Solve the boundary value problem (a , l , b constant…

Solve the boundary value problem ($a$, $l$, $b$ constant parameters) $y^{(4)} - a^4y = 0$, with $y(0)=0,\qquad y'(0)=0,\qquad y''(l)=0,\qquad y'''(l)=b$.

Step 1: Characteristic equation and fundamental system

Substituting $y=e^{rx}$: \[r^4-a^4=0 \;\Longrightarrow\; (r^2-a^2)(r^2+a^2)=0 \;\Longrightarrow\; r=\pm a,\ \pm ai.\]

Four distinct roots (assuming $a\neq0$), giving the general solution in hyperbolic/trigonometric form: \[y = A\cosh(ax) + B\sinh(ax) + C\cos(ax) + D\sin(ax). \tag{1}\]

The three derivatives needed are: $y' = a\bigl[A\sinh(ax)+B\cosh(ax)-C\sin(ax)+D\cos(ax)\bigr]$, $y'' = a^2\bigl[A\cosh(ax)+B\sinh(ax)-C\cos(ax)-D\sin(ax)\bigr]$.

\[ y''' = a^3\bigl[A\sinh(ax)+B\cosh(ax)+C\sin(ax)-D\cos(ax)\bigr]. \tag{2} \]

Step 2: Conditions at $x=0$

Since $\cosh0=\cos0=1$ and $\sinh0=\sin0=0$: \[y(0)=A+C=0 \;\Longrightarrow\; C=-A, \qquad y'(0)=a(B+D)=0 \;\Longrightarrow\; D=-B.\]

The solution now has only two free constants: \[y = A\bigl[\cosh(ax)-\cos(ax)\bigr] + B\bigl[\sinh(ax)-\sin(ax)\bigr]. \tag{3}\]

Step 3: Conditions at $x=l$: a linear system for $A$, $B$

Introduce the shorthand

\[ P=\cosh(al)+\cos(al), \qquad Q=\sinh(al)+\sin(al), \qquad R=\sinh(al)-\sin(al). \]

Substituting $C=-A,\ D=-B$ into $y''(l)=0$ (from (2)): \[a^2\bigl[A\cosh(al)+B\sinh(al)+A\cos(al)+B\sin(al)\bigr]=0 \;\Longrightarrow\; AP+BQ=0. \tag{4}\]

Substituting $C=-A,\ D=-B$ into $y'''(l)=b$ (from (2)): \[a^3\bigl[A\sinh(al)+B\cosh(al)-A\sin(al)+B\cos(al)\bigr]=b \;\Longrightarrow\; a^3\bigl[AR+BP\bigr]=b. \tag{5}\]

Step 4: Solve the system

From (4): $A=-BQ/P$. Substitute into (5): \[a^3\left[-\frac{BQR}{P}+BP\right]=b \;\Longrightarrow\; a^3B\cdot\frac{P^2-QR}{P}=b.\]

Compute $P^2-QR$: $P^2=\cosh^2(al)+2\cosh(al)\cos(al)+\cos^2(al), \qquad QR=\sinh^2(al)-\sin^2(al)$.

\[ P^2-QR=\bigl[\cosh^2(al)-\sinh^2(al)\bigr]+\bigl[\cos^2(al)+\sin^2(al)\bigr]+2\cosh(al)\cos(al) = 2\bigl[1+\cosh(al)\cos(al)\bigr], \tag{6} \]

using $\cosh^2-\sinh^2=1$ and $\cos^2+\sin^2=1$. Hence: \[B=\frac{bP}{a^3(P^2-QR)}=\frac{bP}{2a^3\bigl[1+\cosh(al)\cos(al)\bigr]}, \qquad A=-\frac{BQ}{P}=-\frac{bQ}{2a^3\bigl[1+\cosh(al)\cos(al)\bigr]}.\]

Writing $K=\dfrac{b}{2a^3\bigl[1+\cosh(al)\cos(al)\bigr]}$, the four constants are: \[A=-KQ, \qquad B=KP, \qquad C=-A=KQ, \qquad D=-B=-KP.\]

Step 5: Particular solution

Substituting $A,B,C,D$ back into (1) and grouping the $P$- and $Q$-terms: \[y = -KQ\cosh(ax)+KP\sinh(ax)+KQ\cos(ax)-KP\sin(ax) = K\Bigl[Q\bigl(\cos(ax)-\cosh(ax)\bigr)+P\bigl(\sinh(ax)-\sin(ax)\bigr)\Bigr].\]

Step 6: Verification

At $x=0$: $\cos(a\cdot0)-\cosh(a\cdot0)=1-1=0$ and $\sinh(a\cdot0)-\sin(a\cdot0)=0-0=0$, so both bracket terms vanish and $y(0)=0\ \checkmark$; the same cancellation makes $y'(0)=0\ \checkmark$ (both terms in $y'$ carry a factor of $\sinh(a\cdot0)$ or $1-\cosh$-type combinations that vanish at $x=0$, consistent with step 2).

For the conditions at $x=l$, use $y=K[Q(\cos(ax)-\cosh(ax))+P(\sinh(ax)-\sin(ax))]$ with $A=-KQ,\ B=KP,\ C=KQ,\ D=-KP$ substituted into (2): \[y''(l)=a^2\bigl[A\cosh(al)+B\sinh(al)-C\cos(al)-D\sin(al)\bigr] = a^2K\bigl[-Q\cosh(al)+P\sinh(al)-Q\cos(al)+P\sin(al)\bigr].\]

Grouping: $=a^2K\bigl[P\bigl(\sinh(al)+\sin(al)\bigr)-Q\bigl(\cosh(al)+\cos(al)\bigr)\bigr]=a^2K[PQ-QP]=0\ \checkmark$.

\[ y'''(l)=a^3\bigl[A\sinh(al)+B\cosh(al)+C\sin(al)-D\cos(al)\bigr] = a^3K\bigl[-Q\sinh(al)+P\cosh(al)+Q\sin(al)+P\cos(al)\bigr]. \]

Grouping: $=a^3K\bigl[P\bigl(\cosh(al)+\cos(al)\bigr)-Q\bigl(\sinh(al)-\sin(al)\bigr)\bigr]=a^3K\bigl[P^2-QR\bigr]=a^3K\cdot2\bigl[1+\cosh(al)\cos(al)\bigr]$, using (6) again. Since $K=b/\bigl(2a^3[1+\cosh(al)\cos(al)]\bigr)$, this equals exactly $b\ \checkmark$. All four boundary conditions and the differential equation are satisfied.