Sentence examples for boundary value problem left from inspiring English sources

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Consider the periodic boundary value problem left { textstylebegin{array}l} ddot{u}(t)+ r u t-tau)+nabla F(t, u t-tau))=0, quad mbox{a.e.

Consider the boundary value problem left { textstylebegin{array}{l} -mathscr{D}_{mathbf{t}}^{beta}v(t)=r(t), quad tin 0,1), v(0)=v(1)=0.

Let (e=e(x)) be the solution of the following boundary value problem: left { textstylebegin{array}{l} L varphi(x)=1, quad xinOmega, Bvarphi=0, quad xinpartial Omega.

Consider the periodic boundary value problem left { textstylebegin{array}{ll} ddot{u}(t)+r u t-tau)+nabla F(t, u t-tau))=0, quadmbox{a.e.

The Lyapunov inequality states that a necessary condition for the boundary value problem left { begin{array}{l} y"(t) + q(t y(t) = 0,quad a < t < b, y(a)=0=y(b) end{array} right.

Consider the following Neumann boundary value problem: left { textstylebegin{array}{l} Delta^{2} u t-1)+g(u t-1=h(t), quad tin[1,T]_{mathbf{Z}}, Delta u(0)=Delta u(T)=0, end{array}displaystyle right.

Consider the following second order three-point boundary value problem: left { textstylebegin{array}{l} -u t)=f u), quad tin[0,1], u(0)=0,qquad u(1)=frac{1}{2}u(frac{1}{2}), end{array}displaystyle right.

We consider the boundary value problem left { textstylebegin{array}{l} Delta_{mathbf{g}} u +u^{p}= 0 quadmbox{in } Omega_{R}, u=0 quad mbox{on } partialOmega_{R}, end{array}displaystyle right.

For any (y in C[0,1]), the unique solution of the boundary value problem left { textstylebegin{array}{l} u"'(t)=y(t), u(0)=0, qquad u'(p)=0, qquad u"(1)=lambda[u"] end{array}displaystyle right.

Then, for the solution of the nonlocal boundary value problem left { begin{array}l} -u^{prime prime }+Au(t)=f(t),quad 0< t

Given (rin L^{1} 0, 1)), the boundary value problem left { textstylebegin{array}{l} -mathscr{D}_{mathbf{t}}^{alpha}x t)=r(t),quad 0< t< 1, x 0)=x(1)=0, end{array}displaystyle right.

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