For a real parameter \(\lambda\), find all solutions on \(0\le x\le1\) of \[ \varphi(x)=e^x+\lambda\int_0^x e^{x-y}\varphi(y)\,dy \] and \[ \psi(x)=e^x+\lambda\int_0^1 e^{x-y}\psi(y)\,dy. \]
Solution
1. The first integral equation is equivalent, after setting\[ u(x)\coloneqq e^{-x}\varphi(x), \]to\[ u(x)=1+\lambda\int_0^x u(y)\,dy. \]
Factor \(e^x\) from the kernel: \[ \begin{aligned} \varphi(x) &= e^x+\lambda e^x\int_0^x e^{-y}\varphi(y)\,dy. \end{aligned} \] Divide by \(e^x\) and use the definition of \(u\).
2. The first equation has exactly one solution, namely\[ \boxed{\varphi(x)=e^{(1+\lambda)x}}. \]
Any solution of step 1 is continuous, since its right-hand side is an indefinite integral plus a constant. The fundamental theorem of calculus therefore gives \[ u'(x)=\lambda u(x), \qquad u(0)=1. \] The unique solution of this initial-value problem is \[ u(x)=e^{\lambda x}. \] Thus \[ \varphi(x)=e^xu(x)=e^{(1+\lambda)x}. \] Conversely, direct substitution verifies that this function satisfies the original integral equation, so there are no additional solutions.
3. For a solution of the second equation, define\[ C\coloneqq\int_0^1 e^{-y}\psi(y)\,dy. \]Then necessarily\[ \psi(x)=e^x(1+\lambda C). \]
The integral in the second equation is over the fixed interval \([0,1]\), so \[ \int_0^1e^{x-y}\psi(y)\,dy = e^x \int_0^1e^{-y}\psi(y)\,dy = e^xC. \] Substitution gives the displayed formula.
4. The scalar \(C\) from step 3 must satisfy\[ (1-\lambda)C=1. \]
Insert the formula from step 3 into the definition of \(C\): \[ \begin{aligned} C &= \int_0^1 e^{-y}e^y(1+\lambda C)\,dy\\ &= \int_0^1(1+\lambda C)\,dy\\ &= 1+\lambda C. \end{aligned} \] Rearranging yields the claim.
5. If \(\lambda\neq1\), the second equation has exactly one solution:\[ \boxed{\psi(x)=\frac{e^x}{1-\lambda}}. \]
Step 4 gives \[ C=\frac1{1-\lambda}. \] Then step 3 gives \[ \psi(x) = e^x \left( 1+\frac{\lambda}{1-\lambda} \right) = \frac{e^x}{1-\lambda}. \] The scalar equation determines \(C\) uniquely, so no other solution is possible. Direct substitution verifies the displayed function.
6. If \(\lambda=1\), the second equation has no solution.
For \(\lambda=1\), step 4 becomes \[ 0\cdot C=1, \] which is impossible.
7. Hence all solutions are\[ \boxed{ \varphi(x)=e^{(1+\lambda)x} \quad\text{for every }\lambda\in\RR, } \]and\[ \boxed{ \psi(x)= \begin{cases} \dfrac{e^x}{1-\lambda},&\lambda\neq1,\\ \text{no solution},&\lambda=1. \end{cases} } \]
Combine steps 2, 5, and 6.
8. Q.E.D.
Step 7 is the complete solution set for both equations.