Q^2=1/2 \implies Q = 1/\sqrt{2} = \sqrt{2}/2$. (The other equation $6Q^2=BP$ gives $6(1/2) = 6(1/2) \implies 3=3$, which is consistent.) So the solution for $f(\xi)$ is: $f(\xi) = \frac{1}{2} \mathrm{sech}^2\left(\frac{\xi}{\sqrt{2}}\right)$ Since $\xi=x-ct$ and $c=\pm\sqrt{3}$, the travelling wave solutions are:

  $$$$
• GPT-o3 incorrect

  \boxed{\dfrac12 sech^2 ((x-\sqrt3 t)/(\sqrt2) ); \dfrac12 sech^2 ((x+\sqrt3 t)/(\sqrt2) )}